EEE

St. PETER’S UNIVERSITY St. Peter’s Institute of Higher Education and Research

(Declared Under Section 3 of the UGC Act, 1956) AVADI, CHENNAI – 600 054

TAMIL NADU

STUDY MATERIAL

B. Tech. PROGRAMME (Code No. – 501 - 516)

(Effective From 2009 – 2010)

II SEMESTER

209BTT08 – BASIC ELECTRICAL AND ELECTRONICS ENGINEERING

St. PETER’S INSTITUTE OF DISTANCE EDUCATION (Recognized by Distance Education Council and Joint Committee of

UGC-AICTE-DEC, New Delhi. Ref. F. No. DEC/SPU/CHN/TN/Recog/09/14 dated 02.04.2009 and Ref. F.No.

DEC/Recog/2009/3169 dated 09.09.2009)

Copyright © Laxmi Publications (P) Ltd.

Published by: Laxmi Publications Pvt Ltd., 113, Golden House, Daryaganj, New Delhi-110 002. Tel: 43532500, E-mail: [email protected]

No part of this publication which is material protected by this copyright notice may be reproduced or transmitted or utilized or stored in any form or by any means now known or hereinafter invented, electronic, digital or mechanical, including photocopying, scanning, recording or by any information storage or retrieval system, without prior written permission from the publisher.

Information contained in this book has been published by Laxmi Publications (P) Ltd. and has been obtained by its authors from sources believed to be reliable and are correct to the best of their knowledge. The University has edited the study material to suit the curriculum and distance education mode. However, the publisher/university and its author shall in no event be liable for any errors, omissions or damages arising out of use of this information and specifically disclaim any implied warranties or merchantability or fitness for any particular use.

PREFACE

St. Peter’s University has been recognized by the Distance Education

Council and Joint Committee of UGC-AICTE-DEC, New Delhi, for offering

various programmes including B.Tech., D.Tech., MBA, MCA and other

programmes in Humanities and Sciences through Distance Education

mode.

The Methodology of distance education includes self-instructional study

materials in print form, face-to-face counselling, practical classes, virtual

classes in phased manner and end assessment.

The basic support for distance education students lies on the self-

instructional study materials. Keeping this in mind, the study materials

under distance mode are prepared. The main features of the study

materials are (1) learning objectives (2) self explanatory study materials

unitwise (3) self tests (4) list of references for further studies. The material

is prepared in simple English and graded in terms of technical content. It is

built upon the pre-requisite knowledge.

Students are advised to study the materials several times and get

benefitted. The face-to-face session in the counseling centre will help them

to clear their doubts and difficult concepts which they would have faced

during the learning process.

Students should remember that self study and sustained motivation are

the two important requirements for a successful learning under the

distance education mode.

We wish the students to put forth their best efforts to become

successful in their chosen field of learning.

Registrar St. Peter’s University

CONTENTS Page No.

Scheme of Examinations (vi)

Syllabus of Basic Electrical and Electronics Engineering (vii)

Model Question Paper (ix)

Unit 1: Electrical Circuits and Measurements 1-84

Unit 2: Electrical Mechanics 85-155 Unit 3: Semiconductor Devices and Applications 156-202 Unit 4: Digital Electronics 203-253

Unit 5: Fundamentals of Communication Engineering 254-282

Scheme of Examinations

I Semester

Code No. Course Title Credit Marks

Theory EA Total109BTT01 Technical English – I 1 100 100 109BTT02 Mathematics – I 3 100 100 109BTT03 Engineering Physics – I 3 100 100 109BTT04 Engineering Chemistry – I 3 100 100 109BTT05 Engineering Graphics – I 3 100 100

109BTT06 Fundamentals of Computing and programming 3 100 100

109BTP01 Computer Practices Laboratory – I Record 1 90

10 100

109BTP02 Engineering Practices Laboratory – I Record 1 90

10 100

109BTP03 *Physics & Chemistry Laboratory - I - - - Total 18 800 800

*The Practical Examinations of First Semester will be held only along with those of the second semester.

II Semester

Code No. Course Title Credit Marks

Theory EA Total209BTT01 Technical English – II 2 100 100 209BTT02 Mathematics – II 3 100 100 209BTT03 Engineering Physics – II 2 100 100 209BTT04 Engineering Chemistry – II 2 100 100

209BTT05 209BTT06 209BTT07

Engineering Mechanics (For non-circuit branches) Circuit Theory (For branches under Electrical Faculty) Electric Circuits and Electron Devices (For branches under I & C Faculty)

3

100

100

209BTT08 209BTT09

Basic Electrical & Electronics Engineering (For non-circuit branches) Basic Civil & Mechanical Engineering (For circuit branches)

3

100 100

209BTP01 Computer Practices Laboratory – II Record 1 90

10 100

209BTP02 Physics & Chemistry Laboratory – II Record 1 90

10 100

209BTP03 209BTP04 209BTP05

Computer Aided Drafting and Modelling Laboratory(For non-circuit branches) Electrical Circuits Laboratory (For branches under Electrical Faculty) Circuits and Devices Laboratory (For branches under I & C Faculty) Record

1

90 90 90 10

100 100 100

Total 18 900 900

SYLLABUS

209BTT08 BASIC ELECTRICAL AND ELECTRONICS ENGINEERING UNIT I ELECTRICAL CIRCUITS & MEASURMENTS OhmÊs Law – KirchoffÊs Laws – Steady State Solution of DC Circuits – Introduction to AC Circuits – Waveforms and RMS Value – Power and Power factor – Single Phase and Three Phase Balanced Circuits. Operating Principles of Moving Coil and Moving Iron Instruments (Ammeters and Voltmeters), Dynamometer type Watt meters and Energy meters. UNIT II ELECTRICAL MECHANICS Construction, Principle of Operation, Basic Equations and Applications of DC Generators, DC Motors, Single Phase Transformer, single phase induction Motor. UNIT III SEMICONDUCTOR DEVICES AND APPLICATIONS Characteristics of PN Junction Diode – Zener Effect – Zener Diode and its Characteristics – Half wave and Full wave Rectifiers – Voltage Regulation. Bipolar Junction Transistor – CB, CE, CC Configurations and Characteristics – Elementary Treatment of Small Signal Amplifier. UNIT IV DIGITAL ELECTRONICS Binary Number System – Logic Gates – Boolean Algebra – Half and Full Adders – Flip-Flops – Registers and Counters – A/D and D/A Conversion (single concepts) UNIT V FUNDAMENTALS OF COMMUNICATION ENGINEERING Types of Signals: Analog and Digital Signals – Modulation and Demodulation: Principles of Amplitude and Frequency Modulations. Communication Systems: Radio, TV, Fax, Microwave, Satellite and Optical Fibre (Block Diagram Approach only). TEXT BOOKS: 1. V.N. Mittle „Basic Electrical Engineering‰,Tata McGraw Hill Edition, New Delhi, 1990. 2. R.S. Sedha, „Applied Electronics‰ S. Chand & Co., 2006. REFERENCES: 1. Muthusubramanian R, Salivahanan S and Muraleedharan K A, „Basic Electrical, Electronics

and Computer Engineering‰,Tata McGraw Hill, Second Edition, (2006). 2. Nagsarkar T K and Sukhija M S, „Basics of Electrical Engineering‰, Oxford press (2005). 3. Mehta V K, „Principles of Electronics‰, S.Chand & Company Ltd, (1994). 4. Mahmood Nahvi and Joseph A. Edminister, „Electric Circuits‰, SchaumÊ Outline Series,

McGraw Hill, (2002). 5. Premkumar N, „Basic Electrical Engineering‰, Anuradha Publishers, (2003).

p-2\d\L-sirsa7\Beeg-mod.pmd 22-5-2010

MODEL QUESTION PAPERB.Tech. Degree Examinations

Second Semester

209BTT06 - Basic Electrical and Electronics Engineering

(Regulations 2009)

Time: 3 Hours Maximum: 100 marks

Answer ALL the questions

PART – A (10 × 2 = 20 MARKS)

1. State Kirchhoff’s Laws.

2. Define R.M.S. value.

3. List the main constituents used in construction of a DC machine.

4. In a particular d.c. machine, if P = 8, Z = 400, N = 300 rpm and φ = 100 mWb, calculate generatedemf if winding is connected in lap fashion.

5. Explain the effect of temperature on diode parameters.

6. What are the advantages of full wave rectifier?

7. What do mean universal gates?

8. Minimize the Boolean expression AB + ABC + AB (D + E) = AB(1 + C + D + E).

9. Define modulation. Write the types of modulation.

10. Specify the three basic sections in satellite.

PART – B (5 × 16 = 80 MARKS)

11. (a) Apply kirchhoff’s laws to the circuit shown in figure. Indicate the various branch currents. Writedown the equations relating the various branch currents. Solve these equations to find the val-ues of these currents. Is the sign of any of the calculated currents negative? If yes, explain thesignificance of the negative sign.

+

–

+

–100 V50 V

15 30

20

(OR)

(b) Describe the construction and working of PMMC instruments. Also drive the equation for de-flection in spring controlled PMMC instrument.

12. (a) (i) Explain the principle of operation of DC generator clearly.

(ii)Derive the expression for generated emf in a dc machine both for lap and wave winding.

(OR)

(b) With a neat sketch explain the different types single-phase induction motor.

p-2\d\L-sirsa7\Beeg-mod.pmd 22-5-2010

13. (a) Draw and Explain zener diode characteristics. Also explain the breakdown action in zener diode.

(OR)

(b) Explain the input and output characteristics of a transistor in CB configuration.

14. (a) Explain the half adder and full adder with logic diagram, truth table and K-map simplificationfor carry and sum.

(OR)

(b) Explain the types of shift register according to the data movement.

15. (a) Explain the principle of Amplitude modulation and derive equation for AM wave.

(OR)

(b) Explain with a neat sketch a fiber optic communication system. State functions of each block.

Self-Instructional Material 1

NOTES

Electrical Circuits andMeasurements

STRUCTURE

1.1 Basic Concepts and Ohm’s Law1.2 Ohm’s Law1.3 Kirchhoff’s Laws1.4 Introduction to Alternating Current1.5 Generation and Equations of Alternating Voltages and Currents1.6 Alternating Voltage and Current1.7 Single Phase Circuits1.8 Three Phase Balanced Circuit1.9 Generation of Three-Phase Voltages

1.10 Phase Sequence and Numbering of Phases1.11 Inter-connection of Three Phases1.12 Electrical Principles of Operation1.13 Electrical Indicating Instruments1.14 Moving-iron Instruments (Ammeters and Voltmeters)1.15 Moving-coil Instruments1.16 Wattmeters1.17 Integrating Meters (Energy Meters)

• Summary• Glossary

• Review Questions• Further Readings

OBJECTIVES

After going through this unit, you should be able to:

• define Ohm’s law and Kirchhoff’s laws with their limitations andapplications.

• give an introduction to A.C. circuits in terms of waveform and RMS value,power and power factor.

U N I T

1ELECTRICAL CIRCUITS AND

MEASUREMENTS

2 Self-Instructional Material

NOTES

Basic Electrical andElectronics Engineering

• describe single phase and three phase circuit.• define operating principles of moving coil and moving iron instruments of

ammeter and voltmeter, dynamometer type wattmeter and energy meters.

1.1 BASIC CONCEPTS AND OHM’S LAW

Electricity

It is not easy to define electricity.

Electricity may be defined as a form of energy. It involves making andusing energy.

It may also be defined as a way in which materials behave.Sometimes people use the term ‘electricity’ as the name for a material that

flows through a solid wire, motion of this strange material is called electric current.

Electron Theory

An element is defined as a substance which cannot be decomposed intoother substances. The smallest particle of an element which takes part inchemical reaction is known as an atom.

All matter is composed of atoms which are infinitesimally small. The atom, itself, is composed of electrons, protons and neutrons. The number

and arrangement of these particles determines the type of atom: oxygen,carbon, copper, lead or any other element.

Weight, colour, density, and all other characteristics of an element aredetermined by the structure of the atom. Electrons from lead would be thesame as electrons from any other element.

The ‘electron’ is a very light particle that spins around the centre of theatom. Electrons move in an orbit. The number of electrons orbiting aroundthe centre or nucleus of the atom varies from element to element. Theelectron has a negative (–) electric charge.

The ‘proton’ is a very large and heavy particle in relationship to the electron.One or more protons will form the centre or nucleus of the atom. Theproton has a positive (+) electrical charge.

The ‘nucleon’ consists of neutron and proton bound tightly together inwhich the neutron is electrically neutral; it has no electrical charge.

Normally, the atoms are electrically neutral, that, the number of electronsand protons are the same, cancelling out each other’s electrical force. Atoms‘‘stay together’’ because unlike electrical charges attract each other. Theelectrical force of the protons holds the electrons in their orbits. Likeelectrical charges repel each other so negatively charged electrons will notcollide with each other.


NOTES


++

–

–

Protons

NeutronsNucleus

Electrons

Electronorbits

Fig. 1.1. Atomic structure: Electron, proton and neutron.

Electric Current

In order to have electric current, electrons must move from atom to atom(Fig. 1.2). There are quite a few substances in which it is relatively easyfor an electron to jump out of its orbit and begin to orbit in an adjoining ornearby atom. Substances which permit this movement of electrons arecalled conductors of electricity (e.g., copper, aluminium, silver, etc.)

–

––

–

–

–

–

NucleusFree electron–

–

–

Fig. 1.2. Current: Flow of electrons within a conductor.

The controlled movement of electrons, (or drift) through a substance iscalled current. Current occurs only when a difference of potential (e.m.f.or voltage) is present. For example, we can get a difference of potential byconnecting a battery to the ends of a length of copper wire. The pressurefrom the battery will then move the electrons.

Refer Fig. 1.3 (a). The electrons of copper are free to drift in random fashionthrough the copper. If an imaginary line is set up in the copper wire, it will befound that the same number of electrons will cross the line from both the directions.This random movement however will not result in electric current.


NOTES


Fig. 1.3. (a) With no voltage applied: Electrons drift aimlessly.

Refer Fig. 1.3 (b). One end of the wire attracts electrons because it is connectedto the battery terminal which has a positive charge (lack of electrons). The electronsin the copper wire drift towards this positive charge. As electrons leave the copperwire and enter the positive terminal, more electrons enter the other end of thecopper wire. These electrons are taken from the negative terminal of the battery.

_ +

_ +

Flow of electronsin one direction

Fig. 1.3. (b) When voltage is applied to the wire:A controlled drift of electrons takes place.

Current is the rate at which electrons move. One ampere (unit of current)represent 6.28 × 1018 electrons passing a point each second (1 coloumb past apoint in 1 second).

A moving stream of positive charges also constitutes an electric current andin the case of flow through ionized gases and electrolytes, the current consists partlyof positively charged particles moving in one direction and partly of negativelycharged particles moving in the opposite direction. (Fig. 1.4). In all other cases theelectric current consists of solely of moving electrons.

–

+

–

+

–

+

–

+

–

+

–

Fig. 1.4

One ampere is that constant current, which if maintained in two straightparallel conductors of infinite length, of negligible circular cross-section and placed


NOTES


one metre apart in vacuum, would produce between these conductors a force equalto 2 × 10–7 newton per metre of length.

Note. Those substances whose atoms have their outermost orbit incomplete actas good conductors of electricity i.e., they permit an easy detachment of their outermostelectrons and offer very little hinderance to their flow through atoms. Such substancesare known as good conductors. But substances whose electrons are rigidly attached totheir atoms are termed as bad conductors. Materials like germanium, silicon, and siliconcarbide whose resistances at ordinary temperatures lie in between those of typical metalsand typical insulators are called semi-conductors.

Electromotive Force and Potential

Electromotive force (e.m.f.) is the force that causes a current of electricity to flow.

If we have two compressed-air storage tanks, one of which is at a higherpressure than the other, and we connect the two tanks together by meansof a pipe, air will flow along this pipe from the higher pressure tank to thelow-pressure tank. The force that causes this current of air to flow is thepressure difference between the two tanks.

Similarly, in the zinc-copper voltaic cell, the zinc and copper electrodesboth contain vast quantities of electrons, but owing to chemical action thezinc electrode is charged to a greater electron pressure than the copperelectrode. The electron-pressure difference is the electromotive force thatcauses the flow of electrons from the zinc electrode to copper electrodethrough the external connecting circuit.

The term electron pressure, althoughdescriptive, has no social standing. The wordpotential had been used to express the sameidea for several years before the developmentof the electron theory. Two conducting bodiesare said to be at the same potential if there isno flow of electric current between them whenthey are joined together by a conducting wire.

The potential difference (p.d.) V, between two points in a circuit is theelectrical pressure or voltage required to drive the current between them (Fig. 1.5).

The volt is unit of potential difference and electromotive force. It is defined asthe difference of potential across a resistance of 1 ohm carrying a current of 1 ampere.

Resistance

Some materials have an abundance of free electrons, which require a lowpressure to move them from atom to atom, and establish a high current. Suchmaterials are known as good conductors. Other materials have few free electrons.In these the same electric pressure can move only a few electrons from atom toatom, establishing a low current. These are considered poor conductors. Theprogressive motion of free electrons is hindered in all materials, because they collidewith atoms of the substance used. The opposition to flow of electrons (due to bonds

V

I I

Fig. 1.5. Current and p.d. in asimple circuit.


NOTES


between protons and electrons, as well as to collisions) is called electricalresistance (R).

Resistance may also be defined as ‘‘The property of the electric circuit whichopposes the flow of current’’.

Resistance is analogous in most of its aspects to friction in mechanics orhydraulics and, like friction, results in heat generation. The heating of an electriciron or stove is due to the resistance of the heat-unit conductor materials. Forsafety the material and cross-section of the conductors must be such that thetemperature is kept well below a value which would result in damage to theconductor or its protective coating of insulating material.

The practical unit of electric resistance is ohm (Ω). It (ohm) is defined as theresistance in which a constant current of 1 ampere generates heat at the rate of 1watt. One volt applied across 1 ohm will produce 1 ampere.

1 Mega-ohm (M Ω) = 106 Ω1 kilo-ohm (k Ω) = 103 Ω1 milli-ohm (m Ω) = 10–3 Ω1 micro-ohm (µ Ω) = 10–6 Ω .

Laws of Resistance

The resistance of a conductor, such as a wire, of uniform cross-section dependson the following factors:

(i) Length (l). Varies directly as its length, l.(ii) Cross-section (A). Varies inversely as the cross-section, A of the conductor.

(iii) Nature of the material.(iv) Temperature of the conductor.Neglecting the last factor (iv) for the time being, we can say that

R ∝ lA

or R = ρ lA

...(1.1)

where, l = Length of the conductor,

A = Area of cross-section of the conductor, and

ρ = A constant depending on the nature of the material of the conductorand is known as its specific resistance or resistivity.

If in eqn. (1.1), l = 1 metre, A = 1 metre, then R = ρ.

Hence specific resistance or resistivityof a material may be defined as ‘‘theresistance between the opposite faces of ametre cube of that material ’’ (Fig. 1.6).

Unit of Resistivity. From eqn. (1.1),we have

ρ = RAl

CurrentA = 1 metre

2

CurrentR = ρ

l = 1 metre

Fig. 1.6


NOTES


In S.I. system of units

ρ = R Al

RAl

ohm mm× =

2 ohm-m

Hence the unit of resistivity is ohm-metre (Ω-m).

Volume Resistivity

We know that R = ρlA

= ρ ρl AA A A

××

= V2 ...(i)

∵ V lVl

A

=

=

L

N

MMM

O

Q

PPP

Awhere = volume,

length, and= uniform cross-sectional area of a conductor.

Also, R = ρ ρ× ×

×=l l

A llV

2

...(ii)

i.e., R = ρ ρV2A

lV

=2

[from (i) and (ii)] ...(1.2)

The above eqn. signifies that for a given volume:

1. Resistance of a conductor varies inversely as square of its cross-section.

2. Resistance varies as square of its length.

The values of resistivity and temperature coefficients for some materialsare given in Table 1.1.

Table 1.1 Resistivities and Temperature Coefficients

Material Resistivity in ΩΩΩΩΩ-m Temperatureat 20° C coefficient at 20° C

Copper 1.59 × 10–8 0.00428

Aluminium 2.8 × 10–8 0.0020

Silver 1.52 × 10–8 0.00377

Platinum 11 × 10–8 0.00340

Iron (9 to 15) × 10–8 0.0070

Mercury 19.9 × 10–8 0.00089

German silver (4 Cu; 2Ni; 1 Zn) 20.2 × 10–8 0.00027

Constantan or Eureka 49 × 10–8 – 0.00004 to + 0.00001

Carbon 7000 × 10–8 – 0.0005


NOTES


Conductance (G)

Conductance GR

Al

= =FHG

IKJ

1ρ

is the reciprocal of resistance and therefore be

utilised as a measure of the inducement to current flow offered by the circuit. Theconductance of circuit is directly proportional to the cross-section of the conductorand inversely proportional to its length.

The unit of conductance is reciprocal ohm, or mho ( ).

Conductivity (σ)

The reciprocal of specific resistance σρ

=FHG

IKJ

1 of a material is called its

conductivity. The unit of conductivity σ =FHG

IKJ

GlA

is mho/metre ( /m).

In practice, conductivities are used chiefly to compare one conductor materialwith another. It is therefore convenient to choose one material as the standard andto compare others with it. Accordingly, pure annealed copper has been chosen asthe standard and its conductivity is taken as 100 percent.

Example 1.1. Find the resistance of copper wire at 20°C whose cross-sectionalarea is 0.02 cm2 and length 400 metres. Take resistivity of copper at 20°C as 1.7 ×10–8 Ω-m. What is the conductivity of wire?

Solution. Resistivity of copper at 20° C, ρ20 = 1.7 × 10–8 Ω-m

Cross-sectional area, A = 0.02 cm2 = 0.02 × 10–4 m2

Length of wire, l = 400 m

Resistance of copper wire at 20° C, R20:

Using the relation, R = ρla

or R20 = ρ20la

= 1.7 10 400

0 02 10

8

4× ×

×

−

−. = 3.4 ΩΩΩΩΩ . (Ans.)

Conductivity of wire, σ:

Conductivity (σ) = 1 1

7 10 8ρ=

× −1.

= 58.8 × 106 mho/metre. (Ans.)

1.2 OHM’S LAW

Ohm’s law can be stated as follows:

‘‘The ratio of potential difference between any two points of a conductor to thecurrent flowing through it, is constant provided the physical conditions (i.e.,temperature etc.) do not change.”


NOTES


In other words, Ohm’s law states that in a closedcircuit (Fig. 1.7), when a voltage V is applied across aconductor, then current I flowing through it is directlyproportional to the applied voltage,

i.e., I ∝ V orVI

= constant

∴ VI

= R ohm ...(1.3)

or V = IR volt ...[1.3 (a)]

or I = VR

amp. ...[1.3 (b)]

where,V = Voltage between two points,

I = Current flowing in the circuit, and

R = The resistance of the conductor.

The units of voltage, current and resistance are volt (V), ampere (A) and ohm(Ω) respectively.

The linear relationship (I ∝ V) does not apply to all non-metallic conductors.For example, for silicon carbide, the relationship is given by:

V = k Ix

where, k and x are constants and x is less than unity.

Conditions for Applying Ohm’s Law

Ohm’s law is applicable under the following conditions:

1. Ohm’s law can be applied either to the entire circuit or to a part of thecircuit.

2. When Ohm’s law is applied to a part of a circuit, part resistance and thepotential difference across that part resistance should be used.

3. Ohm’s law can be applied to both D.C. and A.C. circuits. However, in case ofA.C. circuits, impedance Z, is used in place of resistance.

Then, I = VZ

= Applied voltageImpedance in the circuit

Electrical Power. It is the rate of doing work. In other words, the amountof work done in one second is called ‘‘power’’.

or P = Wt

...(1.4)

where P = The power in watt,

W = The work done in joule, and

t = Time in second.

I R

V

Fig. 1.7


NOTES


Power is equal to the product of voltage (V) and current (I) in a particularcircuit.

i.e., P = V × I ...(1.5)

The following relations hold good:

(i) P = VI = I2 R = VR

2

where, power in watt,voltage in volt,current in ampere, andresistance in ohm

PVI

R

====

R

S|

T|

U

V|

W|

.

(ii) I = PV

PR

=

(iii) R = PI

VP2

2

=

(iv) V = PI

PR=

Power is expressed in terms of kW (kilowatt = 1000 W)

or MW (megawatt = 1000 kW or 106 W).

Electrical Energy. It is the total amount of work done in an electric circuit.In other words, it is measured by the product of power and time.

i.e., W = P × t ...(From eqn. (1.4)]or W = VI t

= VQ joules

where Q = The quantity of electricity passing through the circuit in coulombs.

The unit of electrical energy is joule or watt-sec.

It is expressed in kWh (kilowatt hour)

1 kWh (commercial unit) = 1 kW × 1 hour = 1000 watt-hour

= 1000 × 60 × 60 watt-sec

= 3.6 × 106 watt-sec. or joule.

Linear and Non-linear Resistors

A linear resistor is one which obeys ohm’s law. A circuit which contains onlylinear components is called a linear circuit.

Such elements in which V/I (volt-ampere) plots are not straight lines but curvesare called non-linear resistors or non-linear elements.

Examples of non-linear elements: Filaments of incandescent lamps, diodes,thermistors and varistors.

‘‘Varistor (Non-linear resistor)’’:

It is a voltage-dependent metal-oxide material whose resistance decreasessharply with increasing voltage.


NOTES


The zinc oxide-based varistors are primarily used for protecting solid-statepower supplies from low and medium size voltage in the supply line.

Silicon carbide varistors provide protection against high-voltage surgescaused by lightning and by discharge of electromagnetic energy stored inthe magnetic fields of large coils.

Ohm’s law does not apply under the following conditions:

1. Electrolytes where enormous gases are produced on either electrode.

2. Non-linear resistors like vacuum radio valves, semiconductors, gas filledtubes etc.

3. Arc lamps.

4. Metals which get heated up due to flow of current.

5. Appliances like metal rectifiers, crystal detectors, etc., in which operationdepends on the direction of current.

Resistances in Series

Fig. 1.8 shows three resistance connected in series. Obviously current flowingthrough each resistance will be same but voltage drop across each of them will varyas per value of individual resistance.

R1 R2

II

R3

V

V1 V2 V3

I

Fig. 1.8. Resistances in series.

Also the sum of all the voltage drops (V1 + V2 + V3) is equal to the appliedvoltage (V).

i.e., V = V1 + V2 + V3

IR = IR1 + IR2 + IR3 (Using Ohm’s law: V= IR)

i.e., R = R1 + R2 + R3 ...(1.6)

where R is the equivalent resistance of series combination.

Also 1 1 1 1

1 2 3G G G G= + + ...[1.6(a)]

As seen from above, the main characteristics of a series circuit are:

1. Same current flows through all parts of the circuit.

2. Different resistors have their individual voltage drops.

3. Voltage drops are additive.

4. Applied voltage equals the sum of different voltage drops.

5. Resistances are additive.

6. Powers are additive.


NOTES


Resistances in Parallel

Refer Fig. 1.9. In this case voltage across each resistance will be same butcurrent will be different depending upon the value of the individual resistance.

i.e., I = I1 + I2 + I3

I

I1R1

R2

R3

I2

I3

V

Fig. 1.9. Resistances in parallel.

VR

VR

VR

VR

= + +1 2 3

1 1 1 1

1 2 3R R R R= + + ...(1.7)

where R is the equivalent resistance of the parallel combination.

R = R R R

R R R R R R1 2 3

1 2 2 3 3 1+ + ...[1.7(a)]

G = G1 + G2 + G3 ...(1.8)

The main characteristics of a parallel circuit are:

1. Same voltage acts across all parts of the circuit.

2. Different resistors have their individual current.

3. Branch currents are additive.

4. Conductances are additive.

5. Powers are additive.

Example 1.2. In the Fig. 1.10 is shown a combination of resistances. If thevoltage across L and M is 75 volt find:

(i) The effective resistance of the circuit.

(ii) Voltage drop across each resistance.


NOTES


Solution. Refer to Fig. 1.10.

L

2 Ω

4 Ω

4 Ω

6 Ω 10 Ω 8 ΩN P M

75 V

Fig. 1.10

(i) Effective resistance of the circuit:

Resistance between L and N,

1 12

14

16

6 3 212

1112RLN

= + + = + + = Ω

∴ RLN = 1211

Ω ...(Fig. (1.10)

Resistance of branch LNP,

RLNP = 1211

+ 10 = 12211

= 11.09 Ω ...(Fig. 1.11)

L

4 Ω

10 Ω 8 Ω

N P M

75 V

1211

Ω

75 V

4 Ω

L P

11.09 Ω

M

8 Ω

Fig. 1.11 Fig. 1.12

Resistance between L and P

1 1

11.0914RLP

= + [As there are two parallel paths between points

L and P of resistances 11.09 Ω and 4 Ω]

L

2.94 Ω

P

8 Ω

M

75 V

L M

10.94 Ω

75 V

Fig. 1.13 Fig. 1.14


NOTES

Basic Electrical andElectronics Engineering =

4 11.0944 36+

.

or RLP = 44 3615 09

.

. = 2.94 Ω ...(Fig. 1.12)

Effective resistance of the circuit,

RLM = RLP + RPM

= 2.94 + 8 = 10.94 ΩHence effective resistance of the circuit = 10.94 Ω. (Ans.)

Total current through the circuit = VRLM

= 7510 94.

= 6.85 A

Current through 4 Ω resistor = 6.85 × 11.09

11.09 4+ = 5.03 A

Current in branch LNP = 6.85 × 4

11.09 4+ = 1.82 A

∴ Voltage drop across 2, 4 and 6 Ω resistors

= 1.82 × RLN = 1.82 × 1211

= 1.98 V. (Ans.)

Voltage drop across 10 Ω resistor = 1.82 × 10 = 18.2 V (Ans.)



Limitations of Ohm’s Law

In a series circuit or in any branch of a simple parallel circuit the calculationof the current is easily effected by the direct application of Ohm’s law. But such asimple calculation is not possible if one of the branches of a parallel circuit containsa source of e.m.f., or if the current is to be calculated in a part of a network in whichsources of e.m.f. may be present in several meshes or loops forming the network.The treatment of such cases is effected by the application of fundamental principlesof electric circuits. These principles were correlated by Kirchhoff many years agoand enunciated in the form of two laws, which can be considered as the foundationsof circuit analysis. Other, later, methods have been developed, which when appliedto special cases considerably shorten the algebra and arithmetic computationcompared with the original Kirchhoff’s method.

1.3 KIRCHHOFF’S LAWS

For complex circuit computations, the following two laws first stated by GutsavR. Kirchhoff (1824–87) are indispensable.


NOTES


Kirchhoff’s Point Law or Current Law (KCL)

It states as follows:

The sum of the currents entering ajunction is equal to the sum of the currentsleaving the junction. Refer to Fig. 1.15.

If the currents towards a junction areconsidered positive and those away from thesame junction negative, then this law statesthat the algebraic sum of all currents meet-ing at a common junction is zero.

i.e., Σ Currents entering = Σ Currentsleaving

I1 + I3 = I2 + I4 + I5 ...[1.9 (a)]

or I1 + I3 – I2 – I4 – I5 = 0 ...[1.9 (b)]

Kirchhoff’s Mesh Law or Voltage Law (KVL)

It states as follows:The sum of the e.m.fs (rises of potential) around any closed loop of a circuit

equals the sum of the potential drops in that loop.Considering a rise of potential as positive (+) and a drop of potential as negative

(–), the algebraic sum of potential differences (voltages) around a closed loop of acircuit is zero:

ΣE – ΣIR drops = 0 (around closed loop)i.e., ΣE = ΣIR ...[1.10 (a)]or Σ Potential rises = Σ Potential drops ...[1.10 (b)]

To apply this law in practice, assume an arbitrary current direction for eachbranch current. The end of the resistor through which the current enters, is thenpositive, with respect to the other end. If the solution for the current being solvedturns out negative, then the direction of that current is opposite to the directionassumed.

In tracing through any single circuit, whether it is by itself or a part of anetwork, the following rules must be applied:

1. A voltage drop exists when tracing through a resistance with or in the samedirection as the current, or through a battery or generator against their voltage,that is from positive (+) to negative (–). Refer to Fig. 1.16.

M

M

V+ –L

L

Fall in voltage (–V)

Fall in voltage (–E)

+ –E

M

M

V+–L

L

Rise in voltage (+V)

Rise in voltage (+E)

+–

Fig. 1.16 Fig. 1.17

I1 I2

I5

I4

I3

Junction

Fig. 1.15


NOTES


2. A voltage rise exists when tracing through a resistance against or in oppositedirection to the current or through a battery or a generator with their voltage thatis from negative (–) to positive (+). Refer to Fig. 1.17.

Illustration. Consider a circuitshown in Fig. 1.18.

Considering the loop ABEFA, we get

– I1R1 – I3R3 + E1 = 0

or E1 = I1R1 + I3R3 (where I3 = I1 + I2) ...(i)

Considering the loop BCDEB, wehave

I2R2 – E2 + I3R3 = 0 ...(ii)

or E2 = I2R2 + I3R3

If E1, E2, R1, R2 and R3 are known, then I1, I2 and I3 can be calculated fromeqns. (i) and (ii).

Applications of Kirchhoff’s Laws

Kirchhoff’s laws may be employed in the following methods of solving network :

1. Branch-current method

2. Maxwell’s loop (or mesh) current method

3. Nodal voltage method.

Branch-Current Method

For a multi-loop circuit the following procedure is adopted for writingequations:

1. Assume currents in different branch of the network.

2. Write down the smallest number of voltage drop loop equations so as toinclude all circuit elements; these loop equations are independent.

If there are n nodes of three or more elements in a circuit, then write (n–1)equations as per current law.

3. Solve the above equations simultaneously.

The assumption made about the directions of the currents initially is arbitrary.In case the actual direction is opposite to the assumed one, it will be reflected as anegative value for that current in the answer.

The branch-current method (the most primitive one) involves more labourand is not used except for very simple circuits.

Example 1.3. In the circuit of Fig. 1.19, find the current through each resistorand voltage drop across each resistor.

F E DI1 I2

E1 E2

+

+

+ +

–

– –

–

A +R1 R2B C

–

R

I3

I1 I2

Fig. 1.18


NOTES

Electrical Circuits andMeasurements12 W

10 W12 V 10 V

6 W

12 W 6 W

10 W

B

( + )I I1 2

12 V 10 V

A C

F E D

++

+ +

+

–

–

– –

–

I1I2

Fig. 1.19 Fig. 1.20

Solution. Let the currents be as shown in Fig. 1.20.

Applying Kirchhoff’s voltage law to the circuit ABEFA, we get

– 12I1 – 10(I1 + I2) + 12 = 0

– 22I1 – 10I2 + 12 = 0or + 11I1 + 5I2 – 6 = 0 ...(i)

Circuit BCDEB gives, 6I2 – 10 + 10(I1 + I2) = 0

10I1 + 16I2 – 10 = 05I1 + 8I2 – 5 = 0 ...(ii)

Multiplying eqn. (i) by 5 and eqn. (ii) by 11 and subtracting, we get 55I1 + 25I2 – 30 = 0

55I1 + 88I2 – 55 = 0

– – +

– 63I2 + 25 = 0

i.e., I2 = 0.397 A

Substituting this value in eqn. (i), we get

11I1 + 5 × 0.397 – 6 = 0

i.e., I1 = 0.365 A

Hence, Current through 12 Ω resistor, I1 = 0.365 A (Ans.)

Current through 6 Ω resistor, I2 = 0.397 A (Ans.)

Current through 10 Ω resistor, I1 + I2 = 0.762 A (Ans.)

The voltage drop across:

12 Ω resistor = 0.365 × 12 = 4.38 V (Ans.)

6 Ω resistor = 0.397 × 6 = 2.38 V (Ans.)

10 Ω resistor = 0.762 × 10 = 7.62 V (Ans.)

Example 1.4. A battery having an e.m.f. of 110 V and an internal resistance of0.2 Ω is connected in parallel with another battery with e.m.f. of 100 V and a resist-ance of 0.25 Ω. The two in parallel are placed in series with a regulating resistanceof 5 ohms and connected across 220 V mains. Calculate:

(i) The magnitude and direction of the current in each battery.(ii) The total current taken from the mains supply.


NOTES



(I + I )1 2(I + I )1 2

L

220 V

T5 Ω

0.2 Ω

0.25 Ω

N

S

I1

I2

Q

M

100 V

110 V

P

Fig. 1.21

(i) I1, I2:

Let the directions of flow of currents I1 and I2 be as shown in Fig. 1.21.

Applying Kirchhoff’s voltage law to LMNPSQL, we get

110 + 0.2I1 – 0.25I2 – 100 = 0

0.2I1 – 0.25I2 = – 10or I1 – 1.25I2 = – 50 ...(i)

Circuit LMNPTL gives,110 + 0.2I1 – 220 + 5(I1 + I2) = 0

5.2I1 + 5I2 = 110I1 + 0.96I2 = 21.15 ...(ii)

Subtracting (ii) from (i), we get – 2.21I2 = – 71.15

∴ I2 = 32.19 A (Ans.)

and I1 = – 9.75 A (Ans.)

Since I1 turns out to be negative, its actual direction of flow is opposite tothat shown in Fig. 1.21. In other words it is not a charging current but adischarging one. However, I2 is a charging current.

(ii) (I1 + I2):

The total current taken from the mains supply,

I1 + I2 = – 9.75 + 32.19 = 22.44 A (Ans.)

1.4 INTRODUCTION TO ALTERNATING CURRENT

A.C. means alternating current—The current or voltage which alternatesits direction and magnitude every time. Now-a-days 95% of the total energy isproduced, transmitted and distributed in A.C. supply.

The reasons are the following :

(i) More voltage can be generated (upto 33000 V) than D.C. (650 V only).


NOTES


(ii) A.C. voltage can be increased and decreased with the help of a staticmachine called the ‘transformer’.

(iii) A.C. transmission and distribution is more economical as line material(say copper) can be saved by transmitting power at higher voltage.

(iv) A.C. motors for the same horse power as of D.C. motors are cheaper, lighterin weight, require less space and require lesser attention in operation andmaintenance.

(v) A.C. can be converted to D.C. (direct current) easily, when and whererequired but D.C. cannot be converted to A.C. so easily and it will not beeconomical.

However, D.C. entails the following merits and hence finds wide applications.

(i) D.C. series motors are most suitable for traction purposes in tramway,railways, crains and lifts.

(ii) For electroplating, electrolytic and electrochemical processes (batterycharging etc.), D.C. is required.

(iii) Arc lamps for search lights and cinema projectors work on D.C.

(iv) Arc welding is better than on A.C.

(v) Relay and operating time switches, etc., and circuit-breakers, D.C. worksmore efficiently.

(vi) In rolling mills, paper mills, colliery winding, etc., where fine speed controlof speeds in both directions is required, D.C. motors are required.

1.5 GENERATION AND EQUATIONS OF ALTERNATING

VOLTAGES AND CURRENTS

Generation of Alternating Voltages and Currents

Alternating voltages may be generated in the following two ways:

1. By rotating a coil in a stationary magnetic field, as shown in Fig. 1.22.

2. By rotating a magnetic field within a stationary coil, as shown in Fig. 1.23.

+–

Axisof

rota

tion

Current

Flux

Motion

Externalconnections

S N

Cut awaypole

Backcoil

N

S

Rotor

Stator

Fig. 1.22. Rotating a coil in a stationary Fig. 1.23. Rotating a magnetic fieldmagnetic field. within a stationary coil.

NOTES



The value of the voltage generated in each case depends upon the followingfactors:

(i) The number of turns in thecoils ;

(ii) The strength of the field;(iii) The speed at which the coil or

magnetic field rotates. Out of the above two methods the

rotating-field method is mostly used inpractice.

Equations of AlternatingVoltages and Currents

Fig. 1.24 shows a rectangular coilof N turns rotating clockwise with anangular velocity ω radian per second ina uniform magnetic field.

Since by Faraday’s law, the voltage is proportional to the rate at which theconductor its across the magnetic field or to the rate of change of flux linkages, theshape of the wave of voltage applied to the external circuit will be determined bythe flux distribution in the air gap. For a uniform field between the poles it isevident that maximum flux will link with the coil when its plane is in verticalposition i.e., perpendicular to the direction of flux between the poles. Also it isobvious that when the plane of coil is horizontal no flux will link with the coil.

If the position of the coil with reference to the vertical axis be denoted by θ theflux linking with the coil at any instant, as the coil rotates may be determinedfrom the relation,

φ = φmax cos θ = φmax cos ωt ...(i) (∵ θ = ωt)

where, φmax = Maximum flux which can link with the coil, and t = Time taken by the coil to move through an angle θ from vertical

position.Using Faraday’s law to eqn. (i), in order to determine the voltage equation,

e = − Nddt

φ(where e is the instantaneous value of the induced

e.m.f.)

= − Nddt

tmax( cos )φ ω = ωN φmax sin ωt

or e = ωNφmax sin θ ...(ii)

As the value of e will be maximum when sin θ = 1,

∴ Emax = ω Nφmax

The eqn. (ii) can be written in simpler form as

e = Emax sin θ ...(iii)

X′ XO

Y

Y′

φ θmax sinφ θmax sin

φ θmax cosφ θmax cos

θθ

φ maxφ max

θω

=t

θω

=t

coil

coil

t= 0t= 0

φmaxφmax

ω

Fig. 1.24. A coil rotating in a magnet field.


NOTES


Similarly the equation of induced alternating current (instantaneous value) is

i = Imax sin θ (if the load is resistive) ...(iv)

Waveforms

A waveform (or wave-shape) is the shape of a curve obtained by plotting theinstantaneous values of voltage or current as ordinate against time as abscissa.

e, i

0+

–

p 2pt

(a)

e, i

0p 2p

t

(b)

e, i

0p 2p

t

(e)

e, i

0p 2p

t

(c)

e, i

0p 2p

t

(d)

e, i

0p 2p

t

(f)

Fig. 1.25. Waveforms.

Fig. 1.25 (a, b, c, d, e) shows irregular waveforms, but each cycle ofcurrent/voltage is an exactly replica of the previous one. Alternating e.m.fs andcurrents produced by machines usually both have positive and negative half waves,the same shape as shown. Fig. 1.25(f) represents a sine wave of A.C. This is thesimplest possible waveform, and alternators are designed to give as nearly as pos-sible a sine wave of e.m.f.

In general, an alternating current or voltage is one the circuit direction ofwhich reverses at regularly recurring intervals.

The waves deviating from the standard sine wave are termed as distortedwaves.

Complex waves are those which depart from the ideal sinusoidal form. Allalternating complex waves, which are periodic and have equal positive andnegative half cycles can be shown to be made up of a number of pure sinewaves, having different frequencies but all these frequencies are integralmultiples of that of the lowest alternating wave, called the fundamental(or first harmonic). These waves of higher frequencies are called harmonics.

1.6 ALTERNATING VOLTAGE AND CURRENT

Modern alternators produce an e.m.f. which is for all practical purposessinusoidal (i.e., a sine curve), the equation between the e.m.f. and time being

NOTES



e = Emax sin ωt ...(1.11)

where, e = Instantaneous voltage; Emax = Maximum voltage;

ωt = Angle through which the armature has turned from neutral.

Taking the frequency as f hertz (cycles per second), the value of ω will be 2πf,so that the equation reads

e = Emax sin (2πf)t.

The graph of the voltage will be as shown in Fig. 1.26.

+

–

Vol

tsE

.M.F

.

1 Cycle

Time

Emax+

–

Fig. 1.26. The graph of the sinusoidal voltage.

1. Cycle. One complete set of positive and negative values of an alternatingquantity is known as a cycle. A cycle may also sometimes be specified in terms ofangular measure. In that case, one complete cycle is said to spread over 360° or 2πradian.

2. Amplitude. The maximum value, positive or negative, of an alternatingquantity, is known as its amplitude.

3. Frequency (f). The number of cycles/second is called the frequency of thealternating quantity.

Its unit is hertz (Hz).

4. Time Period (T). The time taken by an alternating quantity to completethe cycle is called its time period. For example, a 50 hertz (Hz) alternating current

has a time period of 150

second.

Time period is reciprocal of frequency,

i.e., T = 1 1f

fT

or =FHG

IKJ ...(1.12)

5. Root Mean Square (R.M.S.) Value. The r.m.s. (or effective) value of analternating current is given by that steady (D.C.) current which when flowingthrough a given circuit for a given time produces the same heat as produced by thealternating current when flowing through the same circuit for the same time.

R.M.S. value is the value which is taken for power purposes of any description.This value is obtained by finding the square root of the mean value of the squaredordinates for a cycle or half-cycle (See Fig. 1.26).


NOTES


This is the value which is used for all power, lighting and heating purposes, asin these cases the power is proportional to the square of the voltage.

Refer to Fig. 1.26.

The equation of sinusoidal alternating current is given as:

i = Imax sin θThe mean of squares of the instantaneous values of current over half cycle is

I2 = i d2

0 0θ

π

π

( )−z

I2 = 1 12

0

2

0πθ

πθ θ

π πi d I dmaxz z= ( sin )

= 1 1 2

22 2

0

2

0πθ θ

πθ θ

π πI d

Idmax

maxsincos

z z= −FHG

IKJ

= I

dImax max

2

0

2

021 2

22

2πθ θ

πθ θπ π

( cos )sin− = −z

= I Imax max

2 2

2 2ππ× = or I = I Imax max

2

2 2=

or I = 0.707 Imax. ...(1.13)Note. While solving problems, the values of given current and voltage should always

be taken as the r.m.s. values, unless indicated otherwise.

6. Average or Mean Value. The average value of an alternating current isexpressed by that steady current which transfers across any circuit the same chargeas is transferred by that alternating current during the same time.

The mean value is only of use in connection with processes where the resultsdepend on the current only, irrespective of the voltage, such as electroplating orbattery charging.

Refer to Fig. 1.27.

The value of instantaneous currentis given by

i = Imax sin θRefer to Fig. 1.27. The value of

instantaneous current is given by:

i = Imax sin θ [θ = ωt]

Iav = 1

0 0( )πθ

π

− z id

Limits are taken from 0 to π,since only first half cycle isconsidered. For whole cycle,the average value of sinewave is zero.

O

Cur

rent

+ Imax

Imax

Iav.

–

π 2π

Fig. 1.27

L

N

MMMMMM

O

Q

PPPPPP

NOTES



= 1 1

00π

θ θπ

θπ

π. . sin . cosI d Imax max= −z

= 1π

. Imax [1 – (– 1)] = 2π . Imax

or Iav = 0.637 Imax. ...(1.14)

Note. In case of unsymmetrical alternating current viz. half-wave rectified currentthe average value must always be taken over the whole cycle.

7. Form and Peak Factors

Form Factor. The ratio of r.m.s. (or effective) value to average value is theform factor (Kf ) of the waveform. It has use in voltage generation and instrumentcorrection factors.

Peak Factor. The ratio of maximum value to the r.m.s. value is the peakfactor (Kp) of the waveform.

Reasons for using alternating current (or voltage) of sinusoidal form:

An alternating current (or voltage) of sinusoidal form is normally used becauseof the following reasons:

(i) Mathematically, it is quite simple.

(ii) Its integrals and differentials both are sinusoidal.

(iii) It lends itself to vector representation.

(iv) A complex waveform can be analysed into a series of sine waves of variousfrequencies, and each such component can be dealt with separately.

(v) This waveform is desirable for power generation, transmission andutilisation.

8. Phase and Phase Angle. The‘phase’ of an A.C. wave may be definedas its position with respect to a referenceaxis or reference wave and ‘phase angle’as the angle of lead or lag with respect tothe reference axis or with respect to an-other wave.

Examples. The phase of current at

point L is T4

second where T is the time

period or expressed in terms of angle θ, it is π2

radian (Fig. 1.28). Similar phase of

the rotating coil at the instant shown in Fig. 1.29 is ωt which is therefore called itsphase angle.

The e.m.fs. induced in both the coils (Fig. 1.29) will be of the same frequencyand of sinusoidal shape, although the values of instantaneous e.m.f. induced will

Cur

rent

π/2 π 3 /2π 2π

T

t = 00

T4

L

Fig. 1.28


NOTES


be different. However, the alternating e.m.fs. would reach their maximum andzero values at the same time as shown in Fig. 1.29 (b). Such alternating voltages orcurves are said to in phase with each other.

Y

θ ω= t

XX′

Y′

M

(a)

O

L

L

M

t (Second)

(b)

Fig. 1.29

Refer to Fig. 1.30. M lags behind L by β and N lags behind L by (α + β) becausethey reach their maximum later.

wt

LM

Na

b

Y

Y

(a)

OX X

L M N

b( + )a b

t (Second)

(b)

Fig. 1.30

Example 1.5. (a) What is the equation of a 25 cycle current sine wave havingr.m.s. value of 30 amps?

(b) A 60 cycle engine-driven alternator has a speed of 1200 r.p.m. How manypoles are there in the alternator?

Solution. We know that,

i = Imax sin ωt

= Imax sin 2πft (∵ ω = 2πf )

= 30 × 2 . sin (2π × 25 × t) ∵ R.M.S. valueMax. value

=FHG

IKJ

1

2

= 42.42 sin 157 t. (Ans.)

NOTES


Basic Electrical andElectronics Engineering Using the relation, f =

Np120

where, f = frequency, N = speed in r.p.m., and

p = No. of poles

∴ 60 = 1200

120p

or p = 6. (Ans.)

Example 1.6. The graph in Fig. 1.31 shows the variation of voltage with time.Use the graph to calculate the average and r.m.s. value of the voltage. What is thefrequency of the voltage? What would be the r.m.s. value of sine wave having thesame peak value?


As the graph is symmetrical about time axis, considering only the positivehalf cycle.

Average value,

Vav = 0 10 20 40 100 120 100 40 20 10

10+ + + + + + + + +

= 46 V (Ans.)R.M.S. value,

V = 0 10 20 40 100 120 100 40 20 10

10

2 2 2 2 2 2 2 2 2 2+ + + + + + + + +

= 0 100 400 1600 10000 14400 10000 1600 400 100

10+ + + + + + + + +

= 3860010

3860= = 62.1 V (Ans.)

Fig. 1.31


NOTES


Since the time period ‘T ’ is 20 milli sec.

∴ Frequency ‘f ’ = 1T

= 1

20 10 3× − = 50 Hz (Ans.)

R.M.S. value of a sine wave of the same peak value

= 0.707 × 120 = 84.84 V (Ans.)

1.7 SINGLE PHASE CIRCUITS

The study of circuits involves three basic types of units (R, L, C i.e., resistance,reactance and capacitance respectively) and four possible series combination ofthem. The latter, in turn, may be arranged in many kinds of parallel, series-parallel,parallel-series or other complex circuits.

A.C. Through Pure Ohmic Resistance Alone

The circuit containing a pure resistance R is shown in Fig. 1.32 (a). Let theapplied voltage be given by the equation,

v = Vmax sin θ = Vmax sin ωt ...(i)

(b)

iv

v = V sin tmax wi = sin tImax w

(c)

iv pP

0

+ +

Powerwave

p2p

i = sin tImax wv = V sin tmax w

V2

max maxIV2

max maxI

Vmax maxIVmax maxI

t

Fig. 1.32. A.C. through pure ohmic resistance alone.

Then the instantaneous value of current flowing through the resistance Rwill be,

i = vR

V tR

max=sin ω

...(ii)

NOTES



The value of current will be maximum

when sin ωt = 1 or (ωt = 90°)

∴ Imax = V

Rmax

Substituting this value in eqn. (ii), we get

i = Imax sin ωt ...(iii)

Comparing (i) and (iii), we find that alternating voltage and current are inphase with each other as shown in Fig. 1.32 (b), also shown vectorially in Fig. 1.35 (c).

Power. Refer to Fig. 1.32 (c)

Instantaneous power,

p = vi = Vmax sin ωt × Imax sin ωt = Vmax Imax sin2 ωt

= V I

tV I

tmax max max max

22

21 22× = −sin ( cos )ω ω

= V I V I

tmax max max max

2 2 2 22. . cos− ω

(Constant part) (Fluctuating part)

For a complete cycle the average of V Imax max

2 2. cos 2 ωt is zero.

Hence, power for the whole cycle,

P = V I

V Imax maxr m s r m s

2 2. .. . . . .=

or P = VI watt

where V = R.M.S. value of applied voltage, and

I = R.M.S. value of the current.

It may be observed from the Fig. 1.32 (c) that no part of the power cycle at anytime becomes negative. In other words the power in a purely resistive circuit neverbecomes zero.

Hence in pure resistive circuit we have:

1. Current is in phase with the voltage.

2. Current I = VR

where I and V are r.m.s. values of current and voltage.

3. Power in the circuit, P = VI = I2R.

A.C. Through Pure Inductance Alone

Fig. 1.33 (a) shows the circuit containing a pure inductance of L henry.


NOTES


(a)v = V sin tmax ω

L

I

V

i

(b)

iv

v = V sin tmax ω

i = I (sin t – )max ω π2

1tπ

2

iv p

0

––

+

+v

Powerwave

i′p

(c)

π 2πt

V I2

max maxV I2

max max

Fig. 1.33 (a) (b), (c). A.C. through pure inductance alone. Resultant power zero.

Let the alternating voltage applied across the circuit be given by the equation,

v = Vmax sin ωt ...(i)

Whenever an alternating voltage is applied to a purely inductive coil, a backe.m.f. is produced due to the self-inductance of the coil. This back e.m.f. opposesthe rise or fall of the current through the coil. Since there is no ohmic drop in thiscase, therefore, the applied voltage has to overcome this induced e.m.f. only. Thusat every step,

v Ldidt

=

or Vmax sin ωt = Ldidt

or di = V

Lmax sin ωt dt

Integrating both sides, we get

diV

Lt dtmaxz z= sin ω

or i = V

Ltmax −F

HGIKJ

cos ωω

= V

Ltmax

ωω

πsin −L

NM

O

QP2

or i = VX

tmax

L

sin ωπ

−L

NM

O

QP2

...(ii)

NOTES



where XL = ωL (opposition offered to the flow of alternating current by a pureinductances) and is called Inductive reactance. It is given in ohm if L is in henryand ω is in radian/second.

The value of current will be maximum when sin ω πt −F

HGIKJ

=2

1

∴ Imax = VXmax

L


i = Imax sin ωπ

t −FHG

IKJ2

...(iii)


Instantaneous power, p = vi = Vmax sin ωt × Imax sin ωπ

t −FHG

IKJ2

= – Vmax Imax sin ωt . cos ωt

= −V Imax max

2 × 2 sin ωt cos ωt

= −V Imax max

2 2. . sin 2ωt

∴ Power for the whole cycle, P = − =zV I

tmax max

2 22 0

0

2sin ωπ

Hence average power consumed in a pureinductive circuit is zero.

Hence in a pure inductive circuit, we have:

1. Current I = VX

VL

VfLL

= =ω π2

amp.

2. Current always lags behind the voltage by90°.

3. Average power consumed is zero.

Variation of XL and f:

Since XL = ωL = 2πfL, and here if L is constant, then

XL ∝ f

Fig. 1.34, shows the variation. As frequency is increased XL increases and thecurrent taken by the circuit decreases.

A.C. Through Pure Capacitance Alone

The circuit containing a pure capacitor of capacitance C farad is shown inFig. 1.35 (a). Let the alternating voltage applied across the circuit be given by theequation,

Fig. 1.34. Variation ofXL with f.


NOTES


(b)

iv

v = V sin tmax ω

i = I sin ( t + )max ω π2

tπ2

Fig. 1.35. A.C. through pure capacitance alone. Resultant power is zero.

v = Vmax sin ωt ...(i)

Charge on the capacitor at any instant,

q = C v

Current through the circuit,

i = dqdt

ddt

= (CVmax sin ωt) = ω CVmax cos ωt

or i = VC

max

1/ω sin ω

πt +F

HGIKJ2

∴ i = VX

tmax

C

sin ωπ

+FHG

IKJ2

...(ii)

The denominator XC = 1

ωC (opposition offered to the flow of alternating current

by a pure capacitor) is known as capacitive reactance.

It is given in ohms if C is in farad and ω in radian/second.

The value of current will be maximum when sin ωπ

t +FHG

IKJ

=2

1

∴ Imax = VXmax

C

NOTES




i = Imax sin ωπ

t +FHG

IKJ2

...(iii)


Instantaneous power,

p = vi = Vmax sin ωt × Imax sin ωπ

t +FHG

IKJ2

= Vmax Imax sin ωt cos ωt = V Imax max

2 2. sin 2ωt

Power for the whole cycle = V I

tmax max

2 22 0

0

2. sin ω

π

z =

This fact is graphically illustrated in Fig. 1.35 (c). It may be noted that, duringthe first quarter cycle, what so ever power or energy is supplied by the source isstored in the electric field set-up between the capacitor plates. During the nextquarter cycle, the electric field collapses and the power or energy stored in the fieldis returned to the source. The process is repeated in each alternation and thiscircuit does not absorb any power.

Hence in a pure capacitive circuit, we have

1. I = VXC

= V × 2 πfC amps.

2. Current always leads the applied voltage by 90°.

3. Power consumed is zero.

Variation of XC and f:

Since XC = 1

2πfC and if C is kept constant, than

X fC ∝ 1

Fig. 1.36 shows the variation. As the frequency increases XC decreases, so thecurrent increases.

A.C. Series Circuits

Under this heading, we shall discuss R-L, R-C and R-L-C series circuits.

R-L Circuit (Resistance and Inductance in Series)

Fig. 1.37 (a) shows a pure resistance R and a pure inductive coil of inductanceL connected in series. Such a circuit is known as R-L circuit (usually met a cross inpractice).

Fig. 1.36


NOTES


(a) Circuit diagram

v = V sin tmax ω

I

I

90°

VL

IVR

VV

V = IRRV = IRR V = IXL LV = IXL L

R L

(c) Impedance triangle

A

Z

C

B

X = LL ω

R

φ

I

V

φ I cos φ

I sinφ

(e) Resolution of I

Fig. 1.37. R-L circuit (resistance and inductance in series).

Let V = R.M.S. value of the applied voltage,

I = R.M.S. value of the resultant current,

VR = IR = Voltage drop across R(in phase with I), and

VL = IXL = Voltage drop across L (coil), ahead of I by 90°.The voltage drop VR and VL and shown in voltage triangle OAB in Fig. 1.37 (b),

I being taken as the reference vector in the phasor diagram. Vector OA representsohmic drop VR and AB represents inductive drop VL. Vector OB represents theapplied voltage V which is the vector sum of the two (i.e., VR and VL).

∴ V = V V IR IXR L L2 2 2 2+ = +( ) ( ) = I ( )R XL

2 2+

or I = V

R X

VZ

L2 2+

=( )

where Z = R XL2 2+ (total opposition offered to the flow of alternating current by

R-L series circuit) is known as impedance of the circuit.

NOTES



As seen from the “impedance triangle” ABC [Fig. 1.37 (c)],Z2 = R2 + XL

2

i.e., (Impedance)2 = (Resistance)2 + (Inductive reactance)2

From Fig. 1.37 (b) it is evident that voltage V leads the current by an angleφ such that,

tan φ = VV

IXIR

XR

LR

L

R

L L= = =ω =

Inductive reactanceResistance

∴ φ = tan–1 XR

LFHG

IKJ

The same is illustrated graphically in Fig. 1.37 (d).In other words, I lags V by an angle φ.

Power factor, cos φ = RZ

[From Fig. 1.37 (c)]

Thus, if the applied voltage is given by v = Vmax sin ωt, then current equationis given as,

i = Imax sin (ωt – φ),

where Imax = V

Zmax

In the Fig. 1.37 (e), I has been shown resolved into two components, I cos φalong V and I sin φ in quadrature (i.e., perpendicular) with V.

Mean power consumed by the circuit= V × I cos φ (i.e., component of I which is in phase with V

i.e., P = V I cos φ (= r.m.s. voltage × r.m.s. current × cos φ)

The term ‘cos φ’ is called the power factor =FHG

IKJ

RZ

of the circuit

It may noted that:— In A.C. circuit the product of r.m.s. volts and r.m.s. amperes gives

volt-ampere (i.e., VA) and not true power in watt. True power (W)= volt-ampere (VA) × power factor

or Watt = VA (Apparent power) × cos φ— The power consumed is due to ohmic resistance only since pure inductance

consumes no power.

i.e., P = V I cos φ = V I × RZ

VZ

= IR = I × IR = I2R, watts

(∵ cos φ = R/Z and VZ

= I)

This shows that power is actually consumed in resistance only; the inductordoes not consume any power.

The power consumed in R-L circuit is shown graphically in Fig. 1.37 ( f ).

Thus in R-L circuit we have:

1. Impedance, Z = R XL2 2+ (where XL = ωL = 2π × f L)


NOTES

Electrical Circuits andMeasurements2. Current, I =

VZ

3. Power factor, cos φ = RZ

= =FHG

IKJ

True powerApparent power

WVA

[or angle of lag, φ = cos–1 (R/Z)]

4. Power consumed, P = V I cos φ = × × =FHG

IKJ

IZ IRZ

I R2

Symbolic Notation: Z = R + j XL

The numerical value of impedance vector = R XL2 2+

The phase angle with the reference axis, φ = tan–1 (XL/R).

In polar form: Z Z= ∠φ°.

Apparent, Active (True or real) and Reactive Power:Every circuit current has two components: (i) Active component and

(ii) Reactive component.“Active component” consumes power in the circuit while “reactive component”

is responsible for the field which lags or leads the main current from the voltage.In Fig. 1.38 active component is Iactive = I cos φ, and reactive component is

Ireactive = I sin φ

VI cos ff True power

Reactive

VI sin fApparentVI

Fig. 1.38. Active and reactive Fig. 1.39. Apparent, true and components of circuit current I. reactive power.

So, I = ( ) ( )I Iactive reactive2 2+

Refer to Fig. 1.39.(i) Apparent Power (S). It is given by the product of r.m.s. values of applied

voltage and circuit current.∴ S = V I = (I × Z) . I = I2Z volt-amperes (VA)

(ii) Active or True or Real Power (P or W). It is the power which is actuallydissipated in the circuit resistance.

P = I2R = V I cos φ watts

(iii) Reactive Power (Q). A pure inductor and a pure capacitor do not consumeany power, since in a half cycle what so ever power is received from the source bythese components the same is returned to the source. This power which flows back

NOTES



and forth (i.e., in both directions in the circuit) or reacts upon itself is called “reactivepower.”

It may be noted that the current in phase with the voltage produces active ortrue or real power while the current 90° out of phase with the voltage contributes toreactive power.

In a R-L circuit , reactive power which is the power developed in the inductivereactance of the circuit, is given as:

Q = I2 XL = I2Z sin φ = I . (IZ) sin φ= VI sin φ volt-ampere-reactive (VAR)

These three powers are shown in Fig. 1.39

Relation between VA, W and VR

W = VA cos φ ...(i)

VAR = VA sin φ ...(ii)

∴ VA = W

cos φ...[From (i)]

and, VA = VARsin φ

...[From (ii)]

Power factor (p.f.) = WVA

= True powerApparent power

The larger bigger units of apparent, true and reactive power are kVA (or MVA),kW(or MW) and kVAR (or MVAR) respectively.

The power factor depends on the reactive power component. If it is made equalto the active power component, the power factor becomes unity.

Example 1.7. A coil takes 2.5 amp. when connected across 200 volt 50 Hzmains. The power consumed by the coil is found to be 400 watts. Find the inductanceand the power factor of the coil.

Solution. Current taken by the coil, I = 2.5 A

Applied voltage, V = 200 volt

Power consumed, P = 400 W

We know that P = VI cos φ

or 400 = 200 × 2.5 × cos φ or cos φ = 400

200 25× . = 0.8

Hence power factor of coil is 0.8 (Ans.)

Impedance of the coil, Z = VI

=20025.

= 80 Ω

Also XZ

L = sin φ

∴ XL = Z sin φ

= 80 sin φ = 80 1 2− cos φ

= 80 1 0 82− . = 80 × 0.6 = 48 Ω


NOTES


But XL = 2πfL

∴ L = X

fL

248

2 50π π=

× = 0.1529 H (henry) (Ans.)

Example 1.8. A voltage v = 100 sin 314t – 50 cos 314t, is applied to a circuithaving R = 20 Ω in series with C = 100 µF. Obtain expression for instantaneouscurrent, r.m.s. value of current and the power in the circuit.

Solution. Given: v = 100 sin 314t – 50 cos 314t; R = 20 Ω; C = 100 µF.The R-C circuit and the phasor diagram for the given instantaneous voltage

are shown in Figs. 1.40 and 1.41 respectively.

Fig. 1.40 Fig. 1.41

Resultant voltage, Vmax(R) = ( ) ( )100 502 2+ = 111.8 V

Phase angle with the horizontal, θ = tan–1 −FHG

IKJ

50100 = – 26.56°

∴ v = 111.8 sin (314t – 26.56°)Now, ωt = 314 t or ω = 314

∴ Capacitive reactance, XC = 1 1

314 100 10 6ωC=

× × − = 31.85 Ω

Circuit impedance, Z = R XC2 2 2 220 31 85+ = +( ) ( . ) = 37.6 Ω

Maximum value of current, Imax = V

Zmax =

111 837 6

..

= 2.97 A

Phase difference between voltage and current,

φ = cos–1 RZFHGIKJ = cos–1

2037 6.FHG

IKJ = 57.86° (leading)

∴ Instantaneous value of currenti = 2.97 sin (314t – 26.56° + 57.86°) = 2.97 sin (314t + 31.3°)

or i = 2.97 [sin 314t . cos 31.3° + cos 314t . sin 31.3°]or i = 2.54 sin 314t + 1.54 cos 314t (Ans.)

RMS value of the current, I = Imax

2

2 97

2=

. = 2.1 A (Ans.)

Power in the circuit, P = VI cos φ

= Vmax R( )

2 × I cos φ =

111 8

2

. × 2.1 × cos (57.86°) = 88.32 W (Ans.)

NOTES



R-L-C Circuit (Resistance, Inductance and Capacitance in Series)

Fig. 1.42 shows a R-L-C circuit.

V = ZI (V – V )L C

IV = RR I

VC

VL

Z

R

(X – X )L Cf

Phasor diagram Impedance triangle(b) X > XL C

RI I

I

VR

VL

VC

L C

VL VCVR

= RII I

= XI L = XI C

V

(a) R-L-C Circuit

Z

R

(X – X )C L

fV = ZI

(V – V )C L

I

V = RR I

VC

VL

f

Phasor diagram Impedance triangle(c) X > XC L

Fig. 1.42. Resistance, inductance and capacitance in series.

Important Formulae:

1. Impedance, Z = R X XL C2 2+ −( )

where in and in faradX fL L henry XfC

CL C= =L

NM

O

QP2

12

ππ

, ,

2. Current, I = VZ

3. Power factor, cos φ = RZ

angle of lag when or lead when( ) ( ), cosX X X XRZL C C L> > =L

NM

O

QP

−φ 1

4. Power consumed = VI cos φ (= I2R)

Resonance in R-L-C Circuits

Refer to Fig. 1.42 (a).

Fig. 1.43. Reactance (X) Vs frequency (f). Fig. 1.44. Current in R-L-Ccircuit Vs frequency.


NOTES


The frequency of the voltage which gives the maximum value of the current inthe circuit is called resonant frequency, and the circuit is said to be resonant.

At resonance, XL = XC (i.e., Z = R)

i.e., 2 πfr L = 1

2π f Cr

fr = 1

2π LC...(1.15)

where fr = Resonance frequency in Hz; L = Inductance in henry; and C =Capacitance in farad.

Fig. 1.43 shows variation of XL, XC and X (total reactance = XL – XC) withvariation of frequency f.

Fig. 1.44 shows the variation of current (I) with frequency (f).

At series resonance, it is seen that:

1. Net reactance of the circuit is zero i.e., XL – XC = 0 or X = 0.

2. The impedance of the circuit is minimum and equal to the resistance (R) of

the circuit i e IVR

. ., =FHG

IKJ

. Consequently circuit admittance is maximum.

3. The current drawn is maximum (i.e., Ir = Imax).

4. The phase angle between the current and voltage is zero; the power factoris unity.

5. The resonant frequency is given by fr =1

2π LC; if the frequency is below the

resonant frequency the net reactance in the circuit is capacitive and if the frequencyis above the resonant frequency, the net reactance in the circuit is inductive.

6. Although VL = VC, yet Vcoil is greater than VC because of its resistance.

Example 1.9. A resistance 12 Ω, an inductance of 0.15 H and a capacitance of100 µF are connected in series across a 100 V, 50 Hz supply. Calculate :

(i) The current.

(ii) The phase difference between current and the supply voltage.

(iii) Power consumed.

Draw the vector diagram of supply voltage and the line current.

Solution. Given : R = 12 Ω, L = 0.15 H or XL = 2πfL

= 2π × 50 × 0.15 = 47.1 ΩC = 100 µF = 100 × 10–6F

or XC = 1

21

2 50 100 10 6π πfC=

× × × − = 31.8 Ω

NOTES



φV

V = RR I

( X – X )I IL C

IXC

I

R L C

100 V, 50 Hz.

12 Ω 0.15 H 100 Fµ

(a) R-L-C circuit (b) Vector/phasor diagramFig. 1.45

(i) The current, I: Z = R X XL C2 2+ −( )

= 12 47 1 31 82 2+ −( . . ) = 19.43 Ω

Current, I = VZ

= 10019 43.

= 5.15 A. (Ans.)

(ii) Phase difference, φ:

φ = cos–1 RZ

X XR

L Cor tan− −L

NM

O

QP

1

= cos–1 12

19 4315 312 0

1

...

or tan−L

NM

O

QP = 52° (lag)

Hence current lags supply voltage by 52°. (Ans.)

(iii) Power consumed, P:

P = VI cos φ = 100 × 5.15 × cos 52° = 371.1 W. (Ans.)

Fig. 1.45 (a), (b) show the circuit and vector/phasor diagrams respectively.

1.8 THREE PHASE BALANCED CIRCUIT

Generation, transmission and heavy-power utilisation of A.C. electricenergy almost invariably involve a type of system or circuit called apolyphase system or polyphase circuit. In such a system, each voltagesource consists of a group of voltages having relative magnitudes andphase angles. Thus, a m-phase system will employ voltage sources which,conventionally, consist of m voltages substantially equal in magnitudeand successively displaced by a phase angle of 360°/m.

A 3-phase system will employ voltage sources which, conventionally,consist of three voltages substantially equal in magnitude and displacedby phase angles of 120°. Because it possesses definite economic andoperating advantages, the 3-phase system is by far the most common, andconsequently emphasis is placed on 3-phase circuits.

NOTES



Advantages of Polyphase Systems

The advantages of polyphase systems over single-phase systems are:

1. A polyphase transmission line requires less conductor material than a single-phase line for transmitting the same amount of power at the same voltage.

2. For a given frame size a polyphase machine gives a higher output than asingle-phase machine. For example, output of a 3-phase motor is 1.5 times theoutput of single-phase motor of same size.

3. Polyphase motors have a uniform torque where most of the single-phasemotors have a pulsating torque.

4. Polyphase induction motors are self-starting and are more efficient. On theother hand, single-phase induction motors are not self-starting and are less efficient.

5. Per unit of output, the polyphase machine is very much cheaper.6. Power factor of a single-phase motor is lower than that of polyphase motor

of the same rating.7. Rotating field can be set up by passing polyphase current through stationary

coils.8. Parallel operation of polyphase alternators is simple as compared to that of

single-phase alternators because of pulsating reaction in single-phase alternator.It has been found that the above advantages are best realised in the case of

three-phase systems. Consequently, the electric power is generated and transmittedin the form of three-phase system.

1.9 GENERATION OF THREE-PHASE VOLTAGES

Let us consider an elementary 3-phase 2-pole generator as shown inFig. 1.46. On the armature are three coils, ll′, mm′, and nn′ whose axesare displaced 120° in space from each other.

Rotation

N

S

l

l¢

n¢

n m

m¢

Armaturewinding

Field windingexcited by directcurrent throughslip rings

Field pole producedby direct currentin field winding

Field structureor rotor

Armaturestructureor stator

Fig. 1.46. Elementary 3-phase 2-pole generator.

When the field is excited and rotated, voltages will be generated in thethree phases in accordance with Faraday’s law. If the field structure is so

NOTES



designed that the flux is distributed sinusoidally over the poles, the fluxlinking any phase will vary sinusoidally with time and sinusoidal voltageswill be induced in three-phases. These three waves will be displaced 120electrical degrees (Fig. 1.47) in time as a result of the phases beingdisplaced 120° in space. The corresponding phasor diagram is shown inFig. 1.48. The equations of the instantaneous values of the three voltages(given by Fig. 1.47) are:

030° 60° 90° 120°150°180°210°240°270°300°330°360°

Emax.Emax.ee

l l¢ em m¢ en n¢

q w= ( t)

Fig. 1.47. Voltage waves generated in windings of Fig. 1.46.

el′l = Emax. sin ωt

em′m = Emax. sin (ωt – 120°)

en′n = Emax. sin (ωt – 240°)

The sum of the above three e.m.fs. is always zero as shown below:

Resultant instantaneous e.m.f.

= el′l + em′m + en′n

= Emax. sin ωt + Emax. sin (ωt – 120°)

+ Emax. sin (ωt – 240°)

= Emax. [sin ωt + (sin ωt cos 120°

– cos ωt sin 120° + sin ωt cos 240°

– cos ωt sin 240°)]

= Emax. [sin ωt + (– sin ωt cos 60°

– cos ωt sin 60° – sin ωt cos 60°+ cos ωt sin 60°)]

= Emax. (sin ωt – 2 sin ωt cos 60°)

= Emax. (sin ωt – sin ωt) = 0.

1.10 PHASE SEQUENCE AND NUMBERING OF PHASES

By phase sequence is meant the order in which the three phases attaintheir peak or maximum.

120°

120°

120°

En

El

Em

Fig. 1.48. Phasor diagram ofgenerated voltages.

NOTES



In the generation of three-phase e.m.fs. in Fig. 1.47 clockwise rotation of thefield system in Fig. 1.49 was assumed. This assumption made the e.m.f. of phase‘m’ lag behind that of ‘l’ by 120° and in a similar way, made that of ‘n’ lag behindthat of ‘m’ by 120° (or that of l by 240°). Hence, the order in which the e.m.fs. ofphase l, m and n attain their maximum value is lmn. It is called the phase order orphase sequence l → m → n. If now the rotation of field structure of Fig. 1.46 isreversed i.e., made counter-clockwise, then the order in which three phases wouldattain their corresponding maximum voltages would also be reversed. The phasesequence would become l → n → m. This means that e.m.f. of phase ‘n’ would nowlag behind that of phase ‘l’ by 120° instead of 240° as in the previous case.

The phase sequence of the voltages applied to a load, in general, is determinedby the order in which the 3-phase lines are connected. The phase sequence can bereversed by interchanging any pair of lines. (In the case of an induction motor,reversal of sequence results in the reversed direction of motor rotation).

The three-phases may be numbered l, m, n or 1, 2, 3 or they may be giventhree colours (as is customary).

The colours used commercially are red, yellow (or sometimes white) and blue.In this case sequence is RYB.

Evidently in any three-phase system, there are two possible sequences, inwhich three coils or phase voltages may pass through their maximum value i.e.,red → yellow → blue (RYB) or red → blue → yellow (RBY).

By convention:

RYB ...... taken as positive.

RBY ...... taken as negative.

1.11 INTER-CONNECTION OF THREE PHASES

Each coil of three phases has two terminals [one ‘start’ (S) and another ‘finish’(F)] and if individual phase is connected to a separate load circuit, as shown inFig. 1.49, we get a non-interlinked 3-phase system. In such a system each circuitwill require two conductors, therefore, 6 conductors in all. This makes the whole

ER Load

F

S

EY Load

F

S

EB Load

F

S

S = StartF = Finish

Fig. 1.49. Non-interlinked 3-phase system.

NOTES



system complicated and expensive. Hence the three phases are generallyinterconnected which results in substantial saving of copper.

The general method of inter-connections are:

1. Star or Wye (Y) connection.

2. Mesh or delta (∆) connection.

Star or Wye (Y) Connection

In this method of inter-connection the similar ends either the ‘start’ or‘finish’ are joined together at point N. This common point N [Fig. 1.50 (a)]is called star point or neutral point. Ordinarily only three wires are carriedto the external circuit giving 3-phase, 3-wire start connected system butsometimes a fourth-wire, known as neutral wire is carried to the neutralpoint of the external load circuit giving 3-phase, 4 wire star connectedsystem.

IR

N

IYIB

ERER

EYEY

ERYERY

EBREBR

EYBEYBEBEB

Y

B

Neutral wire

R

E = E = E = EE = E = E = E

(a)

R Y B ph

RY YB BR LI = I

E = EL ph

L ph3

(b)

EBIB

f

f

f ER

IY

EY

IR120°

120°

120°

Fig. 1.50. Star-connected three-phase network.

NOTES



The voltage between any line and the neutral point (i.e., voltage across thephase winding) is called the ‘phase voltage’ (Eph); while the voltageavailable between any pair of terminals (or outers) is called the ‘linevoltage’ (EL).

In star connection, as is evident in Fig. 1.50 (a) there are two-phasewindings between each pair of terminals, but since their similar endshave been joined together, they are in opposition. Obviously, theinstantaneous value of potential difference between any two terminals isthe arithmetic difference of the two-phase e.m.fs. concerned. However,the r.m.s. value of this potential difference is given by the vector differenceof the two-phase e.m.fs.

Fig. 1.50 (b) shows the vector diagram for phase voltages and currents ina star connection where a balanced system has been assumed. [A balancedsystem is one in which (i) the voltages in all phases are equal in magnitudeand differ in phase from one another by equal angles, in this case, theangle = 360/3 = 120°, (ii) the currents in the three phases are equal inmagnitude and also differ in phase from one another by equal angles. Athree-phase balanced load is that in which the loads connected across thethree-phases are identical]. Thus, we have

ER = EY = EB = Eph (phase e.m.f.)

Line voltage, ERY(= EL) = Vector difference of ER and EY

= ER – EY

Line voltage, EYB = EY – EB

Line voltage, EBR = EB – ER.

(a) Relation between Line Voltages and Phase Voltages. Refer to Fig. 1.51.

30°30°

120°120°

120°120°12

0°12

0°

60°60°

EBEBR

– ER

–EY E (= E )RY L

E (= E )R ph

EY–EB

EYB

Rotation ofvectors

Fig. 1.51. Vector diagram for star connected network.

NOTES



The potential difference between outers R any Y is

ERY = ER – EY [vector difference]

or ERY = ER + (– EY) [vector sum]

Hence, ERY is found by compounding ER and EY reversed and its value isgiven by the diagonal of the parallelogram (Fig. 1.52). Obviously the angle betweenER and EY reversed is 60° and the value of

ERY (or EL) = E E E ER Y R Y2 2 2 60+ + °cos

= E E E Eph ph ph ph2 2 1

22+ + × × = 3Eph

Similarly, EYB(= EL) = EY – EB = 3 Eph

and EBR(= EL) = EB – ER = 3Eph

i.e., ERY = EYB = EBR = EL = 3 Eph

Hence, EL = 3 Eph ...(1.16)

(i.e., Line voltage = 3 phase voltage)

(b) Relation between Line Currents and Phase Currents. Since in star-connected system each line conductor is connected to separate phase, so the currentflowing through the line and phase are same.

Current in outer (or line) R = IR

Current in outer Y = IY

Current in outer B = IBSince IR = IY = IB = say, Iph—the phase current∴ Line current, IL = Iph ...(1.17)(c) Power. If the phase current has a phase difference of φ with phase voltage,Power per phase = EphIph cos φTotal power (true), P = 3 × power per phase

P = 3 × EphIph cos φ ...(1.18)

Now Eph = EL

3 and Iph = IL

Hence in terms of line values, the above expression becomes

P = 3 × EL

3 IL cos φ or P = 3ELIL cos φ ...(1.19)

(Apparent power = 3 ELIL).

In a balanced star-connected net work the following points are worthnoting:

(i) Line voltages are 3 times the phase voltages.

(ii) Line currents are equal to phase currents.

(iii) Line voltages are 120° apart.

(iv) Line voltages are 30° ahead of the respective phase voltages.

(v) The angle between line currents and the corresponding line voltages is(30° ± φ), +ve for lagging currents, –ve for leading currents.

NOTES


Electrical Circuits andMeasurements(vi) True power = 3 ELIL cos φ, where φ is the angle between respective phase

current and phase voltage, not between the line current and line voltage.

(vii) Apparent power = 3 ELIL.

(viii) In balance system, the potential of the neutral or star point is zero.

∴ Potential at neutral (or star) point =ENR + ENY + ENB = 0.

Example 1.10. A balanced star connected load of (8 + j6) Ω/phase is connectedto a 3-phase, 230 volts, 50 Hz supply. Find the current, p.f., power, volt ampere andreactive power. Draw the phasor diagram for the above circuit.

8 W

8 W8 W

6 W

6 W6 W

E = 230 VLO

f = 36.87°

Eph

Iph

(a) (b)

Fig. 1.52

Solution. Given: R = 8 Ω; XL = 6 Ω; EL = 230 volt, f = 50 Hz.

The circuit is shown in Fig. 1.52(a),

Phase Voltage, Eph = EL

3230

3= = 132.8 V

Impedence, Z = R XL2 2 2 28 6+ = + = 10 Ω

Current, Iph = IL = E

Zph = 132.8

10 = 13.28 A. (Ans.)

Power factor, cos φ = RZ

= 810

= 0.8. (Ans.)

(∴ φ = cos–1 (0.8) = 36.87°)

Power, P = 3 EL IL cos φ

= 3 × 230 × 13.28 × 0.8 = 4342.3 W. (Ans.)

Apparent power = 3ELIL = 3 × 230 × 13.28 = 5290.4 VA. (Ans.)

Reactive power = 3 ELIL sin φ = 3 × 230 × 13.28 × sin (36.87°)

= 3174 VAR. (Ans.)

The phasor diagram is shown in Fig. 1.52(b).

NOTES



Example 1.11. Three identical coils are connected in star to a 400 V (linevoltage), 3-phase A.C. supply and each coil takes 300 W. If the power factor is 0.8(lagging). Calculate:

(i) The line current; (ii) Impedance; and(iii) Resistance and inductance of each coil.Solution. Line voltage, EL = 400 V

Power taken by each coil, Pph = 300 W

Power factor, cos φ = 0.8 (lagging)

IL; Z; Rph; Lph

Phase voltage, Eph = EL

3400

3= V

Also Pph = EphIph cos φ

300 = 400

3 × Iph × 0.8

∴ Iph = 300 3400 0 8

×× .

= 1.62 A

(i) Line current, IL = phase current, Iph

∴ IL = 1.62 A. (Ans.)

(ii) Coil impedance, Zph = E

Iph

ph

=

4003

1.62 = 142.5 Ω

∴ Zph = 142.5 Ω (Ans.)

(iii) Rph = Zph cos φ = 142.5 × 0.8 = 114 ΩCoil reactance, Xph = Zph sin φ = 142.5 × 0.6 = 85.5 Ω (Ans.)

But Xph = 2πfLph

∴ Lph = X

fph

285 5

2 50π π=

×.

= 0.272 H

Hence, Rph = 114 Ω. (Ans.) and Lph = 0.272 H (Ans.)

Delta (∆∆∆∆∆) or Mesh Connection

In a delta or mesh connection the dissimilar ends of the three-phase windingsare joined together i.e., the ‘starting’ end of one phase is joined to the ‘finishing’end of the other phase and so on as shown in Fig. 1.53. In other words, the threewindings are joined in series to form a closed mesh. Three leads are taken out fromthe junctions as shown and outward directions are taken as positive.

(a) Relation between Line Voltages and Phase Voltages

Since in delta or mesh connected system, only one phase is included betweenany pair of line outers, therefore, potential difference between the line outers, calledthe line voltage, is equal to phase voltage.

i.e., Line voltage, EL = phase voltage, Eph.

NOTES



(b) Relation between Line Currents and Phase Currents

From Fig. 1.53 it is obvious that line current is the vector difference of phase

currents of two phases concerned.I

(I)

RB

ph

ERYERY EBREBR

EYBEYB

R

I(I

)

YRph

I (I )BY ph

I (= I )R L

I (= I )Y L

I (= I )B L

E = E = E = E = E

I = I = I = I = ( 3 I )RY YB BR ph L

R Y B L ph

Fig. 1.53. Delta or mesh connected diagram.

Thus, line current,

IR = IYR – IRB (Vector difference)

= IYR + (– IRB) (Vector sum)

Similarly,

IY = IBY – IYR and IB = IRB – IBY

Refer to Fig. 1.54. Since phase angle between phase current IYR and – IRB is 60°.

∴ IR = I I I IYR RB YR RB2 2 2 60+ + °cos

30°

30°

30°

30°

120°

120°120°

120°

120°

60°

60°

IY

–IYR

IRB

–IBY

IB

IBY –IRB

IYR

IR

Fig. 1.54. Vector diagram for delta connected network.

Assuming the delta connected system or network be balanced, the phasecurrent in each winding is equal and let each be equal to Iph (i.e., IYR = IBY = IRB = Iph).

∴ IR (= IL) = I I I Iph ph ph ph2 2 2 60+ + °cos = 2 22 2 1

2I Iph ph+ × = 3 Iph

NOTES


Basic Electrical andElectronics Engineering Similarly, IY = IB = 3 Iph

Hence, IL = 3 Iph ...(1.20)

(i.e., line current = 3 phase current).(c) Power

Power/phase = EphIph cos φTotal power (true) P = 3EphIph cos φ

But Eph = EL and Iph = IL

3Hence, in terms of line values, the above expression for power becomes

P = 3 × EL × EL

3 cos φ or P = 3ELIL cos φ ...(1.21)

where φ = the phase power factor angle.

(Apparent power = 3 ELIL)

In case of delta or mesh connected system the following points are worth-noting:

(i) Line voltages are equal to phase voltages.

(ii) Line currents are 3 times phase currents.

(iii) Line currents are 120° apart.

(iv) Line currents are 30° behind their respective phase currents.

(v) The angle between line currents and corresponding line voltages is (30° ± φ)as in the star system.

(vi) True power = 3 ELIL cos φ, where φ is the phase angle between respectivephase current and phase voltage.

(vii) Apparent power = 3 ELIL.

(viii) In balanced system, the resultant e.m.f. in the closed circuit will be zero.

i.e., ERY + EYB + EBR = 0.

Hence, there will no circulating current in the mesh if no-load is connected tothe lines.

Example 1.12. A delta-connected balanced 3-phase load is supplied from a3-phase, 400 V supply. The line current is 30 A and the power taken by the load is12 kW. Find:

(i) Impedance in each branch; and(ii) The line current, power factor and power consumed if the same load is

connected in star.Solution. Delta Connection:

Eph = EL = 400 V

IL = 30 A

∴ Iph = IL

330

3= = 17.32 A.

NOTES



(i) Impedance per Phase

Zph = E

Iph

ph

= 40017 32.

= 23.09 ΩΩΩΩΩ. (Ans.)

Now P = 3ELIL cos φ

12000 = 3 × 400 × 30 × cos φ

or cos φ (Power factor) = 12000

3 400 30× × = 0.577

(ii) Star Connection

Eph = EL

3400

3= = 231 V

IL = Iph = E

Zph

ph

= 23123 09.

= 10 A. (Ans.)

Power factor, cos φ = 0.577 (Since impedance is same)

Power consumed = 3ELIL cos φ

= 3 × 400 × 10 × 0.577 = 3997.6 W. (Ans.)Example 1.13. Three 50 ohm non-inductive resistances are connected in (i)

star, (ii) delta across a 400 V, 50 Hz, 3-phase mains. Calculate the power takenfrom the supply system in each case. In the event of one of the three resistancesgetting opened, what would be the value of the total power taken from the mains ineach of the two cases.

Solution. Star Connection:

Phase voltage, Eph = EL

3400

3= = 231 V

Phase current, Iph = E

Rph

ph

= 23150

= 4.62 A

Power consumed, P = 3Iph2 Rph = 3 × 4.622 × 50 = 3200 W. (Ans.)

[or P = 3ELIL cos φ = 3 × 400 × 4.62 × 1 = 3200 W]Delta Connection:Phase voltage, Eph = EL = 400 V

Phase current, Iph = E

Rph

ph

= 40050

= 8 A

Power consumed, P = 3Iph2 Rph = 3 × 82 × 50 = 9600 W. (Ans.)

When one of the resistances is disconnected:(i) Star Connection. Refer to Fig. 1.55.

When one of the resistances is disconnected, the circuit is no longer 3-phasebut converted into single-phase circuit, having two resistances each of 50 ohmconnected in series across supply of 400 V.

NOTES



Hence line current,

IL = ER

L

ph2400

2 50=

× = 4 A

400 V400 V

400 V400 V

400 V400 V

50 W

4 A

50 W Removed

Removed400 V400 V

400 V400 V

400 V400 V

50 W

8 A

Rem

oved

Rem

oved

50W

Fig. 1.55 Fig. 1.56

Power consumed,

P = 42 (50 + 50) = 1600 W. (Ans.)

[or P = VI cos φ = 400 × 4 × 1 = 1600 W].

(ii) Delta Connection. Refer to Fig. 1.56.Potential difference across each resistance, EL = 400 V

Current in each resistance = 40050

= 8 A

Power consumed in both resistances

= 2 × 82 × 50 = 6400 W. (Ans.)

[or P = 2 × Eph Iph cos φ = 2 × 400 × 8 × 1 = 6400 W].

1.12 ELECTRICAL PRINCIPLES OF OPERATION

All electrical measuring instruments depend for their action on any of manyphysical effects of electric current or potential. The following are the effects generallyused in the manufacture :

(i) Magnetic effect. ...... Voltmeters, ammeters, wattmeters, power factormeters etc.

(ii) Thermal effect. ...... Ammeters, voltmeters, maximum demand metersetc.

(iii) Chemical effect. ...... D.C. ampere hour meters (integrating meters).(iv) Electrostatic effect. ...... Voltmeters which can indirectly be used as

ammeters and wattmeters.(v) Electro-magnetic induction effect. ...... Voltmeters, ammeters,

wattmeters and integrating meters used in A.C. only.


NOTES

Electrical Circuits andMeasurements1.13 ELECTRICAL INDICATING INSTRUMENTS

Almost invariably an indicating instrument is fitted with a pointer whichindicates on a scale the value of the quantity being measured. The moving systemof such an instrument is usually carried by a spindle of hardened steel, having itsends tapered and highly polished to form pivots which rest in hollow-groundbearings, usually of saphine, set in steel screws. In some instruments, the movingsystem is attached to thin ribbons of spring material such as beryllium-copperalloy, held taut by tension springs mounted on the frame of movement. Thisarrangement eliminates pivot friction and the instrument is less susceptible todamage by shock or vibration.

Essential Features

Indicating instruments possess three essential features.

(i) Deflecting device. ...... whereby a mechanical force is produced by theelectric current, voltage or power.

(ii) Controlling device. ...... whereby the value of deflection is dependentupon the magnitude of the quantity being measured.

(iii) Damping device. ...... to prevent oscillation of the moving system andenable the latter to reach its final position quickly.

Deflecting Device

A deflecting device produces a deflecting torque which is caused by anyone ofthe previously mentioned effects (i.e., thermal effect, chemical effect, electrostaticeffect etc.); with the help of this deflecting torque the needle or the pointer movesfrom zero position to the final position. The arrangement of the deflecting devicewith each type of instrument will be discussed individually.

Controlling Devices

There are two types of controlling devices:

(i) Spring control

(ii) Gravity control.

(i) Spring Control. Fig. 1.57 shows a commonly used spring controlarrangement. It utilises two spiral hair springs, 1 and 2, the inner ends of whichare attached to the spindle S. The outer end of spring 2 is fixed while that of 1 isattached to a lever, the adjustment of which gives zero adjustments. The two springs1 and 2 are wound in opposite directions so that when the moving system is deflected,one spring winds up while the other unwinds, and the controlling torque is due tothe combined torsions of the springs.


NOTES


JB

B

P

1

S

2

P = PointerB = Balance weightS = SpindleJB = Jewelled bearing1,2 = Springs

Fig. 1.57. Spring control.

Since the torsional torque of a spiral spring is proportional to the angle oftwist, the controlling torque (Tc) is directly proportional to the angular deflection ofthe pointer (θ).

i.e., Tc ∝ θ.

The spring material should have the following properties:

• It should be non-magnetic.• It must be of low temperature co-efficient.• It should have low specific resistance.• It should not be subjected to fatigue.(ii) Gravity Control. With gra-

vity control, weights L and M areattached to the spindle S [Fig. 1.58 (a)],the function of L being to balance theweight of the pointer P. Weight Mtherefore provides the controllingtorque. When the pointer is at zero, Mhangs vertically downwards. When P isdeflected through angle θ, thecontrolling torque is equal to (weight ofM × distance d) and is thereforeproportional to the sine of the angulardeflection [Fig. 1.58 (b)],i.e., Tc ∝ sin θ.

The degree of control is adjusted by screwing the weightup or down the carrying system.

It may be seen from Fig. 1.58 (b) that as θ approaches90°, the distance 1-2 increases by a relatively small amountfor a given change in the angle that when θ is just increasingfrom its zero value. Hence gravity-controlled instrumentshave scales which are not uniform but are cramped orcrowded at their lower ends.

0

1

P

S

L

M

dd

(a)

P = PointerS = SpindleL = Balance weightM = Control weight

= Angle of deflection

Fig. 1.58. Gravitycontrol.

W

W sin W

cos 2

1

(b)


NOTES


Advantages:

1. The gravity controlled instrument is cheaper than corresponding spring-controlled instrument.

2. It is not subjected to fatigue.

3. It is unaffected by temperature.

Disadvantages:

1. Gravity control gives a cramped scale.

2. The instrument must be levelled before use.

1.14 MOVING-IRON INSTRUMENTS (AMMETERS AND

VOLTMETERS)

Moving-iron instruments are commonly used in laboratories and switch boardsat commercial frequencies because they are very cheap and can be manufacturedwith required accuracy.

Moving-iron instruments can be divided into two types:

1. Attraction type ...... in which a sheet of soft iron is attracted towards asolenoid.

2. Repulsion type ...... in which two parallel rods or strips of soft iron,magnetised inside a solenoid, are regarded as repelling each other.

Attraction Type

Fig. 1.59 shows the sectional front and an end elevation of the attracted-irontype instrument. It consists of a solenoid (or coil) C and oval shaped soft-iron discD in such a way that it can move in or out of the solenoid. To this iron a pointer Pis attached so that it may deflect along with the moving iron over a graduatedscale. The soft-iron disc is made of sheet metal specially shaped to give a scale asnearly uniform as possible.

When the current to be measured (or a definite fraction of the current to bemeasured or proportional to the voltage to be measured) is passed through thesolenoid, a magnetic field is set up inside the solenoid, which in turn magnetisesthe iron. Thus the soft-iron disc is attracted into the solenoid/coil, causing thespindle and the pointer to rotate. Damping is provided by vane V attached to thespindle and moving in an air chamber, and control is by hair spring.

P

VJB JB

S

D

C

DS

Section of currentcarrying coil

Direction of forceon the discC

S = SpindleP = PointerD = Soft iron discC = SolenoidJB = Jewelled bearingV = Vane

Fig. 1.59. Attraction-type moving-iron instrument.


NOTES


Repulsion Type

Repulsion-type moving-iron instrument is shown in Fig. 1.60. Here there aretwo irons, one fixed (A) and the other mounted on a short arm fixed (B) to theinstrument spindle. The two irons lie in the magnetic field due to a solenoid/coil C.When there is no current in the coil the two iron pieces (moving one and fixed one)are almost touching each other and the pointer rests on zero position. When

S

B

AC

P

B AS

JB

C

VPS

JB

A = Fixed ironB = Moving ironC = Solenoid / CoilJB = Jewelled bearing

S = SpringV = VaneP = Pointer

Fig. 1.60. Repulsion-type moving-iron instrument.

the current to be measured (or a definite fraction of it or proportional to the voltageto be measured) is passed through the solenoid, a magnetic field is set up insidethe solenoid and the two iron pieces are magentised in the same direction.This sets up a repulsive force so moving iron piece, is repelled by fixed iron piece,thereby results in the motion of the moving iron piece, carrying the pointer. Thepointer comes to rest in a deflected position when equilibrium is attained betweenthe repulsive forces of the working elements and the controlling force.

Such instruments are commonly provided with spring control and air frictiondamping.

In commercial instruments, it is usual for the moving-iron B to be in the formof a thin curved plate and for the fixed iron A to be a tapered curved sheet. Thisconstruction can be arranged to give a longer and more uniform scale than is possiblewith the rods shown in Fig. 1.60.

Deflecting Torque in Moving-iron Instruments

In both the attraction and repulsion type moving-iron instruments it is foundthat for a given position of the moving system, the value of the deflecting torque isproportional to the square of the current, so long as the iron is working belowsaturation. Hence, if the current waveform is as shown in Fig. 1.61, the variationof the deflecting torque is represented by the dotted wave. If the supply frequencyis 50 Hz, the torque varies between zero and a maximum 100 times a second, so


NOTES


that the moving system (due to its inertia) takes up a position corresponding to themean torque, where

Meantorque

0

Current

Time

Deflectingtorque

Fig. 1.61. Deflecting torque in a moving-iron instrument.

mean torque ∝ mean value of the square of the current

= kI2

where k = a constant for a given instrument

and I = r.m.s. value of the current.

Thus the moving-iron instrument can be used to measure both direct currentand alternating current, and in the latter case the instrument gives the r.m.s.value of the current.

Owing to the deflecting torque being proportional to the square of the current,the scale divisions are not uniform, being cramped at the beginning and open at theupper end of the scale.

Note. For both types of instruments (attraction-type and repulsion type) the necessarymagnetic field is produced by the ampere-turns of a current-carrying coil.

• In case the instrument is to be used as an ammeter, the coil hascomparatively few turns of thick wire so that the ammeter has lowresistance because it is connected in series with the circuit.

• In case it is to be used as a voltmeter, the coil has high impedance so as todraw as small a current as possible since it is connected in parallel withthe circuit. As current through the coil is small it has large number ofturns in order to produce sufficient ampere-turns.

Advantages and Disadvantages of Moving-Iron Instruments

Advantages:

1. Can be used both in D.C. as well as in A.C. circuits.

2. Robust and simple in construction.

3. Possess high operating torque.

4. Can withstand overload momentarily.

5. Since the stationary parts and the moving parts of the instrument are simpleso they are cheapest.


NOTES


6. Suitable for low frequency and high power circuits.

7. Capable of giving an accuracy within limits of both precision and industrialgrades.

Disadvantages:

1. Scales not uniform.

2. For low voltage range the power consumption is higher.

3. The errors are caused due to hysteresis in the iron of the operating systemand due to stray magnetic field.

4. In case of A.C. measurements, change in frequency causes serious error.

5. With the increase in temperature the stiffness of the spring decreases.

Sources of Errors

A. Errors with both D.C. and A.C.(i) Errors due to Hysteresis. This source of error is due to hysteresis in the

soft iron moving part, due to which too high values are recorded by the instrument,when the current is increasing and too low readings are liable to be indicated whenthe current is decreasing.

(ii) Errors due to Stray Fields. External stray magnetic fields are liable toaffect adversely the accurate functioning of the instrument. Magnetic shielding ofthe working parts is obtained by using a covering case of cast iron.

B. Errors with A.C. only

Errors may be caused due to change in frequency because change in frequencyproduces (i) change in impedance of the coil and (ii) change in magnitude of eddycurrents. The error due to the former is negligible in ammeters, as the coil currentis determined by the external circuit and the error due to the latter can normallybe made small.

1.15 MOVING-COIL INSTRUMENTS

The moving-coil instruments are of the following two types:

1. Permanent-magnet type......can be used for D.C. only.

2. Dynamometer type......can be used both for A.C. and D.C.

Permanent-Magnet Moving-Coil Type (PMMC) Instruments

A permanent-magnet moving-coil type instrument works on the principle that‘‘when a current-carrying conductor is placed in a magnetic field, it is acted upon bya force which tends to move it to one side and out of the field’’.

Construction

The instrument consists of a permanent magnet M and a rectangular coilC which consists of insulated copper wire wound on light aluminium framefitted with polished steel pivots resting in jewel bearings. The magnet is


NOTES


made of Alnico and has soft-iron pole-pieces PP which are bored outcylindrically.

The rectangular coil C is free to move in air gaps between the soft-ironpole pieces and a soft-iron cylinder A (central core), supported by a brassplate (not shown).

AA CPP

M

SN

D

X X

P P

BC

B

M = Permanent magnetPP = Soft iron pole pieces

A = Soft iron cylinder(central core)

C = Rectangular coilB = Spiral springsD = PointerA

Fig. 1.62. Permanent-magnet moving-coil instrument.

The functions of the central core A are:

(i) To intensify the magnetic field by reducing the length of air gap acrosswhich the magnetic flux has to pass.

(ii) To give a radial magnetic flux of uniform density, thereby enabling thescale to be uniformly divided.

• The movement of the coil is controlled by two phosphor bronze hair springsBB (one above and one below), which additionally serve the purpose ofleading the current in and out of the coil. The two springs are spiralled inopposite directions for neutralizing the effects of changes in temperature.

• The aluminium frame not only provides support for the coil but alsoprovides damping by eddy currents induced in it.

Deflecting Torque. Refer to Fig. 1.63. When current is passed through thecoil, forces are set up on its both sides which produce deflection torque. If I ampereis the current passing through the coil, the magnitude of the force (F) experiencedby each of its sides is given by

F = BIl newton

where B = flux density in WB/m2, and

l = length or depth of coil in metre.


NOTES


N SF

F

tt

Fig. 1.63

For N turns, the force on each side of the coil is

= NBIl newton

∴ Deflecting torque (Td)

= force × perpendicular distance

= NBIl × b = NBI (l × b) = NBIA N-m

where b = breadth of the coil in metres, and

A = face area of the coil.

If B is constant, then

Td ∝ I (i.e., current passing through the coil)

= kI where k is a constant for a given instrument.

Since such instruments are invariably spring controlled, the controlling torque(Tc) of the spiral springs ∝ angular deflection

i.e., Tc ∝ θor Tc = c θwhere c = a constant for given springs, and

θ = angular deflection.

For a steady deflection,

Controlling torque (Tc) = deflecting torque (Td)

Hence cθ = kI

∴ θ = kc

I

i.e., the deflection is proportional to the current and the scale is therefore uniformlydivided.

Advantages and Disadvantages. The moving-coil permanent-magnet typeinstruments have the following advantages and disadvantages:

Advantages:(i) Low power consumption.

(ii) Their scales are uniform.(iii) No hysteresis loss.(iv) High torque/weight ratio.(v) They have very effective and efficient eddy-current damping.

(vi) Range can be extended with shunts or multipliers.(vii) No effect of stray magnetic field as intense polarised or unidirectional

field is employed.


NOTES


Disadvantages:(i) Somewhat costlier as compared to moving-iron instruments.

(ii) Cannot be used for A.C. measurements.(iii) Friction and temperature might introduce errors as in case of other

instruments.(iv) Some errors are set in due to the ageing of control springs and the

permanent magnets.Ranges:

D.C. Ammeters

(i) Without shunt......0/5 micro-amperes up to 0/30 micro-amperes.

(ii) With internal shunts......upto 0/2000 amperes.

(iii) With external shunts......upto 0/5000 amperes.

D.C. Voltmeters

(i) Without series resistance......0/100 milli-volts.(ii) With series resistance......upto 20000 or 30000 volts.Moving-coil permanent-magnet instruments can be used as:(i) Ammeters......by using a low resistance shunt.

(ii) Voltmeters......by using a high series resistance.(iii) Flux-meters......by eliminating the control springs.(iv) Basllistic galvanometers......by making control springs of large moment of

inertia.Extension of Range

The following devices may be used for extending the range of instruments:(i) Shunts (ii) Multipliers (iii) Current transformers and (iv) Potential transformers.Use of ammeter shunts and voltage multipliers is discussed below.

(i) Ammeter Shunts. An ammeter shuntis merely a low resistance that is placed inparallel with the coil circuit of the instrumentin order to measure fairly large currents. Thegreater part of the current in the main circuitis then diverted around the coil through theshunt. The connection diagram for a shunt andmilli-ammeter for measuring large currents isshown in Fig. 1.64.

The shunt is provided with four terminals, the milliammeter being connectedacross the potential terminals. If the instrument were connected across the currentterminals, there might be considerable error due to the contact resistance at theseterminals being appreciable compared with the resistance of the shunt.

Let I = current of the circuit to be measured,

Im = current passing through the ammeter,

Is = current passing through the shunt,

Rs = resistance of the shunt, and

Rm = resistance of the ammeter (plus its leads to the potential terminals).

Fig. 1.64. Use of a shunt formeasuring large currents

Ammeter

+ –Rm

Im

IsIShunt

Rs


NOTES


Thus I = Im + Is ... (i)

Also, since the voltage drop across the shunt and the instrument in same,

∴ Is Rs = Im Rm

or Rs = IIm

s

Rm

Substituting Is = I – Im from equation (i), we get

Rs = I

I Im

m( )− . Rm

or Rs = R

II

m

m

−FHG

IKJ

1...(ii)

The ratio of I

Im

is known as ‘multiplying power’ of the shunt. Denoting, the

ratio by N, we have shunt resistance,

Rs = R

Nm

− 1...(iii)

The shunts are made of a material such as manganin (copper, manganeseand nickel), which has a negligible temperature coefficient of resistance. The material,is employed in the form of thin strips, the ends of which are soldered to two largecopper blocks. Each copper block carries two terminals-one current terminal andother potential terminal. The strips which form the shunt are spaced from eachother to promote a good circulation of air and thus efficient cooling.

Note. A ‘swamping’ resistor r, of material having negligible temperatureco-efficient of resistance, is connected in series with the instrument (moving coil). Thelatter is wound with copper wire and the function of r is to reduce the error due tovariation of resistance of the instrument with variation of temperature.

(ii) Voltmeter Multipliers. The range of the instrument, when used as avoltmeter can be extended or multiplied by using a high non-inductive seriesresistance R connected in series with it as shown in Fig. 1.65.

+

–Rv

R

VV Non-inductiveresistance

Load

Fig. 1.65

Let I = full scale deflection current of voltmeter,

V = voltage of the circuit to be measured,


NOTES


Rv = resistance of the voltmeter, and

R = external series resistance.

Now, voltage across supply leads

= voltage drop across the voltmeter + voltage drop across external resistance

∴ V = IRv + IR

or R = V IR

Iv−

= VI

– Rv

The voltage multipliers to be used for D.C. measurements should satisfy thefollowing requirements:

(i) The resistance should not change with time of usage.(ii) The temperature co-efficient of resistance must be very low.Note. The frequency error introduced by the inductance of the instrument coil can be

compensated by shunting R by a capacitor.

Example 1.14. A milliammeter of 2.5 ohms resistance reads upto 100 milliamperes. What resistance is necessary to enable it to be used as:

(i) A voltmeter reading upto 10 V.(ii) An ammeter reading upto 10 A.Draw the connection diagram in each case.

Solution. Resistance of the milli-ammeter, Rm = 2.5 ΩMaximum current of the milli-ammeter, Im = 100 mA = 0.1 A.

(i) Voltage to be measured, V = 10 voltResistance to be connected in series.

R = VIm

– Rm = 100 1.

– 2.5 = 97.5 Ω (Ans.)

Connection diagram is shown in Fig. 1.66.

(ii) Current to be measured, I = 10 A

R

10 V10 V Load

0.25 V0.25 V

9.75 V9.75 V

I = 0.1 Am

MilliammeterR =2.5

m

Fig. 1.66

Multiplying power of the shunt, N = I

Im

= 100 1.

= 100

∴ Resistance to be connected in parallel,

Rs = R

Nm

−=

−12 5

100 1.

= 0.02525 Ω. (Ans.)


NOTES


SupplySupply

Load

+

–

R = 2.5m

I =9.9 A

s

I =

0.1 A

mI = 10 A

Rs

Milliammeter

10 A

Fig. 1.67

Connection diagram is shown in Fig. 1.67.

Example 1.15. If the moving coil of a voltmeter consists of 100 turns woundon a square former which has a length of 30 mm and the flux density in the air gapis 0.09 Wb/m2, calculate the turning moment on the coil when it is carrying a currentof 10 mA.

Solution. Number of turns, N = 100

Length of each side, l = 30 mm = 0.03 m

Flux density, B = 0.09 Wb/m2

Current through the coil, I = 10 mA = 0.01 A

We know that the force on each side of the coil,

F = NBIl newton

∴ Turning moment (i.e., deflecting torque),

T = F × breadth = F × l = NBIl2 N-m

= 100 × 0.09 × 0.01 × (0.03)2 N-m

= 8.1 × 10–5 N-m. (Ans.)

Electrodynamic or Dynamometer Instruments

In an electrodynamic instrument the operating field is produced by anotherfixed coil and not by permanent magnet. This instrument can be used as an ammeteror as voltmeter but is generally used as a wattmeter.

Refer to Fig. 1.68 (a), (b).

These instruments essentially consist of fine wire moving coil placed in themagnetic field produced by another fixed coil when carrying currents. The coils areusually air cored to avoid hysteresis, eddy currents and other errors when theinstrument is used on A.C. The fixed coil FC is divided into two halves placed closetogether and parallel to each other in order to provide a fairly uniform field withinthe range of the movement of the moving coil.


NOTES


P

AFC FC

MC

SXX

A

FC FC

MCS

FC = Field coils (divided into two halves)MC = Moving coils

S = SpindleA = Spiral hair springs

(a)

I1FC FC

MC

I2

I = Current in FC

I = Current in MC

(b)

1

2

Fig. 1.68. Electrodynamic or dynamometer instrument.

The upper diagram in Fig. 1.68 (a) shows a sectional elevation through fixedcoil FC and the lower diagram represents a sectional plan on XX. The moving coilMC is carried by a spindle S and the controlling torque is exerted by spiral hairsprings A, which may also serve to lead the current into and out of MC.

Deflecting Torque. The deflecting torque is due to interaction of the magneticfields produced by currents in the fixed and moving coils.

Fig. 1.69 (a) shows the magnetic field due to current flowing through FC(I1) in the direction indicated by the dots and cross.

Fig. 1.69 (b) shows the magnetic field due to current (I2) in MC. Fig. 1.69 (c) shows the combined effect of the above magnetic fields. By

combining these magnetic fields it will be seen that when currents(I1 and I2) flow simultaneously through FC and MC, the resultant magneticfield is distorted and effect is to exert a clockwise torque on MC.

Since MC is carrying current (I2) at right angles to the magnetic field producedby FC,

deflecting torque, Td ∝ I1 × I2

or Td = KI1I2 where K is a constant.

Since the instrument is spring-controlled, the restoring of control torque (Tc)is proportional to the angular deflection θ.


NOTES


+ + + +FC FC

MC

(a)

+ +

FC FC

MC

(b)

FC

MC

(c)

Fig. 1.69. Magnetic fields due to fixed and moving coils.

∴ Tc ∝ θ or Tc = K′θThe two torques (Td and Tc) are equal and opposite in the final deflected

position.

∴ Td = Tc

or KI1I2 = K′ θor θ ∝ I1I2

Use of the Instrument as an Ammeter. When the instrument is used asan ammeter then same current passes through both moving coil (MC) and fixed

coils (FC) as shown in Fig. 1.70. In this case, I1 = I2 = I, hence θ ∝ I2 or I ∝ θ . Theconnections of Fig. 1.70. are used when small currents are to be measured.

FCMCFCI I

Supply Load

Fig. 1.70. Measurement of small currents.

In the case of heavy currents, a shunt is used to limit current through themoving coil as shown in Fig. 1.71.

FCMCFCI I

Supply LoadShunt

Fig. 1.71. Measurement of heavy currents.


NOTES


Use of the Instrument as Voltmeter. When the instrument is used as avoltmeter, the fixed and moving coils are used in series along with a high resistanceas shown in Fig. 1.72.

SupplySupply LoadLoad

FC

FC

MC

R

Fig. 1.72. Use of the instrument as a voltmeter

Here again I1 = I2 = I,

where I = VR

...... in D.C. circuits

and I = VZ

...... in A.C. circuits

∴ θ ∝ V × V or θ × V2

or V = θ

Thus whether the instrument is used as an ammeter or voltmeter its scale isuneven through the whole of its range and is cramped or crowded near the zero inparticular.

Note. When the dynamometer instrument is used to measure an alternating currentor voltage, the moving coil-due to its inertia-takes up a position where the average deflectingtorque over one cycle is balanced by the restoring torque of the spiral springs. For thatposition, the deflecting torque is proportional to the mean value of the square of current orvoltage, and the instrument scale can therefore be calibrated to read the r.m.s. value.

In these instruments the damping is pneumatic (i.e., air damping). Eddycurrent damping is admissible owing to weak operating field.

Ranges:

Ammeters. (i) With fixed and moving coils in series......0/0.01 A–0/0.05 A(ii) With moving coil shunted or parallel connections......upto 0/30 A.Voltmeters. Upto 0–750 volt.

Advantages and Disadvantages

Advantages:

(i) Can be used on both D.C. as well as A.C. systems.(ii) They are free from hysteresis and eddy current errors.

(iii) It is possible to construct ammeters upto 10 A and voltmeters upto 600 Vwith precision grade accuracy.


NOTES


Disadvantages:(i) Since torque/weight ratio is small, such instruments have low sensitivity.

(ii) The scale is not uniform because θ ∝ I .(iii) Cost of these instruments is higher in comparison to those of moving iron

instruments. So, these are only used as voltmeters and ammeters for precisionmeasurements.

(iv) Higher friction losses.

1.16 WATTMETERS

A wattmeter is a combination of an ammeter and a voltmeter and, thereforeconsists of two coils known as current coil and pressure coil. The operating torqueis produced due to interaction of fluxes on account of currents in current and pressurecoils.

There are following three types of wattmeters:

1. Dynamometer wattmeter

2. Induction wattmeter

3. Electrostatic wattmeter.

We shall discuss here only the first and second type.

Dynamometer Wattmeter

In Fig. 1.73, the dynamometer is connected as a wattmeter. This is one of theadvantages of this type of meter. If the coils are connected so that a value of currentproportional to the load voltage flows in one, and a value of current proportional tothe load current flows in the other, the meter may be calibrated directly in watt.This is true because the indication depends upon the product of the two magnetic

F.C.

V

F.C.

M.C.

Voltagecoil Series

resistor

Watts

P

0

2

4 6 8

10

To Load

F.C. = Fixed coils(current coils)

M.C. = Moving coil(voltage coil)

P = Pointer

Fig. 1.73. Connection of dynamometer for measuring power.


NOTES


fields. The strength of the magnetic fields depends upon the values of currents flowingthrough the coils. If one current is proportional to load voltage and other current isthe load current, then the meter can be calibrated in terms of watts or true powerconsumed by the load.

Let v = supply voltage,

i = load current, and

R = resistance of the moving coil circuit.

Current through fixed coils, if = i.

Current through the moving coil, im = vR

Deflecting torque, Td ∝ if × im ∝ ivR

.

— For a D.C. circuit the deflecting torque is thus proportional to the power.— For any circuit with fluctuating torque, the instantaneous torque is

proportional instantaneous power. In this case due to inertia of movingparts the deflection will be proportional to the average torque i.e., thedeflection will be proportional to the average power. For sinusoidalalternating quantities the average power is VI cos φ, where

V = r.m.s. value of voltage, I = r.m.s. value of current, and φ = phase angle between V and I.

Hence an electrodynamic instrument, when connected as shown in Fig. 1.73,indicates the power, irrespective of the fact it is conected in an A.C. or D.C. circuit.

Scales of such wattmeters are more or less uniform because the deflectionis proportional to the average power and for spring control, controllingtorque is proportional to the deflection, hence θ ∝ power. Damping ispneumatic.

Errors

The error may creep in due to the inductance of the moving or voltage coil.However, the high non-inductive resistance connected in series with coilswamps, to a great extent, the phasing effect of the voltage coil inductance.

There may be error in the indicated power due to the following:(i) Some voltage drop in the current circuit.

(ii) The current taken by the voltage coil.

This error, however, in standard wattmeters may be overcome by having anadditional compensating winding connected in series with the voltage coil but is soplaced that it produces a field in opposite direction to that of the fixed or currentcoils.

Ranges(i) Current Circuit. 0.25 to 100 A without employing current transformers.

(ii) Potential Circuit. 5 to 750 V without employing potential transformers.


NOTES


Advantages:(i) The scale of the instrument is uniform (because deflecting torque is pro-

portional to true power in both the cases i.e., D.C. and A.C. and the instrument isspring controlled.)

(ii) High degree of accuracy can be obtained by careful design, hence theseare used for calibration purposes.

Disadvantages:(i) The error due to the inductance of pressure coil at low power factor is very

serious (unless special features are incorporated to reduce its effect).(ii) Stray field may effect the reading of the instrument. To reduce it, magnetic

shielding is provided by enclosing the instrument in an iron case.

Induction Wattmeter

Induction wattmeters can be used on A.C. circuit only (in contrast withdynamometer wattmeters can be used both on D.C. and A.C. circuits) and areuseful only when the frequency and supply voltage are constant.

The operation of all induction instruments depends on the production of torquedue to reaction between a flux φ1 (whose magnitude depends on the current or voltageto be measured) and eddy currents induced in a metal disc or drum by another fluxφ2 (whose magnitude also depends on the current or voltage to be measured). Sincethe magnitude of eddy currents also depends on the flux producing them, theinstantaneous value of the deflecting torque is proportional to the square of thecurrent or voltage under measurement and the value of mean deflecting torque isproportional to the mean square of the current or voltage.

Fig. 1.74 shows an induction wattmeter. It has two laminated electromagnetsone is excited by the current in the main circuit—its exciting winding being joinedin series with the circuit, hence it is also called series magnet.

The other electromagnet is excited by current which is proportional to thevoltage of the circuit. Its exciting coil is joined in parallel with the circuit, hencethis magnet is sometimes referred to as shunt magnet.

I2

I1

SupplySupply LoadLoad

Series orcurrent magnet

Aluminiumdisc

Coppershadingbands

Shunt or voltagemagnet

Fig. 1.74. Induction wattmeter.


NOTES


A thin aluminium disc is mounted in such a way that it cuts the fluxes of boththe magnets. Hence two eddy currents are produced in the disc. The deflectiontorque is produced due to the interaction of these eddy currents and the inducingfluxes. Two or three copper rings are fitted on the central limb of the shunt magnetand can be so adjusted as to make the resultant flux in the shunt magnet lag behindthe applied voltage by a suitable angle.

This instrument is spring controlled, the spring being fitted to the spindle ofthe moving system which also carries a pointer. The scale is uniformly even andextends over 300°.

Such wattmeters can handle current upto 100 A. For handling greater currentsthey are used in conjunction with current transformers.

Advantages:(i) Fairly long scale (extending over 300°).

(ii) Free from the effects of stray fields.(iii) Good damping.(iv) Practically free from frequency errors.Disadvantages:

Sometimes subjected to serious temperature errors.

1.17 INTEGRATING METERS (ENERGY METERS)

Integrating or energy meters are used to measure the quantity of electric energysupplied to a circuit in a given time. They give no direct indication of power i.e., asto the rate at which energy is being supplied because their registrations areindependent of the rate at which at given quantity of electric energy is beingconsumed.

The main difference between an energy meter and a wattmeter is that theformer is fitted with some type of registration mechanism whereby all theinstantaneous readings of power are summed over a definite period of time whereasthe latter indicates the value at a particular instant where it is read.

Essential Characteristics of Energy Meters

The essential characteristics of energy meters are given below:

1. They must be simple in design and must not contain any parts which mayrapidly deteriorate.

2. The readings may be given directly by the dials and must avoid anymultiplying factors.

3. The casing of the meter should be dust, water and insect proof.

4. Permanency of calibration is a prime requisite and to attain it, the frictionat the pivots etc., and retarding torque of the magnetic brakes must remain constant.The magnets should be so placed that they are not affected in their strength by themagnetic field of the current coil.


NOTES


5. The friction losses should be minimum (being unavoidable) and must remainpractically constant over long periods of time. This requires that the moving partsshould be light, and the jewels and pivots should be of best quality and kept ingood order.

6. There should be minimum possible friction in the counter device. The torqueof the meter should be high so that the unavoidable irregularities in friction maynot cause inaccuracies.

7. The energy meter should maintain its accuracy under reasonably varyingconditions of voltage and load.

8. The energy loss in the meter itself must be small.

Types of Energy Meters

Energy meters are generally of the following three types:

1. Electrolytic meters

2. Motor meters

3. Clock meters.

Here only motor meters will be discussed.

Motor Meters

The motor meters can be used in D.C. as well as in A.C. circuits. In principlethe motor meter is a small motor of D.C. or A.C. type whose instantaneous speed ofrotation is proportional to the circuit current in case of an ampere-hour meter andto the power of the circuit in case of a watt-hour meter.

The following are the essential parts of the motor meters:

(i) An Operating Torque System. It produces a torque and causes the movingsystem to rotate continuously.

(ii) A Braking Device. It is usually a permanent magnet, known as brakemagnet. This brake magnet induces current in some part of the moving systemwhich in turn produces the braking torque. Thus the braking torque is proportionalto the induced currents whereas the induced currents are proportional to the speedof the moving system (and hence the braking torque is proportional to the speed ofthe moving system (disc). When the braking torque is equal to the driving torquethe moving system attains a steady speed.

(iii) Revolution Registering Device. This device is obtained by having aworm cut on the spindle of the instrument. The worm engages with a pinion andthus drives the train of wheels and registers ampere-hours and watt-hours directly.

Types of Motor Meters. The various types of motor meters are:

(i) Mercury motor meters(ii) Commutator motor meters

(iii) Induction motor/energy meters.• Mercury motor meters and commutator meters are used on D.C. circuits

• Induction meters are used on A.C. circuit.


NOTES


Motor-driven Meter—Watt-hour Meter

The motor-driven meter shown in Fig. 1.75 can be used on direct or alternatingcurrent. It contains a small motor and an aluminium retarding disc. The fieldwinding is connected in series with the load, and the field strength is proportionalto the load current.

The armature is connected across the source, and the current in the armatureis proportional to the source or line voltage. The torque produced in the armature isproportional to the power consumed by the load.

The armature shaft drives a series of counters that are calibrated in watt-hours.The power that is used can be read directly from the dials.

The aluminium disc attached to the armature is used to control the armaturespeed. The disc turns in a magnetic field produced by the permanent magnets, andthe retarding force increases as the rotation increases and stops when the discstops. The retarding force is produced by the aluminium conductor cutting throughthe lines of force of the permanent magnets. This is a form of magnetic damping.

To counter

Armature

Aluminiumdisc Load

Permanent magnets

Powersource

Fieldcoils

Seriesresistor

Fig. 1.75. Motor-driven meter (watt-hour meter)designedto operate on direct or alternating current.

The meter must overcome the friction of the bearings and indicators at verylight loads. A portion of the field is produced by the armature current (coil in serieswith the armature winding). This coil is (called as compensating coil) wound to aidthe field and is adjusted to the point where it just overcomes the meter friction.

Induction Type Watt-hour Meter

This is the most commonly used meter on A.C. circuits for measurement ofenergy.

Advantages:(i) Simple in operation

(ii) High torque/weight ratio(iii) Cheap in cost(iv) Correct registration even at very low power factor(v) Unaffected by temperature variations


NOTES


(vi) More accurate than commutator type energy meter on light loads (owingto absence of a commutator with its accompanying friction).

Induction Type Single Phase Energy Meters

These are, by far, the most common form of A.C. meters met with in every-day domestic and industrial installations. These meters measure electric energyin kWh.

The principle of these meters is practically the same as that of induction wattmeters. Instead of the control spring and the pointer of the watt-meter, the watt-hour meter, (energy meter) employs a brake magnet and a counter attached to thespindle. Just like other watt-hour meters, the eddy currents induced in thealuminium disc by the brake magnet due to the revolution of the disc, are utilisedto control the continuously rotating disc.

Construction. The construction of a typical meter of this type is shown inFig. 1.76. The brake-magnet and recording wheel-train being omitted for clearances.It consists of the following:

(i) Series magnet M1(ii) Shunt magnet M2

(iii) Brake magnet(iv) A rotating disc.The series electromagnet M1 consists of a number of U-shaped iron laminations

assembled together to form a core, wound with a few turns of a heavy gauge wire.

LineLine

LoadLoad

I2 M2

PC

C

CC

M1

D

I1M = Series or current magnet

M = Shunt or voltage magnet

D = DiscC = Copper shading bands

CC = Current coilPC = Pressure coil

1

2

Fig. 1.76. Induction type single phase energy meter.


NOTES


This wound coil is known as current coil and is connected in one of the lines and inseries with the load to be metered. The series electromagnet is energized and setsup a magnetic field cutting through the rotating disc, when load current flowsthrough the current coil C.C. The rotating disc is an aluminum disc mounted on avertical spindle and supported on a sapphire cup contained in a bottom screw. Thebottom pivot, which is usually removable, is of hardened steel, and the end, whichis hemispherical in shape, rests in the sapphire cup. The top pivot (not shown)merely serves to maintain the spindle in a vertical position under working conditionand does not support any weight or exert any appreciable thrust in any direction.

The shunt magnet M2 consists of a number of M shaped iron laminationsassembled together to form a core. A core having large number of turns of fine wireis fitted on the middle limb of the shunt magnet, this coil is known as pressure coilP.C. and is connected across the supply mains.

The brake magnet consists of C shaped piece of alloy steel bent round to forma complete magnetic circuit, with the exception of a narrow gap between the poles.This magnet is mounted so that the disc revolves in the air gap between the polarextremities. The movement of the rotating disc through the magnetic field crossingthe air gap sets up eddy currents in the disc which react with the field and exerts abraking effect. The speed of the rotating disc may be adjusted by changing theposition of the brake magnet or by diverting some of the flux there from.

Working. The shunt electromagnet produces a magnetic field which is ofpulsating character; it cuts through the rotating disc and induces eddy currentsthere in, but normally does not in itself produce any driving force. Similarly serieselectromagnet induces eddy currents in the rotating disc, but does not in itselfproduce any driving force. In order to obtain driving force in this type of meter,phase displacement of 90° between the magnetic field set up by shunt electromagnetand applied voltage V is achieved by adjustment of copper shading band C (alsoknown as power factor compensator or compensating loop). The reaction betweenthese magnetic fields and eddy currents set up a driving torque in the disc.

Sources of Errors. The various sources of errors in an induction-type energymeter are given below:

(i) Incorrect magnitude of the fluxes. These may arise from abnormal voltagesand load currents.

(ii) Incorrect phase relation of fluxes. These may arise from defective lagging,abnormal frequencies, changes in the iron losses etc.

(iii) Unsymmetrical magnetic structure. The disc may go on rotating while nocurrent is being drawn but pressure coils alone are excited.

(iv) Changes in the resistance of the disc. It may occur due to changes intemperature.

(v) Changes in the strength of the drag magnets. It may be due to temperatureor ageing.

(vi) Phase-angle errors due to lowering of power factor.

(vii) Abnormal friction of moving parts.


NOTES


(viii) Badly distorted waveform.

(ix) Changes in the retarding torque due to the disc moving through the fieldof the current coils.

Example 1.16. A 5 A, 230 V meter on full load unity power factor test makes60 revolutions in 360 seconds. If the normal disc speed is 520 revolutions per kWh,what is the percentage error?

Solution. Energy consumed in 360 seconds

= VI tcos φ ×

×3600 1000 kWh

= 230 5 1 360

3600 1000× × ×

× = 0.115 kWh

where, t is in seconds.

Energy recorded by the meter

= 60

520 = 0.11538 kWh

∴ % age error = 0 11538 0 115

0 115. .

.−

× 100 = 0.33% (fast). (Ans.)

Example 1.17. The constant of a 230 V, 50 Hz, single phase energy meter is185 revolutions per kWh. The meter takes 190 seconds for 10 revolutions whilesupplying a non-inductive load of 4.5 A at normal voltage. What is the percentageerror of the instrument?

Solution. Energy consumed in 190 seconds

= VI cos φ

1000 × t =

230 4 5 11000× ×.

× 1903600

= 0.0546 kWh

[cos φ = 1, since load supplied is non-inductive]

Energy registered by the meter = 10185

= 0.054 kWh

∴ % age error = 0 054 0 0546

0 054. .

( . )−

= 0.06% (slow). (Ans.)

SUMMARY

1. Electricity may be defined as a form of energy. It involves making andusing energy.

2. The controlled movement of electrons (or drift) through a substance is calledcurrent. Current is the rate at which electrons move.

3. Electromotive force (e.m.f.) is the force that causes a current of electricity toflow. The volt is a unit of potential difference and electromotive force.


NOTES


4. Ohm’s law may be stated as :

‘‘For a fixed metal conductor, the temperature and other conditions remainingconstant, the current (I) through it is proportional to the potential difference(V) between its ends’’.

5. A linear resistor is one which obeys ohm’s law. Such elements in which theV/I (volt-ampere) plots are not straight lines but curves are called non-linearresistors or non-linear elements.

6. In a series combination : R = R1 + R2 + R3

where R1, R2 and R3 are the resistance in series and R is the equivalentresistance of the combination.

7. In a parallel combination :1 1 1 1

1 2 3R R R R= + +

where R1, R2 and R3 are the resistances connected in parallel and R is theequivalent resistance of the combination.

8. Kirchhoff’s law :

First law : Σ Currents entering = Σ Currents leaving

Second law : Σ potential rises = Σ potential drops.

9. Applications of Kirchhoff’s laws include the following :

(i) Branch-current method

(ii) Maxwell’s loop (or mesh) current method

(iii) Nodal voltage method

10. Modern alternators produce an e.m.f. which is for all practical purposessinusoidal (i.e. a sine curve), the equation between the e.m.f. and time being

e = Emax sin ωtwhere e = instantaneous voltage ; Emax = maximum voltage;

ωt = angle through which the armature has turned from neutral.

11. The r.m.s. value of an alternating current is given by that steady (D.C.)current which when flowing through a given circuit produces the same heatas is produced by the alternating current when flowing through the samecircuit for the same time.

Irms = 0.707 Imax.12. The average or mean value of an alternating current is expressed by that

steady current which transfers across any circuit the same charge as istransferred by that alternating current during the same time.

Iav = 0.637 Imax

Form factor is the ratio of r.m.s. value to average value of the wave form.

Peak factor is the ratio of maximum value to the r.m.s. value of the waveform.


NOTES


13. In a star-connected system,

Eph = EL

3

Iph = IL , P = 3 ELIL cos φ.14. In a single-phase as well as in a three-phase system, the kVA is directly

proportional to the current I. The disadvantage of a lower power factor isthat the current required for a given power is very high, which fact leads tomany undesirable results. The power factor may be improved by thefollowing :

— Static capacitors

— Phase advancers

— Synchronuous capacitors

— Capacitors boosters

— High power factor motors.

15. Essential features of indicating instruments are:

(i) Deflecting device

(ii) Controlling device

(iii) Damping device.

16. Moving iron instruments are of the following two types:

(i) Attraction type

(ii) Repulsion type.

17. Moving-coil instruments are of the following two types:

(i) Permanent magnet type

(ii) Dynamometer type.

GLOSSARY

• Resistance: The property of the electric current which oppose the flow ofcurrent is called resistance, denoted by (R).

• Conductance: Reciprocal of resistance is called conductance, denoted by(G).

• Alternating current: The current or voltage which alternate its directionand magnitude every time.

• Amplitude: The maximum value, positive or negative of an alternatingquantity.

• Frequency: The number of cycle per second of an alternating quantitydenoted by (f).

• Time period: The time taken by an alternating quantity to complete thecycle. It is denoted by (T).

• Reactive power (Q): The power which flow back and forth (i.e., in bothdirections in the circuit) or reacts upon itself is called reactive power.


NOTES


• Apparent power (S): It is the product of r.m.s. values of applied voltageand current circuit.

S = V.I. = (I × z) = I2z volt-amp.• Wattmeter: It is a combination of an ammeter and a voltmeter, consist of

two coils known as current coil and pressure coil. The operating torque isproduced due to interaction of fluxes on account of currents in both coil.

REVIEW QUESTIONS

1. Define the term ‘electricity’.

2. Write a short-note on ‘Electron theory’ ?

3. What is an electric current ?

4. What is the difference between electromotive force and potential difference ?

5. What do you mean by the term ‘resistance’ ? On what factors does it depend ?

6. Define the terms ‘conductance’ and ‘conductivity’.

7. What is temperature co-efficient of resistance ?

8. Derive the relation α2 = 1

1

12 1α

+ −( )t t where α1 and α2 are the temperature

co-efficients of resistance at temperatures t1 and t2°C respectively.

9. State and explain Ohm’s law.

10. What is the difference between linear and non-linear resistors ?

11. Derive an expression for the equivalent resistance in the following cases :

(a) When the resistances are connected in series.

(b) When the resistances are connected in parallel.

12. How does a conductor differ from an insulator ?

13. Derive an expression for insulation resistance of a cable.

14. Define the following terms :

Circuit, Electrical network, Active network, Node and Branch.

15. What are the limitations of Ohm’s law ?

16. State and explain Kirchhoff’s laws.

17. Discuss briefly application of Kirchhoff’s laws.

18. Calculate the resistance of a manganin wire having the following particulars :

Length of the wire = 100 m

Uniform cross-sectional area = 0.1 mm2

Resistivity of the material = 50 × 10–8 Ω m. [Ans. 500 Ω]


NOTES


19. Calculate the area of cross-section and diameter of a copper wire to have aresistance of 0.13 ohm per kilometre ; given that resistivity for copper is 1.7× 10–8 Ω-m. [Ans. 12.9 mm]

20. Find the area of cross-section of a cable of 1 km length to transmit 500 amperesso that total drop in voltage along the cable may not exceed 25 volts.Assume : ρ = 1.7 × 10–8 Ω-m. [Ans. 3.4 cm2]

[Hint. R = VI

=25

500 = 0.05 Ω . ]

21. Find the resistance between the points A and B in the series-parallel net-work shown in Fig. (A).

60 Ω

12 Ω

M NL

100 Ω

S B10 Ω

180 Ω

90 Ω

20 Ω

A

Fig. (A)

22. A resistance of 10 ohms is connected in series with a combination of tworesistances arranged in parallel each of value 20 ohms. Determine theresistance R3 which should be shunted across the parallel combination sothat current drawn by the circuit is 1.5 A with applied voltage of 20 V.

[Ans. 5 ohms]23. A series parallel resistor circuit is shown in Fig. (B). Determine

(i) Equivalent resistance across the battery terminals.(ii) Current supplied by the battery.

(iii) Current through 4 Ω resistor.(iv) Equivalent resistance of the circuit with an open circuit at point L.(v) Equivalent resistance of the circuit with an open circuit at point M.

(vi) Equivalent resistance of the circuit with points N and P short circuited.(vii) Equivalent resistance of the circuit with points L and N short circuited.

[Ans. (i) 12 Ω (ii) 2 A (iii) 1 A (iv) Infinite (v) 12.66 Ω (vi) 8 Ω (vii) 10 Ω]

4 Ω

8 Ω

8 Ω

6 Ω 4 ΩP

L NM

24 V

Fig. (B)


NOTES


24. Find the resistance of the circuit shown in Fig. (C). [Ans. 4.2 Ω]

2 Ω

4 Ω

4 Ω4 Ω

6 Ω

8.4 Ω

5 Ω

Fig. (C)

25. State the maximum power transfer theorem and explain its importance.

26. Define the following terms as applied to an alternating current :

Cycle, frequency, time period, amplitude.

27. What do you mean by the term “Phase difference” ?

28. Explain the following terms relating alternating current :

(i) R.M.S. value (ii) Average value

(iii) Form factor (iv) Peak factor.

29. Explain briefly the following as applied to A.C. series and parallel circuits :

(i) Resonance frequency (ii) Q-factor.

30. An alternating current of frequency 60 Hz has a maximum value of 120 A.Write down the equation for its instantaneous value. Reckoning time fromthe instant the current is zero and is becoming positive, find :

(i) The instantaneous value after 1360

second ;

(ii) The time taken to reach 96 A for the first time.

[Ans. 103.9 A, 0.00245 second]

31. An alternating current of frequency 50 Hz has a maximum value of 100 A.Calculate :

(i) Its value 1/600 second after the instant the current is zero and its valuedecreasing thereafter wards.

(ii) How many seconds after the instant the current is zero (increasingthereafter wards) will the current attain the value of 86.6 A ?

[Ans. – 50 A, 1/300 s]

32. Calculate the r.m.s. value, the form factor of a periodic voltage having thefollowing values for equal time intervals changing suddenly from one valueto the next : 0, 5, 10, 20, 50, 60, 50, 20, 10, 5, 0, 5, 10 V etc. What would be ther.m.s. value of sine wave having the same peak value ?

[Ans. 31 V ; 23 V ; 1.35 ; (app.) ; 42.2 V]


NOTES


33. A sinusoidally varying alternating current has an average value of 127.4 A.When its value is zero, then its rate of change is 62,800 A/s. Find the analyticalexpression for the sine wave. [Ans. i = 200 sin 100 πt]

34. A resistance, a capacitor and a variable inductance are connected in seriesacross a 200 V, 50 Hz supply. The maximum current which can be obtainedby varying the inductance is 314 mA and the voltage across the capacitor isthen 300 V. Calculate the capacitance of the capacitor and the values of theinductance and resistance. [Ans. 3.33 µF, 3.04 H, 637 Ω]

35. A circuit consisting of a coil of resistance 12 Ω and inductance 0.15 H inseries with a capacitor of 12 µF is connected to a variable frequency supplywhich has a constant voltage of 24 V.

Calculate : (i) The resonant frequency, (ii) The current in the circuit atresonance, (iii) The voltage across the capacitor and the coil at resonance.

[Ans. (i) 153 Hz, (ii) 2 A, (iii) 224 V]

36. A resistance of 24 Ω, a capacitance of 150 µF and an inductance of 0.16 H areconnected in series with each other. A supply at 240 V, 50 Hz is applied tothe ends of the combination. Calculate (i) the current in the circuit (ii)the potential differences across each element of the circuit (iii) the frequencyto which the supply would need to be changed so that the current would beat unity power-factor and find the current at this frequency.

[Ans. (i) 6.37 A (ii) VR = 152.8 V, VC = 320 V, VL = 123.3 V (iii) 32 Hz ; 10 A]

37. State the advantages of A.C. polyphase supply system over single-phasesystem.

38. Why is the number of phases in a polyphase system always three ratherthan any other number ?

39. Explain clearly what is meant by ‘phase sequence’ of 3-phase voltages.

40. What are the two systems in which three-phases can be connected ? Whatare the advantages and disadvantages of each system ?

41. Deduce an expression for power in a 3-phase balanced load circuit. Showthat it is the same irrespective of the load being connected in star or delta.

42. Compare the star and delta connections in a 3-phase system.

43. Three equal impedances each having a resistance of 20 ohm and reactanceof 15 ohm are connected in star to a 400 V, 3-phase, 50 Hz system. Calculate :

(i) The line current, (ii) The power factor,

(iii) The power consumed. [Ans. 9.24 A ; 0.8 (lag) ; 5120 W]

44. Three equal impedances are star-connected to a 3-phase, 50 Hz supply. Ifthe resistance and reactance of each branch are 25 ohm and 38 ohmrespectively, calculate :

(i) The line current

(ii) The power consumed. [Ans. 5.28 A ; 2086 W]


NOTES


45. Three identical coils connected in delta across 400 V, 50 Hz, 3- phase supplytakes a line current of 17.32 A at power factor 0.5 lagging. Determine :(i) The current in each phase,

(ii) Resistance, inductance and impedance of each phase winding.[Ans. 10 A ; 20 Ω ; 0.11 H ; 40 Ω]

46. A 220 V, 3-phase voltage is applied to a balanced delta-connected 3-phaseload of phase impedance (15 + j20) Ω. Find :(i) The phasor current in each line.

(ii) What is the power consumed per phase ?(iii) What is the phasor sum of the three line currents ? Why does it have this

value ? [Ans. 15.24 A ; 1161.6 W ; zero]47. Calculate (i) line current and (ii) the total power absorbed when three coils

each having a resistance of 10 Ω and reactance of 7 Ω are connected (a) instar and (b) in delta across a 400 V, 3-phase supply.

[Ans. 18.93 A, 10750 W ; 56.7 A, 32250 W]

48. What are the essential features of an indicating instrument?

49. What are the advantages and disadvantages of gravity controlledinstruments?

50. Explain the construction and working of a dynamometer type instrument.How it can be used as an ammeter and a voltmeter.

51. What is a wattmeter?

52. Enumerate types of wattmeter.

53. Discuss the construction working of a dynamometer wattmeter with the helpof a neat diagram.

54. Draw a neat sketch of an induction wattmeter and explain its working. Alsostate its advantages and disadvantages.

55. What is an integrating or energy meter?

56. What are the essential characteristics of energy meters?

57. With the help of a neat sketch explain the construction and working of aninduction type single phase energy meter. Also discuss the sources of errorsprevalent in this type of energy meter.

58. A moving-coil instrument gives full-scale deflection with 15 mA and has aresistance of 5 Ω. Calculate the resistance to be connected:

(i) in parallel to enable the instrument to read upto 1 A.

(ii) in series to enable it to read upto 100 V. [Ans. (i) 0.076 Ω (ii) 6661.7 Ω]

59. A moving-coil instrument has a resistance of 10 ohms and gives a full-scaledeflection when carrying 50 mA. Show how it can be adopted to measurevoltage upto 750 volts and current upto 100 amperes.

[Ans. R = 14990 Ω, Rs = 0.005 Ω]


NOTES


60. The total resistance of a moving-iron voltmeter is 1000 Ω and coil has aninductance of 0.765 H. The instrument is calibrated with a full-scale deflectionof 50 V D.C. Calculate the percentage error when the instrument is used on(i) 25 Hz supply, (ii) 50 Hz supply, the applied voltage being 50 V in eachcase. [Ans. (i) 0.72%, (ii) 36%]

FURTHER READINGS

• Nagasarkar TK and Sukhija MS, “Basics of Electrical Engineering”, OxfordPress (2005).

• Mahmood Nahvi and Joseph A. Edminister, “Electric Circuits”, SchaumOnline Series, Mchraw Hill (2002).

• Yaduvir Singh, Mandhir Varma, “Fundamental of Electrical Engineering,”University Science Press.


NOTES

Electrical Mechanics

STRUCTURE

2.1 Introduction2.2 Principle of Energy Conversion2.3 Faraday’s Laws of Electromagnetic Induction2.4 Construction of D.C. Machines2.5 F.M.F. Equation of a Generator2.6 Types of D.C. Generators2.7 Direct Current Motor2.8 Basic Definitions: Transformer2.9 Working Principle of a Transformer2.10 Transformer Ratings2.11 Kinds of Transformers2.12 Transformer Construction2.13 Transformer Windings, Terminals, Tapping and Bushings2.14 Transformer Cooling2.15 Single Phase Transformer2.16 Induction Machines2.17 Classification of A.C. Motors2.18 Constructional Details2.19 Theory of Operation of an Induction Motor2.20 Slip2.21 Frequency of Rotor Current2.22 Rotor E.M.F. and Rotor Current2.23 Torque and Power2.24 Induction Generator2.25 Single Phase Motors

• Summary

• Glossary• Review Questions

• Further Readings

U N I T

2ELECTRICAL MECHANICS

NOTES


Basic Electrical andElectronics Engineering OBJECTIVES


• discuss about construction, principle of operation, basic equations andapplications regarding D.C. generators.

• define D.C. motors.• give details about transformer, its construction, working principle in

addition to single phase transformer.• classify A.C. motors and their constructional details.• illustrate the concepts of single phase induction motor.

2.1 INTRODUCTION

The main advantage of electric energy over other forms of energy is thatit can be transmitted over long distances with ease and high efficiency. Itsmain use is in the form of a transmitting link for transporting other formsof energy, e.g., mechanical, sound, light etc. from one physical location toanother.

An electromechanical energy conversion device is one whichconverts electrical energy into mechanical energy and mechanical energyinto electrical energy. Electromechanical energy conversion takes placevia the medium of a magnetic or electric field. The magnetic field beingmost suited for practical conversion devices. The conversion process isbasically a reversible one though practical devices may be designed andconstructed to particularly suit one mode of conversion or other.

— From the view-point of magnitudes of energy involved rotating or linearelectrical machines are the most important energy converters. Theseare primarily used for bulk energy conversion and utilisation.

Examples: Motors and generator (continuous energy-conversion equipment).

— A “second category” includes devices for measurement and control, fre-quently referred to as transducers. These generally operate under linear-output conditions and with relatively small signals.

Examples: Microphones, pictures, sensors, and loudspeakers.

— A “third category” of devices encompasses force-producing devices.

Examples: Solenoids, relays and electromagnets.

2.2 PRINCIPLE OF ENERGY CONVERSION

For converting energy from one form to another the “principle ofconservation of energy” (which states that energy can neither be created nordestroyed, it can merely be converted from one form to another) can be invoked. In


NOTES

Electrical Mechanicsan energy conversion device, out of the total input energy, a larger part of theenergy is converted into useful output energy, some energy is stored and rest isconverted to heat (called energy loss). Thus energy balance, which should includethe above terms, can be written for a motor and a generator as follows:

Motor (action):

Total electricalenergy inputL

NMO

QP=

Mechanical energyoutputL

NMO

QP

+Total energy stored(in magnetic field)

Energy convertedinto heat

L

NMO

QP+ LNM

O

QP

...(2.1)Eqn. (2.1) is written so that the electrical and mechanical energy terms

have positive values for motor action. The equation applies well to generatoraction; these terms then simply have negative values.

Generator (action):

Total mechanicalenergy input

Electrical energyoutput

L

NMO

QP= LNM

O

QP

+ Total energy stored(in magnetic field)

Energy convertedinto heat

L

NMO

QP+ LNM

O

QP

...(2.2)In the above cases, the sign of the heat generation term(s) is such that heat

generation within the system results in a flow of thermal energy out of the system.In the systems, to be considered here, the conversion of energy into heat occurs bysuch mechanisms as ohmic heating due to current flow in the windings of theelectric terminals and mechanical friction due to the motion of the systemcomponents forming the mechanical terminals.

Thus, energy balance equation may be written as:

dW dWinput outputz z= + +z zdW dWmagnetic field heat ...(2.3)

Fig. 2.1 shows the flow of energy in electromechanical energy conversion viaa coupling field.

Electricalsource

Couplingfield

Mechanicalsink

Net mechanicaloutput

Gross mechanicaloutput

Gross electricalinput

Net electricalinput

Electrical losses(ohmic and iron)

Mechanicallosses

(a)

NOTES



Mechanicalsource

Couplingfield

Electricalsink

Net mechanicalinput

Gross mechanicalinput

Gross electricaloutput

Net electricaloutput

Electricallosses

Mechanicallosses

(b)

Fig. 2.1. Energy flow in electromechanical energy conversion via a coupling field.

Owing to the inertia associated with mechanical components, electrome-chanical energy conversion devices are slow-moving. Therefore, the cou-pling field necessary for energy conversion must vary slowly and as suchthis field is quasi-static in nature. Electromagnetic radiation from thecoupling field is almost negligible.

2.3 FARADAY’S LAWS OF ELECTROMAGNETIC

INDUCTION

“The phenomenon whereby an e.m.f. and hence current is induced in anyconductor which is cut across or is cut by a magnetic flux is known aselectromagnetic induction”.

Faraday’s First Law. It states as follows:Whenever the magnetic flux linked with a circuit changes, an e.m.f. is always

induced in it”.Or

“Whenever a conductor cuts magnetic flux, an e.m.f. induced in that conductor”.

Conductor – +

N N NS S S

+ –

(a) (b) (c)

(a) Voltage induced (b) Voltage induced (c) No voltage is inducedacross a wire moving across a wire moving in a wire moving parallel

downward. upward. to the field.

Fig. 2.2. When a conductor is moved across a magneticfield a voltage is induced in the conductor.


NOTES

Electrical MechanicsFaraday’s Second Law. It states as follows:

“The magnitude of induced e.m.f. is equal to the rate of change of flux-linkages”.

Mathematically, e Nddt

= − φvolt ...(2.4)

where, e = induced e.m.f.,ddt

φ = rate of change of flux, and

N = number of turns of the coil.

[Usually, a minus sign is given to the right-hand side expression to signifythe fact that the induced e.m.f. sets up current in such a direction that magneticeffect produced by it opposes the very cause producing it.]

Induced e.m.f.:

Induced e.m.f. may be of the following two types:

1. Dynamically induced e.m.f.

2. Statically induced e.m.f.

Dynamically Induced e.m.f.:

Refer to Fig. 2.2. The e.m.f. induced (e) in the conductor is given by:

e = Blv volt ...(2.5)

where, B = flux density of the magnetic field in tesla,

l = length of the conductor in metres, and

v = velocity of the conductor in m/s.

If the conductor moves at an angle θ with the direction of flux then the inducede.m.f.

e = Blv sin θ volt ...(2.6)

The direction of the induced e.m.f. is given by Fleming’s Right hand rule.

Statically Induced e.m.f.:

The e.m.f. induced by variation of flux is termed as “statically induced e.m.f.”.

Statically induced e.m.f. can be further subdivided as follows:(i) Self-induced e.m.f.

(ii) Mutually induced e.m.f.

2.4 CONSTRUCTION OF D.C. MACHINES

A D.C. machine consists of two main parts:

(i) Stationary Part. It is designed mainly for producing a magnetic flux.

(ii) Rotating Part. It is called the armature, where mechanical energy isconverted into electrical (electrical generator), or conversely, electrical energy intomechanical (electric motor).


NOTES


Typical line ofmagnetic flux

Mainframe

FieldpoleN

S Fieldpole

Shaft

Armature

Armature

magnetic

structuremagnetic

structure

Typical slotto allow forwinding coilspace needs(one of many)

Air gap betweenfaces of pole andarmature surface

Air gap

Fig. 2.3. Generator or motor magnetic structure.

The stationary and rotating parts are separated from each other by anair-gap.

The stationary part of a D.C. machine consists of main poles, designed tocreate the magnetic flux, commutating poles interposed between the mainpoles and designed to ensure sparkless operation of the brushes at thecommutator (in very small machines with a lack of space commutatingpoles are not used); and a frame/yoke.

The armature is a cylindrical body rotating in the space between the polesand comprising a slotted armature core, a winding inserted in the armaturecore slots, a commutator and brush gear.

Fig. 2.3 shows generator or motor magnetic structure.

Description of parts of D.C. machines are given as follows:

Frame

Fig. 2.4 shows the sectional view of four pole D.C. machine.

The frame is the stationary part of a machine to which are fixed the mainand commutating poles and by means of which the machine is bolted to itsbed plate.

The ring-shaped portion which serves as the path for the main and com-mutating pole fluxes is called the ‘yoke’.

Cast iron used to be the material for the frame/yoke in early machines butnow it has been replaced by cast steel. This is because cast iron is saturated by aflux density of about 0.8 Wb/m2 while saturation with cast steel is at about1.5 Wb/m2. Thus the cross-section of a cast iron frame is about twice that of a caststeel frame for the same value of magnetic flux. Hence, if it is necessary to reducethe weight of machine, cast steel is used. Another disadvantage with the use of


NOTES

Electrical Mechanics(+) (–)Loadterminals

Shunt pole Field winding

Pole shoe

Interpole

Commutator

Frame

Fieldrheostat

Armatureconductors

S

S

NN NN

NN

NN

SS

SS

Rotation

Fig. 2.4. Sectional view of a four pole D.C. machine.

cast iron is that its mechanical and magnetic properties are uncertain due to thepresence of blow holes in the casting. Lately, rolled steel yokes have been developedwith the improvements in the welding techniques. The advantages of fabricatedyokes are that there are no pattern charges and the magnetic and mechanicalproperties of the frame are absolutely consistent.

It may be advantageous to use cast iron for frames but for medium andlarge sizes of frames, usually rolled steel is used.

If the armature diameter does not exceed 35 to 45 cm, then, in addition tothe poles, end shields or frame-heads which carry the bearings are alsoattached to the frame. When the armature diameter exceeds 1 m, it iscommon practice to use pedestal-type bearings, mounted separately, onthe machine bed plated outside the frame.

The end shield bearings, and sometimes the pedestal bearings, are of ballor roller type. However, more frequently plain pedestal bearings are used.

In machines with large diameter armatures a brush-holder yoke is fre-quently fixed to the frame.

Field Poles

Formerly the poles were cast integral with the yoke. This practice is stillbeing followed for small machines. But in present day machines it is usual


NOTES


to use either a completely laminated pole, or solid steel poles with laminatedpole shoes.

Laminated construction is necessary because of the pulsations of fieldstrength that result when the notched armature rotor magnetic structurepasses the pole shoe. Variations in field strength result in internal eddycurrents being generated in a magnetic structure. These eddy currentscause losses; they may be largely prevented by having laminated magneticstructures. Laminated structures allow magnetic flux to pass along thelength of the laminations, but do not allow electric eddy currents to passacross the structure from one lamination to another. The assembled stackof laminations is held together as a unit by appropriately placed rivets.The outer end of the laminated pole is curved to fit very closely into theinner surface of the main frame.

Fig. 2.5 shows the constructional details of a field pole. The pole shoe actsas a support to the field coils and spreads out the flux in the air gap andalso being of larger cross-section reduces the reluctance of the magneticpath.

Holes forholding bolts

Rivet heads

Assumed directionof lines of force

Poleshoe

Wire ends

Frame for holding wire ends

Wire woundon frame

Completeconstruction

Fig. 2.5. Constructional details of a field pole.

Different methods are used for attaching poles to the yoke. In case ofsmaller sizes, the back of the pole is drilled and tapped to receive polebolts (see Fig. 2.6). In larger sizes, a circular or a rectangular pole bar isfitted to the pole. This pole bar is drilled and tapped and the pole boltspassing through laminations screw into the tapped bar (see Fig. 2.7).

Yoke

Laminations

Fig. 2.6. Fixing pole to the yoke. Fig. 2.7


NOTES

Electrical MechanicsCommutating Poles

A commutating pole (also called interpole) is similar to a main pole andconsists of core terminating in a pole shoe, which may have various shapes,and coil mounted on the core.

The commutating poles are arranged strictly midway between the mainpoles and are bolted to the yoke.

Commutating poles are usually made of solid steel, but for machines oper-ating on sharply varying loads they are made of sheet steel.

Armature

The armature consists of core and winding. Iron being the magnetic ma-terial is used for armature core. However, iron is also a good conductor ofelectricity. The rotation of solid iron core in the magnetic field results ineddy currents. The flow of eddy currents in the core leads to wastage ofenergy and creates the problem of heat dissipation. To reduce the eddycurrents the core is made of thin laminations.

The armature of D.C. machines (see Fig. 2.8) is built up of thin laminationsof low loss silicon steel. The laminations are usually 0.4 to 0.5 mm thickand are insulated with varnish.

Laminations

SlotKey way

Air holes

Fig. 2.8. Armature of a D.C. machine.

The armature laminations, in small machines, are fitted directly on tothe shaft and are clamped tightly between the flanges which also act assupports for the armature winding. One end of flange rests against ashoulder on the shaft, the laminations are fitted and other end is pressedon the shaft and retained by a key.

The core (except in small size) is divided into number of packets byradial ventilation spacers. The spacers are usually I sections welded tothick steel laminations and arranged to pass centrally down each tooth.


NOTES


Fig. 2.9. Drum armature stamping with axial flow ventilation system.

— For small machines the laminations are punched in one piece (see Fig. 2.9).These laminations are built up directly on the shaft. With such anarrangement, it is necessary to provide axial ventilation holes so that aircan pass into ventilating ducts.

— The armature laminations of medium size machines (having more thanfour poles) are built on a spider. The spider may be fabricated. Lamina-tions up to a diameter of about 100 cm are punched in one piece and aredirectly keyed on the spider (see Fig. 2.10).

End flange

Armature core

Cast spider

Fig. 2.10. Clamping of an armature core.

In case of large machines, the laminations of such thin sections are difficultto handle because they tend to distort and become wavy when assembledtogether. Hence circular laminations instead of being cut in one piece arecut in a number of suitable sections or segments which form part of acomplete ring (see Fig. 2.11). A complete circular limination is made up offour or six or even eight segmental laminations. Usually two keywaysare notched in each segment and are dove-tailed or wedge shaped to makethe laminations self-locking in position.

The armature winding is housed in slots on the surface of the armature.The conductors of each coil are so spaced that when one side of the coil isunder a north pole, the opposite is under a south pole.


NOTES


Air holes

Slots

Key waySegmentallamination

Fig. 2.11. Segmental stampings.

Fig. 2.12 shows the arrangement of conductors and insulation in a slot.

In D.C. machines two layer winding with diamond shaped coils is used.The coils are usually former wound. In small machines, the coils are heldin position by band of steel wire, wound under tension along the corelength. In large machines, it is useful to employ wedges of fibre or wood tohold coils in place in the slots. Wire bands are employed for holding theoverhang. The equilizer connections are located under the overhang onthe side of the commutator. Fig. 2.13 shows a typical arrangement forequilizers. The equilizers can be accommodated on the other end of thearmature also.

Low

erco

ilsi

deU

pper

coil

side Hardwood or

Fibre wedge

Copperconductor

Varnishedcambric

Micawrapper

Cotton tape

Armaturecoils

Equalizerrings

Fig. 2.12. Cross-section of an armature slot. Fig. 2.13. Ring type equilizers.

Commutator

A commutator converts alternating voltage to a direct voltage.

A commutator is a cylindrical structure built up of segments made ofhard drawn copper. These segments are separated from one another andfrom the frame of the machine by mica strips. The segments are con-nected to the winding through risers. The risers have air spaces betweenone another so that air is drawn across the commutator thereby keepingthe commutator cool.


NOTES


Fig. 2.14 shows the components of a commutator. The general appearance ofa commutator when completed is as shown in Fig. 2.15 (a). The commutator andarmature assembly is shown in Fig. 2.15 (b).

Shaft

Hub combined with end ring

SegmentSegment

Micanitevee rings

Riser orlug

End clampring

Clampbolt

Riser or lug

Segment

Micaniteseparator

Fig. 2.14. Commutator components.

Insulatedcoppersegments

Endclamp

Commutatorlugs

Fig. 2.15. (a) General appearance of a commutator after assembly.

Spacingrings

CommutatorSpider

Key rings

Key

Laminations

Armature

Key

Fan

Shaft

Fig. 2.15. (b) Commutator and armature assembly.


NOTES

Electrical MechanicsBrush Gear

To collect current from a rotating commutator or to feed current to it use ismade of brush-gear which consists of:

(i) Brushes (ii) Brush holders(iii) Brush studs or brush-holder arms (iv) Brush rocker(v) Current-collecting busbars.Brushes. The brushes used for D.C. machines are divided into five classes:

(i) Metal graphite (ii) Carbon graphite(iii) Graphite (iv) Electro-graphite(v) Copper. The allowable current density at the brush contact varies from 5 A/cm2 in

case of carbon to 23 A/cm2 in case of copper.

The use of copper brushes is made for machines designed for large currentsat low voltages. Unless very carefully lubricated, they cut the commutatorvery quickly and in any case, the wear is rapid. Graphite and carbongraphite brushes are self lubricating and, are, therefore, widely used. Evenwith the softest brushes, however, there is a gradual wearing away of thecommutator, and if mica between the commutator segments does not weardown so rapidly as the segments do, the high mica will cause the brushesto make poor contact with segments, and sparking will result andconsequent damage to commutator. So to prevent this, the mica isfrequently ‘undercut’ to a level below the commutator surface by meansof a narrow milling cutter.

Brush Holders. Box type brushholders are used in all ordinary D.C.machines. A box type brush holder isshown in Fig. 2.16. At the outer end ofthe arm, a brush box, open at top andbottom is attached. The brush is pressedon to the commutator by a clock spring.The pressure can be adjusted by a leverarrangement provided with the spring.The brush is connected to a flexibleconductor called pig tail. The flexibleconductor may be attached to the brushby a screw or may be soldered.

The brush boxes are usually made of bronze casting or sheet brass. In lowvoltage D.C. machines where the commutation conditions are easygalvanised steel box may be used.

Some manufacturers use individual brush holders while others usemultiple holders, i.e., a number of single boxes built up into one longassembly.

Brush Rockers. Brush holders are fixed to brush rockers with bolts. Thebrush rocker is arranged concentrically round the commutator. Cast iron is usu-ally, used for brush rockers.

Fig. 2.16. Box type brush holder.

Pressureadjustinglever

SpringPigtail

Brush

Brushbox

Commutator


NOTES


Armature Shaft Bearings

With small machines roller bearings are used at both ends. For larger machines roller bearings are used for driving end and ball

bearings are used for non-driving (commutator) end. The bearings are housed in the end shields. For large machines pedestal bearings are used.

Armature Windings

The armature winding is very important element of a machine, as it directlytakes part in the conversion of energy from one form into another. The requirementswhich a winding must meet are diverse and often of a conflicting nature. Amongthese requirements the following are of major importance:

The winding must be designed with the most advantageous utilisation ofthe material in respect to weight and efficiency.

The winding should provide the necessary mechanical, thermal andelectrical strength of the machine to ensure the usual service life of16–20 years.

For D.C. machines proper current collection at the commutator (i.e.,absence of detrimental sparking) must be ensured.

According to the degree of closure produced by winding, armature windingsare of the following two types:

1. Open coil winding 2. Closed coil winding.The closed armature windings are of two types:

(i) Ring winding (ii) Drum windingIn general there are two types of drum armature windings:(i) Lap winding (ii) Wave winding.

“Lap winding” is suitable for comparatively low voltage but high currentgenerators whereas “wave of winding” is used for high voltage, low currentmachines.

— In ‘lap winding’ the finish of each coil is connected to the start of the nextcoil so that winding or commutator pitch is unity.

— In ‘wave winding’ the finish of coil is connected to the start of another coilwell away from the fixed coil.

2.5 E.M.F. EQUATION OF A GENERATOR

Let p = number of poles,

φ = flux/pole, webers (Wb),

Z = total number of armature conductors,

= number of slots × number of conductors/slot,

N = rotational speed of armature, r.p.m.,


NOTES

Electrical Mechanicsa = number of parallel paths in armature, and

Eg = generated e.m.f. per parallel path in armature.

Average e.m.f. generated per conductor = ddt

φ volt

Now, flux cut per conductor in one revolution, dφ = pφ Wb

Number of revolutions/second = N60

∴ Time for one revolution, dt = 60N seconds

Hence, according to Faraday’s laws of electromagnetic induction,

E.m.f. generated per conductor = p Nφ60

volt

For a lap wound generator:

Number of parallel paths, a = p

Number of conductor (in series) in one path = Zp

∴ E.m.f. generated per path = φ φpN Z

pZN

60 60× = volt

For a wave wound generator:

Number of parallel paths, a = p

Number of conductor (in series) in one path = Z2

∴ E.m.f. generator per path = p N Z p ZNφ φ60 2 120

× = volt

In general, generated e.m.f.

Eg = φ φZN p

ap ZN

a60 60× FHGIKJ

=volt ...(2.7)

where a = p ...... for lap winding = 2 ...... for wave winding.

Example 2.1. A six pole lap wound D.C. generator has 720 conductors, a fluxof 40 m Wb per pole is driven at 400 r.p.m. Find the generated e.m.f.

Solution. Number of poles, p = 6

Total number of conductors, Z = 720

Flux per pole, φ = 40 m Wb = 40 × 10–3 Wb

Speed of rotation, N = 400 r.p.m.

Number of parallel paths, a = p = 6 [Since the generator is lap wound.]


NOTES


Generated e.m.f. Eg:

Using the relation, Eg = p ZN

aφ60

6 40 10 720 40060 6

3

=× × × ×

×

−

= 192 V.

Hence, generated e.m.f. Eg = 192 V. (Ans.)

Example 2.2. The armature of a 6-pole D.C. generator has a wave windingcontaining 650 conductors. Calculate the generated e.m.f. when the flux per pole is0.055 Wb and the speed is 300 r.p.m.

Calculate speed at which the armature must be driven to generate an e.m.f. of550 V if the flux per pole is reduced to 0.05 Wb.


Total number of conductors, Z = 650

Flux per pole, φ = 0.055 Wb

Speed of rotation, N = 300 r.p.m.

E.m.f. generated, Eg = ?

Generated e.m.f. (2nd case) = 550 V

Flux per pole (2nd case) = 0.05 Wb

Speed of rotation, N = ?

Case I. E.m.f. generated, Eg:

Using the relation, Eg = p ZN

aφ60

= 6 0 055 650 300

60 2× × ×

×.

[∵ a = 2, as the generator is wave wound]

= 536.25 V.

Hence, e.m.f. generated = 536.25 V. (Ans.)

Case II. Speed of rotation, N:

Eg = p ZN

aφ60

550 = 6 0 05 650

60 2× × ×

×. N

N = 550 60 2

6 0 05 650× ×

× ×. = 338.46 r.p.m.

Hence, speed of rotation = 338.46 r.p.m. (Ans.)

2.6 TYPES OF D.C. GENERATORS

The power stations of modern design generate practically only three-phasealternating current. A large part of this power is used in the form of


NOTES

Electrical Mechanicsalternating current in industry, for lighting and domestic needs. Whenindustrial needs make it necessary too or when it is of greater advantageto use direct current (for chemical and metallurgical plants, electrictraction, etc.) it is generally obtained by converting A.C. to D.C. with thehelp of converters of ionic or machine types. In the latter case wide use ismade of such installations as motor generator sets in which A.C. motor iscoupled to a D.C. generator on a common shaft.

As primary sources of power, D.C. generators are mainly used inself-contained plants such as automobiles and air planes, for electric arcwelding, train car lighting, in sub-marines, etc.

Classification

According to method of excitation D.C. generators are classified as follows:

1. Separately excited generators, 2. Self-excited generators.

Separately Excited Generators

These are those generators whose field magnets are energised from anindependent external source of D.C. current. Such a generator is shown in Fig. 2.17.

Load Armature

+

–

+

–

I

Ia

If

Field

R

Separate sourceof excitation

If Fieldcurrent

=

I Loadcurrent

=

Ia Armaturecurrent

=

Fig. 2.17. Separately excited generator.

Self-excited GeneratorsThese are those generators whose field magnets are energised by the current

produced by the generators themselves. Due to residual magnetism, there is alwayspresent some flux in the poles. When the armature is rotated, some e.m.f. andhence some induced current produced which is partly or fully passed through thefield coils thereby strengthening the pole flux.

Self-excited generators can be divided, in accordance with how the fieldwinding is connected into generators, as follows:

1. Shunt wound generators2. Series wound generators3. Compound wound generators:

(i) Short shunt

(i) Long shunt

1. Shunt Wound Generators: Refer Fig. 2.18. In these generators the fieldwindings are connected across or in parallel with the armature conductors, andhave the full voltage of the generator across them.


NOTES


Important Relations: Refer to Fig. 2.18.

(i) Ish = V

Rsh

(ii) Ia = Ish + I

(iii) V = Eg – IaRa

(iv) Power developed = EgIa

(v) Power delivered = VI

where Ish = shunt field current,

Ia = armature current,

I (or Il) = load current, Ra = armature resistance,

Rsh = shunt field resistance, Eg = generated e.m.f., and

V = terminal voltage

2. Series Wound Generators: Refer toFig. 2.19. In this case, the field windings are joined inseries with armature conductors. As they carry full loadcurrent, they consist of relatively few turns of thick wireor strip. The use of such generators is limited to specialpurposes (as boosters etc.).

Important Relations. (see Fig. 2.19):

(i) Ia = Ise = I (Ise = series field current)

(ii) V = Eg – I(Ra + Rse) (Rse = series field resist-ance)

(iii) Power developed = EgI

(iv) Power delivered = VI.

3. Compound Wound Generators: It is a combination of a few series anda few shunt windings and be either short shunt or long shunt as shown in Figs. 2.20and 2.21 respectively.

Important Relations:

(a) Short Shunt Compound Wound. (see Fig. 2.20):

(i) Ise = I

(ii) Ish = V I R

Rse se

sh

+

(iii) Ia = I + Ish

(iv) V = Eg – IaRa – IseRse

(v) Power developed = EgIa

(vi) Power delivered = VI.

Armature

+

–

Ish

Ia LoadVV

I

Shuntfield

Fig. 2.18. Shunt woundgenerator.

Armature

+

–

Ia

LoadVV

I = I = Ise a

Seriesfield

Fig. 2.19. Series woundgenerator.


NOTES


ArmatureIa

LoadVV

I

I

IseIsh

IaVV

IIsh

Armature

Ise

Rse

+

–

Shuntfield

Load

Fig. 2.20. Short shunt compound Fig. 2.21. Long shunt compound wound generator. wound generator.

(b) Long Shunt Compound Wound. (see Fig. 2.21):

(i) Ish = V

Rsh

(ii) Ia = Ise = I + Ish

(iii) V = Eg – IaRa – IseRse = Eg – Ia(Ra + Rse)

(iv) Power developed = EgIa

(v) Power delivered = VI.

2.7 DIRECT CURRENT MOTOR

General Aspects

The electric motor is a machine which converts electric energy into mechanicalenergy. It depends for its operation on the force which is known to exist on a con-ductor carrying current while situated in a magnetic field.

Construction. A D.C. motor is similar in construction to a D.C. generator. Asa matter of fact any D.C. generator will run as a motor when its field and armaturewindings are connected to a source of direct current. The field winding producesthe necessary magnetic field. The flow of current through the armature conductorsproduces a force which rotates the armature.

Though the essential construction of D.C. motor is identical to that of agenerator, the external appearance of a motor may be somewhat different from thatof a generator. This is mainly due to the fact that the frame of a generator may bepartially open because it is located in relatively clean environment and only skilledoperators are present in its vicinity. A motor, on the other hand, may be operatingin a rather dusty environment and only unskilled operators may be working in itsvicinity. Therefore, frames of motors are to a large extent closed.

The body of D.C. mill motors is made in two halves bolted together for easyaccess to the field windings and inter-poles.


NOTES


Applications. Because of their inherent characteristics D.C. motors findextensive application in:

(i) Steel plants (ii) Paper mills

(iii) Textile mills (iv) Printing presses

(v) Cranes (vi) Winches

(vii) Excavators etc. where precise and accurate speed control over a wide rangeis required.

Advantages. The D.C. motors possess the following advantages:

(i) High starting torque.

(ii) Speed control over a wide range, both below and above the normal speed.

(iii) Accurate stepless speed control with constant torque.

(iv) Quick starting, stopping, reversing and accelerating.

Disadvantages. The disadvantages of D.C. motors are:

(i) High initial cost.

(ii) Increased operating and maintenance costs because of the commutatorsand brushgear.

Principle of Operation of D.C. Motor

The principle of motor action can be stated as follows:

“Whenever a current carrying conductor is placed in a magnetic field, itexperiences a force whose direction is given by Fleming’s left hand rule”.

Fig. 2.22 illustrates this principle.Fig. 2.22 (a) shows the field set up by the poles.Fig. 2.22 (b) shows the conductor field due to flow of current in theconductor.

Fig. 2.22 (c) shows the resultant field produced when the current carryingconductor wire of Fig. 2.22 (b) is inserted in the air gap of Fig. 2.22 (a) with the axisof the conductor at right angles to the direction of the flux.

On the upper side of the conductor in Fig. 2.22 (c) the magnetizing forces ofthe field and of the current in the conductor are additive while on the lower sidethese are subtractive. This explains why the resultant field is strengthened aboveand weakened below the conductor (wire).

The above experiment shows that the wire in Fig. 2.22 (c) has a force on itwhich tends to move it downward. Thus the force acts in the direction of the weakerfield. When the current in the wire is reversed, the direction of the force is alsoreversed, as in Fig. 2.22 (d).

The force (F) developed in the conductor is given by the relation,

F = BIl Newton

where B = flux density, T (Wb/m2),

I = current in conductor, A, and

l = exposed length of conductor, m.


NOTES


N S

(a) Pole field (b) Conductor field

+

(c) force (d)

N S+ N S

Force

Fig. 2.22. The principle of motor action.

Now consider the magnetic field of a D.C. motor in which there is no currentin the armature conductors; the lines of force will be distributed as shown inFig. 2.23.

NN SS

Conductors Armature

Fig. 2.23. Distribution of lines of force in a motor due to magnetic field only.

If, now, the armature carries current, each of its conductors will produce amagnetic field which, when super-imposed on the main field, causes a distributionof magnetic lines as shown in Fig. 2.24.

NN SS

+

+

+

+

+

A

B

Fig. 2.24. Distribution of lines of force in a motor,on load, due to the armature and magnetic field.

The magnetic field is said to be distorted, since the lines of force no longerfollow approximately straight paths.


NOTES


These lines of force have the property of tending to shorten themselves, so thatthey may be regarded as being in tension. Each conductor in Fig. 2.24 will experiencea force like that exerted on a stone in a catapult. Since these conductors are embeddedin slots in the armature, the latter is caused to rotate in a clockwise direction.

Back or Counter E.M.F.

Refer to Fig. 2.25. In a D.C. motor when thearmature rotates, the conductors on it cut the lines offorce of magnetic field in which they revolve, so thatan e.m.f. is induced in the armature as in a generator.The induced e.m.f. acts in opposition to the current inthe machine and, therefore, to the applied voltage, sothat it is customary to refer to this voltage as the ‘backe.m.f’. That this is so can be deduced by Lenz’s law,which states that the direction of an induced e.m.f. issuch as to oppose the change causing it, which is, ofcourse, the applied voltage.

The magnitude of the back or counter e.m.f. canbe calculated by using formula for the induced e.m.f.in a generator, and it is important in the case of the motor, to appreciate that thisis proportional to the product of the flux and the speed. Thus if Eb denotes the backe.m.f., φ the flux and N the speed, we may write,

Eb = k φ N

where k is a number depending on nature of armature winding.

The value of back e.m.f. (Eb) is always less than the applied voltage, althoughdifference is small when the machine is running under normal conditions. It is thedifference between these two quantities which actually drives current through theresistance of the armature circuit. If this resistance is represented by Ra, the backe.m.f. by Eb and the applied voltage by V, then we have

V = Eb + IaRa

where Ia is the current in the armature circuit.

Types of D.C. Motors

There are three main types of motors characterised by the connection of thefield winding in relation to the armature. These are:

1. Shunt wound motor or the shunt motor, in which the field winding isconnected in parallel with the armature.

2. Series motor, in which the armature and field windings are connected inseries.

3. Compound motor, which has two field windings, one of which is connectedin parallel with the armature and the other in series with it.

Shunt Wound Motor. Fig. 2.26 shows the connections of a shunt motor.From these connections it may be observed at once that the field current is constant,

N

+Force

Motion

– +

IaArmature

Eb

Fig. 2.25. Motoringoperation.


NOTES

Electrical Mechanicssince it is connected directly to the supply which is assumed to be at constantvoltage. Hence the flux is approximately constant and, since also the back e.m.f. isalmost constant under normal conditions the speed is approximately constant. Thisis not strictly true, but nevertheless, it is usual for all practical purposes to regardthe shunt motor as a constant speed machine. It is, therefore, employed in practicefor drives, the speeds of which are required to be independent of the loads. Thespeed can, of course, be varied when necessary and this is done by the inclusion ofa variable resistor in series with the field winding, as shown in Fig. 2.26.

VVArmature

Rsh

Resistor

VVArmature

I

Divertor

Rse

I

Fig. 2.26. Connections of a shunt motor. Fig. 2.27. Connections of a series motor.

Series Motor. Fig. 2.27 shows the connections for the series motor. The currentpassing through the field winding is the same as that in the armature, since thearmature current increases as the mechanical load on the shaft increases, so alsodoes the field current. The resultant increase in magnetic flux causes a reductionin the speed, as can be observed from a consideration of the formula:

Eb = p ZN

aφ60

or Eb = kφN

where k = pZa60

being constant

or N = Ek

b

φ.

This is a useful property for many drives in which it is desirable that a heavyincrease in the load should automatically bring about a compensating reduction inspeed. As with the shunt motor, the speed may also be varied independently of theload by the inclusion of a variable resistor in the field circuit. In this case, however,it is connected in parallel with the series winding as shown in Fig. 2.27, and iscalled a divertor compound motor.

Refer to Fig. 2.28. The compound motor has a shunt field winding in additionto the series winding so that the number of magnetic lines of force produced byeach of its poles is the resultant of the flux produced by the shunt coil and that dueto the series coil. The flux so produced depends not only on the current and number


NOTES


of turns of each coil, but also on the windingdirection of the shunt coil in relation to that ofthe series coil. When the two fluxes assist eachother the machine is a cumulativecompound motor, while if they oppose eachother, it is said to be a differentialcompound motor.

Fig. 2.29 shows the field windings andinterpole connections of a differentialcompound wound motor. The shunt coil ismade up of many turns of fine wire, whilst the series coil comprises relatively fewturns of thickwire.

Fig. 2.30 shows the field windings of a cumulative compound motor.

The flow of currents in the shunt and series coils is worth noting in Figs. 2.29and 2.30.

+

–

N

N

S

S

Shuntcoil

Seriescoil

N

+ –

N

S

S

S S

Fig. 2.29. Field windings of a Fig. 2.30. Field windings anddifferential compound motor. interpole connections of a cumulative

compound wound motor.

Speed of a D.C. Motor

We know that the voltage equation of a motor is given by

V = Eb + IaRa

or Eb = V – IaRa

orp ZN

aφ60

= V – IaRa

∴ N = ( )V I R a

Zpa a− ×

φ60

But V – IaRa = Eb

∴ N = E a

Zpb

φ× 60

VVArmatureRsh

Rse

+

–Fig. 2.28. Connections of a com-

pound motor.


NOTES


or N = k′ Eb

φ ...(2.8)

where k′ = 60aZp

= constant.

The equation (2.8) shows that speed is directly proportional to back e.m.f. andinversely to the flux

or N ∝ Eb

φ .

For series motor:

Let N1 = speed in the first case,

Ia1 = armature current in the first case, and

φ1 = flux in the first case.

N2, Ia2 and φ2 = corresponding quantities in the second case.

Using the above relation i e NEb. ., ∝

FHG

IKJφ

, we get

N1 ∝ Eb1

1φ

and N2 ∝ Eb2

2φ

where, Eb1 = V – Ia1Ra

and, Eb2 = V – Ia2Ra

∴NN

EE

b

b

2

1

2

1

1

2

= ×φφ ...(2.9)

Prior to saturation of poles:

φ ∝ Ia

∴NN

EE

II

b

b

a

a

2

1

2

1

1

2

= × ...(2.10)

For shunt motor:

Applying the same equation in this case also, we get

NN

EE

b

b

2

1

2

1

1

2

= × φφ

If φ1 = φ2

NN

EE

b

b

2

1

2

1

= . ...(2.11)


NOTES


Speed Regulation

The speed regulation of a D.C. motor is defined as follows:

“The change in speed when the load on the motor is reduced from rated valueto zero, expressed as percent of the rated load speed.”

∴ Percent speed regulation = no load speed full load speed

full load speed−

.

Motor Characteristics

The properties of all motors and, in particular, D.C. motors are defined as atotality of the following characteristics:

1. Starting; 2. Operating and mechanical;3. Braking; and 4. Regulation.

Starting Characteristics. The starting characteristics determine the opera-tion of starting from the moment the motor begins running to the moment whensteady-state operation is established and include:

The starting current Istart generally determined by the ratioIIstart

run;

The starting torque Tstart, determined by the ratioTT

start

run;

The duration of starting tstart; The economy of operation determined by the amount of energy consumed

in starting; and The cost and reliability of the starting equipment.

Operating Characteristics. The operating characteristics are those thatgive the relation between speed, torque and efficiency as functions of the usefulpower or the armature current for V = constant and constant resistances in thearmature and field circuit.

Mechanical Characteristics. Of major importance for industrial drivemechanisms are the mechanical characteristics, which are the relation N = f(T)(where N and T stand for speed and torque respectively) for conditions of constantvoltage and resistances in the armature and field circuits. These also include thebraking characteristics.

Regulation Characteristics. These characteristics determine the proper-ties of motors when their speed is controlled. These include:

The regulation range determined by the ratioNN

max

min;

The efficiency of regulation from the point of view of the initial cost of theequipment and maintenance;

The nature of regulation—continuous or stepped; and

The simplicity of the control apparatus and methods.


NOTES

Electrical MechanicsThe D.C. motors possess versatile and diverse regulation characteristics, andfor this reason are indispensable in installations where wide-range control of speedis necessary.

The characteristic curves of a motor are those curves which show relationbetween the following quantities:

1. Torque and armature current i.e., Ta/Ia characteristic. This is also knownas electrical characteristic.

2. Speed and armature current i.e., N/Ia characteristic.

3. Speed and torque i.e., N/Ta characteristic. This is also known as mechani-cal characteristic.

This can be obtained from (1) and (2) above.

Following relations are worth keeping in mind while discussing motor char-acteristics:

N ∝ Eb

φ and Ta ∝ φIa.

Torque-current Characteristics Shunt Motor

When running on no-load, a small armature current flows to supply thefield and to drive the machine against the fric-tion and other losses in it.

As the load is applied to the motor, and isincreased, the torque rises almost propor-tionally to the increase in current. This is notquite true, because the flux has been assumedto be constant, whereas it decreases slightlyowing to armature reaction. The effect of thisis to cause the top of the curve connectingtorque and line current to bend over as shownin Fig. 2.31.

The starting torque of a motor is determinedby the starting resistance, which in turn,governs the initial current through the machine when the main switch isclosed. At this moment the speed is zero, so that the back e.m.f. is zeroand the starting current is given by I = V/R, where V is the supply voltageand R is the total resistance, which includes the armature and startingresistance.

If the starting current is limited by heating considerations to twice the fullload current, then with normal supply voltage the starting torque of a shunt motoris twice the full load torque. If, however, the supply voltage is below normal, theflux is also less than twice full load torque. The importance of this will be appreciatedwhen the starting torque of a series motor is compared with that of the shuntmotor.

Tor

que

Line current

Fig. 2.31. Torque-currentcharacteristic of a shunt

motor.


NOTES


Series Motor

In a series motor the torque (Ta ∝ φIa) increases much more than does thearmature current. This is because the flux itself increases with the arma-ture current, though, owing to the magnetic saturation, the two are notstrictly proportional. Nevertheless, for all but the heavy loads which tendto produce saturation of the field system, it may be said that the torque isapproximately proportional to the square of the load current.

Fig. 2.32 shows the relationship between torqueand current. Here the current commences at the no-loadvalue, rises parabolically at first, but increases moreslowly as the effects of armature reaction and magneticsaturation becomes appreciable.

This property of a series motor, by virtue ofwhich a heavy current gives rise to a very hightorque, also influences its starting charac-teristics. In a case of a shunt motor, it hasalready been seen, that the current at themoment of starting may be as high as twicethe full-load value; if we allow for thearmature of the magnetisation characteristicand for weakening effect of armature reactionand assume that the flux is increased to 1.5 times its full-load value, thenit is obvious that the starting torque of a series motor is three times thefull-load torque.

Further more, if the supply voltage falls, the starting current may still bemaintained at twice full-load value by cutting out some of the startingresistance, so that the high value of starting torque may still be main-tained.

This type of motor (series motor) is superior to shunt motor for drives in whichmachines have to be started and accelerated from rest when fully loaded, as is thecase with traction equipment.

Compound Motors

(i) Differential Compound Motor. Refer Fig. 2.29. In this type of motor the twofield windings (shunt and series) oppose each other. On light loads, such a machineruns as a shunt motor, since the series field winding, carrying only a small current,has relatively little effect.

On heavy loads, the series coils strengthen and since they are in opposition tothe shunt winding, cause a reduction in the flux and a consequence decrease intorque.

On heavy overloads or when starting up on load, the series winding may be-come as strong as shunt, or it may even predominate, in which the torque will bereduced to zero or may even be reversed. In the latter case, the motor would tend tostart-up in the wrong direction.

Fig. 2.32. Torque-currentcharacteristic of a series

motor.

Tor

que

Current (I )a


NOTES

Electrical MechanicsIt is obvious that such characteristics may cause dangerous results, so thatdifferential compound motors have only very limited applications in practice.

Fig. 2.33 shows the torque-current characteristic of a differential compoundmotor.

Tor

que

Current

Tor

que

Current

Fig. 2.33. Torque-current characteristic of Fig. 2.34. Torque-current characteristic ofa differential compound motor. a cumulative compound motor.

(ii) Cumulative Compound Motor. In this type of motor the two field windingsassist each other as shown in Fig. 2.30. The flux on no-load is that due to the shunt-winding, while on load the flux and torque rise with the load current. The torque,therefore, increase more rapidly than in the case of the shunt machine, and onheavy loads it resembles the characteristic of the series motor.

Fig. 2.34 shows the torque-current characteristic of a cumulative compoundmotor.

The torque-current characteristics of shunt, series and compound (differentialand cumulative) motors are showing in Fig. 2.35 from which their properties, maybe compared, as far as torque is concerned.

Tor

que

Current

Differential

Shunt

Series

Cumulative

Fig. 2.35. Torque-current characteristics of shunt, series and compound motors.

Speed-current Characteristics

The speed-current characteristics of various motors can be deduced from thefollowing motor equation,


NOTES

Basic Electrical andElectronics Engineering N ∝

Eb

φor N ∝

V IR−φ

.

Shunt Motor

In the shunt motor, the field circuit is connected to the supply terminalsso that the exciting current remains constant as long the temperature ofthe machine is constant, and field regulator is not adjusted. Actually asthe machine warms up, the field resistance increases and the excitingcurrent decreases by about 4% for every 10°C rise in temperature.Neglecting this effect and also due to armature reaction, it is seen thatthe speed of a shunt machine falls slightly as the load increases. The fallin speed is proportional to the volt drop IR in the armature circuit. If,however, we consider the effect ofarmature reaction, an increase of loadcauses a slight decrease in flux, unlessthe machine is fitted with compensatingwindings. This weakening of the fieldtends to raise the speed, so that theactual fall in speed is less than thatcalculated by a consideration of the voltdrop in the armature.

On the whole, the shunt motor may beregarded as one in which the speed is approx-imately constant, falling slightly as the loadincreases (see Fig. 2.36).

The speed of a shunt machine can be increased by inserting resistance inthe field by means of a field regulator. This weakens the field and causesthe motor to run faster in order to generate the necessary back e.m.f. Ofcourse, it is impossible to reduce the speed by this method below that atwhich it runs with no field resistance in the circuit.

Series Motor

In case of a series motor the flux does not remain constant, or evenapproximately constant, because the field winding is in series with theload, so that as the load increases so also does the strength of the magneticfield. At first the flux increases approximately in proportion to the load,but as the field approaches saturation, owing to the heavier loads, theincrease is not so rapid. The effects of temperature changes and ofarmature reaction may be neglected (in comparison with the abovementioned effect).

It will be appreciated, while considering motor equation, that the backe.m.f. decreases as the armature current increases, as in shunt motor; inthe latter, however, the decrease is due to the volt drop in the armature,while in the series machine the loss in volts occurs in the field as well asin the armature, since they are in series. The back e.m.f. in a series motor,therefore, decreases more rapidly than in a corresponding shunt machine.The speed, however, is proportional to the back e.m.f. divided by the flux;

Spe

ed

Armature current

No-loadspeed

Fig. 2.36. Speed-current charac-teristic of a shunt motor.


NOTES

Electrical Mechanicsthe former decreases, while the latter increases with increasing load sothat the speed decreases rapidly as the armature current increases. Thisproperty is a valuable feature in a drive of which the speed is requiredautomatically to adjust to compensate for changes in load.

The speed-current characteristic of a series motor are shown in Fig. 2.37 (a).

On very low current, a series motor runs at very high speeds, or tends torace, as it is termed. This is dangerous, since the machine may be destroyedby the centrifugal forces set up in the rotating parts. For this reason,when installing a series motor it must be positively connected to its loadby gearing or by direct connection and never by belting. Moreover, theminimum load should be great enough to keep the speed within safe limits,as is the case, for example, with railway motors, hoists and rolling mills.

Spe

ed

Load current

Spe

ed

Current

Differential

Cumulative

Fig. 2.37. (a) Speed-current characte- Fig. 2.37. (b) Speed-current characteristicristic of a series motor. of compound motors.

Compound Motor

A compound motor runs on ‘no-load’ at a speed determined by its shuntwinding since the series field contributes little to the total flux in thiscondition.

In a cumulative compound motor at ‘no-load’, the series field strengthensthe shunt winding so that the speed falls as in a series machine. Sincethe flux at any load is equal to the shunt and series fluxes, the speed isless than it would be if running on either field alone. The speed-currentcharacteristic of such a machine is shown in Fig. 2.37 (b). Such acharacteristic has two important advantages. These are:

(i) The machine has the compensating action of reducing its speed onheavy loads, as is the case with series machine.

(ii) The maximum speed on no-load is limited by the shunt winding, sincethis produces a maximum flux even on no-load.

This type of motor, therefore, is suitable for driving machines which operateon a cycle consisting of a power or working stroke followed by a return or idle stroke.The series winding produces a fall in speed on the working stroke, while the shuntwinding permits the return stroke to be completed at a high, but safe, speed. A fly-wheel is also provided to act as a load equilizer in such a drive.


NOTES


In case of a differential compound motor, since the series winding opposes theshunt, the resultant flux decreases as the load increases; thus the machine runs ata higher speed than it would do as a shunt motor. If the series windings wererelatively weak, this reduction in flux might be just sufficient for the fall in speed,brought about by the volt drop in the machine. Such a motor would have a usefulapplication in driving loads at a constant speed. If the series field were strong,however, an increase in load would result in a decrease in the magnetic flux and arise in speed would take place, the heavier the load, the faster would the motortend to run. This is the property which may have dangerous consequences, since aheavy overload would result in such a high speed that the motor would destroyitself.

Speed-Torque (or Mechanical) Characteristics

The speed-torque characteristics of the four types, i.e. shunt, series, cumulativeand differential of motors drawn on the same diagram are shown in Fig. 2.38 forthe purpose of comparison.

The main properties of individual motors, from this diagram, may besummarised as under:

(i) Shunt Motor. As the load torqueincreases the speed falls somewhat, but themachine may be regarded as an approximatelyconstant speed motor.

The shunt motor is used:

When the speed is required to remainapproximately constant from no-load tofull-load.

When the load has to be driven at anumber of speeds, any one of which isrequired to remain approximatelyconstant.

(ii) Series Motor. As the load torque increases the speed falls rapidly. At lowtorque the speed becomes very high and machine tends to race.

The series motors are used:

When large starting torque is required (as in traction motors).

When the load is subject to heavy fluctuations, and a reduced speed isdesired to compensate for the high torque, provided that there is nopossibility of the machine ‘losing’ its load.

(iii) Cumulative Compound Motor. In this type of motor the speed fallsappreciably as the torque increases, but on low torques the maximum speed is limitedto a safe value. These motors are used:

When a large starting torque is required but when the load may fall so lowthat a series motor would race.

When the load is of a fluctuating nature and a reduced speed is desirableon the heavy loads.

Fig. 2.38. Speed-torquecharacteristic of D.C. motors.

Spe

ed

Torque

Differential

Cumulative

Shunt

Series


NOTES

Electrical MechanicsIn such a case a flywheel is usually fitted so that when speed is so reduced thekinetic energy stored in the flywheel at high speeds is given up to assist the motorin driving the heavy load.

When the supply voltage is subject to fluctuations (as in traction systems).(iv) Differential Compound Motor. The speed at low torque is limited by

the shunt winding, as in the cumulative compound machine. At high torques, thespeed may be arranged to remain constant or, with a stronger series field, thespeed may rise with increasing load.

On very heavy loads the machine may tend to race.Its use is usually restricted to applications which require a constant speed.

Industrial Applications of D.C. Motors

1. Shunt Motors:

(i) Drills and milling machines(ii) Line-shaft drives

(iii) Boring mills(iv) Grinders and shapers(v) Spinning and weaving machines

(vi) Wood working machines(vii) Small printing presses

(viii) Light machine tools generally.

2. Series Motors:

(i) Traction drives generally(ii) Tram cars and railway cars

(iii) Cranes, derricks, hoists, elevators and winches(iv) Fans and air compressors(v) Vacuum cleaners, hair driers, sewing machines

(vi) Universal machines generally.

3. Cumulative Compound Motors:

(i) Punching, shearing and planing machines(ii) Lifts, haulage gears and mine hoists

(iii) Pumps and power fans(iv) Rolling mills, stamping presses and large printing presses(v) Trolley buses.

4. Differential Compound Motors:

(i) Battery boosters

(ii) Experimental and research work.

2.8 BASIC DEFINITIONS: TRANSFORMER

A transformer is a static electromagnetic device designed for the transfor-mation of the (primary) alternating current system into another (second-ary) one of the same frequency with other characteristics, in particulars,other voltage and current.


NOTES


As a rule a transformer consists of a core assembled of sheet transformersteel and two or several windings coupled electromagnetically, and in thecase of autotransformer, also electrically.

A transformer with two windings is called double-wound transformer; atransformer with three or more windings is termed a triple wound ormulti-winding one.

According to the kind of current, transformers are distinguished as single-phase, three-phase and poly-phase ones. A poly-phase transformer windingis a group of all phase windings of the same voltage, connected to eachother in a definite way.

Primary and secondary windings. The transformer winding to whichthe energy of the alternating current is delivered is called the primarywinding; the other winding from which energy is received is called thesecondary winding.

In accordance with the names of the windings, all quantities pertainingto the primary winding as, for example, power, current, resistance etc.,are also primary, and those pertaining to the secondary winding secondary.

h.v. and l.v. windings. The winding connected to the circuit with thehigher voltage is called the high-voltage winding (h.v.), the windingconnected to the circuit with the lower voltage is called the low-voltagewinding. (l.v.). If the secondary voltage is less than the primary one, thetransformer is called a step-down transformer and if more-a step-uptransformer.

A tapped transformer is one whose windings are fitted with special tapsfor changing its voltage or current ratio.

Oil and dry transformers. To avoid the deterimental effect of the airon the winding insulation and improve the cooling conditions of thetransformer its core together with the windings assembled on it isimmersed in a tank filled with transformer oil. Such transformers arecalled oil transformers. Transformers not immersed in oil are calleddry transformers.

2.9 WORKING PRINCIPLE OF A TRANSFORMER

A transformer operates on the principle of mutual inductance, between two(and sometimes more) inductively coupled coils. It consists of two windings in closeproximity as shown in Fig. 2.39. The two windings are coupled by magneticinduction. (There is no conductive connection between the windings). One of thewindings called primary is energised by a sinusoidal voltage. The second winding,called secondary feeds the load. The alternating current in the primary windingsets up an alternating flux (φ) in the core. The secondary winding is linked by mostof this flux and e.m.fs are induced in the two windings. The e.m.f. induced in thesecondary winding drives a current through the load connected to the winding.Energy is transferred from the primary circuit to the secondary circuit throughthe medium of the magnetic field.


NOTES

Electrical MechanicsIn brief, a transformer is a device that:

(i) transfers electric power from one circuit to another;

(ii) it does so without change of frequency; and

(iii) it accomplishes this by electromagnetic induction (or mutual inductance).

ff

Primarywinding

Secondarywinding

Laminated core

Fig. 2.39. Two-winding transformer.

2.10 TRANSFORMER RATINGS

The rated quantities of a transformer, its power, voltage, frequency, etc., aregiven in Manufacturer’s name plate, which should always be arranged so as to beaccessible. But the term ‘rated’ can also be applied to quantities not indicated onthe name plate, but relating to the rated duty, as for example, the rated efficiency,rated temperature conditions of the cooling medium, etc.:

The rated duty of a transformer is determined by the quantities given inthe name plate.

The rated power of the transformer is the power at the secondaryterminals, indicated in the name plate and expressed in kVA.

The rated primary voltage is the voltage indicated in the transformername plate; if the primary is provided with taps, the rated tapped voltageis specially noted.

The rated secondary voltage is the voltage across the transformer secondaryterminals at no-load and with the rated voltage across the primaryterminals; if the secondary winding has taps, then their rated voltage isspecially indicated.

The rated currents of the transformer, primary and secondary, are thecurrents indicated in the name plate of the transformer and calculatedby using the corresponding rated values of power and voltage.


NOTES

Basic Electrical andElectronics Engineering 2.11 KINDS OF TRANSFORMERS

The following kinds of transformers are the most important ones:

1. Power transformers. For the transmission and distribution of electricpower.

2. Auto-transformers. For converting voltages within relatively small limitsto connect power systems of different voltages, to start A.C. motors etc.

3. Transformer for feed installations with static convertors. (Mercuryarc rectifiers, ignitrons, semi-conductor valves, etc.) When converting A.C. into D.C.(rectifying) and convering D.C. into A.C. (inverting).

4. Testing transformers. For conducting tests at high and ultra-high voltages.

5. Power transformers for special applications. Furnace, welding etc.

6. Radio-transformers. It is used in radio engineering etc.Note. Distribution transformers should be designed to have maximum efficiency at a

load much lower than full-load (about 50 per cent).

Power transformers should be designed to have maximum efficiency at or nearfull-load.

2.12 TRANSFORMER CONSTRUCTION

All transformers have the following essential elements:

1. Two or more electrical windings insulated from each other and from thecore (except in auto-transformers).

2. A core, which in case of a single-phase distribution transformers usuallycomprises cold-rolled silicon-steel strip instead of an assembly of punchedsilicon-steel laminations such as are used in the larger power-transformer cores.The flux path in the assembled core is parallel to the directions of steel’s grain or‘orientation’. This results in a reduction in core losses for a given flux density andfrequency, or it permits the use of higher core densities and reduced size oftransformers for given core losses.

Other necessary parts are:

A suitable container for the assembled core and windings.

A suitable medium for insulating the core and its windings from eachother and from the container.

Suitable bushings for insulating and bringing the terminals of thewindings out of the case.

The two basic types of transformer construction are:

1. The core-type.

2. The shell-type.


NOTES

Electrical MechanicsThe above two types differ in their relative arrangements of copper conductorsand the iron cores. In the ‘core-type’, the copper virtually surrounds the iron core,while in the ‘shell-type’, the iron surrounds the copper winding.

Core-type T ransformer

The completed magnetic circuit of the core-type transformer is in the shape ofthe hollow rectangle, exactly as shown in Fig. 2.40 in which I0 is the no-load currentand φ is the flux produced by it. N1 and N2 are the number of turns on the primaryand secondary sides respectively.

ff

V2V2A.C. SupplyA.C. Supply V1V1

I0I0

Primary

Secondary

N1 N2

Fig. 2.40. Magnetic circuit of a core-type transformer.

The core is made up of silicon-steel laminations which are, either rectangularor L-shaped. With the coils wound on two legs the appearance is that of Fig. 2.41.If the two coils shown were the respective high and low-side coils as in Fig. 2.41,the leakage reactance would be much too great. In order to provide maximum linkagebetween windings, the group on each leg is made up of both high-tension and low-tension coils. This may be seen in Fig. 2.42, where a cross-sectional cut is takenacross the legs of the core. By placing the high-voltage winding around thelow-voltage winding, only one layer of high-voltage insulation is required, thatbetween the two coils. If the high-voltage coils were adjacent to the core, an additionalhigh-voltage insulation layer would be necessary between the coils and the ironcore.

Primary andsecondary

coils

Primary andsecondarycoils

Fig. 2.41. Core-type transformer.

CoreLow voltage insulationLow voltage windingHigh voltage insulationHigh voltage winding

Fig. 2.42. Cross-section of a core-type transformer.


NOTES


Fig. 2.43 shows the coils and laminations of a core-type transformer with acruciform core and circular coils.

Fig. 2.43 shows the different types of cores used in core-type transformers.

Rectangular cores [Fig. 2.44 (a)] with rectangular cylindrical coils can be usedfor small size core-type transformers. For large size transformers it becomeswasteful to use rectangular cylindrical coils and socircular cylindrical coils are preferred. For suchpurposes, ‘square cores’ may be used as shown inFig. 2.44 (b) where circles represent the tubularformer carrying the coils. Evidently a considerableamount of useful space is still wasted. A commonimprovement on the square core is to employ a‘cruciform core’ [Fig. 2.44 (c)] which demands,atleast, two sizes of core strips. For very largetransformers, further core stepping is done as inFig. 2.44 (d) where atleast three sizes of core platesare necessary. Core stepping not only gives highspace factor but also results in reduced length ofthe mean turn and the consequent I2R loss. Threestepped core is the most commonly used althoughmore steps may be used for very large transformersas shown in Fig. [2.44 (e)].

(a)

dd

0.71d

0.71d0.71d

(b)

0.53d

0.85d

0.53

d

0.85

d

(c)

(d)

0.42

d

0.70

d

0.90

d

0.42d

0.70d

0.90d

(e)

0.36

d

0.60

d

0.78

d

0.92

d

0.36d

0.60d

0.78d

0.92d

Fig. 2.44. Various types of cores.

Fig. 2.43. Coils and lamina-tions of a core-type trans-

former.


NOTES

Electrical MechanicsShell-type T ransformer

In the shell-type construction the iron almost entirely surrounds the copper(Fig. 2.45). The core is made up of E-shaped or F-shaped laminations which arestacked to give a rectangular figure eight. All the windings are placed on the centreleg, and in order to reduce leakage, each high-side coil is adjacent to a low-side coil.The coils actually occupy the entire space of both windows, are flat or pencake inshape, and are usually constructed of strip copper. Again, to reduce the amount thehigh-voltage insulation required, the low-voltage coils are placed adjacent to theiron core.

Lowvoltagewinding

Highvoltagewinding

f f

Insulation

Fig. 2.45. Shell-type transformer.

Fig. 2.45 shows the coils and laminations of a typical shell-type transformer.

Choice of Core- or Shell-type Construction. In general, the core-type hasa longer mean length of core and a shorter mean length of coil turn. The core-typealso has a smaller cross-section of iron and so will need a greater number of turnsof wire, since, in general, not as high a flux may be reached in the core. However,core-type is better adopted for some high-voltage service since there is more room forinsulation. The shell-type has better provision for mechanically supporting andbracing the coils. This allows better resistance to the very high mechanical forcesthat develop during a high-current short circuit.

Fig. 2.46. Coils and laminations of a shell-type transformer.

The choice of core- or shell type construction is usually one of cost, for similarcharacteristics can be obtained with both types.

Both core and shell forms are used, and selection is based upon many factorssuch as voltage rating, kVA rating, weight, insulation stress, mechanical stress,and heat distribution.


NOTES


Spiral-core T ransformer

The typical spiral core is shown in Fig. 2.47.The core is assembled either of a continuous stripof transformer steel wound in the form of a circularor elliptical cylinder or of a group of short stripsassembled to produce the same elliptical-shapedcore. By using this construction the core fluxalways follows along the grain of the iron. Cold-rolled steel of high silicon content enables thedesigner to use higher operating flux densities withlower loss per kg. The higher flux density reducesthe weight per kVA.

2.13 TRANSFORMER WINDINGS, TERMINALS,

TAPPINGS AND BUSHINGS

T ransformer Windings

The most important requirements of transformer windings are:

1. The winding should be economical both as regards initial cost, with a viewto the market availability of copper, and the efficiency of the transformer in service.

2. The heating conditions of the windings should meet standard requirements,since departure from these requirements towards allowing higher temperaturewill drastically shorten the service life of the transformer.

3. The winding should be mechanically stable in respect to the forces appearingwhen sudden short circuit of the transformer occur.

4. The winding should have the necessary electrical strength in respect to overvoltages.

The different types of windings are classified and briefly discussed below:

1. Concentric windings 2. Sandwich windings

(i) Cross-over

(ii) Helical

(iii) Disc.

Concentric Windings

Refer to Fig. 2.48. These windings are used for core-type transformers. Eachlimb is wound with a group of coils consisting of both primary and secondary turnswhich may be concentric cylinders. The l.v. winding is placed next to the core andh.v. winding on the outside. But the two windings can be sub-divided, and interlacedwith high tension and low tension section alternately to reduce leakage reactance.These windings can be further divided as follows:

Fig. 2.47. Spiral-coretransformer.

Wound cores


NOTES


h.v.

l.v.

Core

l.v.

h.v.

Fig. 2.48. Concentric coils.

(i) Cross-over Windings. Cross-over windings are used for currents up to 20A and so they are suitable for h.v. winding of small transformers. The conductorsare either cotton covered round wires or strips insulated with paper. Cross-overcoils are wound over formers and each coil consists of a number of layers witha number of turns per layer. The complete winding consists of a number of coilsconnected in series. Two ends of each coil are brought out, one from inside andone from outside. The inside end of a coil is connected to the outside end of theadjacent coil.

(ii) Helical Winding. A helical winding consists of rectangular strips woundin the form of a helix. The strips are wound in parallel radially and each turnoccupies the total radial depth of winding.

Helical coils are well suited for l.v. windings of large transformers. They canalso be used for h.v. windings by putting extra insulation between layers inaddition to insulation of conductors.

(iii) Continuous Disc Winding. This type of winding consists of a number offlat strips wound spirally from inside (radially) outwards. The conductor is usedin such lengths as are sufficient for complete winding or section of windingbetween tappings. The conductor can either be a single strip or a number ofstrips in parallel, wound on the flat. This gives a robust construction for eachdisc. The discs are wound on insulating cylinders spaced from it by strips alongthe length of cylinder. The discs are separated from each other with press boardsectors attached to the vertical strips. The vertical and horizontal spacers provideducts for free circulation of oil which is in contact with every turn.

Sandwich Coils

Sandwich coils (Fig. 2.49) are employed in transformers of shell-type. Bothhigh and low voltage windings are split into a number of sections. Each highvoltage section lies between the low voltage sections.

The advantage of sandwich coils is that their leakage can be easily controlledand so any desired value of leakage reactance can be had by the division ofwindings.


NOTES


l.v.

h.v.

l.v.

h.v.

l.v.

Core

Fig. 2.49. Sandwich coils.

T erminals and Leads

The connection to the windings are of insulated copper rods or bars. The shapeand size of leads is important in high voltage transformers owing to dielectric stressand corona which are caused at bends and corners. Connections from windings aredirectly taken to the busbars in the case of air-cooled transformers while they aretaken to insulated bushings in the case of oil-cooled transformers.

T appings

In a supply network the voltage can be controlled by changing thetransformation ratio. This can be done by tapping the winding in order to alter thenumber of turns. The change in number of turns may be effected when thetransformer is out of circuit (known as off load tap changing) or when on load(known as on load tap changing). The tappings are provided on the high voltagewinding because a fine voltage variation is obtained owing to large number of turns.It is difficult to obtain voltage variation within close percentage limits in low voltagewinding as there are few turns and voltage per turn is a large percentage of the totalvoltage.

In transformers, the tappings can be provided at:

(i) phase ends; and

(ii) neutral point or in the middle of the windings.

The advantage of providing tappings at phase ends is that the number ofbushing insulators is reduced, this is important where the cover space islimited. Some transformers have reinforced insulation at the phase ends.It is essential that in such cases either the tapping should not be providedat end turns or the reinforcement should be carried beyond the lower tap.

When the tappings are made at the neutral point the insulation betweenvarious parts is small. This arrangement is economical especially in thecase of high voltage transformers.


NOTES

Electrical MechanicsBushings

The bushings are employed for insulating and bringing out terminals of thewinding from the container to the external circuit. For low-voltage transformersthis is accomplished by employing bushings of porcelain around the conductor atthe point of entry. For high voltages it is necessary to employ bushings of largersizes. In modern transformers the problem is met by using large porcelain orcomposition bushings for voltages as high as 33 kV, above that oil filled or condensertype bushings are used.

2.14 TRANSFORMER COOLING

Cooling Methods

The transformers get heated due to iron and copper losses occurring in them.It is necessary to dissipate this heat so that the temperature of the windings iskept below the value at which the insulation begins to deteriorate. The cooling oftransformers is more difficult than that of rotating machines because the rotatingmachines create a turbulent air flow which assists in removing the heat generateddue to losses. Luckily the losses in transformers are comparatively small.Nevertheless the elaborate cooling arrangements have been devised to deal withthe whole range of sizes.

As far as cooling methods are concerned, the transformers are of followingtwo types:

1. Dry type. 2. Oil immersed type.

Dry Type Transformers. Small transformers upto 25 kVA size are of the drytype and have the following cooling arrangements:

(i) Air Natural. In this method the natural circulation of surrounding air isutilized to carry away the heat generated by losses. A sheet metal enclosure protectthe winding from mechanical injury.

(ii) Air Blast. Here the transformer is cooled by acontinuous blast of cool air forced through the core andwindings (Fig. 2.50). The blast is produced by a fan.The air supply must be filtered to prevent accumulationof dust in ventilating ducts.

Oil Immersed Transformers. In general mosttransformers are of oil immersed types. The oil providesbetter insulation than air and it is a better conductor ofheat than air. Mineral oil is used for this purpose.

Oil immersed transformers are classified asfollows:

(i) Oil Immersed Self-cooled Transformers. The transformer is immersed inoil and heat generated in cores and windings is passed to oil by conduction. Oil incontact with heated parts rises and its place is taken by cool oil from the bottom.

Core

Fan

Coi

ls

Fig. 2.50


NOTES


The natural oil transfers its heat to thetank walls from where heat is taken awayby the ambient air. The oil gets cooler andfalls to the bottom from where it isdissiptated into the surroundings. Thetank surface is the best dissipator of heatbut a plain tank will have to be excessivelylarge, if used without any auxiliary meansfor high rating transformers. As both spaceand oil are costly, these auxiliary meansshould not increase the cubic capacity ofthe tank. The heat dissipating capacity canbe increased by providing (i) corrugations,(ii) fins, (iii) tubes (Fig. 2.51) and (iv) radiator tanks.

The advantages of ‘oil natural’ cooling is that it does not clog the ducts and thewindings are free from effects of moisture.

(ii) Oil Immersed Forced Air-cooled Transformers. In this type of cooling, airis directed over the outer surfaces of the tank of the transformer immersed in oil.

(iii) Oil Immersed Water-cooled Transformers. Heat is extracted from the oilby means of a stream of water pumped through a metallic coil immersed in the oiljust below the top of the tank. The heated water is in turn cooled in a spray pond ora cooling tower.

(iv) Oil Immersed Forced Oil-cooled Transformers. In such transformers heatis extracted from the oil by pumping the oil itself upward through the winding andthen back by way of external radiators which may themselves be cooled by fans.The extra cost of oil pumping equipment must of course be economically justifiedbut it has incidentally the advantage of reducing the temperature difference betweenthe top and bottom of enclosing tank.

Transformertank Windings

Conservator

Buchholz relay

Radiator

Pump

Fan

Aircirculation

Fig. 2.52. Air blast cooling radiator.

Transformerbody

Coolingtubes

Fig. 2.51. Transformer withcooling tubes.


NOTES

Electrical MechanicsFig. 2.52 shows the cooling of transformers having capacities from 10000 kVAand higher. In such cases air blast cooling of radiator is used.

Transformer Oil

It is a mineral oil obtained by refining crude petroleum. It serves the followingpurposes:

• Provides additional insulation.

• Carries away the heat generated in the core and coils.

• Protects the paper from dirt and moisture.

The transformer oil should possess the following properties:

1. High dielectric strength.

2. Low viscosity to provide good heat transfer.

3. Good resistance to emulsion.

4. Free from inorganic acid, alkali and corrosive sulphur.

5. Free from sludging under normal operating conditions.

6. High flash/fire point.

Conservator and Breather

Conservator. The oil should not be allowed to come in contact withatmospheric air as it may take up moisture which may spoil its insulating properties.Also air may cause acidity and sludging of oil. To prevent this, many transformersare provided with conservators. The function of a conservator (Fig. 2.52) is to takeup contraction and expansion of oil without allowing it to come in contact withoutside air. The conservator consists of an air tight metal-drum fixed above thelevel of the top of the tank and connected with it by a pipe. The main tank iscompletely filled with oil when cold. The conservator is partially filled with oil. Sothe oil surface in contact with air is greatly reduced. The sludge thus formed remainsin the conservator itself and does not go to the main tank.

Breather. When the temperature changes, the oil expands or contracts andthere is a displacement of air. When the transformer cools, the oil level goes down,and air is drawn in. This is known as breathing. The air, coming in, is passedthrough an apparatus called breather for the purpose of extracting moisture. Thebreather consists of a small vessel which contains a drying agent like silica getcrystal impregnated with cobalt crystal.

Note. Sludging means the slow formation of solid hydrocarbons due to heatingand oxidation. The sludge deposit itself on the windings and cooling ducts producingoverheating. This makes transformer still hotter producing more sludge. This processmay continue till the transformer becomes unusable due to overheating. So the contactof oil with air should be avoided as the air contains oxygen.


NOTES

Basic Electrical andElectronics Engineering 2.15 SINGLE PHASE TRANSFORMER

Elementary Theory of an Ideal Transformer

The basic theory of a transformer is not difficult to understand. To simplifymatters as much as possible, let us first consider an ideal transformer, that is, onein which the resistance of the windings is negligible and the core has no losses.

Let the secondary be open (Fig. 2.53), and let a sine wave of potential differencev1 (Fig. 2.54) be impressed upon the primary. The impressed potential differencecauses an alternating current to flow in the primary winding. Since the primaryresistance is negligible and there are no losses in the core, the effective resistanceis zero and the circuit is purely reactive. Hence the current wave im lags the

ff

V1V1 V2V2E1 E2

Primarywinding

(N1 turns)

Secondarywinding

(N2 turns)

Induced emfin secondary = E2

Induced emfin primary = E1

Fig. 2.53. Elementary diagram of an ideal transformerwith an open secondary winding.

impressed voltage wave v1 by 90 time degrees, as shown in Fig. 2.55. The reactanceof circuit is very high and the magnetizing current is very small. This current inthe N1 turns of the primary magnetizes the core and produces a flux φ that is at alltimes proportional to the current (if the permeability of the circuit is assumed tobe constant), and therefore in time phase with the current. The flux, by its rate ofchange, induces in the primary winding E1 which at every instant of time is equalin value and opposite in direction to V1. It is called counter e.m.f. of the primary.The value which the primary current attains must be such that the flux which itproduces in the core is of sufficient value to induce in the primary the requiredcounter e.m.f.

V1V1

0 1m

f

u1

e1

e2

2pp

V1

90°90°

90°90°

IIm

E2

f

E1

Fig. 2.54, 2.55. Current, voltage and flux curvesof an ideal (no loss) transformer.


NOTES

Electrical MechanicsSince the flux also threads (or links) the secondary winding a voltage e2 isinduced in the secondary. This voltage is likewise proportional to the rate of changeof flux and so is in time phase with e1, but it may have any value depending uponthe number of turns N2 in the secondary.

E.M.F. Equation of a Transformer

Let N1 = number of turns in primary,

N2 = number of turns in secondary,

φm = maximum flux in the core, Wb.

= Bm × A, [where Bm is the maximum flux density in the core andA is the core area], and

f = frequency of a.c. input, Hz.

fmax.fmax.

–fmax.–fmax.T4

14f

=

T 1f

=

CycleCycle

Flu

x,f

Time, t

Fig. 2.56

Refer Fig. 2.56. Since the flux increases from its zero value to maximum value

φm in one quarter of the cycle i.e., in T

orf4

14

second (T being time-period of the

cycle),

∴ Average rate of change of flux = φ

φm

f

f1

4

4= max Wb/s or volt

If flux φ varies sinusoidally, then r.m.s. (root mean square) value of inductede.m.f. is obtained by multiplying the average value with form factor.

But, form factor = r.m.s. value

average value = 1.11

∴ r.m.s. value of e.m.f./turn= 1.11 × 4fφmax = 4.44fφmax volt

Now, r.m.s. value of induced e.m.f. in the whole of primary winding, E1 = 4.44fφmax N1 ...(2.12)

Similarly r.m.s. value of induced e.m.f. in secondary is, E2 = 4.44fφmax N2 ...(2.13)

In an ideal transformer on no-load V1 = E1 and V2 = E2 (Fig. 2.53).


NOTES


Voltage Transformation Ratio (K)

The transformation ratio is defined as the ratio of the secondary voltage toprimary voltage. It is denoted by the letter K.

From eqns. (2.12) and (2.13), EE

NN

2

1

2

1

= = K ...(2.14)

If N2 > N1 i.e., K > 1, then transformer is called step-up transformer.

If N2 < N1 i.e., K < 1, then transformer is called step-down transformer.

Again for an ideal transformer

Input (VA) = Output (VA)

V1I1 = V2I2 or E1I1 = E2I2

orII

EE

NN K

2

1

1

2

1

2

1= = = ...(2.15)

i.e., Primary and secondary currents are inversely proportional to their respectiveturns.

Example 2.3. A 40 kVA, single phase transformer has 400 turns on the primaryand 100 turns on the secondary. The primary is connected to 2000 V, 50 Hz supply.Determine:

(i) The secondary voltage on open circuit.

(ii) The current flowing through the two windings on full-load.

(iii) The maximum value of flux.

Solution. Rating = 40 kVA

Primary turns, N1 = 400

Secondary turns, N2 = 100

Primary induced voltage, E1 = V1 = 2000 V

(i) Secondary voltage on open, circuit V2:

Using the relation, EE

NN

2

1

2

1

=

E2 = E1 × NN

2

1

∴ E2 = V2 = 2000 × 100400

= 500 V

Hence, V2 = 500 V. (Ans.)

(ii) Primary current, I1:

Secondary current, I2:

Primary full-load current, I1 = kVA 1000

V40 1000

20001

× = × = 20 A (Ans.)


NOTES


Secondary full-current, I2 = kVA 1000

V40 1000

5002

× = × = 80 A (Ans.)

(iii) Maximum value of flux, φmax:

Using e.m.f. equation, E1 = 4.44fφmax N1

∴ 2000 = 4.44 × 50 × φmax × 400

∴ φmax = 2000

4 44 50 400. × × = 0.0225 Wb.

Hence, φmax = 0.0225 Wb. (Ans.)

Example 2.4. The voltage per turn of a single-phase transformer is 1.1 V.When the primary winding is connected to a 220 V, 50 Hz A.C. supply, the secondaryvoltage is found to be 550 V. Find:

(i) Primary and secondary turns.

(ii) Core area if the maximum flux density is 1.1 T.

Solution. Voltage per turn = 1.1 V

Primary, E1 = 220 V

Secondary, E2 = 550 V

Max. flux density, Bmax = 1.1 T

(i) Primary turns, N1 = E1

1.12201.1

= = 200. (Ans.)

Secondary turns, N2 = E2

1.15501.1

= = 500. (Ans.)

(ii) Core area A:

Using the relation, E1 = 4.44fφmax N1

220 = 4.44 × 50 × φmax × 200

∴ φmax = 220

4.44 50 200× × = 0.004955 Wb

Core area, A = φmax

maxB= 0.004955

1.1 0.004504 m2

= 45.04 cm2. (Ans.)

Transformer Losses

The losses in a transformer are classified as follows:

1. Iron losses (or core losses).

2. Copper losses.

Iron or Core Losses. It includes hysteresis loss and eddy current loss.

(i) Hysteresis Loss. Since the flux in a transformer core is alternating, poweris required for the continuous reversals of the elementary magnets of which theiron is composed. This loss is known as hysteresis loss.


NOTES


Hysteresis loss = Khf B1.6max ...(2.16)

where, f is the frequency in Hz, Bmax is the maximum flux density in core and Khis a constant.

(ii) Eddy Current Loss. This is due to the flow of eddy currents in the core.Thin laminations, insulated from each other, reduce the eddy current loss to smallproportion.

Eddy current loss = Kef2 B2

max ...(2.17)

where Ke is a constant.

Iron or core loss is found from open circuit test. The input of the transformerwhen on no-load measures the core loss.

Copper Losses. These losses are due to the ohmic resistance of the transformerwindings.

Total copper loss = I12R1 + I2

2R2 = I12R01 + I2

2R02

These losses, as is evident, are proportional to square of the current (or kVA)2.

The value of copper losses is found from the short-circuit test.

Transformer Efficiency

The efficiency of a transformer at a particular load and power factor is definedas the ratio of power output to power input.

∴ Efficiency = outputinput

= output

output + lossesoutput

output + cu loss + iron loss= ...(2.18)

or Efficiency = input losses

input1

lossesinput

− = − ...(2.19)

It may be noted that efficiency is based on power output in watts and not involt-amperes, although losses are proportional to volt-ampere. Hence at anyvolt-ampere load, the efficiency depends on power factor, being maximum at unitypower factor.

Efficiency can be calculated by determining core losses from open-circuit testand copper losses from short-circuit test.

Condition for Maximum Efficiency:

Iron losses, Pi = hysteresis loss + eddy current loss

= Ph + Pe

Copper losses,Pc = I12R01 or I2

2R02

Considering Primary Side:

Input to primary = V1I1 cos φ1

Efficiency, η = V I

V IV I I R P

V Ii1 1 1

1 1 1

1 1 1 12

01

1 1 1

coscos

coscos

φφ

φφ

− = − −losses


NOTES


= 1 – I R

VP

I Vi1 01

1 1 1 1 1cos cosφ φ=

Differentiating both sides w.r.t. I1, we get

ddI

RV

P

V Iiη

φ φ1

01

1 1 1 12

1

0= − +cos cos

For η to be maximum,

ddI

η

1 = 0. Hence the above equation reduces to

RV

P

V Ii01

1 1 1 12

1cos cosφ φ+

or Pi = I12R01 or I2

2R02 ...(2.20)

or Copper losses = Iron losses

The output current corresponding to maximum efficiency is

I2 = P

Ri

02...(2.21)

By proper design it is possible to make the maximum efficiency occur at anydesired load.

2.16 INDUCTION MACHINES

Introduction. An induction motor is simply an electric transformer whosemagnetic circuit is separated by an air gap into two relatively movable portions,one carrying the primary and the other, the secondary winding. Alternating currentsupplied to the primary winding from an electric power system induces an opposingcurrent in the secondary winding, when latter is short-circuited or closed throughan external impedance. Relative motion between the primary and secondarystructures is produced by the electromagnetic forces corresponding to the powerthus transferred across the air gap by induction.

The essential feature which distinguishes the induction machine from othertypes of electric motors is that the secondary currents are created solely by induction,as in a transformer instead of being supplied by a D.C. exciter or other externalpower source, as in synchronous and D.C. machines.

Advantages. Three-phase induction motor is the most commonly used motorin industrial applications because of the advantages listed below:

1. Simple design

2. Rugged construction

3. Reliable operation

4. Low initial cost

5. Easy operation and simple maintenance


NOTES


6. High efficiency

7. Simple control gear for starting and speed control.

Applications. Induction motors are available with torque characteristics suit-able for a wide variety of loads as follows:

(i) The standard motor has a starting torque of about 120 to 150 per cent offull-load torque. Such motors are suitable for most applications.

(ii) For starting loads such as small refrigerating machines or plunger pumpsoperating against full pressure or belt conveyors, high torque motors with a startingtorque of twice normal full-load torque, or more, are used.

(iii) For driving machines that use large flywheels to carry peak loads, such aspunch presses and shears, a high-torque motor with a slip at full-load up to 10 percent is available. The high slip permits enough change in speed to make possible theproper functioning of the flywheel.

(iv) By the use of a wound-rotor with suitable controller and externalresistances connected in series with the rotor winding, it is possible to obtain anyvalue of starting torque up to the maximum breakdown torque. Such motors arewell adapted as constant-speed drives for loads that have large friction loads toovercome at starting.

2.17 CLASSIFICATION OF A.C. MOTORS

Different A.C. motors may be classified as follows:

1. According to the ‘type of current’:

(i) Single-phase (ii) Three-phase.

2. According to ‘speed’:

(i) Constant speed (ii) Variable speed

(iii) Adjustable speed.

3. According to ‘principle of operation’:

(A) Synchronous motors(i) Plain (ii) Super.

(B) Asynchronous motors (a) Induction motors:

(i) Squirrel cage—single—double

(ii) Slip-ring (external resistance)(b) Commutator motors:

(i) Series

—single-phase

—universal

NOTES


Electrical Mechanics(ii) Compensated

—conductively

—inductively.

4. According to ‘structural features’:

(i) Open (ii) Enclosed

(iii) Semi-enclosed (iv) Ventilated

(v) Pipe-ventilated (vi) Riveted frame eye

(vii) Splash proof (viii) Totally enclosed fan-cooled

(ix) Explosion proof (x) Water proof.

2.18 CONSTRUCTIONAL DETAILS

The Stator

The stator frame consists of a symmetrical and substantial casting, havingfeet cast integral with it. The stator core, consisting of high grade, lowloss electrical sheet-steel stampings, is assembled in the frame underhydraulic pressure. The thickness of stampings/laminations is usuallyfrom 0.35 to 0.6 mm. The stator laminations are punched in one piece forsmall induction motor (Fig. 2.57). In induction machines of large size thestator core is assembled from a large number of segmental laminations.

The slots are sometimes of the ‘open type’(i.e., having parallel walls) for the accommo-dation of former wound coils. But the usualpractice is to have practically ‘enclosed slots’ inorder to reduce the effective length of air-gap.

The stator windings are given theutmost care to make themmechanically and electrically sound,so as to ensure long life and highefficiency. After the winding is inposition it is thoroughly dried outwhilst still hot and is completelyimmersed in a high grade syntheticresin varnish. It is then acid, alkali,moisture and oil proof.

For small motors working at ordinary voltages, single layer mushwinding is used. For medium size machines double layer lap windingwith diamond shaped coils is used. Single layer concentric windings areused for large motors working at high voltages.

Frames of electrical machines house the stator core. Frames of small andmedium sizes of induction motors have hollow cylindrical form and thatof larger motors have the shape of a circular box. In small induction motors,

Fig. 2.57. Stator stamping.


NOTES


Fig. 2.58. Rotor stamping.

having a frame diameter of up to about 150 cm, the frame also supportsthe end shields. The frame should be strong and rigid as rigidity is veryimportant in the case of induction motors of large dimensions. This isbecause the length of the air gap is very small and if the frame is not rigid,it would create an irregular air gap around the machine resulting inproduction of unbalanced magnetic pull. Frames for small machines aremade as a single unit and are usually cast. The frames of medium andlarge sized machines are fabricated from rolled steel plates.

The Rotor

The rotors are of two types:

1. Squirrel-cage;

2. Wound rotor.

Squirrel-cage. The squirrel-cage rotor is made up of stampings (Fig. 2.58)which are keyed directly to the shaft. The slots are partially closed and the windingconsists of embedded copper bars to which the short-circuited rings are brazed.The squirrel cage rotor is so robust that it is almost indestructible.

The great majority of present dayinduction motors are manufactured withsquirrel-cage rotors, a common practice beingto employ winding of cast aluminium. In thisconstruction the assembled rotor lamin-ations are placed in a mould after whichmolten aluminium is forced in, underpressure, to form bars, end rings and coolingfans as extension of end rings. This is knownas die cast rotor and has become very popularas there are no joints and thus there is nopossibility of high contact resistance.

In this type of rotor, it may be noted thatslots are not made parallel to the shaft butthey are ‘skewed’ to serve the following pur-poses:

(i) To make the motor run quietly by reducing the magnetic hum.(ii) To reduce the locking tendency of the rotor.Wound Rotor. The wound rotor has also slotted stampings and the windings

are former wound. The wound rotor construction is employed for induction motorsrequiring speed control or extremely high values of starting torque. The wound rotorhas completely insulated copper windings very much like the stator windings. Thewindings can be connected in star or delta and the three ends are brought out atthe three slip rings. The current is collected from these slip rings with carbonbrushes from which it is led to the resistances for starting purposes. When themotor is running, the slip rings are short-circuited by means of a collar, which ispushed along the shaft and connects all the slip rings together on the inside. Usuallythe brushes are provided with a device for lifting them from the slip rings when themotor has started up, thus reducing the wear and the frictional losses.

NOTES



Terminal boxStatorwinding

Statorpunchings

Ballbearings

Ballbearing

EndshieldsAir deflector andinner dust cap

RotorBearinghousing

Outer dust capOuter dust capTerminal box

cover

Fig. 2.59. Component parts of a small squirrel-cage induction motor.

The number of slots in the rotor should never be equal to the number of slots inthe stator. If they are, there would be a variation of reluctance of the magnetic pathfrom maximum, when teeth are opposite slots, to minimum when teeth are oppositeteeth. The resulting flux pulsations would have a high frequency, since the periodictime would be the interval period for a tooth to occupy similar positions oppositetwo successive teeth. This will not only cause extra iron loss but the rotor will tendto lock with the stator if at the time of starting teeth are opposite teeth. The bestplan is to make the number of the stator and the rotor teeth prime to each other.

Skewed rotor

Slip-rings

Fig. 2.60. Induction motor with phase-wound rotor showing thethree slip rings on the rotor shaft.

Figs. 2.61 and 2.62 show squirrel-cage and phase-wound induction motorsrespectively.


NOTES


Stator Mainswitch

Mainswitch

Slip-rings

Startingresistance

Fig. 2.61. Squirrel-cage motor. Fig. 2.62. Phase-wound motor connected to athree-phase star-connected starting resistance.

Advantages of a squirrel-cage motor over a phase-wound inductionmotor. As compared with a wound rotor a squirrel-cage induction motor entailsthe following advantages:

1. Slightly higher efficiency.2. Cheaper and rugged in construction.3. No slip rings, brush gear, short-circuiting devices, rotor terminals for starting

rheostats are required. The star-delta starter is sufficient for starting.4. It has better space factor for rotor slots, a shorter overhang and consequently

a small copper loss.5. It has a smaller rotor overhang leakage which gives a better power factor

and a greater pull out torque and overload capacity.6. It has bare end rings, a large space for fans and thus the cooling conditions

are better.The major ‘disadvantage’ of squirrel-cage motor is that it is not possible to

insert resistance in the rotor circuit for the purpose of increasing the starting torque.The cage rotor has a smaller starting torque and large starting current as comparedwith wound rotor.

Slip Rings

The slip rings for wound rotor machines are made of either brass or phosphorbronze. They are shrunk on to a cast iron sleeve with moulded silica insulation.This assembly is passed on to the rotor shaft. The slip rings are rotated eitherbetween the core and the bearing or on the shaft extension. In the latter case theshaft is made hollow to allow the three connections from rotor to slip rings to passthrough bearings.

Shaft and Bearings

In an induction motor the air gap is made as small as possible. Therefore theshaft is made short and stiff in order that the rotor may not have any significantdeflection, as even a small deflection would create large irregularities in the air gapwhich would lead to production of an unbalanced magnetic pull. There is also a

NOTES


Electrical Mechanicspossibility of rotor and stator fouling with each other. Ball and roller bearings aregenerally used as with their use, accurate centering is much simpler than withjournal bearings. Also the overall length of machine is reduced. For small motors,a roller bearing may be used at the driving end and a ball bearing at the non-driving end. For large and heavy rotors journal bearings are used.

2.19 THEORY OF OPERATION OF AN INDUCTION

MOTOR

When a three-phase is given to the stator winding a rotating field is set-up.This field sweeps past the rotor (conductors) and by virtue of relative motion, ane.m.f. is induced in the conductors which form the rotor winding. Since this windingis in the form of a closed circuit, a current flows, the direction of which is, by Lenz’slaw, such as to oppose the change causing it.

Now, the change is the relative motion of the rotating field and the rotor, sothat, to oppose this, the rotor runs in the same direction as the field and attempts tocatch up with it. It is clear that torque must be produced to cause rotation, and thistorque is due to the fact that currents flow in the rotor conductors which are situ-ated in, and at right angles to, a magnetic field.

Stator

Rotor

Ns

Force

Np

S p

Rotatingfield

Ns

N – NsN

Rotor

N = Synchoronous speed, r.p.m.N = Rotor speed, r.p.m.S = South poleN = North pole

s

p

p

×

Fig. 2.63. Induction motor action.

Fig. 2.63 shows the induction motor action.

When the motor shaft is not loaded, the machine has only to rotate itselfagainst the mechanical losses and the rotor speed is very close to thesynchronous speed. However, the rotor speed cannot become equal to thesynchronous speed because if it does so, the e.m.f. induced in the rotorwinding would become zero and there will be no torque. Hence the speedremains slightly less than the synchronous speed. If the motor shaft isloaded, the rotor will slow down and the relative speed of the rotor withrespect to the stator rotating field will increase. The e.m.f. induced in therotor winding will increase and will produce more rotor current whichwill increase the electromagnetic torque produced by the motor. Conditionsof equilibrium are attained when the rotor speed has adjusted to a newvalue so that the electromagnetic torque is sufficient to balance themechanical or load torque applied to the shaft. The speed of the motorwhen running under full load conditions is somewhat less than the no-loadspeed.


NOTES

Basic Electrical andElectronics Engineering 2.20 SLIP

As earlier stated, the rotor speed must always remain less than thesynchronous speed. The difference between the synchronous speed and therotor speed is known as ‘slip’. It is usually expressed as a fraction of thesynchronous speed. Thus slip s is

s = N N

Ns

s

−...(2.22)

or N = Ns(1 – s) ...(2.23)

where Ns = synchronous speed (r.p.m.)

N = motor speed (r.p.m.).

In practice the value of slip is very small. At no-load, slip is around 1% or soand at full-load it is around 3%. For large efficient machines the slip at full-loadmay be around 1% only. The induction motor, is therefore, a motor with substantiallyconstant speed and fills the same role as D.C. shunt motor.

When the rotor is stationary (standstill) its speed is zero and s = 1. Therotor cannot run at synchronous speed because then there will be no rotore.m.f. and no rotor current and torque. If the rotor is to run at synchronousspeed an external torque is necessary. If the rotor is driven such thatN > Ns, the slip becomes negative, the rotor torque opposes the externaldriving torque and the machine acts as induction generator.

The induction motor derives its name from the fact that the current in therotor circuit is induced from the stator. There is no external connection tothe rotor except for some special purposes.

If the rotor reactance at standstill is X2 its value at slip ‘s’ becomes sX2. Thisis very desirable, for at no-load the reactance becomes almost negligible and therotor impedance is now all resistance. Further if the rotor resistance is smallthe rotor current is large, so that motor works with a large torque which brings thespeed near to synchronous speed, i.e., the slip is reduced.

2.21 FREQUENCY OF ROTOR CURRENT

At standstill (i.e., when the rotor is stationary), the frequency of the rotorcurrent is the same as the supply frequency (f). But when the rotor starts revolving,then the frequency depends upon the relative speed or slip-speed. If fr is the fre-quency of the rotor current, then

Ns – N = 120f

pr

...(i)

Also Ns = 120f

p...(ii)

NOTES


Electrical MechanicsDividing (i) by (ii), we get

N N

Nff

s

s

r− = or s = ffr

or fr = sf ...(2.24)

2.22 ROTOR E.M.F. AND ROTOR CURRENT

Rotor e.m.f

When the rotor is stationary, an induction motor is equivalent to a 3-phasetransformer with secondary short-circuited. Therefore, the induced e.m.f. per phaseE2 in the rotor at the instant of starting is given as:

E2 = E1 × NN

2

1...(2.25)

where E1 = applied voltage per phase to primary i.e., stator winding,

N1 = number of stator turns, and

N2 = number of rotor turns.

When the rotor starts gaining speed, the relative speed of the rotor with re-spect to stator flux i.e., slip, is decreased. Hence induced e.m.f. in the rotor, whichis directly proportional to the relative speed i.e., slip, is also decreased and is givenby sE2. Hence for slip ‘s’, the induced e.m.f. in the rotor is s times the induced e.m.f.in the rotor at standstill.

Rotor Current

Let R2 = rotor resistance/phase,

L2 = rotor inductance/phase, and

E2 = induced e.m.f. of rotor/phase at standstill.

At Standstill

Induced e.m.f. of rotor/phase = E2

Rotor winding resistance/phase = R2

Rotor winding reactance/phase, X2 = 2πfL2 where f is the supply frequency

Rotor impedance/phase, Z2 = R X22

22+

∴ Rotor current/phase = EZ

E

R X

2

2

2

22

22

=+

.

At Slip ‘s’

Induced e.m.f. of rotor/phase = sE2

Rotor winding resistance = R2

Rotor winding reactance = 2πfrL2 = 2πsfL2 = s(2πfL2) = sX2


NOTES

Basic Electrical andElectronics Engineering Rotor winding impedance/phase = R sX2

22

2+ ( )

∴ Rotor current/phase, I2 = sE

R sX

2

22

22+ ( )

= sE

R s X

E

R s X

2

22 2

22

2

22

22+

=+( / )

...(2.26)

The rotor currrent I2 lags the rotor voltage E2 by rotor power factor angle φ2given by

φ2 = tan–1 sXR

2FHG

IKJ

Power factor of rotor current,

cos φ2 = R

R s X

R s

R s X

2

22 2

22

2

22

22+

=+

/

( / )...(2.27)

2.23 TORQUE AND POWER

The torque of an induction motor (being due to interaction of a rotor andstator fields),

T ∝ φI2 cos φ2

where φ = flux of rotating stator,

I2 = rotor current/phase, and

cos φ2 = rotor power factor.

Since rotor e.m.f./phase at standstill, E2 ∝ φ∴ T ∝ E2I2 cos φ2

or T = kE2I2 cos φ2 where k is any constant ...(2.28)

T = kE2 sE

R s X

R

R s X

2

2 22

2

2

2 22

2+×

+

i.e., T = ksR E

R s X2 2

2

22 2

22+

...(2.29)

Starting Torque

At start slip ‘s’ = 1. Therefore, expression for starting torque may be obtainedby putting s = 1 in eqn. 2.29.

Starting torque, Tst = kR E

R X2 2

2

22

22+

. ...(2.30)

NOTES


Electrical MechanicsCondition for Maximum Torque

The value of torque when motor is running is given by

T = ksR E

R s X2 2

2

22 2

22+

Torque will be maximum when,

sR

R s X2

22 2

22+

orR

Rs

sX

2

22

22+

or RR

sX s R X

2

22 2 22−

FHG

IKJ

+ is maximum, viz.,

R

s2 – X2 s = 0

or s (= smT) = RX

2

2

...(2.31)

(where smT = slip corresponding to maximum torque)

∴ Maximum torque, Tmax = kE

X2

2

22...(2.32)

From the above expression, the following conclusions can be drawn:

Maximum torque is independent of rotor circuit resistance. Maximum torque varies inversely as standstill reactance of the rotor.

Therefore to have maximum torque, standstill reactance (i.e., inductance)of the rotor should be kept as small as possible.

The slip at which the maximum torque occurs depends upon the resistanceof the rotor.

The condition for getting maximum torque at starting can be obtained byputting s = 1 in eqn. (2.32).

Thus, starting torque will be maximum if

RX

2

2 = s = 1 or R2 = X2.

Starting Torque of a Squirrel-cage Motor

The squirrel-cage rotor resistance is fixed and small as compared to its reactancewhich is very large especially at start (because at standstill the frequency of rotorcurrent is equal to that of supply frequency). Hence, the starting current I2 of therotor, though very large in magnitude, lags by a very large angle behind E2;consequently the starting torque per ampere is very poor. It is roughly 1.5 times thefull-load torque although the starting current is 5 to 7 times the full-load current.Thus such motors are not suitable for applications where these have to be startedagainst heavy loads.

Starting Torque of a Slip Ring Motor

In a slip ring motor the torque is increased by improving its power factor byadding external resistance in the rotor circuit from the star- connected rheostat; as


NOTES


the motor gains speed the rheostat resistance is gradually cut out. This additionalresistance, however, increases the rotor impedance and so reduces the rotor current.At first, the effect of improved power factor predominates the current-decreasingeffect of impedance, hence starting torque is increased. But after a certain point,the effect of increased impedance predominates the effect of improved power factorand so the torque starts decreasing.

Torque-Slip and Torque-Speed Curves

The expression for torque is as follows:

T = ksR E

R s X2 2

2

22 2

22+

From the above expression, it is evident, that

Torque is zero when slip s = 0 (i.e., speed is synchronous).

When slip ‘s’ is very low the value of the term sX2 is very small and isnegligible in comparison with R2, therefore torque T is approximatelyproportional to slip ‘s’ if rotor resistance R2 is constant. This means thatat speeds near to synchronous speed the torque-speed and torque-slipcurves are approximately straight lines (Figs. 2.64 and 2.65).

When the slip ‘s’ increases (i.e., as the speed decreases with increase in

load) torque increases and reaches its maximum value when s = RX

2

2.

The maximum torque is also known as ‘pull-out’ or ‘break-down’ torque.

When the slip is further increased the torque decreases. The result isthat motor slows down and eventually stops. The motor operates for thevalue of slip between zero and that corresponding to maximum torque.

With higher slip, R2 becomes negligible as compared to sX2 and torque variesas follows:

T ∝ s

s X s22

2

1∝ if standstill reactance is constant.

0 0.25 Ns 0.5 Ns 0.75 Ns Ns

Tor

que

With large rotorresistance

With medium resistance

With low resistance

Fig. 2.64. Torque-speed curves.

NOTES



0 0.25 0.5 0.75 1.0

Slip

Tor

que

With low rotorresistance

With medium rotor resistance

With large rotor resistance

Fig. 2.65. Torque-slip curves.

This means that speed-torque or slip-torque curves are rectangular hyperbolawith the speed or slip beyond that corresponding to maximum torque. Figs. 2.64and 2.65 show the torque-speed and torque-slip curves for different values of rotorresistance. It is observed that although maximum torque is independent of rotorresistance R2, yet the exact location of Tm is dependent on it. Greater the R2, greateris the value of slip at which maximum torque occurs.

Example 2.5. A 3-phase, 4-pole, 50-Hz induction motor is running at 1440r.p.m. Determine the slip speed and slip.


Frequency, f = 50 Hz

Actual speed of rotor,N = 1440 r.p.m.

Slip speed =?

Slip =?

Synchronous speed,Ns = 120 120 50

4f

p= ×

= 1500 r.p.m.

Slip speed = Ns – N = 1500 – 1440 = 60 r.p.m. (Ans.)

Slip, s = N N

Ns

s

− = −1500 14401500 = 0.04. (Ans.)

Example 2.6. A 3-phase alternator having 12-poles is driven at a speed of 500r.p.m. It supplies power to an 8-pole, 3-phase induction motor. If the slip of themotor at full-load is 4%, calculate the full-load speed of the motor.

Solution. Number of poles of the alternator = 12

Speed of alternator = 500 r.p.m.

Number of poles of induction motor = 8

Percentage slip = 4%

Full-load speed, N:

Supply frequency, f = 12 500

120×

= 50 Hz


NOTES

Basic Electrical andElectronics Engineering Synchronous speed, Ns =

120 508×

= 750 r.p.m.

∴ % slip = N N

Ns

s

− × 100

4 = 750

750− N

× 100

∴ N = 750 – 4 750

100×

= 720 r.p.m.

Hence, full-load speed, N = 720 r.p.m. (Ans.)

2.24 INDUCTION GENERATOR

Any induction motor, if driven above its synchronous speed when connectedto an A.C. power source, will deliver power to the external circuit.

This generator operation is easily visualised for motor-circle diagramcorresponding to the lower half of the circle in which the current vector isdescribed below the OS line. A unique feature is that the power factor ofthe output is fixed in value by the generator characteristics and is alwaysleading, independent of the external circuit. The explanation is that thegenerator draws all its excitation from the system and so must receive adefinite amount of lagging kVA for a given voltage and load currents. Forthis reason, induction generators alone cannot supply a power system butmust always operate in parallel with synchronous machines or withcapacitors. They are, therefore, no more helpful in system stability thanthe addition of parallel reactors with a rating equal to the generatormagnetizing reactance.An induction generator delivers an instantaneous 3-phase short-circuitcurrent equal to the terminal voltage divided by its standstill reactance,but its rate of decay is much faster than that of a synchronous generator ofthe same rating, and its sustained short-circuit current is zero.Such an induction generator must have a laminated rotor, to provide forthe slip frequency rotor magnetic field, its construction is not adapted tohigh speeds as synchronous machines employing solid steel rotors. Forthese various reasons, induction generators have found few practicalapplications, their chief use being perhaps in variable-ratio frequencyconverter sets, where the induction end of the set operates as a motor or agenerator depending on the direction of power flow through the set.

The principle of the induction generator is important, however, in thedynamic braking of induction motors; the machine acts as an overspeeddevice and produces braking action when the motor speed exceedssynchronous speed, such an induction motor will automatically becomean induction generator in that event. It is also important in computingshort-circuit system protection when induction motors are used.

NOTES


Electrical MechanicsAdvantages:

1. It does not hunt or drop out of synchronism.

2. Simple and rugged in construction.

3. Cheaper in cost.

4. Easy maintenance.

5. Induction regulators provide a constant voltage adjustment depending onthe loading of the lines.

Disadvantages:

1. Cannot be operated independently.

2. Can deliver only leading current (whereas most commercial loads requirelagging current).

3. Dangerously high voltages may occur over long transmission lines if thesynchronous machines at the far end become disconnected and the linecapacitance excites the induction machines.

4. The induction generators are not helpful in system stability.

Applications: The induction generator is very useful in the followingcases/applications:

1. When the prime-mover does not run at constant speed (such as in hydro-electric) stations having a variable low-head water supply.

2. For installation in small power stations where it can be operated in paralleland feeding into a common mains without attendant.

3. For braking purposes in railway work.

2.25 SINGLE PHASE MOTORS

General Aspects

• The number of machines operating from single-phase supplies is greaterthan all other types taken in total. For the most part, however, they areonly used in the smaller sizes, less than 5 kW and mostly in the fractionalH.P. range. They operate at lower power-factors and are relativelyinefficient when compared with polyphase motors. Though simplicity mightbe expected in view of the two-line supply, the analysis is quite complicated.

Single phase motors perform a great variety of useful services in the home,the office, the factory, in business establishments, on the farm, and manyother places where electricity is available. Since the requirements of thenumerous applications differ so widely, the motor-manufacturing industryhas developed several types of such machines, each type having operatingcharacteristics that meet definite demands. For example, one type operatessatisfactorily on direct current or any frequency up to 60 cycles; anotherrotates at absolutely constant speed, regardless of load; another developsconsiderable starting torque and still another, although not capable of


NOTES


developing much starting torque, is nevertheless extremely cheap to makeand very rugged.

Types of Single-phase Motor

The single-phase motors may be of the following types:

1. Single-phase Induction Motors:

A. Split-phase motors

(i) Resistance-start motor (ii) Capacitor-start motor(iii) Permanent-split (single-value) capacitor motor(iv) Two-value capacitor motor.B. Shaded-pole induction motor.

C. Reluctance-start induction motor.

D. Repulsion-start induction motor.

2. Commutator-Type, Single-Phase Motors:

A. Repulsion motor. B. Repulsion-induction motor.

C. A.C. series motor. D. Universal motor.

3. Single-phase Synchronous Motors:

A. Reluctance motor. B. Hysteresis motor.

C. Sub-synchronous motor.

Single-phase Induction Motors

Applications and Disadvantages

Applications:

Single phase induction motors are in very wide use in industry especiallyin fractional horse-power field.

They are extensively used for electric drive for low power constant speedapparatus such as machine tools, domestic apparatus and agricultural machineryin circumstances where a three-phase supply is not readily available.

There is a large demand for single-phase induction motors in sizes rangingfrom a fraction of horse-power upto about 5 H.P.

Disadvantages:

Though these machines are useful for small outputs, they are not used forlarge powers as they suffer from many disadvantages and are never used in caseswhere three-phase machines can be adopted.

The main disadvantages of single-phase induction motors are:

1. Their output is only 50% of the three-phase motor, for a given frame sizeand temperature rise.

2. They have lower power factor.

3. Lower-efficiency.

NOTES


Electrical Mechanics4. These motors do not have inherent starting torque.

5. More expensive than three-phase motors of the same output.

Construction and Working

Construction:

A single phase induction motor is similar to a 3-φ squirrel-cage inductionmotor in physical appearance. Its rotor is essentially the same as thatused in 3-φ induction motors. Except for shaded pole motors, the stator isalso very similar. There is a uniform air-gap between the stator and rotorbut no electrical connection between them. It can be wound for any evennumber of poles, two, four and six being most common. Adjacent poleshave opposite magnetic property and synchronous speed equation,

Ns = 120f

p also applies.

The stator windings differ in the following two aspects:— Firstly single phase motors are usually provided with concentric coils.

— Secondly, these motors normally have two stator windings. In motorsthat operate with both windings energised, the winding with theheaviest wire is known as the main winding and the other is calledthe auxiliary winding. If the motor runs with auxiliary windingopen, these windings are usually referred as running and starting.

— In most of motors the main winding is placed at the bottom of the slotsand the starting winding on top but shifted 90° from the runningwinding.

Working:

When the stator winding of a single phase induction motor is connected tosingle phase A.C. supply, a magnetic field is developed, whose axis is always alongthe axis of stator coils. The magnetic field produced by the stator coils is pulsating,varying sinusoidally with time. Currents are induced in the rotor conductors bytransformer action, these currents being in such a direction as to oppose the statorm.m.f. Then the axis of the rotor m.m.f. wave coincides with that of the stator field,the torque angle is, therefore, zero and no torque is developed on starting. However,if the rotor is given a push by hand or by other means in either direction, it willpick-up the speed and continue to rotate in the same direction developing operatingtorque. Thus a single phase induction motor is not inherently self starting andrequires some special means for starting.

The above mentioned behaviour of this type of motor can be explained by anyone of the following theories:

1. Double revolving field theory 2. Cross-field theory.

The results given by both the theories are approximately same.


NOTES

Basic Electrical andElectronics Engineering SUMMARY

1. An electromechanical energy conversion device is one which convertselectrical energy into mechanical energy and mechanical energy into electricalenergy.

2. The phenomenon where by an e.m.f. and hence current is induced in anyconductor which is cut across or cut by a magnetic flux is known aselectromagnetic induction.

3. The examples of singly-excited magnetic field system are electromagneticrelays, reluctance motor.

4. The examples of multiply-excited magnetic field systems are synchronousmotor, alternator on which stator and rotor have A.C. and D.C. excitationrespectively.

5. The expression for mechanical force developed in a current excited system isgiven as:

FW i x

xfldfld=

′∂∂

( , ) .

6. Basic type of D.C. machine is that of commutator type. This is actuallyan alternating current (A.C.) machine, but furnished with a special device, acommutator, which under certain conditions converts alternating currentinto direct current.

7. An electrical generator is a machine which converts mechanical energy (orpower) into electrical energy (or power). It works on the following principle:

“Whenever a conductor cuts magnetic flux, dynamically induced e.m.f. isproduced in it according to Faraday’s Laws of Electromagnetic induction”.

8. The function of a transformer is to transform alternating current energyfrom one voltage into another voltage.

9. A transformer operates on the principle of mutual inductance between two(sometimes more) inductively coupled coils.

10. Both core and shell types of transformers are used, and selection is basedupon many factory such as voltage rating, kVA rating, weight insulationstress, mechanical stress, and heat distribution.

GLOSSARY

• Electromagnetic induction: It is a phenomenon whereby an e.m.f. and hencecurrent is induced in any conductor which is cut across or is cut by amagnetic flux.

• Yoke: It is a ring shaped portion of a D.C. machine which serve as the pathfor the main and commutating pole fluxes.

• Commutator: It is a device which converts alternating voltage to a directvoltage.

NOTES


Electrical Mechanics• Speed regulation: The change in speed when the load on the motor isreduced from rated value to zero, expressed as percent of the rated loadspeed.

• Transformer: A transformer is a static electromagnetic device designedfor the transformation of the (primary) alternating current system intoanother (secondary) one of the same frequency with other characteristics,in particular other voltage and current.

• Transformer efficiency: The efficiency of a transformer at a particular loadand power factor is defined as the ratio of power output to power input.

Efficiency = OutputInput

• Slip: The difference between the synchronous speed (Ns) and the rotorspeed (N) is known as slip. It is usually expressed as a fraction of thesynchronous speed. Thus, slip (s) is

s = N N

Ns

s

−

REVIEW QUESTIONS

1. Describe Faraday’s laws of electromagnetic induction.

2. Explain briefly an electromechanical energy conversion device with the helpof a block diagram.

3. For a singly excited magnetic field system, derive the relation for the magneticstored energy.

4. Show that the torque developed in a doubly excited system is equal to therate of increase of field energy with respect to displacement at constant.

5. Derive an expression for the magnetic force developed in a multiply-excited translational magnetic system.

6. Discuss briefly about torque production in rotating machines.

7. Discuss briefly general analysis of electromechanical system, and derive anexpression for the mechanical force developed in a current excited system.

8. What is a basic type of D.C. machine ?

9. Explain the principle on which a generator works.

10. Explain the construction and working of an elementary generator.11. Describe briefly various parts of a D.C. machine.12. Why is a commutator and brush arrangement necessary for operation of a

D.C. machine ?13. Name the main parts of a D.C. machine and state the materials of which

each part is made.14. Enumerate all parts of a D.C. machine and indicate their functions.


NOTES


15. Draw up the winding table for a 4 pole, wave connected armature having 30coil sides and give a developed diagram of the winding showing the polarityand position of the brushes, the main poles and the direction of motion of thearmature for a D.C. motor.

16. Draw the developed winding diagram of lap winding for 6 poles, 18 slotswith two coil sides/slot, double layer showing therein position of poles,direction of motion, direction of generated e.m.f. and position of brushes.

17. Derive the e.m.f. equation of a D.C. generator.

18. How are D.C. generators classified?

19. What is the difference between a separately excited and a self-excitedgenerator?

20. Sketch the following types of D.C. generators :

(i) Shunt (ii) Series

(iii) Compound.

State with reason(s) where each is used ?21. What is the difference between the short-shunt and long shunt compound

generators ?

22. With the help of a neat diagram show power division in a D.C. generator.

23. What is the working principle of a D.C. motor?

24. “In every D.C. generator motor action occurs and in every D.C. motor agenerator action occurs”. Explain.

25. What is torque ? What is the source of the torque force in a D.C. motor?

26. What is back e.m.f. or counter voltage?

27. Explain speed-current and torque-current characteristics of a series motor?

28. What is meant by a compound motor?

29. What conditions require the use of a compound motor?

30. What is a transformer? How does it transfer electric energy from one circuitto another?

31. Explain the principle of operation of a transformer.

32. What is meant by transformer action? Under what conditions will it takeplace?

33. If an alternating current is impressed on one coil, what will be the frequencyof the induced voltage in another coil with which it is coupled?

34. Derive an expression for induced e.m.f. in a transformer in terms of frequency,the maximum value of flux and the number of turns on the windings.

35. Derive an expression for the e.m.f. of an ideal transformer winding.

36. ‘Losses and efficiency’ of a transformer. Explain briefly.

37. Explain the principle of operation of the polyphase induction motor.

38. Show that a rotating magnetic field can be produced by the use of 3-phasecurrents of equal magnitude.

NOTES


Electrical Mechanics39. What is meant by slip in an induction motor? Develop an expression for thefrequency of rotor currents in it.

40. What are the applications of single-phase motors?

FURTHER READINGS

• Muthusubramanian R., Salivahanan S. and Murleedharan K.A., “BasicElectrical, Electronics and Computer Engineering”, Tata Mchraw Hill,Second Edition, (2006).

• “Basic Electrical and Electronics Engineering”, Laxmi Publications (P) Ltd.• Prem Kumar N., “Basic Electrical Engineering”, Anuradha Publishers,

(2003).

NOTES



STRUCTURE

3.1 Semiconductors3.2 P-N Junction Diode3.3 Zener Diode3.4 Rectifiers3.5 Bipolar Junction Transistor (BJT)3.6 Point-contact Transistor3.7 Junction Transistor3.8 P-N-P and N-P-N Transistors3.9 Transistor Circuit Configuration3.10 Common-base (CB) Configuration3.11 Common-emitter (CE) Configuration3.12 Common-collector (CC) Configuration3.13 Small Signal Amplifiers3.14 Common Emitter Amplifier Design3.15 Capacitor Coupled Two-stage Common Emitter Amplifier3.16 Direct Coupling between Stages Direct Coupled Circuit3.17 Two-stage Circuit with Emitter Follower Output3.18 Small-signal High Frequency Amplifier


• Review Questions


U N I T

3SEMICONDUCTOR DEVICES

AND APPLICATIONS

OBJECTIVES


• determine characteristics of semiconductors, P-N junction diode, Zenereffects and Zener diode.


NOTES

Semiconductor Devicesand Applications

• give concepts about half wave and full wave rectifiers with voltageregulation.

• define Bipolar Junction Transistor (BJT), regarding its characteristics andconfigurations—CB, CE and CC.

• describe small signal amplifiers in a simple conceptual way.

3.1. SEMICONDUCTORS

“Semiconductors are solid materials, either non-metallic elements orcompounds, which allow electrons to pass through them so that they conductelectricity in much the same way as a metal”.

Characteristics of Semiconductors

Semiconductors possess the following characteristics :

1. The resistivity is usually high.

2. The temperature coefficient of resistance is always negative.

3. The contact between semiconductor and a metal forms a layer which has ahigher resistance in one direction than the other.

4. When some suitable metallic impurity (e.g., Arsenic, Gallium etc.) is addedto a semiconductor, its conducting properties change appreciably.

5. They exhibit a rise in conductivity in the increasing temperature, with thedecreasing temperatures their conductivity falls off and at low temperaturessemiconductors become dielectrics.

6. They are usually metallic in appearance but (unlike metals) are generallyhard and brittle.

3.2 P-N JUNCTION DIODE

In an N-type material (Fig. 3.1) the electron is called the majority carrierand the hole of the minority carrier.

In a P-type material (Fig. 3.2) the hole is the majority carrier and the electronis the minority carrier. The N- and P-type materials represent the basic buildingblocks of semiconductor devices.

⊕ – ⊕ – ⊕ –– –

–

+ –⊕– – ⊕ ⊕

⊕ ⊕ ⊕ ⊕⊕+ –

– – –+ –

–

Donor ionsMajoritycarriers

Minoritycarrier

– + – ++ – + – +– + – ++ –

–– –

+– + – +

+++ –

Acceptor ions

Majoritycarriers

Minority carrier

Fig. 3.1. N-type material. Fig. 3.2. P-type material.

NOTES



The semiconductor diode is simply bringing these materials together(constructed from the same base-Ge or Si). At instant the two materials are “joined”the electrons and holes in the region of the junction will combine resulting in alack of carriers in the region near the junction. This region of uncovered positiveand negative ions is called the depletion region due to the depletion of carriers inthis region.

Construction and Types of P-N Junction Diodes

The most extensively used elements in the manufacture of junction diodesare germanium and silicon (although some other materials are also assumingimportance in recent years).

A P-N junction diode (known as a semiconductor or crystal diode) consists ofa P-N junction, formed either in germanium or silicon crystal. The diode has twoterminals namely anode and cathode. The anode refers to the P-type region andcathode refers to the N-type region as shown in Fig. 3.3 (a). Its circuit symbol is asshown in Fig. 3.3 (b).

P N

(a) Construction

Anode Cathode

(b) Symbol

A K

Fig. 3.3. P-N junction diode.

The arrow head, shown in the circuit symbol, points the direction of currentflow, when it is “forward biased” (It is the same direction in which the movement ofholes takes place).

The commercially available diodes, usually have some notations to identifythe P and N terminals or leads. The standard notation consists of type numberspreceded by IN, such as IN 240 and IN 1250. Here 240 and 1250 correspond tocolour bands. In some diodes, the schematic symbol of a diode is painted or thecolour dots are marked on the body.

A(Anode)

K(Cathode)

Colourband

A

K

A

K

Red

Blue

A

K

(a) (b) (c) (d)

(a), (b) = Low current diodes ; (c) = Medium current diode ;(d) = High current or power diode.

Fig. 3.4. Low, medium and high current diodes.


NOTES


Fig. 3.4 shows low, medium and high current diodes.

— Refer Fig. 3.4 (a). The diode shown has a colour band located near one ofthe ends. The end, which is near the colour band, is identified as cathode;and the other end, obviously, is the anode (A).

— Refer Fig. 3.4 (b). The diode has a schematic symbol actually painted atits cathode (K) and the other end as anode.

The diodes of Fig. 3.4 (a) and 3.4 (b) can pass a forward current of 100 mA andare known as low current diodes.

— Refer Fig. 3.4 (c). The diode has colour dots marked on its body. The endlying near the blue dot is a cathode, while the other end is anode. Some-times this diode is shown bigger in size than that of diodes shown in Fig. 3.4(a) and (b). The diodes of this size can pass a forward current of 500 mAand are known as medium current diodes.

— Refer Fig. 3.4 (d). It shows a diode, which can pass a forward current ofseveral amperes. Therefore it known as a power diode or a high currentdiode.

The outstanding property of P-N junction/crystal diode to conduct currentin one direction only permits it to be used as a rectifier.

Potential Barrier and Biasing

A P-N junction diode which consists of P- andN-type semiconductors formed together to make a P-Njunction is shown in the Fig. 3.5. The place dividingthe two zones is known as a ‘‘junction’’.

Potential Barrier

As a result of diffusion some electrons and holesmigrate across the junction there by forming adepletion layer on either side of the junction byneutralisation of holes in the P-regional and of freeelectrons in the N-region. This diffusion of holes andelectrons across the junction continues till a potentialbarrier is developed in the depletion layer which thenprevents further diffusion. By the application of anexternal voltage this potential barrier is either increased or decreased.

The barrier voltage of a P-N junction depends upon three factors namelydensity, electronic charge and temperature. For a given P-N junction, the first twofactors are constant, thus making the value of VB dependent only on temperature.It has been observed that for both germanium and silicon the value of VB decreaseby 2mV/°C. Mathematically, the decrease in barrier voltage, ∆VB = – 0.002 × ∆t,where ∆t is the increase in temperature in °C.

Forward Biasing

The junction is said to be biased in the forward direction when then positivebattery terminal is connected to P-type region and the negative battery terminal

Depletionlayer

Potentialbarrier

Width

Height (V )B

Fig. 3.5

NOTES



to the N-type (Fig. 3.6). This arrangement permits the flow of current across theP-N junction. The holes are repelled by the positive battery terminal and electronsby the negative battery terminal with the result that both holes and electrons will bedriven towards the junction where they will recombine. Hence as long as the batteryvoltage is applied large current flows. In other words, the forward bias lowers thepotential barrier across the depletion layer thereby allowing more current to flowacross the junction.

Potential barrier decreased Potential barrier increased Fig. 3.6. Forward biasing. Fig. 3.7. Reverse biasing.

Reverse Biasing (Zener Diode)

The junction is said to be reversed biased when battery connections to thebattery are reversed as shown in Fig. 3.7. In this arrangement holes are attractedby the negative battery terminal and electrons by the positive battery terminal sothat both holes and electrons move away from the junction. Since there is norecombination of electron-hole pairs, diode current is negligible and the junctionhas high resistance. Reverse biasing increases the potential barrier at the junction,thereby allowing very little current to flow through the junction.

V-I Characteristics of a P-N Junction Diode

The V-I (volt-ampere) characteristic of atypical P-N junction diode with respect tobreakdown voltage (VBR) is shown in Fig. 3.8.

For typical junction concentrations andcurrent densities at a temperature of300 K, forward voltage ranges between0.2 and 0.3 V in germanium and between0.5 and 0.75 V in silicon.

The reverse current is related to minoritycarrier concentration, which dependsupon temperature and the energy gap ofthe material.Reverse current increases exponentially with temperature. It is a limitingfactor in the high-temperature junction of semiconductor junction devices.

The high-frequency response of a semi-conductor diode may be seriouslylimited by charge stored in the depletion region. This charge gives a

Forward

Cur

rent

+

Voltage +

VBR

Reverse

Fig. 3.8. Characteristic of aP-N junction diode.


NOTES


capacitive effect since it changes with voltage. The value of the storedcharge is that of the ionized impurity atoms in the depletion regions oneither side of the junction. The width of the depletion region increases withhigher reverse voltage and higher doping. The result is lower capacitance,as in the case of a parallel-plate capacitor with wider spacing betweenplates.

The maximum reverse voltage of a P-N junction is limited by the field inthe depletion region. The field accelerates carriers, which may gain enoughenergy to create new hole-electron pairs by colliding with atoms of thelattice structure. Each of these carriers may also create a hole-electronpair. As reverse voltage is increased, as avalanche breakdown point isreached at which this multiplicative action causes the current to increaseabruptly. Avalanche breakdown voltage is higher in lightly doped regions,since depletion region is wider, making the internal electric field smallerfor any given voltage.

P-N junction diodes usually made of germanium or silicon, are commonlyused as power rectifiers.

The circuit arrangements for obtaining forward and reverse characteristics ofa P-N junction diode in a laboratory are shown in Fig. 3.9 (a) and (b) and 3.9 (a)and (b) respectively.

For

war

dcu

rren

t (m

A)

B

A

O 0.6 VForward voltage

(b) Forward characteristic.

+

– VAA P

V+ –

Diode

(a) Circuit arrangement.

mA

R

+ –

Fig. 3.9

Refer Fig. 3.9. It may noted that if excessive current is permitted to flowthrough the diode, it may get permanently damaged.

Refer Fig. 3.9. The circuit is similar to that shown in Fig. 3.10 (a), excepttwo changes namely the diode terminals are reversed and milliammeter is

+

– VAA P

V+ –

Diode

(a) Circuit arrangement.

µA

R

+ –

(b) Reverse characteristic.

D

VBR O

C

Reverse voltage

IRS

Rev

erse

cur

rent

(A

)µ

Fig. 3.10

NOTES



replaced by a microammeter. It may be noted that negative terminal ofthe voltage source is connected to the anode of a diode and positive terminalto the cathode. Hence, the diode is reversed.

3.3 ZENER DIODE

A properly doped P-N junction crystal diode which has a sharp breakdownvoltage is known as Zener diode.

The voltage-regulator diode is commonly called a ‘Zener’ diode. It is a voltagelimiting diode that has some applications in common with the oldervoltage-regulator gas tubes but serves a much wider field of application, becausethe devices cover a wide spectrum of voltages and power levels.

Performance/Operations

The electrical performance of a zenerdiode is based on the avalanchecharacteristics of the P-N junction. Whena source of voltage is applied to a diode inthe reverse direction (negative to anode),a reverse current IR is observed (seeFig. 3.11). As the reverse potential isincreased beyond the “Zener knee”avalanche breakdown becomes welldeveloped at zener voltage VZ. At voltageVZ, the high counter resistance drops to alow value and the junction currentincreases rapidly. The current must ofnecessity be limited by an externalresistance, since the voltage VZ developedacross the zener diode remains essentially constant. Avalanche breakdown of theoperating zener diode is not destructive as long as the rated power dissipation of thejunction is not exceeded.

Externally, the zener diode looks much like other silicon rectifying devices,and electrically it is capable of rectifying alternating current.

The following points about the Zener diode are worth noting :(i) It looks like an ordinary diode except that it is properly doped so as to

have a sharp breakdown voltage.(ii) It is always reverse connected i.e., it is always reverse biased.

(iii) It has sharp breakdown voltage, called Zener voltage VZ.(iv) When forward biased, its characteristics are just those of ordinary diode.(v) It is not immediately burnt just because it has entered the breakdown

region (The current is limited only by both external resistance and power dissipationof Zener diode).

The location of Zener region can be controlled by varying the doping levels.An increase in doping, producing an increase in the number of addedimpurities, will decrease the Zener potential.

Rev

erse

curr

ent (

I)

R

Forward bias

Zener knee

Zenervoltage

Reverse bias

For

war

dcu

rren

t

Fig. 3.11


NOTES


Zener diodes are available having Zener potentials of 1.8 to 200 V with

power ratings from 14

to 50 W. Because of its higher temperature and

current capability, silicon is usually preferred in the manufacture of Zenerdiodes.

Applications of Zener Diode

Zener diode serves in the following variety of applications :

Voltage Reference or Regulator Element

The primary use of a zener diode is as a voltage reference or regulator element.Fig. 3.12 shows the fundamental circuit for the Zener diode employed as a shuntregulator. In the circuit, diode element and load RL draw current through the seriesresistance RS. If Ein increases, the current through the Zener element will increaseand thus maintain an essentially fixed voltage across RL. This ability to maintainthe desired voltage is determined by the temperature coefficient and the diodeimpedance of the zener device.

RS+

–

E inZenerdiode

+

–

RL

R = Series resistanceS R = Load resistanceL

Fig. 3.12. Basic Zener-diode regulator circuit.

Shunt Transistor Regulator

The Zener diode may also be used tocontrol the reference voltage of a transistorregulated power supply. An example of this ina shunt transistor regulator is shown inFig. 3.12, where Zener element is used tocontrol the operating point of the transistor.The advantage of this circuit over that shownin Fig. 3.13 are increased power handlingcapability and a regulating factor improved by utilizing the current gain of thetransistor.

Audio or r-f Applications

The Zener diode also finds use in audio or r-f (radio frequency) applicationswhere a source of stable reference voltage is required, as in bias supplies. Frequently,Zener diodes are connected in series package, with, for example, one junctionoperating in the reverse within a single direction and possessing a positivetemperature VZ coefficient ; the remaining diodes are connected to operate in theforward direction and exhibit negative temperature VZ coefficient characteristics.The net result is close neutralization of VZ drift vs. temperature change ; suchreference units are frequently used to replace standard voltage cells.

RS

+

–

Zenerdiode

+

–

Fig. 3.13. Shunt transistorregulator.

NOTES



Computer Circuits

Zener diodes also find use in computer circuits designed for switching aboutthe avalanche voltage of the diode. Design of the Zener diode permits it to absorboverload surges and thereby serves the function of protecting delicate circuitry formovervoltage.

— The usual voltage specifications VZ on Zener diodes are 3.3 to 200 V with± 1, 2, 5, 10 or 20% tolerances.

— Typical power dissipation ratings are 500 mW, 1, 10 and 50 W.

— The temperature co-efficient range on VZ is as low as 0.001% °C.

3.4 RECTIFIERS

A rectifier is a circuit, which uses one or more diodes to convert A.C. voltageinto pulsating D.C. voltage.

A rectifier may be broadly categorized in the following two types :

1. Half-wave rectifier, and

2. Full-wave rectifier.

Half-wave RectifierFig. 3.14 (a) shows a half-wave rectifier circuit. It consists of a single diode in

series with a load resistor. A P-N junction diode can easily be used as a rectifierbecause it conducts current only when forward biased voltage is acting, and doesnot conduct when reverse bias voltage is acting.

The input to the half-wave rectifier is supplied from the 50 Hz A.C. supply,whose wave form is shown in Fig. 3.14 (b).

OperationWhen and A.C. voltage source is connected across the junction diode as shown

in Fig. 3.14 (a) the positive half cycle of the input acts as a forward bias voltage andthe output across the load resistance varies correspondingly. The negative halfcycle of the input acts as a reverse bias and practically no current flows in thecircuit. The output is, therefore, intermittent, pulsating and unidirectional.

It is evident from the above discussion, that as the circuit uses only one-halfcycle of the A.C. input voltage, therefore, it is popularly known as a “half-waverectifier”.

Input A.C.

Output(b)

+–

t

t

RLA.C.

Supply Vout

A

B

Vin

(a)

i

Fig. 3.14. Half-wave rectifier.


NOTES


Disadvantages

The main disadvantages of a half-wave rectifier are :

(i) The A.C. supply delivers power only half the time ; therefore, its output islow.

(ii) The pulsating current in the load contains alternating component whosebasic frequency is equal to the supply frequency. Therefore, an elaborate filteringis required to produce steady direct current.

Efficiency of Half-wave Rectifier

The ratio of D.C. power output to the applied A.C. input power is known asrectifier efficiency.

i.e., Rectifier efficiency, η = D.C. power outputA.C. input power

RLA.C.

Supply Voutv = V sinm θ

i

v

O

i

θ

dθ θO

Fig. 3.15

Consider a half-wave rectifier shown in Fig. 3.15. Let v = Vmax sin θ be thealternating voltage that appears across the secondary winding.

Let, rf = Diode forward resistance, and

RL = Load resistance.

D.C. Power Output

The output current is pulsating direct current. Therefore, in order to findD.C. power, average current has to be found out,

Iav. = Idc = 1

2πθ

πθ

θπ π

0 0

12z z=

+i d

Vr R

df L

max sin( )

= Vr R

dVr Rf L f L

max max

( )sin

( )cos

2 200

πθ θ

πθ

ππ

+=

+−L

NM

O

QPz

= Vr R

Vr Rf L f L

max max

( ) ( )22

1π π π+

× =+

×

= Imax

π∵ I

Vr Rf L

maxmax

( )=

+

L

NMM

O

QPP

∴ D.C. power, Pdc = Idc2 × RL

= Imax

πFHG

IKJ

2

× RL ...(3.1)

NOTES



A.C. Power Input

The A.C. power input is given by,

Pac = Irms2 (rf + RL)

For a half-wave rectified wave, Ir.m.s. = Imax

2

∴ Pac = Imax

2

2FHG

IKJ

× (rf + RL) ...(3.2)

∴ Rectifier efficiency = D.C. output powerA.C. input power

= ( / )

( / ) ( )max

max

I RI r R

L

f L

π 2

22×+

i.e., ηrectifier(half-wave) = 0.406 0.406Rr R r

R

L

f L f

L

+=

+1...(3.3)

The efficiency will be maximum when rf is negligible as compared to RL.

∴ Maximum rectifier (half-wave) efficiency = 40.6%.

This means in half-wave rectification, a maximum of 40.6% of A.C. power isconverted into D.C. power.

Example 3.1. A diode crystal having internal resistance rf = 25 Ω is used forhalf-wave rectification. If the applied voltage v = 60 sin ωt and load resistance =725 Ω, find :

(i) Imax, Idc, Irms. (ii) A.C. input power and D.C. output power.

(iii) D.C. output voltage. (iv) Efficiency of rectification.

Solution. Given : rf = 25 Ω ; RL = 725 Ω ; v = 60 sin ωt.

(i) Imax, Idc, Irms:Now, v = 60 sin ωt ...(Given)

i.e., Vmax = 60 V

∴ Imax = Vr Rf L

max

+=

+60

25 725 = 0.08 A or 80 mA (Ans.)

Idc = Imax

π π=

60 = 25.46 mA (Ans.)

Irms = Imax

2802

= = 40 mA (Ans.)

(ii) A.C. input power, D.C. output power:

Pac = (Irms)2 × (rf + RL) = 40

1000

2FHG

IKJ

× (25 + 725) = 1.2 W (Ans.)

Pdc = (Idc)2 RL =

25 461000

2.FHG

IKJ

× 725 = 0.47 W (Ans.)


NOTES


(iii) D.C. output voltage, Vdc:

Vdc = Idc × RL = 25.461000

× 725 = 18.46 V (Ans.)

(iv) Efficiency of rectification:

Efficiency of rectification = PP

dc

ac

= 0.471.2

= 0.3917 or 39.17% (Ans.)

Full-wave rectifier

A full-wave rectifier is a circuit, which allows a unidirectional current toflow through the load during the entire input cycle. This can be achieved with twodiodes working alternately. For the positive half-cycle of input voltage, one diodesupplies current to the load and for the negative half-cycle, the other diode does so ;current being always in the same direction through the load.

For full-wave rectification the following two circuits are commonly used :1. Centre-tapped full-wave rectifier2. Full-wave bridge rectifier.

Centre-tapped Full-wave RectifierFig. 3.16 shows the circuit of a centre-tapped full-wave rectifier. The circuit

uses two diodes (D1 and D2) which are connected to the centre-tapped secondarywinding AB of the transformer.

Operation: During the positive half-cycle of secondary voltage, the end A of the

secondary winding is positive and end B negative. This makes the diodeD1 forward biased and diode D2 reverse biased. Therefore, diode D1conducts while diode D2 does not. The conventional current flows throughdiode D1, load resistor RL and the upper half of secondary winding asshown by the dotted arrows.

During the negative half-cycle, the end A of the secondary becomes negativeand end B positive. Therefore, D2 conducts while diode D1 does not. Theconventional current flow is through D2, RL and lower half winding asshown by solid arrows.

A.C.supply

A

Vmax

Vmin

B

+D1

+–

RL+

–

O

D2

Input A.C.

D1

t

tD2 D1 D2

Rectifiedoutput

(a) (b) Waveforms of full-wave rectifier.

Fig. 3.16. Centre-tapped full-wave rectifier.

NOTES



It may be noted [Fig. 3.16 (a)] that the current in the load RL is in the samedirection for both the cycles of input A.C. voltage. Therefore, D.C. is obtained fromthe load RL.

Also, Peak inverse voltage (PIV) = Twice the maximum voltage across thehalf-secondary winding

i.e., PIV = 2 Vmax.

Advantages:

1. The D.C. output voltage and load current values are twice than those ofhalf-wave rectifiers.

2. The ripple factor is much less (0.482) than that of half-wave rectifier(see Fig. 3.14).

3. The efficiency is twice that of half-wave rectifier.

For a full-wave rectifier, the maximum possible value of efficiency is 81.2%while that of half-wave rectifier is 40.6%.

Disadvantages:

1. The diodes used must have high peak inverse voltage.

2. It is difficult to locate the centre tap on the secondary winding.

3. The D.C. output is small as each diode utilises only one-half of thetransformer secondary voltage.

Full-wave Bridge Rectifier

It uses four diodes (D1, D2, D3, D4) across the main supply, as shown inFig. 3.17 (a). The A.C. supply to be rectified is applied to the diagonally oppositeends of the bridge through the transformer. Between other two ends of the bridge,the load resistance RL is connected.

D D D D D D1 3 2 4 1 3t

(b)

Vout

O

Secondaryvoltage

t++

O –

Vin

(a)

VoutRL

–

+

B

D4

D3D2

D1

A

+

–

A.C.supply

L

M

Fig. 3.17. Full-wave bridge rectifier.

Operation:

During positive half-cycle of secondary voltage, the end L of the secondarywinding becomes positive and end M negative. This makes D1 and D3forward biased while diodes D2 and D4 are reverse biased. Therefore, onlydioded D1 and D3 conduct. These two diodes will be in series through theload RL as shown in Fig. 3.18 (a). The current flows (dotted arrows) fromA to B through RL.


NOTES


(a)

RL

–

+

+

–

A.C.supply

D3

D1

AB

(b)

RL

–

+

–

+

A.C.supply

D2

AB

D4

Fig. 3.18

During the negative half-cycle of the secondary voltage, end L becomesnegative and M positive. This makes D2 and D4 forward biased whereasdiodes D1 and D3 are reverse biased. Therefore, only diodes D2 and D4conduct. These two diodes will be in series with RL as shown in Fig. 3.18 (b).The current flows (solid arrows) from A to B through RL i.e., in the samedirection as for positive half-cycle. Therefore, D.C. output is obtained acrossRL.

Further it may be noted that peak inverse voltage (PIV) of each diode is equalto the maximum secondary voltage of transformer.

Advantages:

1. It can be used with advantage in applications allowing floating inputterminals i.e., no output terminal is grounded.

2. The transformer is less costly as it is required to provide only half thevoltage of an equivalent centre-tapped transformer used in a full-wave rectifiercircuit.

3. No centre-tap is required on the transformer.

4. The output is twice that of the centre-tapped circuit for the secondary voltage.

Disadvantages:

1. It uses four diodes as compared to two diodes for centre-tapped full waverectifier.

2. Since during each half-cycle of A.C. input two diodes that conduct are inseries, therefore, voltage drop in the internal resistance of the rectifying unit willbe twice as great as in the centre-tapped circuit. This is objectionable whensecondary voltage is small.

NOTES



These days, the bridge rectifiers are so common that manufacturers arepacking them as a single unit with bakelite or some other plasticencapsulation with external connections brought out.

Efficiency of full-wave Rectifier

Fig. 3.19 shows the process of full rectification.

The instantaneous current i is given by

i = v

r RV

r Rf L f L( )sin

( )max

+=

+θ

where rf and RL are the diode forward resistance and load resistance respectively.

D.C. output power (Pdc):

Idc = 2Imax

π

∴ D.C. power output,

Pdc = Idc2 × RL =

2 2Imax

πFHG

IKJ × RL

A.C. input power (Pac):

Pac = (Irms)2 (rf + RL)

For a full-wave rectified wave, we have

Irms = Imax

2

∴ Pac = Imax

2

2FHG

IKJ

(rf + RL) ...(3.4)

∴ Full-wave rectification efficiency,

η = PP

I R

I r Rdc

ac

L

f L

=+

( / )

( / ) ( )max

max

2

2

2

2

π

= 82π

×+

=+

Rr R

Rr R

L

f L

L

f L( ) ( )0.812

i.e., η = 0.812

1 +r

Rf

L

...(3.5)

The efficiency will be maximum if rf is negligible as compared to RL.

∴ ηmax = 81.2%

This is double the efficiency than that of half-wave rectifier. Therefore, a full-wave rectifier is twice as effective as a half-wave rectifier.

vi

O

O dq q

q

Fig. 3.19


NOTES


Example 3.2. A full-wave rectifier uses two diodes, the internal resistance ofeach diode may be assumed constant at 25 Ω. The transformer r.m.s. secondaryvoltage from centre tap to each end of secondary is 60 V and load resistance is 750Ω. Find :

(i) The mean load current.(ii) The r.m.s. value of load current.Solution. Given : rf = 25 Ω ; RL = 750 Ω

(i) The mean load current, Idc:

Vmax = 60 × 2 = 84.85 V

Maximum load current, Imax = V

r Rf L

max

( )+

= 84.85

25 750+ = 0.109 A or 109 mA

∴ Idc = 2 2 109Imax

π π=

× = 69.39 mA. (Ans.)

(ii) The r.m.s. value of load, Irms:

Irms = Imax

2109

2= = 77.1 mA. (Ans.)

Ripple Factor

The output voltage (or load current) of a rectifier consists of two componentsnamely D.C. component and A.C. component. (Any wave which varies in a regularmanner has an A.C. component). The A.C. component present in the output is calleda ripple. As a matter of factor, the ripple is undesirable and accounts for pulsationsin the rectifier output :

The effectiveness of a rectifier depends upon the magnitude of ripple in theoutput. Smaller the ripple, more effective will be the rectifier. Mathematically, theripple factor,

γ = The r.m.s. value of A.C. component of output voltage

The D.C. component of output voltage

= V

V

I

Ir

dc

r

dc

( (r.m.s) r.m.s.)=

where, Vr(r.m.s) = The r.m.s. value of the A.C. component of the output voltage,

Vdc = The average or D.C. value of the output voltage,

Ir(r.m.s) = The r.m.s. value of the A.C. components of current, and

Idc = The average or D.C. value of the load current.

We know that the r.m.s. value of the rectified load current,

Ir.m.s. = ( ) ( )( )I Idc r2 2+ r.m.s.

NOTES



Dividing both sides by Idc, we get

II

I I

II

Idc

dc r

dc

r

dc

r.m.s (r.m.s.) r.m.s.. ( )( ) ( )=

+= +

RST

UVW

L

N

MM

O

Q

PP

2 2 2

1

Squaring and rearranging the above expression,

I

III

r

dc dc

( )r.m.s. r.m.s.=F

HGI

KJ−

2

1

or Ripple factor, γ = IIdc

r.m.s.F

HGI

KJ−

2

1 ...(3.6)

(i) Ripple factor of a half-wave rectification. In half wave rectification,

Ir.m.s. = I

II

dcmax max;2

=π

∴ Ripple factor, γ = ( / )( / )

max

max

II

21

2

πL

NM

O

QP − = 1.21

This indicates that A.C. component exceeds the D.C. component in the outputof a half-wave rectifier. This results in greater pulsations in the output, hencehalf-wave rectifier is not very successful for conversion of A.C. into D.C.

(ii) Ripple factor of a full-wave rectifier. In full-wave rectification,

Ir.m.s. = Imax

2 ; Idc =

2Imax

π

∴ Ripple factor, γ = I

Imax

max

//2

21

2

πF

HGI

KJ− = 0.48

Voltage Regulation

The change in D.C. output voltage from no load to full load with respect to fullload voltage of a power supply is known as its voltage regulation.

Mathematically, % voltage regulation = V V

VNL FL

FL

− × 100

where, VNL = No-load D.C. voltage at the output terminals of the power supply,and

VFL = Full load D.C. voltage at the output terminals of the power supply.

In a well designed power supply percentage regulation should not be morethan 1 per cent.


NOTES

Semiconductor Devicesand Applications3.5 BIPOLAR JUNCTION TRANSISTOR (BJT)

A transistor may be defined as follows:— The word transistor was derived from the two word combination, transfer-

resistance (Transfer + resistor → Transistor). A transistor is a device totransfer a low resistance into a circuit having a high resistance.

— A ‘transistor’ is a semiconductor device in which current flows insemiconductor materials.

— When a thin layer of P-type or N-type semiconductor is between a pair ofopposite types it constitutes a transistor.

The transistor is a solid state device, whose operation depends upon theflow of electric charge carriers within the solid.

A transistor is a semiconductor device having both rectifying and amplifyingproperties.

“The main difference between a vacuum triode and a transistor is that while avacuum triode is a voltage controlled device, a transistor is a current controlleddevice”.

The transistor was invented by a team of three scientists at Bell Laboratories,USA in 1947. Although the first transistor was not a bipolar junction device, yet itwas the beginning of a technological revolution that is still continuing in the twentyfirst century. All of the complex electronic devices and systems developed or in usetoday, are an outgrowth of early developments in semiconductor transistors.

The two basic types of transistors are :

1. Bipolar junction transistor (BJT).

2. Field-effect transistor (FET).

The bipolar junction transistor is used in the following two broad area ofelectronics :

(i) As a linear amplifier to boost an electric signal.(ii) As an electronic switch.Transistors are of the following two types :

1. Point-contact transistors.

2. Junction transistors.

We will discuss about them as follows:

3.6 POINT-CONTACT TRANSISTOR

Refer to Fig. 3.20. A point-contact transistor has a block of P-type germaniummounted on a metal plate with two wire contacts attached to the opposite side ofthe crystal. The two wire contacts are very close together ; one is referred to as theemitter (E) ; the other is collector (C) ; and P-type germanium block is termed thebase (B). The emitter and the base form a point-contact rectifier having a forwarddirection from the wire contact toward the plate contact.

NOTES



P

B

Metal plate

Germaniumblock

CE

Wire contacts

P

B

– + – +

+ 6 V– 0.2 V E C

E = emitter, C = collector, B = base

Fig. 3.20. Basic construction of a Fig. 3.21. A basic point-contact transistor. point-contact transistor.

Fig. 3.21 shows the emitter being acted on by a negative e.m.f. of – 0.2 Vcausing an electron flow of 0.008 A. In the same manner, the collector and the basecombination are acted on by an e.m.f. in the reverse direction. Normally the positive6 volts produce only 0.00024 A of electron flow to the collector due to the smallnumber of free electrons within the P-type crystals. However, with the emitterinjecting a great number of electrons into the crystal very near the collector, theflow of electrons to the collector becomes at least double that to the emitter.

It is important to know that the emitter injects electrons (or holes) into thecrystal to effect a reduction of the reverse resistance between the base and the collector.When the base of the point-contact transistor is of the N-type germanium material,the polarities of electromotive forces will be reversed and the emitter will literallyinject holes.

3.7 JUNCTION TRANSISTOR

A trade transistor consists of two P-N junction diodes placed back to back.The following are the two common types of junction transistors :

1. Grown junction type

2. Alloy-junction type.

Grown Junction Transistor

P N PE C

Emitter CollectorBase

N P NE C

B B

E C

BIndium

N-typeGermanium

(a) (b) (c)

Fig. 3.22

Fig. 3.22 (a) shows a grown P-N-P Junction triode transistor.

Fig. 3.22 (b) shows the form of N-P-N junction transistor.


NOTES


In the manufacture of grown junction transistors the single crystal growingprocess is employed.

— The left hand section or region is called the emitter.

— The right hand section or region is called the collector.

— The middle section is called the base region or base. It is extremely thin ascompared to either the emitter or collector and is lightly doped.

Function of the emitter is to inject majority charge carriers into the base andthat of the collector is to collect or attract these carriers through the base.

Alloy-junction Transistor

Fig. 3.22 (c) shows an alloy-junction transistor which consists of two leads ofindium metal alloyed on the opposite sides of a thin slice of an N-type germanium.Collector is larger in size than the emitter. These transistors may be also of P-N-Ptype or N-P-N type.

Junction Transistors—Illustration of Emitter, Base and Collector :

P N PForwardvoltage

Reversevoltage

Em

itter

Bas

e

Col

lect

or

+–

+

–

Emitter CollectorN

P P

Base

(a)

N P NForwardvoltage

Reversevoltage

Em

itter

Bas

e

Col

lect

or

+–

+

–

Emitter CollectorP

N N

Base

(b)

Fig. 3.23. Junction transistors illustrating emitter, base and collector.

As shown in Fig. 3.23, the transistor consists of a P-N junction and an N-Pjunction, by making either P or N semiconductor between opposite types. Thepurpose is to have the first section supply charges, either holes or electrons, to becollected by the third section, through the middle section. The electrode that suppliescharges is emitter ; the electrode at the opposite end to collect the charges is thecollector. The base in the middle forms two junctions between emitter and collector,to control the collector current.

Emitter

It is more heavily doped than any of other regions because its main functionis to supply majority charge carriers (either electrons or holes) to the base.

The emitter-base junction is biased with forward voltage. (Typicalvalues : Ge = 0.2 V, Si = 0.6 V).

NOTES



— As shown for the P-N-P transistor in Fig. 3.23 (a), the P emitter supplies hole charges to its junction with the base. This direction is indicated bythe emitter arrow for forward hole current in the schematic symbol. Thearrow pointed into the base shows a P-N junction between emitter andbase, corresponding to the symbol for a P-N diode.

— For the N-P-N transistor in Fig. 3.23 (b), the emitter supplies electrons tothe base.

Therefore, the symbol for the N emitter shows the arrow out from the base,opposite to the direction of electron flow.

In the schemic symbols, only the emitter has an arrow. The arrow pointinginto the base means a P-N-P transistor, the arrow out from the base means an N-P-N transistor.

Practically all small transistors for audio and r.f. amplifiers are N-P-N, madeof silicon, with a typical forward bias of 0.6 V between base and emitter collector.

The function of a collector is to remove charges from the junction with the base.In Fig. 3.23 (a), the P-N-P transistor has a P collector receiving hole charges. Forthe N-P-N transistor in Fig. 3.23 (b), the N collector receives electrons. The collectorbase junction always has reverse voltage. Typical values are 4 to 100 V. This polaritymeans no majority charges can flow from collector to base. However, in the oppositedirection, from base to collector, the collector voltage attracts the charges in thebase supplied by the emitter.

Base

The base in the middle separates the emitter and collector. The base-emitterjunction is forward based. As a result, the resistance is very low for theemitter circuit. The base-collector junction is reverse biased, providing amuch higher resistance in the collector circuit.

It is very lightly doped. It is very thin (10–6 m) as compared to either emitter or collector.

Collector Current

The final requirement for transistor action is to have the collector currentcontrolled by the emitter-base circuit. The emitter has heavy doping to supplymajority charges. However, the base has only light doping and is very thin, so thatits charges can move to the collector junction. The collector voltage is relativelyhigh. Because of these factors, practically all the charges supplied by the emitterto the base are made to flow in the collector circuit. Typically, 98 to 99% or more ofthe emitter charges provide collector current (IC). The remaining 1 to 2% or lessbecomes base current (IB).

The key to the transistor action is the lightly doped thin base between theheavily doped emitter and moderately doped collector.

A transistor has two junctions (emitter-base and collector-base junctions),and each of these two junctions, may be forward biased or reverse biased,therefore, there are four possible ways of biasing these two junctions.Accordingly it may operate in different conditions as listed below :


NOTES


Condition Emitter-Base Collector-Base Region of(EB) Junction (CB) Junction Operation

1. Forward-Reverse (FR) Forward-Biased Reverse-Biased Active

2. Forward-Forward (FF) Forward-Biased Forward-Biased Saturation

3. Reverse-Reverse (RR) Reverse-Biased Reverse-Biased Cut-off

4. Reverse-Forward (RF) Reverse-Biased Forward-Biased Inverted

3.8 P-N-P AND N-P-N TRANSISTORS

To understand the basic mechanism of transistor operation the following factsneed to be kept in mind :

1. Since emitter is to provide charge carriers, it is always “forward biased”.

2. First letter of transistor type indicates the polarity of the emitter voltagewith respect to base.

3. Collector’s job is to collect or attract those carriers through the base, hence itis always “reverse-biased”.

4. Second letter of transistor type indicates the polarity of collector voltagewith respect to the base.

The above points apply both to P-N-P and N-P-N transistors.

Working of P-N-P Transistor

Fig. 3.24 shows a P-N-P transistor connected in the common-base (orgrounded-base) configuration (it is so called because both the emitter and collectorare returned to the base terminals). The emitter junction is forward-biased whereasthe collector junction is reverse-biased. The holes in the emitter are repelled by thepositive battery terminal towards the P-N or emitter junction. The potential barrierat the junction is reduced due to the forward-bias, hence holes cross the junctionand enter the N-type base. Because the base is thin and lightly-doped, majority ofthe holes (about 95%) are able to drift across the base without meeting electrons tocombine with. The balance of 5% of holes are lost in the base region due torecombination with electrons. The holes which after crossing the N-P collectorjunction enter the collector region are swept up by the negative collector voltage VC.

Hole flow Hole flow

IE IC

VCBIBVEB

Emitter(E)

Base(B)

Collector(C)

P N PEmitter (E) Collector (C)

Base (B)

IE IC

IB

Fig. 3.24. P-N-P transistor.

NOTES



The following points are worth noting:

1. In a P-N-P transistor majority charge carriers are holes.

2. The collector current is always less than the emitter current because somerecombination of holes and electrons take place.

(IC = IE – IB).

3. The current amplification (α) (or gain of P-N-P transistor) for steadyconditions when connected in common base configuration is expressed as :

α = IIC

E

(collector current)(emitter current)

< 1.

4. Emitter arrow shows the direction of flow of conventional current. Evidently,electron flow will be in the opposite direction.

Working of N-P-N Transistor

Fig. 3.25 shows a N-P-N junction transistor. The emitter is forward-biasedand the collector is reverse-biased. The electrons in the emitter region are repelledby the negative battery terminal towards the emitter or N-P junction. The electronscross over into the P-type base region because potential barrier is reduced due toforward bias. Since the base is thin and lightly doped, most of the electrons (about95%) cross over to the collector junction and enter the collector region where theyare readily swept up by the positive collector voltage VC . Only about 5% of theemitter electrons combine with the holes in the base and are lost as charge carriers.

Electronflow

IE IC

VCVE

Emitter(E)

Base(B)

Collector(C)

N P NEmitter (E) Collector (C)

Base (B)

IC

IB

Electronflow

IE

IB

Fig. 3.25. N-P-N transistor.

The following points are worth noting :

1. In a N-P-N transistor, majority charge carriers are electrons.

2. IC (collector current) is less than IC (emitter current) so that α < I.

3. Emitter arrow shows the direction of flow of conventional current.

The choice of N-P-N transistor is made more often because majority chargecarriers are electrons whose mobility is much more than that of holes.

Note. The junction transistors have been made in power ranges from a few milliwattsto tens of watts. The tiny junction transistor is unparalleled in that it can be made to workat power levels as I microwatt.


NOTES

Semiconductor Devicesand Applications3.9 TRANSISTOR CIRCUIT CONFIGURATIONS

A transistor is a three-terminal device (having three terminals namely emitter,base and collector) but we require four terminals-two for the input and two for theoutput for connecting it in a circuit. Hence one of the terminals of the transistor ismade common to the input and output circuits. Thus there are three types ofconfigurations for operation of a transistor. These configurations are:

(i) Common-base (CB) configuration.(ii) Common-emitter (CE) configuration.

(iii) Common-collector (CC) configuration.The term ‘common’ is used to denote the electrode that is common to the

input and output circuits. Because the common electrode is generally grounded,these modes of operation are frequently referred to as ground-base, ground-emitter and grounded-collector configurations as shown in Fig. 3.26 for a N-P-Ntransistor.

Each circuit configuration has specific advantages and disadvantages. It maybe noted here that regardless of circuit connection, the emitter is always biased inthe forward direction, while the collector always has a reverse bias.

E C

B

Inpu

t

Out

put

(a) CB configuration (b) CE configuration

Inpu

t

Out

put

B

C

E

(c) CC configuration

Inpu

t

Out

put

B

E

C

Fig. 3.26. Different circuit configurations for N-P-N transistor.

3.10 COMMON-BASE (CB) CONFIGURATION

In this circuit configuration, input is applied between emitter and base andoutput is taken from collector and base. Here, base of the transistor is common toboth input and output circuits and hence the name common base configuration. Acommon-base configuration for N-P-N transistor is shown in Fig. 3.27.

CE ICIE

OutputInput

VCCVEE

RC

IB

B

CEOpen

B

ICBO

Fig. 3.27. Common-base N-P-N transistor. Fig. 3.28

NOTES



Current Amplification Factor (α)

It is the ratio of output current to input current. In CB configuration, theinput current is the emitter current IE and output current is the collector currentIC.

The ratio of change in collector current to the change in emitter current atconstant collector-base voltage VCB is known as current amplification factori.e.,

α = ∆∆

II

C

E at constant VCB ...(3.7)

If only D.C. values are considered, then α = II

C

E...(3.8)

α is less than unity. This value can be increased (not more than unity) bydecreasing the base current. This is accomplished by making the base thin anddoping it lightly.

In commercial transistors, practical value of α varies from 0.9 to 0.99.

Collector Current (IC)

Total collector current, IC = αIE + Ileakage

(αIE is the part of emitter current that reaches the collector terminal)

where, IE = Emitter current, and

Ileakage = Leakage current (This current is due to movement of minoritycarriers across base-collector junction on account of it being reversed;it is much smaller than αIE)

When emitter is open (Fig. 3.28) IE = 0, but a small leakage current still flowsin the collector circuit. This Ileakage is abbreviated as ICBO , meaning collector-basecurrent with emitter open.

IC = αIE + ICBO ...(3.9)

∴ IC = α(IC + IB) + ICBO (∵ IE = IC + IB)

or IC(1 – α) = αIB + ICBO

or IC = α

α α1 1−FHG

IKJ

+−

II

BCBO

( )...(3.10)

— In view of improved construction techniques, the magnitude of ICBO forgeneral-purpose and low-powered transistors (especially silicontransistors) is usually very small and may be neglected in calculations.

— For high power calculations, ICBO appears in µA range.

— ICBO is temperature dependent, therefore, at high temperature it must beconsidered in calculations.

Example 3.3. In a common-base configuration, current amplification factoris 0.92. If the emitter current is 1.2 mA, determine the value of base current.


NOTES


Solution. Given : α = 0.94 ; IE = 1.2 mA

We know that, α = II

C

E

or IC = αIE = 0.92 × 1.2 = 1.1 mAAlso, IE = IC + IB

∴ IB = IE – IC

= 1.2 – 1.1 = 0.1 mA. (Ans.)

Characteristics of Common-base Transistor

Curves representing the variation of current with voltage in a transistortriode circuit are called transistor characteristic curves. There are thefollowing two types of characteristic curves :

1. Input characteristic curves of IE versus emitter-base voltage (VEB).2. Output characteristic curves of collector current (IC) versus collector-

base voltage (VCB).Fig. 3.29 shows the circuit of an N-P-N junction triode (common-base) studying

characteristic curve.

B

N-P-NE C

mAmA

VEB VCB Emitter (E) = Forward biasedCollector (C) = Reverse biased

IB

ICIE

Fig. 3.29. Circuit of an N-P-N junction triode.

Input Characteristic Curves— To plot these curves the collector voltage is first put at zero potential

(say), i.e., VCB = 0.— The emitter-base voltage (VEB) is now increased from zero onwards and

emitter current (IE) is recorded.— A graph is plotted between IE and VEB as shown in Fig. 3.30.

Collector-base voltage, VCB

Col

lect

orcu

rren

t,I C

I = 6 mAE

I = 4 mAE

I = 2 mAE

Emitter-base voltage, VEB

Em

itter

curr

ent,

I E

V = 30 voltsCBV = 0CB

Fig. 3.30. Input characteristic curves. Fig. 3.31. Output characteristic curves.

NOTES



— Another similar graph is plotted for VCB = 30 volts (say).

From the graph we observe that :

(i) For a given collector voltage, the emitter current rises rapidly even with avery small increase in emitter potential. It means that the input resistance Ri

=F

HGI

KJ∆∆VI

VEB

ECBat constant of the emitter-base circuit is very low.

(ii) The emitter current is nearly independent of collector-base voltage.

The Output Characteristic CurvesThese curves are obtained by plotting the variation of collector current (IC)

with collector-base voltage (VCB) at different constant values of emitter current(IE).

— These curves shown in Fig. 3.31, indicate that some collector current ispresent even when the collector voltage is zero. To make the collectorcurrent zero, we have to give a certain amount of negative potential to thecollector.

— The curves also indicate that the collector current attains a high valueeven at a very low collector voltage and further increase in collectorvoltage does not produce any appreciable increase in collector current.

It means that the output resistance RVI

IoCB

CE=

F

HGI

KJ∆∆

at constant of the

collector-base circuit is very large.

The collector current is always a little less than the emitter current because ofthe neutralisation of a few holes and electrons within the base due to recombination.

Feedback Characteristic Curves

These curves represent the variation of collector current (IC) with emitter-base voltage (VEB) for a constant-emitter current. A number of emitter currentvalues are selected at which measurements are made. The nature of curves isshown in Fig. 3.32.

= 4

= 3

= 2

= 1

I = 0E

Col

lect

or c

urre

nt, I

C

Emitter-base voltage, VEBEmitter current, IE

= 5 = 4 = 3 = 2 I = 1 AC

Col

lect

or-b

ase

volta

ge, V

CB

Fig. 3.32. Feedback characteristic. Fig. 3.33. Forward characteristic.


NOTES


Forward Characteristic CurvesRefer to Fig. 3.33. This type of curve is a graph between emitter current (IE)

and collector-base voltage at constant value of collector current.

3.11 COMMON-EMITTER (CE) CONFIGURATION

In CE configuration, input is applied betweenbase and emitter and output is taken from thecollector and emitter. Here, emitter of the transistoris common to both input and output circuits andhence the name common-emitter configuration.Fig. 3.34 shows common-emitter N-P-N transistorcircuit:

Base Current Amplification Factor (β)In CE configuration, input current is IB and

output current is IC. The ratio of change in collectorcurrent (∆IC) to the change in base current (∆IB) isknown as “base current amplification factor” i.e.,

β = ∆∆

II

C

B...(3.11)

If D.C. values are considered, β = II

C

B...[3.11 (a)]

In almost every transistor 5% of emitter current flows as the base current.Therefore the value of β is generally greater than 20, β usually varies from 20 to500.

CE configuration is frequently used as it gives appreciable current gain aswell as voltage gain.

Relation between β and α. The relation between β and α is derived as follows:

β = ∆∆

II

C

B...(i)

α = ∆∆

II

C

E...(ii)

Now, IE = IB + IC

or ∆IE = ∆IB + ∆IC or ∆IB = ∆IE – ∆IC

Inserting the value of ∆IB in (i), we get

β = ∆

∆ ∆I

I IC

E C−...(iii)

Dividing the numerator and denominator of R.H.S. by ∆IE, we get

β = ∆ ∆

∆ ∆ ∆ ∆I I

I I I IC E

E E C E

/( / ) ( / )−

=−α

α1 ∵ α =F

HGI

KJ∆∆

II

C

E

RC Output

InputIE

VBB VCC

E

IB

C

B

IC

Fig. 3.34. Common emitterN-P-N transistor.

NOTES


Basic Electrical andElectronics Engineering ∴ β =

αα1 −

...(3.12)

It is evident from the above expression that when α approaches unity, βapproaches infinity. In other words the current gain in CE configuration is veryhigh. It is due to this reason that this circuit arrangement is used is about 90 to 95percent of all transistor applications.

Collector Current

In CE configuration, IB is the input current and IC is the output current :

Now, IE = IB + IC ...(i)

and IC = αIE + ICBO ...(ii)

or IC = α(IB + IC) + ICBO

or IC(1 – α) = αIB + ICBO

or IC = αα α1

11−

+−

IB ICBO ...(iii)

It is evident from (iii) that if IB = 0 (i.e., base circuit is open), the collectorcurrent will be the current to the emitter. This is abbreviated as ICEO meaningcollector-emitter current with base open.

Inserting the value of 1

1 − α ICBO = ICEO in (iii), we get

IC = α

α1 − IB + ICEO

or IC = βIB + ICEO ...(3.13) ∵ β αα

=−

FHG

IKJ1

It may be noted that,

ICEO = (β + 1) ICBO ...(3.14)

Example 3.4. Find the α rating of the transistor shown in Fig. 3.35. Hencedetermine the value of IC using both α and β.


β = α

α1 −...[Eqn. (3.12)]

or β(1 – α) = αor β – αβ = αor β = α(1 + β)

∴ α = β

β149

1 49+=

+ = 0.98

∴ IC = αIE = 0.98 × 10 mA = 9.8 mA. (Ans.)

Also IC = βIB = 49 × 200 µA = 49 × 0.2 mA = 9.8 mA. (Ans.)

IB = 200 Am

IE = 10 mA

IC

b = 49

Fig. 3.35


NOTES


Characteristics of Common-emitter Transistor

Fig. 3.36 shows the circuit of a N-P-N common-emitter junction transistor forthe study of characteristic curves.

µA mA

C

EB

IB

IC

VBEIE

VCE

0.7

4

3

2

1

V=

1 V

CE

V=

10 V

CE

1.4 2.1 V (Volts)BE

I ( A)B µ

Fig. 3.36. Circuit of N-P-N common Fig. 3.37emitter junction transistor.

Input Characteristic Curves

It is the curve between base current IB and the base-emitter voltage VBE atconstant collector-emitter voltage VCE (Refer to Fig. 3.37).

Input resistance, Ri = ∆∆VIBE

B

at constant VCE . Its value is of the order of a few

hundred ohms.

Fig. 3.38 shows the graph of collector current (IC) with base current (IB) atconstant collector-emitter voltage. It may be noted from the curve thatthere is a collector current even when the basic current is zero. This isknown as collector leakage current:

It increases with rise in temperature and also arises due to the reverse biasingbetween base and collector. The value of leakage current ranges from 100 µA to500 µA.

10 20 30Base current, I , A)B m(

Col

lect

orcu

rren

t,(m

A)

I C V = ConstantCE

1

2

3

4

5

2 4 6 8 10 12Collector-emitter voltage, VCE

Col

lect

or c

urre

nt,

(mA

)I C

= 50 Aµ

= 40 Aµ

= 30 Aµ

IB = 20 Aµ

Fig. 3.38 Fig. 3.39

NOTES



Output Characteristic Curves

The collector-emitter voltage (VCE) is varied and the corresponding collectorcurrent (IC) is noted for various fixed values of base current (IB).

The shape of the curves is shown in Fig. 3.39.

Such common-emitter characteristics are widely used for design purpose.

It may be noted IC >> IB

Output resistance Ro = ∆∆VICE

C at constant IB. Its value is of the order of 50 kΩ

(less than that of CB circuit).

3.12 COMMON-COLLECTOR (CC) CONFIGURATION

In this type of configuration, input is appliedbetween base and collector while output is takenbetween the emitter and collector. Here, collectorof the transistor is common to both input andoutput circuits and hence the name common-collector connection. Fig. 3.40 shows the common-collector N-P-N transistor.

Current Amplification Factor (γγγγγ)In CC configuration, the input current is the

base current IB and output current is the emittercurrent IE. The ratio of change in emitter current(∆IE) to the change in base current (∆IB) is knownas “current amplification factor” i.e.,

γ = ∆∆

II

E

B

This circuit provides the same gain as the common-emitter configuration as∆IE ~− ∆IC. However, its voltage gain is always less than one.

Relation between γ and α

γ = ∆∆

II

E

B

...(i)

α = ∆∆

II

C

E...(ii)

Now, IE = IB + IC

or ∆IE = ∆IB + ∆IC or ∆IB = ∆IE – ∆IC

Inserting the value of ∆IB in (i), we get

γ = ∆

∆ ∆I

I IE

E C−

RC

Out

put

IC

VBB VEE

IB

IE

E

B

CACinput

Fig. 3.40. Common-collectorN-P-N transistor.


NOTES


Dividing the numerator and denominator of R.H.S. by ∆IE , we get

γ = ∆ ∆

∆ ∆ ∆ ∆I I

I I I IE E

E E C E

/( / ) ( / )−

=−1

1 α∵ α =F

HGI

KJ∆∆

IIC

E

∴ γ = 1

1 − α...(3.15)

Collector Current

We know that, IC = αIE + ICBO ...(From Eqn. 3.9)

Also, IE = IB + IC = IB + (αIE + ICBO)

or IE(1 – α) = IB + ICBO

or IE = I IB CBO

1 1−+

−α α

or IC , IE = (β + 1)IB + (β + 1)ICBO ...(3.16)

β αα

β αα α

=−

∴ + =−

+ =−

L

NM

O

QP1

11

11

1

Commonly used transistor connection:

Out of the three configuration, the CE configuration is the most efficient. Itis used in about 90 to 95% of all transistor applications. This is due to followingreasons :

1. High current gain ; it may ranges from 20 to 500.

2. High voltage and power gain.

3. Moderate output to input impedance ratio (this ratio is small, to the tune of50). This makes this configuration an ideal one for coupling between varioustransistor stages.

3.13 SMALL SIGNAL AMPLIFIERS

A single-stage circuit may be employed as a small signal amplifier, but twocascaded stages gives much greater amplification. For very high input impedance,a field effect transistor may be used as an input stage, with a bipolar junctiontransistor as the second stage.

Calculation of circuit resistor values merely involves application of Ohm’slaw, after selecting suitable voltage and current levels throughout the circuit. Eachcapacitor is calculated at the circuit low 3 dB frequency in terms of the resistancein series with the capacitor.

3.14 COMMON EMITTER AMPLIFIER DESIGN

Bias circuit design for the common emitter amplifier circuit is shown inFig. 3.41 (a). To design this circuit, it is necessary to work from a specification

NOTES



which might state the supply voltage to be used, the desired voltage gain, thefrequency response, the signal source impedance, and the load impedance.

Selection of IC, RC and RE. Designing for a particular voltage gain requiresthe use of ac negative feedback to stabilize the gain. The circuit in Fig. 3.41 (a) hasno provision for negative feedback ; thus, it is designed to achieve the largest possiblegain. The voltage gain of the equation is given by

Av = – hFe (RC || RL)/hie

+ VCC

VRC

IC

RCR1

C1

I2

R2 VB

RE VEC2

VCE

C4 RL

C3

(a) Common emitter amplifier circuit

hib

E

XC2

VS Zi Vi

XC1 XC3

VoRL

(b) h is in serieswith X

ib

C2

(c) Small signal voltageis potentially dividedacross and ZiXC1

(d) The collector ac voltageis potentially dividedacross X and RC3 L

Fig. 3.41. Circuit for designing a common emitter amplifier.

Since Av is directly proportional to RC || RL, design for the greatest voltagegain normally requires selection of the largest possible value of collector resistanceRC. However, an extremely large value of RC may make the collector current toosmall for satisfactory operation. For most small-signal transistors, IC should benot less than 500 µA. A good minimum to aim for is 1 mA. (Special low noisetransistors operate with much lower collector current levels).

The transistor collector current might be selected high to gives the largest hFevalue, again to achieve the greatest Av. But a high level of IC results in a smallvalues of RS. So a high IC may actually produce a lower voltage gain, although hFemay be relatively large.

For a given level of IC, the largest possible voltage drop rRC gives the greatestvalue of RC, (RC = VRC/IC). To make VRC as large as possible, VCE and VE should beheld to a minimum. The collector-emitter voltage should typically be at least 3 V,


NOTES


to ensure that the transistor operates in it’s active region. This easily allows amaximum output voltage swing of ± 1 V, which is usually quite adequate for smallsignal amplifiers.

For good bias stability, the emitter resistor voltage drop ‘VE’ should be muchlarger than the transistor base-emitter voltage VBE. This is because VE = VB – VBE,and when VE >> VBE, any variation in VBE (due to temperature or other effects) hasonly a slight effect on VE. Consequently, IE and IC remain fairly stable atIC ≈ IE = VE/RE. In most circumstances, a minimum VE of 5 V gives good biasstability. Where the supply voltage is less than 10 V, VE may have to be reduced to3 V to allow for reasonable levels of VCE and VRC. Normally, a voltage much lessthan 3 V is likely to result in poor bias stability.

Only VE, VCE, and IC are decided upon, VRC is determined

VRC = VCC – VCE – VE

Then RC and RE are calculated,

RC = VI

RVI

RC

CE

E

C

, = .

Bias Resistors. The selection of potential divider current I2 as IC/10 givesgood bias stability and high input resistance. Where the input resistance is notimportant, I2 may be made equal to IC for excellent bias stability R1 and R2 arecalculated as follows :

R2 = VI

RV VI I

B CC B

B21

2

, =−

+.

Bypass Capacitor. The bypass and coupling capacitors should be chosen tohave the smallest possible capacitance value, both for economy and to minimizethe physical size of the circuit since each capacitor has it’s highest impedance atthe lowest operating frequency, the capacitor values are calculated at the lowestsignal.

Frequency that the circuit is required to amplify. This frequency is the circuitlower cutoff frequency, or low 3 dB frequency, Fi.

Bypass capacitor C2 in Fig. 3.41 (a) is normally the largest capacitor in thecircuit. The circuit low 3 dB frequency F1 is determined by C2 is developed for acommon emitter circuit with an unbypassed emitter resistor RE. Rewriting this

equation to include XC2 in parallel with RE gives

Av = −

+ +h R R

h h R XFe C L

ie Fe E C

( || )( || )( )1 2

Normally, RE >> XC2. Also, XC2

is capacitive. Therefore

Av ≈ −− +h R R

h j h XFe C L

ie Fe C

( || )( )( )1

2

or | Av | ≈ −

− +

h R R

h h XFe C L

ie Fe C

( || )

( )]2 212

[( )

NOTES



When hie = (1 + hFe) ( XC2),

| Av | ≈ −

+=

h R R

hFe C L

ie

( || ) (mid Frequency gain)21 12 2

= (mid Frequency gain) – 3 dB.

Therefore, at F1, hie = (1 + hFe) ( XC2)

or XC2 =

hhie

e( )1 + F...(3.17)

giving 1

2 1 2πF CF

HGI

KJ = hie/(1 + hFe)

hib = h

hie

Fe1 + = (impedance seen when looking into transistor emitter)

Therefore, at F1, XC2 = (impedance when looking into transistor emitter).

= (impedance in series with XC2):

Equation 3.17 gives the smallest value for the bypass capacitor, when selectinga standard capacitor, the next larger value should be chosen, in order to give an F1slightly lower than the frequency used in the calculation.

Coupling Capacitors. Coupling capacitors should have very little effect on

amplifier frequency response XC2 and Zi constitute a potential divider [see

Fig. 3.41 (c)]. If XC1is too large Vi will be smaller than Vs and will increase with

increasing signal frequency. Similarly XC3 and RL potentially divide the a.c. collector

voltage [Fig. 3.41 (d)]. To minimize the effects of XC1and XC3

, the reactance of each

coupling capacitor at the lowest operating frequency for the circuit is selected to beapproximately equal to one tenth of the impedance in series with it.

XC1 =

Zi

10...(3.18)

XC3 =

RL

10...(3.19)

Once again, these equations give minimum capacitance values. The next largerstandard-size capacitors should be selected.

Equation (3.18) and (3.19) gives an impression that approximately 10% of thesignal and output voltage is lost across C1 and C3, respectively. This would be true ifall the quantities were resistive. However, XC1

and XC3 are capacitive (Zi and RL are

usually resistive). From Fig. 3.41 (d)

| V0 | = V R

R XC C

L C

×

+2 23

with RL = 100 kΩ and XC3 = 10 kΩ,

| V0 | = V k

k kC ×

+

100

100 102 2

Ω

Ω Ω( ) ( ) = 0.995 VC.


NOTES


Thus, loss of ac output voltage across C3 amounts to only 0.5%. Similarly, itcan be shown that approximately 0.5% of the input signal is lost across C1 when Ziis purely resistive and XC1

= Zi/10. Of course, for some circuits, RL and Zi may notalways be purely resistive and XC1

= Zi/10. Of course, for some circuits, RL and Zimay not always be purely resistive and in such cases the foregoing analysis wouldnot be correct.

Shunting Capacitor. Sometimes an amplifier is required to have a particularupper cutoff frequency, or high 3 dB frequency F2. Normally, the transistor mustoperate satisfactorily to a frequency much higher than F2, that is the transistorcutoff frequency must exceed the circuit cutoff frequency.

The upper cutoff frequency for a circuit can be set quite easily by including asmall capacitor to shunt the output terminal to ground. Capacitive impedance XC4

isthen part of the transistor ac load, so that the voltage gain equation becomes

Av = − h R R X

hFe C L C

ie

( || || )4

At low-and midband signal frequencies XC4is very much larger than RC || RL,

so that it has no effect on amplifier gain. At higher frequencies XC4 becomes smaller

and begins to reduce the amplifier gain. Because XC4 is capacitive and RC || RL is

normally resistive, RC || RL || XC4 can be rewritten as

RC || RL || XC4 =

11 1

4/( || ) /R R j XC L C+

or | (RC || RL || XC4) | = 1

1 12 2

4( || )R R XC L C

F

HGI

KJ+F

HG

I

KJ

When XC4 = RC || RL,

| (RC || RL || XC4) | =

11

1 12 2

( || )R RC L+

= R RC L||

2

This gives | Av | = h R R

hFe C L

ie

( || ) (mid frequency gain)22

=

Therefore, at F2, XC4 = RC || RL ...(3.20)

Example 3.5. Calculate suitable resistor values for the common emitteramplifier in Fig. 3.41 (a). The circuit is to use a 2 N 3904 transistor, the supplyvoltage is VCC = 24 V and the external load resistance is RL = 120 kΩ.

Solution. RC << RL

Let RC = RL/10

= 120 kΩ/10 = 12 kΩ (standard value resistor)

Next, let VE = 5 V, VCE = 3 V

Then VRC = VCC – VCE – VE = 24 V – 3 V – 5 V = 16 V

NOTES


Basic Electrical andElectronics Engineering IC =

VR k

RC

C

= 1612

VΩ

= 1.3 mA

RE = VI

E

C

= 5 V1.3 mA

= 3.75 kΩ (use a 3.9 kΩ standard value) I2 = IC/10 = 1.3 mA/10 = 133 µA

R2 = V V

IE BE+ = +

2

5 V 0.7 V133 Aµ

~– 43 kΩ (use 39 kΩ and recalculate I2)Now, I2 = 5.7 V/39 kΩ. ≈ 146 µA

So, R1 ≈ V V

ICC B– –

2

= 24 V 5.7 V146 Aµ

.

≈ 125 kΩ (use a 120 kΩ standard value).

3.15 CAPACITOR COUPLED TWO-STAGE COMMON

EMITTER AMPLIFIER

Circuit Design. A capacitor coupled, two-stage amplifier circuit is shown inFig. 3.42. Each stage is similar to the single stage circuit in Fig. 3.41 (a). Thesimplest approach to the design of this circuit to make each stage identical. Then,when stage 2 has been designed the components for stage 1 are selected to be R1 =R5, R2 = R6, R3 = R7, R4 = R8, C1 = C3 and C2 = C4. Note that C3 is calculated interms of the input resistance to stage 2, which should be identical to that for stage1. If the circuits are otherwise identical C5 is determined by making XC5

<< RL.Because of the presence of C3, stage 1 does not need a capacitor that corresponds toC5.

One way in which design of a two-stage circuit differs from that of a single-stage circuits is in the calculation of capacitors C2 and C4. Consider eqn. (3.17).

XC2 = hie/1 + hFe

This equation was derived to make the voltage gain of a single-stage circuit3 dB down from the mid frequency gain at frequency F1. If the equations is used tocalculate the capacitance of C2 and C4, it is found that both the gain of stage-1 andthe gain of stage-2 are down by 3 dB at F1. This means that the overall gain of thecircuit is down by a total of 6 dB at frequency F1. For a 3 dB total reduction inoverall voltage gain at F1, the bypass capacitors must be calculated to give a 1.5 dBreduction in gain at each stage A 1.5 dB reduction in stage gain at F1 is achieved bymaking the reactance of the emitter bypass capacitor equal to 0.65 times theimpedance in series with the capacitor. Thus, the equation for the bypass capacitorsnow becomes,

XC2 = 0.65

hhie

Fe1 +...(3.21)


NOTES


AC Analysis. AC analysis of a two-stage, capacitor-coupled circuit is similarto analysis of single-stage circuits. The h-parameter equivalent circuit for theamplifier circuit of Fig. 3.42 is drawn in Fig. 3.43. Once again, this equivalentcircuit is produced by replacing the supply voltage and all capacitors with shortcircuit and substituting device h-parameter equivalents for each transistor. Theresultant equivalent circuit simply consists of two single stage h-parameterequivalent circuits. As always, voltage polarities and current directions are indicatedfor an instantaneous positive-going signal voltage. The voltage polarities show a180° phase shift between Vi and VC1

and a further 180° shift from VC1 to V0.

The performance equation for the two-stage circuit are readily derived fromthe equations for a single-stage circuit. Input and output impedance are exactlythe same as for the single-stage circuit.

Zi = R1 || R2 || hie ...(3.22)

Z0 ≈ R7 ...(3.23)

R1 120 k

R= 12 k

3

R 120 k5 R12 k

7

C = 0.15 F5

V = 24 VCC

18 F

18 F

ViR =39 k

2 R4

3.9 k

C2

250 F

Q2 N 3904

1

VC1 R639 k

Q2 N 3904

2

C4

250 F

R8

3.9 k

RL120 k Vo

Vi

VC1

Vo

Fig. 3.42. Two-stage, capacitor coupled common emitter amplifier.

Stage-1 Stage-2IS Ii IC1

rS

VS

ZiVi R1 || R2 hie1

I2

V1CR3 R5 || R6 hie2

IC2

R7

Zo

hFeI2

RLV0

Fig. 3.43. The h-parameter equivalent circuit for a two stagecapacitor coupled amplifier.

The voltage gain equation for the second stage is exactly the same as thesingle-stage gain equation. The capacitor coupled impedance at the collectorterminal of the first stage is the input impedance of the second stage. So the voltagegain equation are,

NOTES


Basic Electrical andElectronics Engineering Av1

= – ( )h R Z

hFe i

ie

1 2

1

3 ||...(3.24)

and Av2 =

– ( )h R R

hFe L

ie

2

2

7 ||...(3.25)

The overall voltage gain is found by multiplying the two individual stage-gains:

Av = A Av v1 2× ...(3.26)

Equations for current gain and power gain can also be derived from the single-stage equations, but these quantities are not of great importance.

3.16 DIRECT COUPLING BETWEEN STAGES DIRECT

COUPLED CIRCUIT

For economical circuit design, the number of components used should be keptto a minimum. The use of direct coupling between stages is one way of eliminatingcomponents. Fig. 3.44 shows a circuit in which the base of transistor Q2 is directlycoupled to the collector of Q1. Comparing this circuit to the capacitor-coupled circuitin Fig. 3.42, it is seen that the two bias resistors for stage 2 and the interstagecoupling capacitors have been eliminated. This is a saving of only three components;however, when several hundred or more similar circuits are to be manufactured,the savings can be considerable.

Circuit Design. The first step in the design of the direct coupled circuit inFig. 3.44 is to determine a suitable level of base bias voltage for Q2. This is done byestimating satisfactory levels of VE1

and VCE1 for transistor Q1. Then,

V V V VB C E CE2 1 1 1= = +

R =68 k

1

C = 471 F

I2

R =47 k

2 R

4.7 k4

C330

2F

VE1

V E1C

Q1, 2 N 3903Q2, 2 N 3903

IB2

5.6 k

VB2

R8.2 k

6

VE2 C3330 F

R53.9 k

RL

C4 = 0.56 F

VCC = 14 V

Fig. 3.44. Two-stage, direct coupled common emitter amplifier.

The emitter voltage of Q2 is

V V VE B BE2 2= −


NOTES


and the voltage across Resistor R5 is V V V VR CC E CE5 2 2

= − − .

As for all amplifier circuits, the resistor at the collector of the output transistor(R5 in Fig. 3.44). Should be much smaller than the external load resistor RL. OnceR5 is selected, IC2

can be calculated using the voltage VR5 already determined. If

IC2 looks too small for satisfactory operation of the transistor, a suitable current

level should be selected and a new value of R5 calculated. The collector current ofQ1 is determined by making IC1

very much greater than the base current for Q2.This is done to ensure that IB2

has a negligible effect on the bias conditions of Q1.Normally just making IC1

equal to IC2 is the simplest way to achieve the desired

effect.AC Analysis. A.C. analysis of a two stage direct-coupled circuit is similar to

analysis of a two-stage capacitor coupled circuit. The h-parameter equivalent forthe circuit of Fig. 3.44 is exactly like Fig. 3.43, except that the bias resistors for thesecond stage are emitted. The voltage gain and impedance equations are asdetermined for the capacitor-coupled circuit. Because the component numbers differslightly for the two circuits, care must be taken in substituting components intothe equation. For example, from eqn. 3.25, the voltage gain of the second stage ofthe circuit in Fig. 3.44 is

Av2 =

– ( )h R R

hFe

ie

2

2

5 6||

Use of Complementary Transistors. The direct-coupled, two stage circuitillustrated in Fig. 3.45 is similar to that in Fig. 3.44, except that transistor Q2 is apnp device Q1 and Q2 are usually selected to have similar characteristics andparameters, although one is npn and the other is pnp. In this case the transistorsare complementary. Suppose the circuit in Fig. 3.45 is to be designed to use a 14 Vsupply, the base voltage for Q2 is then

V V VB R2 36= = and V V V V V VE B BE2 2

6 0 7 5 3= − = − =. .

Hence, V V V VR CC E CE6 2 2= − − = 14 V – 5.3 V – 3 V = 5.7 V.

R68 k

1

C 471 F

R47 k

2 R

4.7 k4

C330

2F

Q1, 2 N 39032 N 3905

V5.6 k

R3

R5.6 k

6

C4

0.56 F

R55.6 k

RL40 k

VCC = 14 V

R3

VB2

VE2 C330

3F

Fig. 3.45. Direct coupled, two-stage circuit using complementary transistor.

NOTES



Thus, using complementary transistors, the voltage across the collector resistorin the second stage is 5.7 V instead of 3.7 V. For a given level of collector current,the larger voltage drop gives a larger resistor, which results in greater voltagegain.

The design procedure for this circuit is very similar to the procedure fordesigning the circuit in Fig. 3.45.

3.17 TWO-STAGE CIRCUIT WITH EMITTER FOLLOWER

OUTPUT

Fig. 3.46 shows another direct-coupled circuit. This time the second-stage is acommon collector circuit, or emitter follower. Stage-2 gives the circuit a very lowoutput impedance, but has unity voltage gain. Stage-1 still has substantial voltagegain.

R68 k

1

C 221 F

R =27 k

2 R

2.7 k4

C180

2F

Q1, 2 N 3903C3

18 FRL =100 k

VCC = 20 V

R5.6 k

3

Q2, 2 N 3903

VB2

R3.3 k

5 VE2

Fig. 3.46. Direct-coupled two stage circuit.

The design procedure for this circuit is similar to the procedure alreadydiscussed. Base bias voltage VB2

is fixed as the collector voltage of the previousstage.

A suitable level of emitter current IE2 is selected and RE is calculated to be

(V VB BE2− )/IE2

. Coupling capacitor C1 is once again calculated from XC1 = Zi/10.

Load resistor RL may be relatively small, which means that coupling capacitor C3must be large. Therefore both C2 and C3 are used to determine a low 3 dB frequencyof the circuit by making

XC2 = 0.65 × hie/(1 + hFe) and XC3

= 0.65 × RL at F1.

It is important to realize that the emitter follower in Fig. 3.46 is a small-signal circuit. In fact, this type of circuit functions best when the amplitude of thea.c. output voltage is much smaller than the d.c. base-emitter voltage of the outputtransistor VBE 2

. With ac output voltages which approach or exceed VB2. The base-

emitter junction of Q2 may be reverse biased when the output goes rapidly in anegative direction. Thus, the circuit does not function correctly, where large outputvoltages and low output impedance are required, a complementary emitter followeris used.


NOTES


With the circuit operating as a small-signal common collector amplifier, emitterresistor R5 in Fig. 3.46 does not have to be very much smaller than external loadRL. However, for satisfactory operation, emitter current IE2

should be greater thanthe peak ac load current ip. The peak output current is calculated by dividing thedesired peak output voltage VP by the load resistance, i.e., xp = VP/RL.

3.18 SMALL-SIGNAL HIGH FREQUENCY AMPLIFIER

Common-base Amplifier. The common base circuit, analysis shows to havevery low input impedance, as well as output impedance similar to that of a commonemitter circuit. The voltage gain of a common base circuit is also shown to be similarto that of a common-emitter circuit, with the important exception that there is nophase inversion between output and input. Because of this absence of phaseinversion, there is no miller effect (or amplification of the input capacitance) witha common base circuit. Consequently a common base circuit is able to operate atmuch higher frequencies than a common-emitter circuit. A practical common-basecircuit is illustrated in Fig. 3.47. The procedure for designing this circuit is exactlyas covered in section 2 for designing a common emitter circuit.

When the a.c. equivalent circuit for Fig. 3.47 is drawn it is seen that theimpedance in series with capacitor C2 is the impedance seen when looking into theemitter of the transistor i.e.,

hib = hie/(1 + hFe)

This low input impedance makes C2 the largest capacitor in the circuit,consequently C2 is used to set the circuit low 3 dB frequency F1 and is calculated inthe same way as the emitter bypass capacitor in a common emitter circuit.

Fig. 3.47. Practical circuit for common-base amplifier.

The impedance in series with capacitor C1 is the impedance hie seen whenlooking into the base of the transistor. At the circuit low 3 dB frequency F1, XC1

should be very much smaller than hie. This is similar to the calculation of C1 in thecommon emitter circuit.

XC1=

hie

10...(3.27)

NOTES



Cascode Amplifier. The cascode amplifier in Fig. 3.48 is one approach tosolve the low input impedance problems of a common base circuit Q1 and it’sassociated components operate as a common emitter input stage, while the circuitof Q2 functions as a common base output stage. This arrangement gives the cascodecircuit the high input impedance of a common emitter amplifier, as well as thegood voltage gain and high frequency performance of a common base circuit.

VCC

C4

R3R1IC2

I2

R2 VB2 C1

Q2I

C1

E2= I

Q1

IE1

R4C2

R6VB1

R5

C3

VCC

RL

Fig. 3.48. Cascode amplifier.

For the dc bias conditions of the circuit, it is seen that emitter current for Q1

is set by VE1 and R4. Collector current IC1

approximately equals IE1 and IE2

is the

same current as IC1. Therefore, IC2

approximately equals IE1. This current remains

constant regardless of the level of VB2 as long as VCE1

remains large enough for

current operation of Q1.

The input impedance to the emitter of Q2 constitutes the a.c. load in thecollector circuit of Q1.

Av1 =

− ×=

− × +h Z to Qh

h h hh

Fe i z

ie

Fe ie Fe

ie

( ) [ /( )]1 ≈ – 1.

With a stage gain of only 1, no miller effect occurs at transistor Q1.

From the above equation, the voltage gain of stage-2 is

Ah R R

hvFb L

ib2

3=× ( || )

Converting to common emitter parameters, the overall voltage gain for thecascode circuit is the same as that for a common emitter amplifier.

Av = – h R R

hFe L

ib

× ( || )3

The design procedure for the cascode circuit is easily determined from thecommon emitter design already covered. The voltage across R4 should typically be


NOTES


VE1 = 5V, as already explained. Also, as previously discussed, the collector-emitter

voltage for each transistor should typically be a minimum of VCE = 3V. The emitter

voltage for Q2 is V V VE CE E2 1 1= + . The transistor base voltages are V V VB E BE1 1

= +

and V V VB E BE2 2= + . Capacitor C1 is calculated using eqn. (3.27), and C2 is

determined by making XC2 = hie/(1 + hFe) at F1. Coupling capacitors C3 and C4 are

determined in the usual way.

SUMMARY

• Semiconductors are solid materials, either non-metallic elements or com-pounds, which allow electrons to pass through them so that they conductelectricity in much the same way as a metal.

• A P-N junction diode (known as a semiconductor or crystal diode) consistsof a P-N junction, formed either in germanium or silicon crystal. The diodehas two terminals namely anode and cathode.

• A properly doped P-N junction crystal diode which has a sharp breakdownvoltage is known as Zener diode.

• A rectifier is a circuit, which uses one or more diodes to convert A.C.voltage into pulsating D.C. voltage.

• A full-wave rectifier is a circuit, which allows a unidirectional currentto flow through the load during the entire input cycle. This can be achievedwith two diodes working alternately.

• The change in D.C. output voltage from no load to full load with respect tofull load voltage of a power supply is known as its voltage regulation.

• The transistor is a solid state device, whose operation depends upon theflow of electric charge carriers within the solid.

• There are three types of configurations for operation of a transistor. Theseconfigurations are:

(i) Common-base (CB) configuration.(ii) Common-emitter (CE) configuration.

(iii) Common-collector (CC) configuration.• A single-stage circuit may be employed as a small signal amplifier, but

two cascaded stages gives much greater amplification. For very high inputimpedance, a field effect transistor may be used as an input stage, with abipolar junction transistor as the second stage.

• The common base circuit, analysis shows to have very low input imped-ance, as well as output impedance similar to that of a common emittercircuit. The voltage gain of a common base circuit is also shown to besimilar to that of a common-emitter circuit, with the important exceptionthat there is no phase inversion between output and input.

NOTES


Basic Electrical andElectronics Engineering GLOSSARY

• Junction: In semiconductor diode, there is a place which divide the twozones is known as junction.

• Ripple factor: In output voltage of a rectifier, the A.C. component presentin the output is called a ripple. As a matter of factor, the ripple isundesirable and accounts for pulsations in the rectifier output.

• Current amplification factor (α): The ratio of change in collector current(∆Ic) to the change in emitter current (∆IE) at constant collector voltage(VCB).

α = ∆∆

II

c

E at constant VCB.

• Base current amplification factor (β): The ratio of change in collector current(∆Ic) to change in base current (∆IB).

β = ∆∆

II

c

B

.

REVIEW QUESTIONS

1. Define a ‘semiconductor’.

2. List the important characteristics of semiconductors.

3. Give examples of semiconducting materials.

4. What is the difference between a semiconductor and an insulator?

5. What is a P-N junction diode ? How its terminals are identified?

6. Draw the V-I characteristics of a junction diode when it is (a) forward biasedand (b) reverse biased.

7. Draw the graphical symbol of a crystal diode and explain its significance.How the polarities of function diode are identified?

8. What are the important applications of a diode?

9. Write a short note on the power and current ratings of a diode.

10. What is a Zener diode? Draw its equivalent circuit

11. Explain briefly the applications of a Zener diode.

12. What do you understand by Zener voltage?

13. Explain why Zener diode is always operated in reverse biasing.

14. Explain how a Zener diode can stabilize the voltage across the load.

15. Explain the process of Zener breakdown.

16. Draw and explain a Zener diode voltage regulator.

17. Define the term ‘Transistor’.

18. What are the various types of transistors?


NOTES


19. Differentiate between P-N-P and N-P-N transistors. Why are collector andemitter currents nearly equal in these transistors?

20. Define α and β of a transistor and derive the relationship between them.

21. Define three basic configurations of N-P-N transistor.

22. Draw input and output characteristics of CB transistor configuration.

23. Draw the circuits of the various transistor configurations. List their importantfeatures. Why CE configuration is mainly used?

24. Derive an expression for the efficiency of a full-wave rectifier.

25. What is a ripple factor? What is its value for a half-wave and a full-waverectifier?

26. The current flowing in a certain P-N junction diode at room temperature is2 × 10–7 A, when large reverse voltage is applied. Calculate the currentflowing, when 0.1 V forward bias is applied at room temperature.

27. Determine the germanium P-N junction diode current for the forward biasvoltage of 0.22 V at room temperature 25°C with reserve saturation currentequal to 1 mA. Take η = 1.

28. For the circuit shown in Fig. A, find the maximum and minimum values ofZener diode current.

IZ

10 kΩ

ILI5 kΩ

80 – 120 V

Fig. A

29. The zener diode shown in Fig. A has VZ = 18 V. The voltage across the loadstays as 18 V as long as IZ is maintained between 200 mA and 2A. Find thevalue of series resistance RS so that Vout remains 18 V while input voltage isfree to vary between 22 V to 28 V. [Ans. 3.33 Ω]

30. In a common-base configuration, the emitter current is 1 mA. If the emittercircuit is open, the collector current is 50 µA. Find the total collector current.Given that α = 0.92. [Ans. 0.97 mA]

31. In a common-base configuration, α = 0.95. The voltage drop across 2 kΩresistance which is connected in the collector is 2 V. Find the base current.

[Ans. 0.05 mA]

32. Calculate IE in a transistor for which β = 50, IB = 20 µA. [Ans. 1.02 mA]

33. For a transistor, β = 45 and voltage drop across 1 kΩ which is connected inthe collector circuit is 1 volt. Find the base current for common-emitterconnection. [Ans. 0.022 mA]

34. A transistor is connected in CE configuration in which collector supply is 8 Vand the voltage drop across resistance RC connected in the collector circuit is0.5 V. The value of RC = 800 Ω. If α = 0.96, determine the collector emittervoltage and base current. [Ans. 7.5 V, 0.026 mA]

NOTES



35. A JFET has the following parameters :

IDSS = 32 mA ; VGS (off) = – 8 V ; VGS = – 4.5 V.Find the value of drain current. [Ans. 6.12 mA]

36. A full-wave rectifier uses two diodes, the internal resistance of each diodemay be assumed constant at 20 Ω. The transformer r.m.s. secondary voltagefrom centre tap of each end of secondary is 50 V and load resistance is 980 Ω.Calculate :

(i) The mean load current.

(ii) The r.m.s. value of load current. [Ans. (i) 45 mA, (ii) 50 mA]

37. Write an equation for calculating the coupling capacitors in a capacitor-coupled common-emitter amplifier. Show that, when this equation is used,the coupling capacitors have very little effect on the performance of the circuit.

38. Show how a capacitor may be used to obtain a desired upper cutoff frequencyin a transistor common-emitter amplifier. Derive the equation for determiningthe capacitor value.

39. Discuss the factors involved in the section of ID, RD, and RS for a single-stagecommon-source FET amplifier circuit using voltage divider bias. Also, explainthe process for determining suitable bias resistors.

40. Sketch the circuit of a two-stage direct-coupled common-emitter amplifierusing npn BJTs. Discuss the advantages of direct coupling between stages.

41. Sketch the h-parameter equivalent circuit for the amplifier. Write equationsfor Zi, Zo, and Av for the circuit.

FURTHER READINGS

• Nagasarkar T.K. and Sukhija M.S., “Basic Electrical Engineering”, OxfordPress (2005).

• Mehta V.K., “Principles of Electronics,” S. Chand & Company Ltd. (1994).• R.K. Rajput, “Basic Electrical and Electronics Engineering”, Laxmi

Publications (P) Ltd.


NOTES

Digital Electronics

STRUCTURE

4.1 Number System4.2 Conversions in Number System4.3 Addition and Subtractions of Number System4.4 Complements4.5 Logic Gates4.6 Universal Gates4.7 Half Adder4.8 Full Adder4.9 Boolean Algebra4.10 Flip-Flop Circuits4.11 Counters4.12 Registers4.13 A/D and D/A Converters: An Introduction4.14 Digital to Analog Conversions4.15 Resistor Divider D/A Converter4.16 Analog to Digital Conversions4.17 Voltage to Time Analog to Digital Converter4.18 A/D Converter Specifications




U N I T

4DIGITAL ELECTRONICS

OBJECTIVES


• define number system, its conversions and mathematical operations suchas addition and subtraction.

NOTES



• illustrate the topics like logic gates, half and full adder, Boolean Algebra,counters and registers.

• describe A/D and D/A conversion (single concepts only).

4.1 NUMBER SYSTEM

Number systems provide the basis for all operations in information processingsystems. In a number system, the information is divided into a group of symbols;for example, 26 English letters, 10 decimal digits etc.

In conventional arithmetic, a number system based upon ten units (0 to 9) isused. However, arithmetic and logic circuits used in computers and other digitalsystems operate with only 0's and 1's because it is very difficult to design circuitsthat require ten distinct states.

The number system with the basic symbols 0 and 1 is called binary i.e., Abinary system uses just two discrete values. The binary digit (either 0 or 1) iscalled a bit.

A group of bits which is used to represent the discrete elements of informationis a symbol. The mapping of symbols to a binary value is known a binary code. Thismapping must be unique. For example, the decimal digits 0 through 9 arerepresented in a digital system with a code of four bits. Thus a digital system is asystem that manipulates discrete elements of information that is representedinternally in binary form.

Decimal Numbers

The decimal number system has been the most important useable numbersystem. The decimal number system uses positional number representation, whichmeans that the value of each digit is determined by its position in a number. Herethe following factors must be kept in mind because these factors are applied to allother number system.

1. The base, also called the radix of a number system is the number of symbolsthat the system contains. For example, the base or radix of decimal numbersystem is 10.

2. The decimal system has ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

3. In other words, it has a base of 10. Each position in the decimal system is10 times more significant than the previous position.

4. The numeric value of a decimal number is determined by multiplying eachdigit of the number by the value of the position in which the digit appearsand then adding the products. Thus the number 2734 is interpreted as

2 × 1000 + 7 × 100 + 3 × 10 + 4 × 1 = 2000 + 700 + 30 + 4

Here 4 is the least significant digit (LSD) and 2 is the most significantdigit (MSD).


NOTES

Digital ElectronicsBinary Numbers

The binary number has a radix of 2. As r = 2, only two digits are needed, andthese are 0 and 1. Like the decimal system, binary is a positional system, exceptthat each bit position corresponds to a power of 2 instead of a power of 10.

In binary n bits can represent N = 2n symbols. e.g., 3 bits can represent up to8 symbols, 4 bits for 16 symbols etc. For N symbols to be represented, the minimumnumber of bits required is the lowest integer n that satisfies the relation givenbelow:

2n ≥ N

For example, if N = 26, minimum n is 5 since 24 = 16 and 25 = 32.

Octal Numbers

Digital systems operate only on binary numbers. Since binary numbers areoften very long, two shorthand notations—octal and hexadecimal, are used forrepresenting large binary numbers.

Octal systems use a base or radix of 8. Thus it has digits from 0 to 7 (r – 1). Asin the decimal and binary systems, the positional valued of each digit in a sequenceof numbers is fixed. Each position in an octal number is a power of 8, and eachposition is 8 times more significant than the previous position.

Hexadecimal Numbers

The hexadecimal numbering system has a base of 16. There are 16 symbols.The decimal digits 0 to 9 are used as the first ten digits as in the decimal system,followed by the letters A, B, C, D, E and F, which represent the values 10, 11, 12,13, 14 and 15 respectively.

Table 4.1 shows the relationship between decimal, binary, octal andhexadecimal number systems.

Table 4.1

Decimal Binary Octal Hexadecimal

0 0000 0 01 0001 1 12 0010 2 23 0011 3 34 0100 4 45 0101 5 56 0110 6 67 0111 7 78 1000 10 89 1001 11 9

10 1010 12 A11 1011 13 B12 1100 14 C13 1101 15 D14 1110 16 E

15 1111 17 F

NOTES


Basic Electrical andElectronics Engineering 4.2 CONVERSIONS IN NUMBER SYSTEM

Conversion of the Base to Decimal

In order to conversion between binary and decimal numbers, we need tounderstand the relationship between the digits of a given number, the position ofthose digits and the base of the number system. Let's take another look at ourexample number 462.15 in the decimal system.

Hundreds Tens Ones Tenths Hundredths

102 101 100 10–1 10–2

4 6 2 . 1 5

The first digit of our number is 4, but because this digit is located in thehundreds column, we know it really represents the value 4 × 102 or 400. Similarly,the 6 in the tens column represents the value 6 × 101 or 60. Continuing this pattern,we can express the number 462.1510 as follows:

4 × 102 = 4 × 100 = 400

6 × 101 = 6 × 10 = 60

2 × 100 = 2 × 1 = 2

1 × 10–1 = 1 × 0.1 = 0.1

5 × 10–2 = 5 × 0.01 = + 0.05462.15

In general, the relationship between a digit, its position, and the base ofthe number system is expressed by the following formula:

DECIMAL NUMBER = DIGIT × BASEPOSITION

In digital systems, the binary number system and other number systemsclosely related to it are used almost exclusively.

Binary to Decimal Conversion

Now let's look at the number 110.112 using the same relationship betweendigit, base, and position number. Recall the formula for computing the value of adigit:

DECIMAL NUMBER = DIGIT × BASEPOSITION

Since the binary system is a base-2 number system, we can make this formulamore specific:

DECIMAL NUMBER = DIGIT × 2POSITION

As we did with the previous decimal example, we can analyze the digits ofour binary number 110.112 in a table.

Fours Twos Ones Halves Fourths

22 21 20 2–1 2–2

1 1 0 . 1 1


NOTES

Digital ElectronicsNotice that the position values in our table are now powers of two instead often. Using formula, we see that the first digit of binary number represents thevalue 1 × 22 or 4 and the second digit represents the value 1 × 21 or 2. Continuingthis pattern, we can express the number 110.112 as follows:

1 × 22 = 1 × 4 = 4

1 × 21 = 1 × 2 = 2

0 × 20 = 0 × 1 = 0

1 × 2–1 = 1 × 0.5 = 0.5

1 × 2–2 = 1 × 0.25 = + 0.25

6.75

To find the decimal value of a binary number, we simply calculate the valueof each binary digit and then sum these values.

Octal to Decimal Conversion

Since the binary system is a base-8 number system, we can make formulamore specific:


Now consider the octal number 1473.28. As you can see in the table below,each column of this number represents a power of eight.

512s 64s Eights Ones Eighths

83 82 81 80 8-1

1 4 7 3 . 2

Using the information from this table, we can find the decimal value for1473.28. We will use the same approach as we used for converting binary numbersto decimal.

1 × 83 = 1 × 512 = 512

4 × 82 = 4 × 64 = 256

7 × 81 = 7 × 8 = 56

3 × 80 = 3 × 1 = 3

2 × 8–1 = 2 × 0.125 = + 0.25

827.25

From our analysis, we see that 1473.28 is equivalent to 827.2510 in decimal.

Hexadecimal to Decimal Conversion

Since the binary system is a base-16 number system, we can make formulamore specific:


NOTES



Now let's look at a hexadecimal number: 33B.416. Since we are using a base-16 number system, the columns of our table are now powers of 16.

256s Sixteens Ones Sixteenths

162 161 160 16–1

3 3 B 4

As we did before, we can find the decimal value of 33B.416 by computer thevalue of each digit and summing them. Notice that we substitute 1110 for thehexadecimal digit B in this conversion.

3 × 162 = 3 × 256 = 768

3 × 161 = 3 × 16 = 48

B × 160 = 11 × 1 = 11

4 × 16–1 = 4 × 0.0625 = + 0.25827.25

Comparing all of our answers, we see that 827.2510 is 1473.28 in octal and33B.416 in hexadecimal.

Decimal to Base Conversion

Decimal to Binary Conversion

If any number has been given in decimal; then to convert in binary, follow thefollowing steps.

(a) Successive division for Integer Part

Steps are as follows:

1. Divide the integer part of given decimal number by the base ‘2’, noting theremainder.

2. Divide the integer part until no further division is possible. At each stepnote the remainder.

3. List the remainder values in reverse order to find the equivalent.

Example 4.1. Convert (105)10 to equivalent binary number.

Solution. As discussed, divide the (105)10 by base 2.remainder

2

1

3

6

13

26

52

105

2

2

2

2

2

1

1

0

1

0

0


NOTES

Digital ElectronicsNow arrange the remainders in reverse order.

(105)10 = (1101001)2

(b) Successive Multiplication for Fractional Part

Follow the steps given below:

1. Multiply the given fractical decimal number by base ‘2’.

2. Record the carry generated by multiplication at most significant position.

3. Multiply only fractional number of the product in step ‘2’ by the base andagain record the carry.

4. Repeat step 3. Until we get a pattern of carry generated repeated.

If not repeating pattern is achieved, follow up to five digits of carry.

5. Now list the carry in order.

Example 4.2. Convert the decimal number (0.42)10 into binary.

Solution. Follow the steps discussed above.

0.84 × 2

0.42 × 2

1.68 × 2

1.36 × 2

0.72 × 2

1.44

= (0.01101)2

=

So, (0.42)10 = (0.01101)2

(c) Conversion of Mixed Number to Binary

If any number has integer and fractional both parts, perform theconversion of both part separately to get equivalent binary.

Decimal to Octal Conversion

To convert a decimal number to octal number, the process is same as discussedfor decimal to octal conversion. But now the

1. Successive division of integer part is performed by base 8.

2. Successive multiplication of fractional part is performed by base 8.

Decimal to Hexadecimal Number Conversion

To convert a decimal number to hexadecimal number, the process is same butnow with base value 16.

Example 4.3. Convert the following:

(257.13)10 = (?)16

(a) Taking integer part

(257)10 = (?)16.

NOTES


Basic Electrical andElectronics Engineering 16

16 16

2571

01

(257)10 = (101)16.

(b) Taking fractional part

(0.13)10 = (?)16.

0.08 × 16

0.13 × 16

1.28 × 16

1.48 × 16

7.68 × 16

1.08

(0.13)10 = (0.21171)16.

So (257.13)10 = (101.21171)16.

Decimal Number to Any Base

As discussed different decimal to base conversion example, it is clear that toconvert any decimal number to any base:

(a) Divide given decimal integer number by base until no further division ispossible.

(b) Multiply given decimal fractional number by base for five places.

Example 4.4. Convert the following:

(a) (2CCD)16 = (?)5 (b) (2 AC 9)16 = (?)7

Solution. As there is no direct conversion method for given problem. So, firstof all base-16 is to be converted to base-10. The base-10 is converted to base-5, ‘or’base-7.

(a) (2CCD)16 → (?)10 → (?)5

So (2CCD)16 = (?)10

= 2 × 163 + C × 162 + C × 161 + D × 160

= 2 × 163 + 12 × 162 + 12 × 161 + 13 × 160

= 131072 + 3072 + 192 + 13

= (134349)10.

Now (134349)10 = (?)5.


NOTES

Digital Electronics

5

8

42

214

1074

5373

26869

5

5

5

5

5

4

4

2

3

3

4

41343495

(134349)10 = (13244344)5

So, finally

(2CCD)16 = (13244344)5

(b) (2AC9)16 = (?)10 = (?)7

So (2AC9)16 = (?)10

(2AC9)16 = 2 × 163 + A × 162 + C × 161 + 9 × 160

= 2 × 163 + 10 × 162 + 12 × 161 + 9 × 160

= (10953)10

Now (10953)10 = (?)7

(10953)10 = (43635)7

So finally

(2AC9)16 = (43635)7

Octal to Binary Conversion

We can see a special relationship between three of these number systems.Binary, octal, and hexadecimal all have bases that are powers of 2. This relationshipallows us to convert between these systems quite easily.

To convert 1473.28 to binary, we simply replace each digit of octal with anequivalent 3-bit binary number.

NOTES



Octal 1473.2

1 4 7 3 . 2

Binary 001 100 111 011 . 010

001100111011.010

We can see the 3 to 1 relationship between octal and binary when we comparetheir bases. Octal has a base of 8 or 23 and binary has a base of 2 or 21.

Hexadecimal to Binary Conversion

As you might guess, binary and hexadecimal have a 4 to 1 relationshipsince the base of hexadecimal is 16 or 24. We can also convert easily between binaryand hexadecimal using this 4 to 1 relationship.

Hexadecimal 33B.4

3 3 B . 4

Binary 0011 0011 1011 . 0100

001100111011.0100

4.3 ADDITION AND SUBTRACTION OF NUMBER

SYSTEM

Decimal Addition

Decimal addition process may help to understand the addition of numbers ofdifferent base. In decimal addition, we follow the steps given below:

Step 1. Add the least significant digit. If this addition is greater than ‘or’equal to base number-10, a carry is generated.

Step 2. If addition is greater than base number-10, then substract 10 from it.Write down reminder at place of addition and forward one carry to next place.

Step 3. Now perform the same operation for next digit (till most significantdigit).

Example 4.5. Add the following:

(256)10 + (629)10

Solution.

(629) 10

(256) 10LSBMSB

Addition of LSB digits is

6 + 9 = 15

that is greater than base = 10

So, Remainder = 15 – 10 = 5


NOTES

Digital ElectronicsThus addition of 6 and 9 produces ‘5’ with carry ‘1’

(629) 10

(256) 10

(885) 10 Result

Carry1

Decimal Substraction

Decimal substraction can be performed in following steps.

Step 1. If minuend (number from which second number is to be substracted)is less than subtrahend (number to be substracted) then we take a borrow. Thevalue of borrow is equal to base (10). When we take borrow, a carry is returned tonext place.

Example 4.6. Substract.

(64)0 – (26)10

So, (6 4)10

– (2 6)10

At LSD, the minuend ‘4’ is less than subtrahend ‘6’ so take a borrow from nextplace and return one carry to that place.

1

6 4

(2 8) 10

Carry

Borrow(10)

– 2 6

Binary Addition

To add two binary number, follow the same process as in decimal addition.Maximum possible four cases of binary additions are:

(A + B) Sum Carry

0 0 0 0

0 1 1 0

1 0 1 0

1 1 1 1

Note: Addition of 1 and 1 is ‘2’ that is equal to base 2. So.

2 – 2 = 0. addition and 1 carry is generated.

Example 4.7. Add the following:

A = (10111)2 and B = (11001)2

NOTES



Solution.

+ 1 1 0 0 1

1 1 1 1 1

(1 0 0 0 0) 2

Carry

1 0 1 1 1

Note that at 5th step 1 + 1 + 1 = 3.

that is greater than base 2.

So, 3 – 2 = 1 is add result with 1 carry generated.

Binary Substraction

For binary substraction, follow the same process as in case of decimal substraction.

Example 4.8. Substract the following:

A = (110 11)2 and

B = (10 11 0)2

Solution :

1

1 1 0 1 1

0 0 1 0 0

Carry

Borrow2

1 0 1 1 0

Result

Note that at third place. 0 – 1 is not possible without borrow. So we havetaken one borrow from next place. Value of borrow is always equal to base.

But a carry = 1 is returned to that place.

Octal Addition

To add two octal number, follow the same process as in decimal addition.

Step 1. Start addition from LSD. (least significant digit).

Step 2. If addition is equal to ‘or’ greater than base ‘8’, substract ‘8’ fromaddition. Remainder is result and one carry is forwarded to next position.

Example 4.9. Perform the following:

(354)8 + (266)8.

Solution.

6 12 10

3 5 4

6 4 2

Carry1 1

+ 2 6 6 If greater 'or'equal to 8–8 –8

(354)8 + (266)8 = (642)8

Note that when addition of digit is equal to ‘or’ greater than ‘8’, 8 is substractfrom addition. The remainder is result and one carry is forwarded to next place.


NOTES

Digital ElectronicsOctal Substraction

To substract two octal number, follow the same process as in decimal substraction.

Step 1. Start substraction from LSD.

Step 2. If minuend is less than substrahend, then take a borrow. The value ofborrow is equal to base-8.

But carry is returned to that place and value of carry is always 1.


(354)8 – (266)8

Solution.

1 1

(3 5 4) 8

6 6

Carry(always 1)

Borrow(always equal to base value 8)

8 8

– (2 6 6) 8

(354)8 – (266)8 = (66)8

Hexadecimal Addition

Process of adding two hexadecimal number is same as decimal.

Step 1. Start addition from LSD.

Step 2. If addition is equal to ‘or’ greater than base ‘16’, substract' 16’ fromaddition. The remainder is result and one carry is forwarded to next place.


(C2)H + (3E)H

Solution.

(16) (16)

(12) (2)

– (16) – (16)

C . 2Carry1 1

(3) (14)

Equal to base value

1 0 0

+ 3 E

(C2)16 + (3E)16 = (100)16

Hexadecimal Substraction

To substract two hexadecimal number, follow the same process as in decimalsubstraction.


Step 2. If minuend is less than subtrahend, then take a borrow. The borrowis equal to base value 16.

But carry is returned to that place and value of carry is always 1.

NOTES



Example 4.12. Perform the following

(C2)16 – (3E)16

Solution.

1

12 2

8 4

C 2Borrow(equal to base)

16

3 14 3 E

Carry(always 1)

(C2)16 – (3E)16 = (84)16

Any Base Addition

On the basis of above example; the final process is:

Step 1. Start addition from LSD.

Step 2. If addition is equal to ‘or’ greater than base, substract base valuefrom addition. Reminder is result and one carry is forwarded to next place.

Example 4.13. Perform the following

(52)7 + (66)7

Solution.

12 8

(5 2) 7

(– 8) (– 7)

Carry1 1

+ (6 6) 7

greater thanbase value

1 4 1

So, (52)7 + (66)7 = (141)7

Any Base Substraction

One the basis of above example, the final process is:


Step 2. If minuend is less than subtrahend, then take a borrow. The borrowis equal to base value.

But carry is returned to that place and value of carry 1.


(52)7 – (36)7

Solution.

1

5 2

1 3

Borrow(equal to base value)

7

– 3 6

Carry(always 1)

(52)7 – (36)7 = (13)7


NOTES

Digital Electronics4.4 COMPLEMENTS

Complements are used for simplifying the substraction. There are two types ofcomplements for each base-r system.

1. The r’s complement

2. The (r–1)’s complement.

r’s and (r–1)’s Complement

r’s Complement

The basic formula for r’s complement of any number ‘N’ is given by.

r’s complement = rn – N

where r = Base of Number

N = Number whose complement is to be taken.

n = Number of digits/Bits in integer part of N.

So for different Number Systems we may have following r’s complements.

Binary, r = 2 ; 2’s Complement.

Octal, r = 8 ; 8’s Complement.

Hexadecimal, r = 16; 16’s Complement.

Decimal; r = 10; 10’s Complement.

For all number systems, the formula remains same.

Example 4.15. Find 10’s complement of (52520)10

Solution. By Formula

10’s Complement = rn – N

But r = 10,

n = 5

N = 52520

So 10’s complement = 105 – (52520)10

= 100000

– 52520 47480

(r – 1)’s Complement

The basic formula for (r – 1)’s complement of any number ‘N’ is given by (r – 1)’scomplement = rn – r–m – N.

where n = Number of digits/Bit is integer part of N.

m = Number of digits in fractional part of N.

N = Number

r = Base of number N.

NOTES



So for different Number Systems, we may have following (r – 1)’s complements.

Binary, r = 2; 1’s complement

Octal, r = 8, 7’s complement

Hexadecimal r = 16, 15’s, complement

Decimal r = 10, 9’s complement

For all the number systems, the formula remains same.

Example 4.16. Find the 9’s complement of (52520)10

Solution. Applying formula = rn – r–m – N.

For 9’s complement r = 10,

n = 5

m = 0

N = 52520

So 9’s complement = 105 – 10–0 – (52520)

= (100000)10–1– (52520)10

= 99999

– 52520

47479

Substraction using r ’s Complement

The substraction of two number (M – N), both of base r, may be done by using r’scomplement as follows:

1. Take r’s complement of substrahend N.

2. Add the minuend M to the r’s complement of subtrahend N.

3. Now

(a) If an end carry (carry at MSD) occurs, discard it.

(b) If an end carry does not occur.

Answer = – (r’s complement of result).

All the steps are identical for

(a) 10’s complement for r = 10.

(b) 8’s complement for r = 8.

(c) 16’s complement for r = 16.

and all other number system.

Example 4.17. Using 10’s complement, substract: 72532 – 3250.

Solution. Follow the steps:

(a) Given that M = (72532)10

N = (3250)10.


NOTES

Digital Electronics(b) Take the 10’s complement of (3250)10.

= rn – N

= 104 – 3250

= (10000)10 – (3250)10

= (96750)10

(c) Add M and 10’s complement of N.

= (72532)10

+ (96750)10

End carry → 1) 69282

(d) As end carry has been generated. So discord it.

So, (72532)10 – (3250)10 = (69282)10

Substraction using (r – 1)’s Complement

The substraction of two number (M – N), both of base ‘r’ may be done by (r – 1)’scomplement as follows:

1. Take (r – 1)’s complement of N.

2. Add the minuend, M to the (r – 1)’s complement of subtrahend N.

3. Now

(a) If end carry (carry at MSD) occurs, add this carry to result of step 2)

Answer = (Result + Carry).

(b) If end carry does not occur;

Answer = –[(r – 1)’s complement of Result].

Example 4.18. Using 9’s complement, substract (72532)10 – (3250)10

Solution. Follow the steps:

(a) Given that M = (72532)10

And N = (3250)10

(b) Take the 9’s complement of (3250)10

= rn – r–m – N

= 104 – 12–0 – (3250)10

= (9999)10 – (3250)10

= (96749)10

(c) Add M with 9’s complement of N.

= (75532)10

+ (96749)10

End carry → 1) 69281

NOTES



(d) End carry has been generated. So

Answer = (Result + Carry)

= 69281

+ 1 Carry added

(69282)10

So (72532)10 – (3250)10 = (69282)10

4.5 LOGIC GATES

General Aspects

A digital circuit with one or more input signals but only one output signal iscalled a logic gate.

Or

A logic gate is an electronic circuit which makes logic decision.

Logic gates are the basic building blocks from which most of the digitalsystems are built up. They implement the hardware logic function basedon the logical algebra developed by George Boolean which is called Booleanalgebra in his honour.— A unique characteristic of Boolean algebra is that variables used in it

can assume only one of the two values i.e., either 0 or 1. Hence, everyvariable is either a 0 or a 1 (Fig. 4.1—limits on TTLIC’s).

Each gate has distinct graphic symbol and its operation can be describedby means of Boolean algebraic function.

Notallowed

5V

2V

0.8 V

0

Logic ‘1’

Logic ‘0’

Fig. 4.1. Voltage assignment in a digital system.

The table which indicates output of gate for all possible combinations ofinput is known as a truth table.

These gates are available today in the form of various IC families. Themost popular families are :

(i) Transistor-transistor logic (TTL)(ii) Emitter-coupled logic (ECL)

(iii) Metal-oxide-semiconductor (MOS)(iv) Complementary metal oxide-semiconductor (CMOS).


NOTES

Digital ElectronicsApplications of Logic Gates

The following are the fields of application of logic gates :

1. Calculators and computers

2. Digital measuring techniques

3. Digital processing of communications

4. Musical instruments

5. Games and domestic appliances, etc.

6. The logic gates are also employed for decision making in automatic controlof machines and various industrial processes and for building more complex devicessuch as binary counters etc.

Positive and Negative Logic

The number symbols 0 and 1 represent in computing systems two possiblestates of a circuit or device. It does not make any difference if these two states arereferred to as ‘ON’ and ‘OFF’, ‘Closed’ and ‘Open’, ‘High’ and ‘Low’, ‘Plus’ and ‘Minus’or ‘True’ and ‘False’, depending upon the situations. The main point is they mustbe symbolized by two opposite conditions. In positive logic a ‘1’ represents : an ‘ONcircuit’ ; a ‘Closed switch’ ; a ‘High voltage’ , a ‘Plus sign’, a ‘True statement’.Consequently, a O represent : an ‘OFF circuit’ ; an ‘Open switch’, a ‘Low voltage’; a ‘Minus sign’, a ‘False statement’.

In negative logic, the just opposite conditions prevail.

Example. A digital system has two voltage levels of 0 V and 5 V. If we say thatsymbol 1 stands for 5 V and symbol 0 for 0 V, then we have positive logic system. Ifon the other hand, we decide that a 1 should represent 0 V and 0 should represent5 V, then we will get negative logic system.

Main point is that in ‘positive logic’ the more positive of the two voltagelevels represents the 1 while in ‘negative logic’ the more negative voltage representsthe 1.

Types of Logic Gates

In the complex circuits, the following six different digital electronics gates areused as basic elements :

1. NOT Gate 2. NAND Gate3. AND Gate 4. OR Gate5. NOR Gate 6. XOR Gate.

— A truth table has 2n rows. It gives in each of its row m outputs for a givencombination of n inputs.

1. NOT gate :

Not operation means that the output is the complement of input. If input islogic ‘1’ , the output is logic ‘0’ and if input is logic ‘0’, the output is logic ‘1’,

Fig. 4.2 shows the symbol of NOT Gate. It is generally represented by atriangle followed by a bubble (or a bubble followed by a triangle).

NOTES



NOT gate is used when an output is desired to be complement of the input. If all inputs of NAND gates are joined it shall act as NOT gate. NOT gate is also called ‘inverting logic circuit’. It is also called a

‘complementing circuit’.

2. NAND gate :

A NAND gate can said to be basic building block of the all digital TTLlogic gates and other digital circuits.

It is represented by the symbol shown in Fig. 4.3. Its unique property is that output is high ‘1’ if any of the input is at low ‘0’

logic level.Let us consider two inputs with the states A and B at the NAND gate. The

answer (output) X = A B. . Bar denotes a NOT log operation on A.B. The meaningof A.B, called AND operation, is given in 3 below.

3. AND gate :

A NAND gate followed by a NOT gate gives us AND gate. It is represented by a symbol in Fig. 4.4. Its symbol differs from NAND

only by omission of a bubble (circle). Its unique property is that its output is ‘0’ unless all the inputs to it are at

the logic 1’s. A two inputs, AND gate has X = A.B. Dot between the two states indicates

‘AND’ logic operation using these.

4. OR gate :

An OR operation means that the output is ‘0’ only if all the input are ‘0’. It is represented by a symbol shown in Fig. 4.5. If any of the inputs is ‘1’ the output is ‘1’ . A two inputs ‘OR’ gate has

X = A + B. Sign + between the two states indicates an ‘OR’ logic operation.

5. NOR gate :

An ‘OR’ circuit followed by a NOT circuit gives a ‘NOR’ gate (Fig. 4.6). Its unique property is that its output is ‘0’ if any of input is ‘1’. A NOR gate is a basic building block for other types of the logic gates than

TTLs. In the TTL circuits, a NOR is fabricated in an IC by the severalNANDs.

A two inputs NOR has X = A B+ .

6. XOR gate :

A XOR gate (Fig. 4.7) is called ‘Exclusive OR’ gate. Its unique property is that the output is ‘1’ only if odd number of the inputs

at it are ‘1’s.

The ‘Exclusive OR’ can be written as : X = A .B + A . B or A ⊕ B. Exclusive OR gate is important in the circuits for addition of two binary

numbers.


NOTES

Digital Electronics7. Coincidence gate :

This gate (Fig. 4.8) can be written as : X = A B. + A . B.

Output available to those states when the inputs are identical.

Tab

le 4

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Boo

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Exp

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ion

AB

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XA

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A B

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A B

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10

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10

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10

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ion

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NOTES



Basic building blocks. AND, OR and NOT gates are called basic buildingblocks or basic gates because they are essential to realize any boolean expression.

Universal gates. NAND and NOR gates are known as universal gates becauseany logic gate can be constructed either by using NAND gates only or by usingNOR gates only.

4.6 UNIVERSAL GATES

NAND and NOR gates are known as universal gates.

The AND, OR, NOT gates can be realized using only NAND or NOR gates.

— Demorgan’s theorem afford a convenient method to use these two gates inlogic design. The entire logic system can be implemented by using any ofthese two gates.

— These two gates are easier to realize and consume less power than othergates.

(i) Realization of logic gates using NAND gates :Fig. 4.9 (a), (b), (c) shows realization of NOT, AND, OR gates using NAND

gates respectively, which is self explanatory.

A X = A

(a) Realization of NOT gate using NAND gate

AX = A.B

B A.B

(b) Realization of AND gate using NAND gates

A

X = A.B = A + B

B

A

B

(c) Realization of OR gate using NAND gates

Fig. 4.9. Realization of NOT, AND and OR gates using NAND gates.

(ii) Realization of logic gates using NOR gates :The realization of NOT, OR and AND gates using NOR gates is shown in

Fig. 4.10 (a), (b), (c) respectively.

X = AA

(a) Realization of NOT gate using NOR gates


NOTES

Digital ElectronicsX = A + B

A

B A + B

(b) Realization of OR gate using NOR gates

A

X = A + B = A.B

BB

A

(c) Realization of AND gate using NOR gates

Fig. 4.10. Realization of NOT, OR and AND gates using NOR gates.

4.7 HALF ADDER

It is a 1-bit adder and carries out binary addition with the help of XOR andAND gates. It has two inputs and two outputs.

It can add 2 binary digits at a time and produce a 2-bit data i.e., SUM andCARRY according to binary addition rules.

The circuit of a half adder is shown in Fig. 4.11. (a). It consists of an Ex-ORgate and AND gate. The outputs of the Ex-OR gate is called the SUM (S), while theoutput of the AND gate is known as CARRY (C). As the AND gate produces a highoutput only when both inputs are high and Ex-OR gate produces a high output ifeither input (not both) is high, the truth table of a half adder is developed by writingthe truth table output of AND gate in the CARRY column and the output truthtable of Ex-OR gate in SUM column. Truth table for half adder is given in table.

A

BCARRY = A.B

SUM = A B+

A

B

C

S

OutputsInputs HA

(a) Logic circuit (b) Logic symbol

Fig. 4.11. Half adder.

Table 4.3. Truth Table for Half Adder

Inputs Outputs

A B C S

0 0 0 00 1 0 11 0 0 11 1 1 0

NOTES



The logical expressions for CARRY and SUM can be written from the truthtable for a half adder as follows :

CARRY, C = A.B

SUM, S = A ⊕ B

This circuit is called half-adder, because it cannot accept a CARRY-INfrom previous additions. Owing to this reason the half-adder circuit canbe used for binary addition of lower most bit only.

For higher-order columns, 3-input adder called full adder are used.

4.8 FULL ADDER

A full adder has three inputs and two outputs. It can add 3 digits (or bits) at atime. The bits A and B which are to be added come from the two registers and thethird input C curves from the ‘carry’ generated by the previous addition. It producestwo outputs, SUM and CARRY-OUT (going to next higher column).

CBA

CARRY = AB + BC + CA

SUM = A B C+ +

(a) Logic circuit

A

C

CARRY

SUM

FAB

(b) Logic symbol

HA

Full adder

A B = D+

+D CI

D.CI

CARRY C0

SUM

A

B

CHA

A.B

(c) Full adder circuit

Fig. 4.12. Full adder.


NOTES

Digital ElectronicsTable 4.4. Truth table for Full Adder

0

0

0

0

1

1

1

1

A

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

0

0

1

0

1

1

1

0

1

1

0

1

0

0

1

B C CARRY SUM

A simple circuit of a full adder is shown in Fig. 4.12 (a), though other designsare also possible. It uses 3 AND gates, one Ex-OR gate and one OR gate. The finalCARRY is given by the OR gate while the final SUM is given out by the Ex-ORgate.

Fig. 4.12 (b) shows the logic symbol for a full adder.

Truth table for full adder for all passible inputs/outputs is given in Table 4.4.Truth table can be checked easily for its validity.

A full adder can be made by using two half adders and an OR gate. The circuitis shown in Fig. 4.12 (c).

The full adder can do more than a million additions per second. Besidesthat, it never get tired or bored or asks for a rest.

Note : Binary additions : Following are the four rules/cases for addition of binarynumbers :

(1) 0 + 0 = 0

(2) 0 + 1 = 1

(3) 1 + 0 = 1

(4) 1 + 1 = 10 (This sum is not ten but one-zero).

4.9 BOOLEAN ALGEBRA

George Boolean in 1854 developed a mathematics now referred as Booleanalgebra. It is the algebra of logic presently applied to the operation of computerdevices. The rules of this algebra are based on human reasoning.

Digital circuits perform the binary arithmetic operations with binary digits 1and 0. These operations are called logic functions or logic operations. The algebraused to symbolically describe logic functions is called Boolean algebra. Booleanalgebra is a set of rules and theorems by which logical operations can be expressedsymbolically in equation form and be manipulated mathematically.

Boolean algebra differs from ordinary algebra in that Boolean constant andvariables can have only two values ; ‘0’ and ‘1’ :

NOTES



In Boolean algebra the following four connecting symbols are used :

1. Equal sign (=). In Boolean algebra the ‘equal sign’ refers to the standardmathematical equality. In other words, the logical value on one side of the sign isidentical to the logical value on the other side of the sign.

Example. We are given two logical variables such that A = B. Then if A = 1,then B = 1 and if A = 0 then B = 0.

2. Plus sign (+). In Boolean algebra the ‘plus sign’ refers to logical ORoperation.

The statement A + B = 1 means A ORed with B equals 1. Consequently, eitherA = 1 or B = 1 or both equal to 1.

3. Multiply sign ( . ). In Boolean algebra the ‘multiply sign’ refers to ANDoperation.

The statement A.B = 1 means A ANDed with B equals 1. Consequently, A = 1and B = 1.

The function A . B often written as AB, omitting the dot for convenience.

4. Bar sign (–). In Boolean algebra the ‘bar sign’ refers to NOT operations.The NOT has the effect of inverting (complementing) the logic value.

Thus if A = 1, then A = 0.

Boolean Laws (For Outputs from Logic Inputs)

The following Laws can said to be associated with Boolean algebra :

1. ‘OR’ Laws

The ‘OR’ Laws are described by the following equations :

A + 1 = 1 ...[4.1(a)]

A + 0 = A ...[4.1(b)]

A + A = A ...[4.1(c)]

A + A = 1 ...[4.1(d)]

An ‘OR’ operation is denoted by plus sign. ‘OR’ Law means :(i) Any number (0 or 1) is a first input to an OR gate and another member at

the second input is 1 then answer is 1,

(ii) If another is 0 then answer is as first input, and

(iii) If two inputs to an OR gate complements then output is ‘1’.

2. ‘AND’ Laws

‘AND’ operation is denoted by the dot sign.

True and true make true True and false make false False and false make false.

A . 1 = A ...[4.2(a)]

A . 0 = 0 ...[4.2(b)]


NOTES

Digital ElectronicsA . A = A ...[4.2(c) ]

A . A = 0 ...[4.2(d) ]

3. ‘NOT’ Laws (Laws of Complementation)

A NOT operation is denoted by putting a bar over a number.

The NOT true means false. The NOT false means true.

1 = 0 ...[4.3(a)]

A = A ...[4.3(b)]

Eqn. [4.3(c)] means that if A is inverted (complemented) and then againinverted, we get the original number.

4. Commutative Laws

These Laws mean that order of a logical operation is immaterial.

A + B = B + A ...[4.4(a)]

A . B = B . A ...[4.4(b)]

5. Associative Laws

These laws allow a grouping of the Boolean variables.

A + (B + C) = (A + B) + C ...[4.5(a)]

A . (B . C) = (A . B) . C ...[4.5(b)]

6. Distributive Laws

These laws simplify the problems in the logic designs.

A . (B + C) = (A . B) + (B . C) ...[4.6(a)]

A + (B . C) = (A + B) . (A + C) ...[4.6(b)]

A + ( A . B) = A + B ...[4.6(c)]

The last two equations are typical to the Boolean algebra, and are not followedin the usual algebra.

De morgan’s Theorems

First theorem shows an equivalence of a NOR gate with an AND gate havingbubbled inputs (Fig. 4.13), and is given by the equation :

A B A B+ = . ...(4.7)

NOTAND

NOT

NOT

A . B

A

B

A

0

1

0

1

B

0

0

1

1

X

1

0

0

0

A

NORB

A + B

Fig. 4.13. De Morgan’s First theorem showing an equivalenceof a NOR gate (same holds for multiple inputs).

NOTES



Second theorem shows an equivalence of a NAND gate with an OR havingbubbled inputs as shown in Fig. 4.14 and is given by the equation :

A B A B. = + ...(4.8)

A

0

1

0

1

B

0

0

1

1

X

1

1

1

0

NOT

OR

NOTA + B

A

B

NAND

A . B

Fig. 4.14. De Morgan’s Second theorem showing the equivalenceof a NAND gate (same holds for the multiple inputs).

In fact with the eqns. (4.7) and (4.8) also hold for the cases of the multiple(more than two) inputs.

A B C+ + + ....... = A B C. . ...[4.9(a)]

A . B . C ....... = A B C+ + ...[4.9(b)]The purpose of these theorems is to enable digital circuit designers to implement

all the other logic gates with the help of either NOR gates only or NAND gates only.For example, a NOT gate is implementable by a NAND or a NOR as shown in theleft part or lower right part of Fig. 4.13 respectively. This theorem finds wide usein the digital logic circuits as these are implementable on one single basic logicgate considered as a basic building unit.

The ‘first statement’ (De Morgan’s) says that the complement of a sumequals the product of the complements. The ‘second statement’ says thatthe complement of a product equals the sum of the complements. In fact, itallows transformation from a sum-of-products form to a product-of-sumform.

The procedure required for taking out an expression from under a NOTsign is as follows :

1. Complement the given expression i.e., remove the overall NOT sign.

2. Change all and ANDs to ORs and all the ORs to ANDs.

3. Complement or negate all individual variables.

Examples : (i) A BC+ = A + BC ...Step 1

= A(B + C) ...Step 2

= A B C( )+ ...Step 3

(ii) ( ) ( ) ( ) ( )A B C A B C A B C A B C+ + + + = + + + +

= ABC ABC+ = A BC AB C+ = ABC AB C+

This process is called demorganization.


NOTES

Digital Electronics— It may be noted that the opposite procedure would be followed to bring anexpression under the NOT sign :

Example : A B C A B C+ + = + + ...Step 3

= A + B + C

= ABC ...Step 2

= ABC ...Step 1

Operator Precedence

For evaluating Boolean expression, the operator precedence is :

(i) parenthesis, (ii) NOT, (iii) AND and (iv) OR. In other words :

— The expression inside the parenthesis must be evaluated before all otheroperations,

— The next operation that holds precedence is the complement,— Then follows the AND, and— Finally the OR.

Example. In the Boolean expression A +B (C + D), and expression inside theparenthesis will be evaluated first, then B will be evaluated, then the results ofthe two [i.e., B and (C + D)] will be ANDed and finally, the result of the productORed with A.

Example 4.19. Prove the following identity :

AC + ABC = AC

Solution. Taking the left hand expression as X, we get

X = AC + ABC = AC(1 + B)

Now, 1 + B = 1 [Eqn. 1 (a)]

∴ X = AC.1 = AC

∴ AC + ABC = AC ...Proved.

Example 4.20. Draw the logic circuit represented by the expression :

X = AB + A B. + A . B . C.

Solution. A circuit using gates can simply be designed by looking at theexpression and finding out the basic gates which can be used to realize the variousterms and then correct these gates appropriately.

In the given expression there are three input logical variables and X is theoutput.

The first term A . B is obtained by ANDing A with B as shown in Fig. 4.15(i).

The second term A B. is obtained by using two INVERTERs and oneAND gate and connecting them as shown in Fig. 4.15 (ii).

The last term is used by using one INVERTER one AND gate andconnecting them as shown in Fig. 4.15 (iii).

Now, the complete logic expression is realised by ORing the three outputs ofthe arrangements explained above i.e., by ORing A.B, A B. and A . B . C. The logicgate implementation for the given expression is shown in Fig. 4.16.

NOTES



A

B

A . B

A . BA

B

A

B

A . B . C

AA

B

C

B

C

X = A . B + A . B

+ A . B . C

A

B

A . B

( )i

A . BA

B

A

B( )ii

A . B . C

AA

B

C( )iii

Fig. 4.15 Fig. 4.16. Logic gate implementation of

expression A . B + A . B + A . B . C.

Example 4.21. Simplify the expression A A + C( )A C+ + AC.

Solution. A A + C( )A C+ + AC

= 0 + C( )A C+ + AC ...[from eqn. 4.2 (d)]

= C( . )A C + AC ...[from eqn. (2.7)]

= C A C + AC

= 0 + AC ...[from eqn.4.2 (d)]

= AC. (Ans.)

4.10 FLIP-FLOP CIRCUITS

The memory elements used in clocked sequential circuits are called flip-flops.These circuits are binary cells capable of storing one bit of information. It has twooutputs, one for the normal value and one for the complement value of the bit storedin it. Binary information can enter a flip-flop in a variety of ways. Hence there aredifferent types of flip-flops.

A number of flip-flops are available in IC form. Some of these are SR (Set-Reset), J-K and D flip-flops. They are widely used as switches, latches, counters,registers and memory cells in computers.

A salient feature of the flip-flop is that output can exist in one of the twostable states, logic 1 and logic 0, simultaneously. This is ensured by theappropriate crossed feedback connections associated with the mostelementary form of the flip-flop known as a latch.

The following flip-flops will be discussed in the following articles :

1. R-S flip-flop.

2. Clocked R-S flip-flop.

3. D flip-flop.


NOTES

Digital Electronics4. J-K flip-flop.

5. T flip-flop.

R-S Flip-flop

Fig. 4.17 shows a R-S flip-flop using NOR gates. There are two inputs to theflip-flops called S (set) and R (reset). The cross-coupled connection from the outputof one gate and input of the other constitutes a feedback path. For that reason, thecircuit is classified as synchronous circuit.

A low R and a high S results in the set state. A high R and a low S give the reset state. If both R and S are high, the output becomes indeterminate and this is

known as ‘race condition’. This condition is avoided by proper design.The truth table is shown in Table 4.5.

R

0

0

1

1

S

0

1

0

1

Q

NC

1

0

Comment

No change

Set

Reset

Race

Table 4.5. Truth Table for NOR Latch

(b) Truth table

R

S

Q

Q

(a) Circuit diagram

Fig. 4.17. R-S flip-flop using NOR gates.

Fig. 4.18 shows a R-S flip-flop using NAND gates. Table 4.6 shows the truthtable. It is seen that the inactive and race conditions are reversed.

When R is low, output Q is high. When R is high, output Q is low. When both R and S are low, we get race condition which must be avoided. When both R and S are high, no change condition.

R

0

0

1

1

S

0

1

0

1

Q

1

0

NC

Comment

Race

Set

Reset

No change

Table 4.6. Truth Table for NAND Latch

(b) Truth table(a) Circuit diagram

R

S

Q

Q

Fig. 4.18. R-S flip-flop using NAND gates.

Clocked R-S Flip-Flop

A large number of flip-flops are used in a computer. In order to coordinatetheir working a square wave signal known as clock is applied to the flip-flop. This

NOTES



clock signal (indicated as CLK) prevents the flip-flop changing state till the rightinstant occurs.

S Q

R Q

CLK

(b) Symbol(a) Circuit diagram

R

S

Q

Q

S

R

CLK

N3 N1

N2N4

N , N , N , N = NAND gates1 2 3 4

Fig. 4.19. Clocked R-S flip-flop.

Fig. 4.19(a) shows a clocked R-S flip-flop using NAND gates (N1 and N2). Thiscircuit uses two NAND gates N3 and N4 to apply CLK signal.

When CLK is low, the flip-flop output Q indicates no change. If S is high and R is low, the flip-flop must wait till CLK becomes high

before Q can be set on 1. If S is low and R is low, the flip-flop must wait for CLK to be high before

Q is reset to low (0).Clocked R-S flip-flop is a synchronous sequential logic circuit because output

state of the circuit changes at discrete clocked instant of time.

Fig. 4.19 (b) shows a symbol for clocked R-S flip-flop.

Level Clocking and Edge Triggering

In a clocked flip-flop, the output can change state when CLK is high. WhenCLK is low, the output remains in the same state. Thus, the output can changestate during the entire half cycle when CLK is high. This may be a disadvantage inseveral situations. It is necessary that the output should change state only at oneinstant in the positive half cycle of the clock. This is known as edge triggeringand the resulting flip-flop is known as edge triggered flip-flop.

Edge triggering can be made feasible by the use of an RC circuit. The timeconstant RC is made much smaller than the width of the clock pulse. Therefore,the capacitor can charge fully when CLK is high. The exponential charging producesa narrow positive voltage strike across the resistor. The input gates are activatedat the instant of this positive strike.

D Flip-Flop

A D flip-flop is an improvement over the R-S filp-flop to avoid race condition.It can be level clocked or edge triggered. The edge triggered one causes the changein output state at a unique instant.

In a clocked R-S flip-flop two input signals are required to drive the flip-flopwhich is a disadvantage with many digital circuits. In some events, both input


NOTES

Digital Electronicssignals become high which is again an undesirable condition. So these shortcomings/drawbacks of clocked R-S flip-flop are overcome in D flip-flop.

D Q

Q

CLK

(b) Symbol

S Q

R Q

CLKCLK

(a) Circuit diagram

InputD

0

1

n

OutputQ

0

1

n+1

(c) Truth-table

Fig. 4.20. D flip-flop.

Fig. 4.20(a), (b), (c) show the circuit diagram, symbol and truth table of D flip-flop respectively. It may be observed that only single data bit, D is required todrive the flip-flop.

— When the clock signal is at low level, data bit D is prevented to reach atoutput Q until clock signal becomes high at next pulse.

— It may be noted from the truth table that when data bit Dn is high, outputQn + 1 gets at high level and when data bit Dn is low, Qn + 1 gets at low level.Thus D flip-flop transfers the data bit D to Q as it is, and Q remains in thesame state until the next pulse of the clock arrives.

The flip-flop is named (D) flip-flop since the transfer of data from the inputto output is delayed.

The D-type flip-flop is either used as a delay device or as a latch to store1-bit of binary information.

Edge Triggered D Flip-Flop

Fig. 4.21(a) shows the circuit diagram and symbols of an edge triggered Dflip-flop. The clock provides the square wave signal. RC circuit converts this signalinto strikes so that triggering occurs at the instant of positive strike. The data bitD drives one of the inputs. Because of inverter, the complement D drives the otheroutput. At the instant of positive strike, input D and its complement D cause theoutput Q to set or reset. Fig. 4.21(b) shows the truth table.

CLK

01

D

×××01

Q

No changeNo changeNo change01

(b) Truth table(a) Circuit diagram

R

S

Q

Q

D

CLK

Squarepulse

N3 N1

N2N4

N , N , N , N = NAND gates1 2 3 4

R

C

NOTES


Basic Electrical andElectronics Engineering D Q

Q

CLK

Positive or leading

10

D Q

Q

CLK

Negative or trailingSymbols

10

Fig. 4.21. Edge triggered D flip-flop.

When CLK is 0 or 1, the D input is not there and there is no change instate of Q.

On the negative edge of the clock (marked ) the output remains in thesame state.

One the positive edge of the clock (marked ) Q changes to 0 if D is 0 andto 1 if D is 1.

Edge Triggered J-K Flip-Flop

— J-K (fip-flop is very versatile and is perhaps the most widely used type offlip-flop.

— The J and K designations for the inputs have no known significance exceptthat they are adjacent letters in the alphabet.

— J-K flip-flop functions identically to R-S flip-flop.

The difference is that the J-K flip-flop has no invalid state as does the R-Sflip-flop.

— It is widely used in digital devices such as counters, registers, arithmeticlogic units, and other digital systems.

Fig. 4.22 (a) shows the circuit diagram of a edge triggered J-K flip-flop used indigital counters. The CLK input is through an RC circuit with a short time constant.The RC circuit converts the rectangular clock pulse to narrow spikes as shown.Due to double inversion through NAND gates, the circuit is positive edge triggered.

(a) Circuit diagram

R

S

Q

Q

J

CLK

N3 N1

N2N4R

C

K


NOTES

Digital Electronics

J Q

Q

CLK

(b) Symbol for positiveedge triggered J.K

flip-flop

K

J Q

Q

CLK

(c) Symbol for positiveedge triggered J.K flip-flop

with present and clear

K

J Q

Q

CLK

K

PR

CLR

PR

CLR

(d) Symbol for negativeedge triggered

J.K flip-flop

Fig. 4.22. Edge triggered J-K flip-flop.

When both inputs J and K are low, the circuit is inactive at all timesirrespective of the presence of CLK pulse.

When J is low (i.e., 0) and K is high (i.e., 1), the circuit will be reset whenpositive CLK edge strikes the circuit and Q = 0. The flip-flop will remainin reset state if it is already in reset state.

When J = 1 and K = 0, the circuit sets at the arrival of next positive clockedge.

When J = 1 and K = 1, the flip-flop will toggle (means to switch to oppositestate) on the next positive CLK edge. The action is illustrated in theTable 4.7 :

Table 4.7. Positive Edge Triggered J-K flip-flop

CLK J K Q

0 × × No change

1 × × No change× × No change

X 0 0 No change

0 1 0 (reset)1 0 1 (set)1 1 toggle

Use of RC circuit for edge triggering is not very convenient for fabrication.Actual circuits use additional NAND gates for edge triggering, such circuits areknown as direct coupled circuit.

— Fig. 4.22(b) shows the symbol for positive edge triggered J-K flip-flop.

— Fig. 4.22(c) shows a positive edge J-K flip-flop with present (PR) and clear(CLR)

— Fig. 4.22(d) shows the symbol for negative edge triggered J-K flip-flopwith PR and CLR. The small bubble at CLR indicates negative triggering.

NOTES



T Flip-flop

T flip-flop is basically a J-K flip-flop, in this circuit input terminals J and Kare connected with each other and this input is named as T.

Fig. 4.23(a) and (b) show the circuit diagram and symbol respectively of atrailing edge triggered T flip-flop.

J Q

Q

CLK

K

(a) Circuit diagram

T Q

Q

CLK

(b) Symbol

Input Tn

0

1

Output Qn+1

Qn

Qn

(c) Truth table

Fig. 4.23. Trailing edge triggered T flip-flop.

When low level signal is applied to the input terminal T, then initial stateof output of flip-flop remains the same.

When high level signal is applied to the input terminal T, then output ofthe flip-flop toggles after arrival of every new clock pulse. So the frequencyof output signal is half of the clock signal frequency.

This flip-flop can be treated as frequency divider or a device which takesthe input frequency at the clock terminal and divide it by two.

4.11 COUNTERS

A counter is a sequential circuit that goes through a prescribed sequenceof states upon the application of input pulses.

The input pulses, called count pulses, may be clock pulse or may originatefrom an external source and may occur at prescribed intervals of true orrandom.

The sequence of states in a counter may follow a binary count or any othersequence of states.

They are used for counting the number of occurrences of an event and areuseful for generating time sequences to control operations in a digitalsystem.

Straight Binary Sequence Counter. It is the simple and most straight forward.An n-bit binary counter has n flip-flops and can count in binary from 0 to 2n – 1.

Binary ripple counter. It consists of a series connections of T flip-flops withoutany logic gates. Each flip-flop is triggered by the output of its preceding flip-flopgoes from 1 to 0. The signal propagates through the counter in a Ripple manner,i.e., the flip-flop essentially changes once at a time in rapid succession. It is themost simplest and most straight forward. It, however, has speed limitations ; anincrease in speed can be obtained by the use of a parallel or a synchronous counter.

Synchronous 3-bit Binary Counter. In this all flip-flops are triggeredsimultaneously by count pulse. The flip-flop is complemented only if its T input isequal to 1.


NOTES

Digital ElectronicsCounter-decoder Circuits :

Counters together with decoders are used to generate timing andsequencing signals that control the operation of digital systems.

The counter-decoder can be designated to give any desired number ofrepeated timing sequence.

Applications of Counters :

The fundamental applications of counters are given below :

1. Measurement of time interval.

2. Direct counting.

3. Measurement of speed.

4. Measurement of frequency.

5. Measurement of distance.

6. Gating a counter.

4.12 REGISTERS

A register is a group of memory elements which work together as one unit.The simple resisters only store a binary word. The other registers modify the storedword by shifting its bits to left or right.

The number of registers differs from processor to processor. A counter is aspecial kind of register to count the number of clock pulses aiming at the input.

The registers can be classified as :

(i) Accumulator(ii) General purpose registers

(iii) Special purpose registers.In a digital computer, the programs are executed in central processing unit

(CPU). The various instructions of the program are executed in proper sequence.The computer has a number of registers to store data temporarily during the executionof the program.

The ALV requires two numbers on which it operates (e.g., adds, subtractsetc.) and produces the results. These two numbers are known as ‘operands’.One of the operands is obtained from memory and the other fromaccumulator (an accumulator is a one word memory). The result is placedback into the accumulator. Thus ‘accumulator’ is the most frequently usedregister.

The ‘general purpose registers’ store data and intermediate results duringprogram execution.

The ‘special purpose registers’ include program counter (to store the addressto the memory location which contains the next instruction to be obtainedfrom the memory), status register (to hold indications like vary sign, parityetc.), index register (for addressing) etc.

NOTES



A “buffer register” is the simplest kind of register, it is used only to store adigital word.

A “shift register” is an array of flip-flops designed to store and shift thedata.

4.13 A/D AND D/A CONVERTERS : AN INTRODUCTION

There are numerous advantages of processing signals using digital systemsand because of these advantages digital systems are widely used for control,communication, computers, instrumentation etc. In many such applications ofdigital systems, the signals are not available in the digital form. Therefore, toprocess these analog signals using digital hardware, they have to be convertedinto digital form. The process of conversion of analog signal to digital signal isreferred as analog to digital conversion. The system that realizes the conversion isreferred as analog to digital converter or A/D Converter or ADC.

The output of the system may be desired to be of analog form. Therefore, theoutput of the digital system is required to be converted back to the analog form.The process of converting the digital signal to analog form is called digital to analogconversion and the system used for this purpose is referred to as digital to analogconverter or D/A converter or DAC.

AnalogSignal

DigitalSignal

AnalogSignal

DigitalSignalAnalog to

DigitalConverter

DigitalProcessor

Digital toAnalog

Converter

Fig. 4.24

In the present trend of technology, the most of the signal processing is basedon the digital systems. But the real world signals are analog, in nature. A/Dconverter and D/A converter are the bridge between the analog world and digitalworld. They find their applications in almost every system of signal processing. Anelementary analog signal processing system with the use of digital processor isillustrated with the block diagram in Fig. 4.24. In this unit, the D/A converter willbe discussed first as it also serves as sub-system of A/D converter.

4.14 DIGITAL TO ANALOG CONVERSIONS

The digital to analog conversions needed in digital data processing requiresinsulation of digital information to an equivalent analog information. Digital toanalog converters are first discussed here as they also find application as feedbackelement in most of analog to digital converters. Digital to analog converters are fedwith a digital information applied to its input and give an analog output in theform of current voltage signal, which is proportional to the numerical value of thedigital input. The advantages of keeping the information in digital form and thenusing digital to analog converters in the analog applications are its independence


NOTES

Digital Electronicsof drift with time and temperature and its better immunity to noise than analoginformation. Digital to analog (D/A) converters, may be used to translate the outputof a digital system into an analog form for the purpose of driving a pen recorder orfor a cathode ray oscilloscope. The D/A converter is commonly referred to as adecoding device, since it is used to decode the digital signals into proportional voltageor current signals for an entry into an analog system.

4.15 RESISTOR DIVIDER D/A CONVERTER

The resistor divider D/A converter converts a digital signal represented inbinary or BCD code into an analog voltage or current signal, which is proportionalto the digital value. Figure 4.25 shows the symbol of a 3-bit resistor dividerD/A converter, having three digital inputs A, B, C, which are derived from theoutput register of a digital system. It has one analog output Va as shown.

A 20

B 21

C 22

Analog outputVa

ResistorDividerD/A converter

Fig. 4.25. Three bit D/A converter with voltage output.

Let the digital input levels be: Low level or ‘0’ = 0V and high level or ‘1’ = 7V.There are 23 = 8 different binary numbers represented by these three bits. In generalfor n digital inputs i.e., for n bit D/A converter, there will be 2n different binarynumbers represented by these n bits. For each digital input number, the output ofD/A converter is unique and here for the digital input voltage levels as supposedabove, the analog output voltage Va is equal in volts to the binary input number.Thus for an input of 001, the analog output will be +1V for input of 010, the analogoutput will be +2V and for input of 011, the analog output voltage will be +3V, andso on, as given in Table 4.8.

Table 4.8

Digital Input Analog output

Voltage in Volts

22 21 20

C B A

0 0 0 00 0 1 + 10 1 0 + 20 1 1 + 31 0 0 + 41 0 1 + 51 1 0 + 61 1 1 + 7

NOTES



Thus in a resistor divider network, every 20 bit is changed to +1V, 21 bit ischanged to +2V, the 22 bit is changed to +4V and so on, whenever the input voltagelevels represented by these bits are high or at ‘1’. Then the voltage representingthe different digital bits are added together to give the equivalent analog outputvoltage.

The circuit diagram of the resistor divider network, which performs the aboveconversion is illustrated in Fig. 4.26. Here RL is much greater than R. The resistorsR, R/2 and R/4 are connected to form the resisters divider network. The resistanceRL represents the load resister through which the divider network is earthed. It isvery large compared to resister R, thus avoiding the possibility of loading the resistordivider network.

R/4 R/2 R

RL

V (Analog output)a

2 V22 2 V1

1 2 V0 0

Fig. 4.26. Resistive divider network.

Now considering the digital input signal 001 applied to this resistor dividernetwork, the equivalent circuit is drawn as illustrated in Fig. 4.27.

R/4 R/2 R

0V2

0V1

1V0 = + 7 V

MV (Analog output)a

Fig. 4.27. Three input resister divider for digital input 001.

The analog output voltage Va can easily be determined by using Millman’stheorem, which states that voltage at any node point in a resistive circuit is equalto the ratio of the sum of the currents entering that node to the sum of theconductances connected to that node

Mathematically, we have

Va = + + ++ + +

1 1 2 2 3 3

1 2 3

/ / / ....1 / 1/ 1/ ....

V R V R V RR R R

For the equivalent circuit in Fig. 4.27, using Millman’s theorem, we get theexpression for voltage Va at node M, as

Va = + + ++ + +

1 1 2 2 3 3

1 2 3

/ / / ....1 / 1/ 1/ ....

V R V R V RR R R


NOTES

Digital Electronics=

+ ++ +

7 / 0 / / 2 0 / / 41/ 2 / 4 /R R R

R R RSince here V0 = ‘1’ (high level) = +7V

V1 = ‘0’ (Low) =0V, V2 = ‘0’ (Low) = 0V for digital input

0 0 1C B A

∴ Va = +7/7 /

RR

= + 1V

Considering the digital input signal 011 applied to the resister divider network,the equivalent circuit is drawn as in Fig. 4.28.

R/2R/4 R

M

Va

0V2

1V2 = +7 V

1V0 = +7 V

Fig. 4.28. Three input resister divider for digital input 011.

For the equivalent circuit of Fig. 4.27, using Milman’s theorem, we get theexpression for voltage Va at mode M, as

Va =+ ++ +

0 1 2/ / / 2 / /41/ 1/ /2 1/ /4

V R V R V RR R R

=+ + +

+ +7 / 7 / / 2 0 / / 4

1/ 2/ 4 /R R R

R R R

=+7 / 14 /

7 /R R

R =

21/7/

RR

= +3V

which is the same as shown in the Table 4.8 of output analog voltages for thedigital input of 011. It can be thus shown that all other input combinations willgive the analog output voltage given in Table 4.8.

Thus the basic problem, we come across in converting a digital input signalinto an equivalent analog signal, is to change the n input digital voltage levels intoone equivalent analog voltage. This is accomplished by designing an appropriatevariable resistor divider network, which changes each of the n digital levels intoan equivalent binary weighted voltage or current. Seven discrete levels can bedefined between digital 000 to digital 111 in a three-bit system. The least significantbit (LSB) here is 2°, which represents the smallest incremental change in the digitalsignal. As has been shown, this bit should cause a change in the analog outputvoltage, which must be equal to 1/7th of the full scale analog output voltage. Thenetwork thus has been designed in such a manner that a ‘1’ in the 20 position willgive 7 × 1/7 = 1V at the output. Since 21 bit number is twice the number represented

NOTES



by 2° bit , therefore a ‘1’ in the 21 position will causes a change of 7 × 2/7 = 2V in theanalog output voltage and so on. Each successive bit has a value which is equal totwice that of the preceding bit. The sum of weights of all the bits of the systemmust be 1. Thus for a three bit digital system, the sum of weights of 20 bit, 21 bitand 22 bit is equal to 1/7 + 2/7 + 4/7 = 7/7 = 1 and for a four-bit system the sum ofweights of 20 bit, 21 bit, 22 bit, 23 bit is equal to 1/15 + 2/15 + 4/15 + 8/15 = 15/15= 1. As a general rule, the binary equivalent weight assigned to the Least Significant

Bit (LSB) of a n bit digital system is −

12 1n

.

The remaining subsequent bits are determined by successive multiplicationby 2.

The variable resistor divider type of D/A converter is an economical and simplemethod of D/A conversion as only one resistor is required per bit. But the methodsuffers from some serious drawbacks, as given below:

(i) Each resistor required in the network is of different value, as such theresistors used are to be chosen from wide range of values. Practically andeconomically it becomes difficult to have precision resistors of widelydifferent values and at the same time obtain high accuracy and stability.

(ii) The resister used for most significant bit MSB is of very small value ascompared to the resistor used for LSB. For example in an eight bit digitalsystem, the resistor used for MSB will be R/128 if the resister used forLSB is R. This means that the resistor used for MSB has to handle a veryhigh current as compared to the LSB resistor. For a eight-bit digital system,the current through the MSB resistor will be 128 times as large as thecurrent through the LSB resistor.

(iii) It becomes difficult and cumbersome to select all the resistors of widelydifferent values to be used in such a way, so that they are all constructedfrom the same material, to have exactly the same resistor temperaturecoeffcients. Thus different resistors used will vary differently, whenD/A converter is operated over a wide range of temperature, giving anerror in the analog output voltage.

(iv) When the number of input bits is large, the resistor used for LSB has to beof very high value, which may be near to the input resistance of theamplifier, thus giving error in the results.

(v) If each higher bit resistor is not exactly half of the previous bit resistor,the step size will change.

Taking into account all these drawbacks of the variable resistor divider networktype of D/A converter, a better type of D/A converter using binary ladder networkhas been developed. It overcomes all the drawbacks discussed above and is describedahead.


NOTES

Digital Electronics4.16 ANALOG TO DIGITAL CONVERSIONS

An analog to digital (A/D) converter, converts an analog input voltage into anequivalent digital output, after a certain time delay. The A/D conversion process ismore complicated and more time consuming than the D/A conversion process. Herethe analog input signal is digitized into n segments or bits. The analog data isgenerally required to be converted into digital form because the outputs of most ofthe transducers are analog signals. Therefore A/D converters are all the moreessential, if information about the physical phenomena is to be processed, storedand displayed as digital data. The data in the digital form can be easily processed,stored recorded and displayed. Therefore A/D converters find applications in digitalprocessing equipments, Instrumentation and process control, computers,communications, etc.

Different important types of A/D converters have been developed and theyuse D/A converter as a part of their circuitry. The timing for their operation isprovided by an input clock signal. Logic circuitry contained in the control unitgenerates a sequence to start the conversion process. A/D converters employ one ormore comparators. An operational amplifier (opamp) is generally used as acomparator. An opamp has two analog inputs and one digital output, which switchesthe states, depending upon which analog input is greater. A/D converters areclassified into various kinds depending upon different requirements and on optimumcombination of the desired parameters like accuracy, precision, stability ofconversion, conversion speed, noise immunity and economic considerations. Eachtype of A/D converter has certain main features and attributes, which are mentionedas each method is discussed.

4.17 VOLTAGE TO TIME ANALOG TO DIGITAL

CONVERTER

The voltage to time, A/D converter offers direct conversion of an input voltageinto the corresponding equivalent pulse width. The pulse width is converted todigital word by using a counter which counts the number of cycles of a referencefrequency, which occurs during the time duration of the pulse width. Figure 4.29(a)illustrates the block diagram of such an A/D converter and Fig. 4.29(b) illustratesthe ramp voltage wave-form.

NOTES



AttenuatorInput

singnal

Ramp generator

Sample rateoscillator

Digital counter

gatecontrolinput

gatecontrolinput

Gate Crystaloscillator

Comparatorea

er e0

+

–

Digital output

Fig. 4.29. (a) Block diagram of voltage to time A/D converter.

This A/D converter operates on the principle of counting the number of cyclesof a fixed constant frequency source for a variable period of time. The voltage totime A/D converter employs the single slope ramp comparison technique.

The input voltage ea here, is compared with a reference ramp voltage er whichis generated by the ramp generator. When the ramp voltage starts its switch agate opens after receiving a signal from the ramp generator. When the ramp voltageer reaches a value equal to ea, the output of the comparator reverses the polarityand sends gate control signal to gate to close the gate. This results in a time interval∆t, which is proportional to the input analog voltage. This time interval ∆t is theneasily converted into a digital word by sending clock pulses into the digital counterfor the duration of this time interval ∆t.

Clock

time

Rampvoltageer

t

Fig. 4.29. (b) Showing wave-form.

The major advantage of this method is that it is very simple and can be realizedby using relatively little electronic equipment to produce a time interval pulse bythis ramp technique. Another advantage is that pulse width of the wave-form canbe easily transferred or transmitted over long lines. The main disadvantages ofthe ramp technique are relatively poor accuracy, typically of the range 0.1 to 1%,conversion process is slow and possibility of large errors due to noise in the inputsignal ea. The problem of error due to noise can however be minimized by usinginput filters.


NOTES

Digital Electronics4.18 A/D CONVERTER SPECIFICATIONS

A/D converters are required to convert analog signals into corresponding,equivalent digital codes. A wide variety of A/D converters with differentspecifications and widely different prices are available Suitable and proper selectionhas to be made depending upon the particular requirement for the specificapplication. In order to make the user familiar with the manufacturer’sspecifications to choose the particular device from the wide variety of A/D convertersavailable, the following terms have been defined.

A/D Resolution. A/D resolution is defined as the change in the input voltagerequired for a one-bit change in the output. The resolution of A/D converter isequal to the resolution of D/A converter, which it contains. It actually depends onthe number of bits in the output digital code. It can also be expressed as a percentage.

A/D Gain. Gain of an A/D converter is defined as the equivalent voltage ofthe digital output divided by the analog input voltage at the linearity referenceline. It is usually not a serious problem because it can be zeroed out.

A/D Drift. Drift of an A/D converter is the quality of a circuit to change theparameters with time. Drift errors of ±½ LSB can cause a maximum error of oneLSB from first transition to the last transition. It becomes difficult to achieve avery low drift as it increases the cost of the A/D converter.

A/D Speed. This is defined as the time required to perform one conversion oras the time taken betweeen successive conversions at the highest possible rate. Itdepends upon the settling time of the various components and the internal speedof the logic.

Quantization Error. The problem associated with conversion of analogsignals to discrete digital approximations is called fidelity. Quantizing is the processof reducing series of complex wave-forms to reasonably accurate straight lineapproximations. Resolution can also be thought of as a built-in error, which isgenerally referred to as quantization error. This built-in error is commonly specifiedas ± 1LSB; which means that the result could be off on either side by the weight ofone LSB. This quantization error can be modified or reduced by increasing thenumber of bits in the digital counter and D/A converter or by centering thecomparator, as is possible in the continuous null balance A / D converter, in whichthe quantization error is modified to ± 1/2 LSB.

Accuracy. The main source of analog error in the A / D converter is thecomparator. The secondary sources of error are variations of resistance of resistorsin the ladder, reference-voltage supply ripple and noise. Therefore the accuracy ofthe A / D converter depends on the accuracy of the circuit components such ascomparator, precision resistors used in D/A converters, supply voltage ripple andnoise and current switches. In an A / D converter, an analog voltage is converted tothe equivalent digital code and then this digital code is again converted back to theanalog voltage, using a D/A converter. It is found that due to certain errors, asmentioned above, these two analog voltages are not exactly the same. The maximumdifference between these two analog voltages is expressed as a fraction of the fullscale output voltage is defined as the accuracy of the A / D converter.

NOTES



Aperture Rate. The aperture rate is the rate at which the discrete pointsalong the wave are analysed and expressed in conversions/second. The higher theaperture rate, the better the fidelity. Fidelity is the problem associated with theconversion of analog signals to reasonably accurate digital approximations.Normally the aperture rate should be at least twice the frequency of the highestharmonic in the wave-form. An easy way to determine the aperture rate is to analysethe given data with a harmonic wave analyser and to find how many harmonicsare necessarily required for the work. To find error due to aperture rate, let usassume the analog input voltage V = A sin ωt.

∴dvdt

= ωA Cos ωt

Thus when ωt = 0, dv/dt = ωA cos 0 = ωA. At this point (wt = 0), the curvechanges very rapidly, because the slope of the curve dv/dt is maximum atwt = 0. If amplitude A is equal to 1 volt peak to peak and w = 2πf, thendv/dt = 2πf × 1 = 2πf.

Replacing dv with ∆V and dt with ∆t, which is the aperture time, the aboverelation can be written as ∆V/∆t = 2πf or ∆V = 20πf. ∆t.

The percentage error in voltage due to aperture time is = 2πf.∆t × 100%.

For a reasonably accurate digital approximation of an analog signal, both theaperture rate and quantizing voltage should be carefully determined.

Example 4.22. Calculate the conversion time of a ten-bit digital ramp A/Dconverter and a ten-bit successive approximation A/D converter, when the clockfrequency used by them is 1000 kHz.

Solution.

Time for one clock cycle = ×1

1000 1000 = 10–6 sec = 1 µ sec

The conversion time for a digital ramp A/D converter is given by

t = (2n – 1) × (one clock cycle time)

= (210– 1) × l µ sec = 1023 µ sec.

The conversion time for a ten-bit successive approximation A / D converter istime for 10 clock cycles

= 10 × 1µsec = 10 µ sec

Thus it is seen that conversion speed is very fast in case of a sucessiveapproximation A/D converter.

Example 4.23. Find the resolution and the percentage resolution of an eight-bit A / D converter, having an input voltage of –5V to +5V.

Solution. Resolution of an A/D converter is the change in input voltagenecessary for one-bit change in the output.

Change in input voltage = E = +5 – (–5) = 10V

∴ Resolution = = =−− −8

10 10256 12 1 2 1n

E


NOTES

Digital Electronics

=10255

= 0.03922 V = 39.22 mV

Percentage resolution = × = ×− −8

1 1100 100

2 1 2 1n

=−

100256 1

=100255

= 0.3922%.

Example 4.24. Find out how bit A/D converter is required to achieve aresolution of 1 mV if the maximum full scale input voltage is 10 volts.

Solution. The n bit A/D converter resolution is given by

Resolution =−2 1n

E

∴ 1 mV =−

102 1n

or 1 × 10–3 =−

102 1n

∴ 2n –1 =−3

1010

= 10 × 103

∴ 2n = 10001

Thus 2n must be equal to or greater than 10001 for n = 13, value of 2n = 8192and for n = 14, value of 2n = 16384.

Therefore value of n must be 14.

Thus a 14-bit A/D converter is required.

Example 4.25. What is the maximum conversion time of a ten-bit counter typeA/D converter if it is driven by an input clock of 500 kHz frequency.

Solution. A ten-bit A/D converter has a maximum of 210 = 1024 counts. Forthe clock of 500 kHz frequency, the counter will advance at the rate of 1 count foreach 2 × 10–6 seconds, because it will count 500000 counts in one second.

Therefore to advance 1024 counts, it will require a conversion time of1024 × 2 × 10–6 seconds.

= 2048 µseconds = 2.048 milliseconds

Example 4.26. An A/D converter has an aperture time of 5 µsec and the highestharmonic present in the wave-form is of 500 Hz. Find the error in voltage and thepercentage error.

Solution. Error in voltage = 2π f.∆t

Where f is the aperture rate, which is equal to twice the frequency of thehighest harmonic present in the wave-form and is = 2 × 500 = 1000 Hz

NOTES



∆t = aperture time = 5 µsec = 5 × 10–6 sec

∴ Error in voltage = ∆v = 2Πf. ∆t

= 2 × 3.14 × 1000 × 5 × 10–6 V

= 3.14 × 1000 × 10 × 10–6 = 0.0314

Percentage error = 2Π f.∆ t × 100%

= 2 × 3.14 × 1000 × 5 × 10–6 × 100

= 3.14%.Example 4.27. Find the aperture time required to digitize a 1000 Hz signal

with a 0.5% accuracy in voltage.Solution.

Percentage accuracy = 2πf. ∆t × 100∴ 0.5 = 2 × 3.14 × 1000 × ∆t × 100

∆t = −= ×× × ×

50.5 0.510

2 3.14 1000 100 6.28

=−× × 60.5 10 10

seconds6.28

=−× 65 10

seconds6.28

= 0.796 µseconds.

Example 4.28. Calculate the quantization error of a 10-bit A/D converter.

Solution. The quantization error of an n bit A/D converter = 1/2n

Therefore quantization error of a ten-bit A/D converter = 1/210 = 1/1024

The quantization error expressed as percentage = 1/1024 × 100 = 0.1%

Example 4.29. An eight-bit successive approximation A/D converter has aresolution of 20 mV. If the analog input voltage is 1.085 V, find out its digitaloutput in binary form.

Solution. A resolution of 20 mV means that the input voltage of the A/Dconverter increases by 20 mV required for analog input voltage of 1.085 V is

= −= = =

× 31.085 1.085 1.085

54.2520 2020 10mV

Thus the 54th step will produce a voltage of 1.08 V and the 55th step will producea voltage of 1.10 V. The successive approximation A/D converter produces a digitaloutput voltage, which is step below analog input voltage. Therefore for input voltageof 1.085 V the digital output will be decimal 54 = (00110110)2.

SUMMARY

1. Number systems provide the basis for all operations in information processingsystems. In a number system, the information is divided into a group ofsymbols; for example, 26 English letters, 10 decimal digits etc.


NOTES

Digital Electronics2. The binary number has a radix of 2. As r = 2, only two digits are needed, andthese are 0 and 1. Like the decimal system, binary is a positional system,except that each bit position corresponds to a power of 2 instead of a power of10.

3. Complements are used for simplifying the substraction. There are two typesof complements for each base-r system.

1. The r’s complement

2. The (r–1)’s complement.

4. A digital circuit with one or more input signals but only one output signal iscalled a logic gate.

In the complex circuits, the following six different digital electronics gates areused as basic elements :

1. NOT Gate 2. NAND Gate3. AND Gate 4. OR Gate5. NOR Gate 6. XOR Gate.

5. The algebra used to symbolically describe logic functions is called Booleanalgebra.

6. The memory elements used in clocked sequential circuits are called flip-flops.These circuits are binary cells capable of storing one bit of information.

7. A counter is a sequential circuit that goes through a prescribed sequence ofstates upon the application of input pulses.

8. A register is a group of memory elements which work together as one unit.The simple resisters only store a binary word. The other registers modify thestored word by shifting its bits to left or right.

9. The process of conversion of analog signal to digital signal is referred as analogto digital conversion. The system that realizes the conversion is referred asanalog to digital converter or A/D Converter or ADC.

10. The process of converting the digital signal to analog form is called digital toanalog conversion and the system used for this purpose is referred to as digitalto analog converter or D/A converter or DAC.

GLOSSARY

• Bits: Binary number system represent either by 0 or 1 binary digit. Thesedigits are called as bits.

• Symbol: A group of bits used to represent the discrete elements ofinformation is known as symbol.

• Truth table: The table which indicates output of gate for all possiblecombination of input.

• Binary ripple counter: It consist of a series connection of flip-flop withoutlogic gate. Each flip-flop is triggered by the output of its preceding flip-flop, goes from 1 to 0.

NOTES



• Buffer register: It is the simplest kind of register used only to store a digitalword.

• Shift register: It is an array of flip-flop designed to store and shift thedata.

REVIEW QUESTIONS

1. Why binary system is preferred in ‘digital system’ ?

2. Discuss the importance of 1’s and 2’s complement numbers.

3. Define the terms ‘logic function’ and ‘logic gate’.

4. What is Boolean algebra ? How is it different from ordinary algebra ?

5. Draw a NOT gate. Write its truth table.

6. Differentiate between OR and NOR gates. NOR is a combination of whichgates ?

7. Which gates combine to form a NAND gate ?

8. What is exclusive OR gate ? Write its Boolean expression.

9. Show symbols of OR, AND, NOT, NAND, NOR, exclusive OR gates.

10. Simplify the Boolean expression : (a) B . 1 + A, (b) A + B + 1, (c) (A + B) . 1

[Ans. (a) A + B, (b) 1, (c) A + B]

11. Draw a network to generate the function Y = A B C. + .

12. Simplify the Boolean expression

Y = A . B . C + A .B C. + A .B . C + A . B . C .[Ans. A]

13. Simplify the following Boolean function and draw a network to generate theoriginal function and simplified function

Y = A B A C. .+ . [Ans. A . B . C]

14. Draw the truth tables for the following functions : (a) X = A .B + A . B,

(b) Y = A . B . C + A . B . Draw circuits to generate above function.

15. Prove that :

(A + B) . ( A + C) = A . C + A B.

A . B . C + A . B .C = A. B.

16. Using only NAND gates draw network to generate Y = A + B . C.

17. If A = 1, B = 0 and C = 1, find the values of

(a) A B. + A . B + B + C (b) A . B .C A+ . B + A . B .

[Ans. (a) 1, (b) 1]

18. Differentiate between combinational and sequential logic circuits.

19. Distinguish between a ‘half adder’ and a ‘full adder’.


NOTES

Digital Electronics20. State De-Morgan’s theorems.

21. What is the importance of binary numbers ? What are the advantages ofdigital signals over the analog signals ?

22. Show that a multiplexer may be used as a sequential data selector.

23. What are demultiplexers ? What is the difference between a demultiplexerand a decoder ? Show connection diagram of a demultiplexer and a decoder.

24. With the aid of a neat sketch explain the operation of a BCD to decimaldecoder.

25. Draw the diagram of a clocked S-R flip-flop and give the truth table.

26. Show that a R-S flip-flop results when two NOR gates are cross-coupled.

27. What is a flip-flop ? Explain the principle of operation of S-R flip-flop withtruth table.

28. With the aid of a neat sketch, explain the operation of J-K flip-flop.

29. Explain the operation of a R-S flip-flop with the help of waveform and truthtable.

How does the master-slave action in a J-K flip-flop improves its operation ?30. Briefly describe R-S, J-K, D- and T-type flip-flops.

31. Find out the binary equivalent weights of each bit in a four-bit resistivedivider D/A converter.

32. Explain D/A accuracy, resolution and monotonicity.

33. What do you understand by quantization error, resolution, accuracy andaperture rate of an A / D converter?

34. Write a short note on A / D converter specification.

FURTHER READINGS

• Muthusubramanian R., Salivahanan S and Muraludharan K.A., “BasicElectrical, Electronics and Computer Engineering”, Tata McGraw Hill,Second Edition, 2006.

• Yaduvir Singh, Mandhir Verma, “Fundamental of Electrical Engineering”,University Science Press.

• Vikramditya Dave, “Electrical and Electronics Engineering”, UniversityScience Press.

NOTES



STRUCTURE

5.1 Communication5.2 Types of Signals5.3 Modulation and Demodulation5.4 Radio Communication Systems5.5 Television Systems5.6 Microwaves5.7 Satellite Communication5.8 Fiber Optic Communications




U N I T

5FUNDAMENTALS OF

COMMUNICATION ENGINEERING

OBJECTIVES


• define signals, its types, modulation and demodulation of signals.• illustrate radio communication systems in terms of AM and FM.• give introduction of television systems.• describe about microwaves, satellite communication, and fiber optic

communications.

5.1 COMMUNICATION

Communication means to share one’s thought with others. It is a bi-directional process.


NOTES

Fundamentals ofCommunication

Engineering

The term communication refers to thesending, processing and reception ofinformation by electrical means.

Depending upon the types of informationto be sent and received, following systems havebeen developed over the years:

(i) Radio

(ii) Telephony

(iii) Telegraphy

(iv) Broadcasting

(v) Radar

(vi) Radio telemetry

(vii) Radio aids to navigation etc.

Fig. 5.1 shows the block diagram for anycommunication system.

Information source. It provides information to be communicated.

Transmitter. It processes the information and makes it fit for transmission.

Receiver. Its functions is that of demodulation or detection or decoding.

Destination. The receiver separates the original information from the highfrequency wave and feeds it to the loudspeaker, punched cards, radar displays orTV picture tube etc., known as destination.

Any communication system must fulfill the following two basicrequirements:

(i) Accurate communication.

(ii) Fast communication.

The ratio S/N =FHG

IKJ

useful signal powernoise signal power

is an important parameter in

evaluating the performance of a system and for accurate transmission,this should be as high as possible.

5.2 TYPES OF SIGNALS

Encoders produce the following two types of electrical signals :

1. Analog Signals

— A telephone, radio broadcast or TV signals are very common types of signalsfor use of general public.

— These are represented by voltage waveforms that have different amplitudesat different instants of time.

Source ofinformation

Transmitter

M

Receiver

Destination

Decodingdemodulation(distortion)

Encodingmodulation(distortion)

NoiseChannel(distortion)

Fig. 5.1. Block diagram ofcommunication system.

NOTES



2. Digital Signals

— These signals comprise of pulses occurring at discrete intervals of time.

— The pulses may occur singly at a definite period of time or as a codedgroup.

— These signals play a very important role in the transmission and receptionof coded messages.

— Simplest types of digital signals are the Telegraph and Teleprinter signalsbut sometimes analog messages are also converted into digital form beforebeing transmitted for certain reasons.

5.3 MODULATION AND DEMODULATION

Several signals out of different types of signals that are generally encounteredin communication systems have frequency spectra that is not suitable for directtransmission especially when atmosphere is used as the transmission channel. Insuch a case, the frequency spectra of the signal may be translated by modulatinghigh frequency carrier wave with signal.

In order to transmit and receive the intelligence (code, voice, music etc.)successfully, the following two processes are essential :

(i) Modulation (ii) Demodulation.

Modulation

Definition

Modulation may be defined as follows :

Modulation is the process of combining the low-frequency signal with avery high-frequency radio wave called ‘carrier wave (CW)’. The resultantwave is called modulated carrier wave. This job is done at the transmittingstation.

Or

Modulation is a process in electronic circuits by which the characteristicsof one waveform (carrier) is modified by the variations in another wave(audio signal).

Or

Modulation is the process of combining an audio-frequency (AF) signalwith a radio frequency (RF) carrier wave. AF signal is called a modulatingwave and the resulting wave produced is called modulated wave. Duringmodulation, some characteristic of the carrier wave is varied in time withthe modulating signal and is accomplished by combining the two.


NOTES


Engineering

RF carrier

AF signal

Modulator

Modulatedwave

Fig. 5.2

Need of Modulation

In electronic communication, the modulation is necessary for the followingreasons :

1. Modulation increases operating range.

2. It reduces the size of transmitting and receiving antennas.

3. It permits transmission without wire.

4. It is extremely difficult to radiate low frequency signals through earth’satmosphere in the form of electromagnetic energy.

Methods of Modulation

For a sinusoidal carrier wave, the mathematical expression is given as

e = Ec sin (ωct + φ) = Ec sin (2πfct + φ) ...(1)

Thus, the waveform can be varied by any of its following three factors orparameters :

(i) Ec – The amplitude ;

(ii) fc – The frequency ;

(iii) φ – The phase.

Accordingly, there are three types of sine-wave modulations known as :

1. Amplitude Modulation (AM)

2. Frequency Modulation (FM)

3. Phase Modulation (PM).

Amplitude Modulation (AM)

The process by which the amplitude of a carrier wave is varied in accordancewith the modulating signal is called amplitude modulation.

The process of amplitude modulation is shown graphically in Fig. 5.3.

— For the sake of simplicity, the AF/message signal has been assumedsinusoidal [Fig. 5.3 (a)].

— The carrier wave by which it is desired to transmit the AF signal is shownin [Fig. 5.3 (b)].

— The resultant wave called modulated wave is shown in [Fig. 5.3 (c)].

The function of the modulation is to mix these two waves in such a way that(a) is transmitted along with (b).

NOTES



— The fluctuations in the amplitude ofthe carrier wave depend on the signalamplitude and the rate at whichthese fluctuations take place dependson the frequency of the audio signal.

All stations broadcasting on thestandard broadcast band (550 – 1550kHz) use AM modulation.

Percent modulation indicates thedegree to which the AF signalmodulates the carrier wave.

Methods of amplitude modulationare :

(i) Amplifier modulation.

(ii) Oscillator modulation.

Limitations of Amplitude Modulation

Following are the limitations ofamplitude modulation :

1. Smaller operating range.

2. Poor efficiency.

3. Poor audio quality.

4. Noisy reception.

Frequency Modulation (FM)

The process by which the frequency of a carrier wave is varied in accordancewith the modulating signal is called frequency modulation.

The process of frequency modulation is shown graphically in Fig. 5.4.

The three waves shown in the figure are : (a) Signal ; (b) Carrier ; (c) Theresultant frequency modulated wave.

When a signal of frequency fs is modulated with carrier wave of frequency fc,a resultant modulated wave is produced.

The following points are worth noting :

(i) The amplitude of the modulated wave is the same as that of the carrierwave.

(ii) The frequency of the modulated wave varies in accordance with the messagesignal.

Advantages

1. Better audio quality.

2. High transmission efficiency.

3. Noiseless reception.

Fig. 5.3. Amplitude modulation.

es

0t

Message (AF) signal

(a) AF signal

ec

(b) Carrier wave

0

Carrier wave

t

e

0

(c) Modulated wave

ResultantAM wave

Ec(min)

Envelope

Ec(max)

t


NOTES


Engineering

es

W

P0 Q

Y Signal

R S Ut

ZX

(a) Signal

ec

0

Carrier

t

(b) Carrier

ev

t

P W Q X R Y S Z U

(c) AM wave

0

Fig. 5.4. Frequency modulation.

Limitations

1. Smaller area of reception.

2. Wider channel is needed.

3. Equipment used is more complex and costly.

Pulse Modulation

Pulse modulation is a technique of modulating the analog signal andconverting it into corresponding values. In this process the instantaneous voltage ofthe analog signal is sampled at regular intervals and transmitted during thesesampling periods only.

— In analog modulation, the amplitude or frequency or phase (any oneparameter) of carrier is varied according to the instantaneous voltageof signal.

— In pulse modulation, either amplitude or width or position of thepulse is varied according to the instantaneous voltage of signal (exceptfor PCM).

NOTES



Pulse modulation is classified as follows :Pulse modulation(a) Analog

(i) PAM

(ii) PTM

PDM, PWM

PPM

(b) Digital(i) PCM, (ii) Delta-modulation.

Sampling Theorem : This theorem was developed by Nyquist.It states : “If sampling rate in any pulse modulating system exceedstwice the maximum signal frequency, the original signal can beconstructed at the receiver with minimum distortion.”

— If fm is the maximum signal frequency then sampling rate is greaterthan 2fm.

— The minimum samplings rate is called the Nyquist rate.— The reciprocal of the sampling rate (or sampling frequency) is called

sampling line or Nyquist interval.

Pulse Amplitude Modulation (PAM)— PAM can be generated by using an AND gate.— PAM can be demodulated by passing through low pass filter with cut-

off frequency as the highest signal frequency.— Generally FM is used to modulate the pulses so that it can be

transmitted ; such a system is called PAM-FM.

Pulse Width Modulation (PWM)— It is also called Pulse Duration Modulation (PDM) or Pulse Length

Modulation (PLM).— It can be generated using a monostable multivibrator.— It can be demodulated by feeding the PWM signal to an integrating

circuit.— In PWM the width of the pulse is changed in accordance with the

instantaneous value of the modulating signal with amplituderemaining constant.

— The most oftenly used integrating circuit is the loudspeaker itself.

Pulse Position Modulation (PPM)— It can be generated in the similar way as PWM but the pulse width is

kept constant from the starting point of occurrence of pulse.— It can be demodulated by converting into PWM using a flip-flop.— In PPM the position of the pulse or the time of occurrence of the pulses

is changed in accordance with the instantaneous magnitude of themodulating signal.

Pulse Code Modulation (PCM)— In this type of modulation, digital equivalents of the instantaneous

voltage levels of the signals are transmitted in the form of pulses.


NOTES


Engineering

— The total amplitude range by which the signal may occupy is dividedinto a number of standard levels ; this process is called Quantizing.The level actually sent is the nearest standard level. The quantizationlevels depend upon number of bits per sample.

— PCM encoder functions similar to A/D converter ; PCM decoder func-tions similar to D/A convertor.

Digital Modulation Techniques

In digital communications, the modulating signal consists of binary data oran M-array encoded version of it. This data is used to modulate a carrier wave(usually sinusoidal) with fixed frequency. In fact the input data may represent thedigital computer outputs or PCM waves, generated by digitizing voice or videosignals. The channel may be a telephone channel, microwave radio link, satellitechannel or an optical fiber.

In digital communication, the modulation process involves switching orkeying the amplitude, frequency or phase of the carrier in accordance with the inputdata. Thus, there are three basic modulation techniques for the transmission ofdigital data ; they are known as :

(i) Amplitude-shift keying (ASK).(ii) Frequency-shift keying (FSK).

(iii) Phase-shift keying (PSK).The above techniques can be viewed as special cases of amplitude modulation,

frequency modulation and phase modulation respectively.

Types of Digital Modulation Techniques

Various types of digital modulation techniques are of two types :

I. Coherent digital modulation techniques :

1. Coherent binary modulation techniques.

2. Coherent binary amplitude shift keying or on-off keying.

3. Coherent demodulation of binary ASK.

4. Binary phase shift keying (BPSK).

5. Coherent binary frequency shift keying (BFSK).

II. Non-coherent binary modulation techniques :

1. Non-coherent binary ASK.

2. Non-coherent detection of FSK.

3. Differential phase shift keying (DPSK).

4. Quadrature phase shift keying (QPSK).

Demodulation or Detection

The process of extracting the low frequency modulating signal from themodulated wave is known as demodulation or detection.

NOTES



The demodulation of an ‘AM wave’ involves two operations :(i) Rectification of the modulated wave ;

(ii) Elimination of the RF component of the modulated wave. The demodulation of an ‘FM wave’ involves three operations :

(i) Conversion of frequency changes produced by modulating signal intocorresponding amplitude changes ;

(ii) Rectification of the modulating signal ;(iii) Elimination of RF component of the modulated wave.

5.4 RADIO COMMUNICATION SYSTEMS

Radio Transmitter

A radio transmitter is a device that transmits information by means of radiowaves.

The signal intelligence is translated in terms of a high frequency wavecommonly termed as Carrier wave and the process of intelligence translation intohigh frequency is termed as Modulation. All radio transmitters use one form ofmodulation or the other for transmission of intelligence.

All radio transmitting systems must have the following :

1. A section for generation of high frequency carrier wave ;

2. A section for converting information into electrical impulses and amplifyingthem to the required level ; and

3. A section for modulating thecarrier with signal intelligenceamplification stages for increasing thelevel of the modulated wave to the desiredpower and antenna system fortransmitting these signals into freespace.

A basic radio transmitter system isshown in Fig. 5.5.

Transmitters are usually nameddepending upon:

(i) The type of signal to betransmitted ;

(ii) The type of modulationemployed ;

(iii) The carrier frequency used ; and

(iv) The type of radio waves radiatedby the system.

— A transmitter, may, therefore, be named as broadcast transmitter,telephony or telegraphy transmitter depending upon whether the signal isan entertainment programme, speech or code signal or a picture signal.

Signaltransducer

Modulatingamplifier

Modulator

Carrier waveamplifier

Carrier wavegenerator

Poweramplifier

Antenna

Fig. 5.5. A basic radio transmissionsystem.


NOTES


Engineering

— It may be termed as AM (amplitude-modulated) or FM (frequency-modulated) transmitter, depending upon the modulation process employed.

— Similarly, it may be termed as Medium wave, ‘Short wave’, VHF, UHF orMicrowave transmitter depending upon the carrier frequency employed.

— Lastly, a transmitter may be termed as long distance transmitter or a lineof sight transmitter depending upon whether the transmission is by skywaves or space waves.

AM Transmitters

These transmitters are generally employed for radio broadcasts over long,medium or short waves, point-to-point communication systems using radiotelephony/telegraphy signals over short waves or VHF waves and picturetransmission over the VHF or UHF ranges.

These transmitters employ one or the other method of producing modulatedwaves such that the depth of modulation is directly proportional to themagnitude of the modulation signal.

Antenna systems for AM transmitters are large and must be located atsome point remote from the studio operations. All the studio signaloperations are performed at relatively low levels and transmitted to themain transmitter location either over telephone wire lines or a radio linksuch as a microwave system.

FM Transmitters

Direct FM

In frequency modulation, the amplitude of r-f carrier remains constantbut its frequency is continuously varied in accordance with theinstantaneous amplitude of an audio signal. Thus the original sound wavesare converted into frequency deviation of the r-f carrier frequency that isproportional to the intensity of the sound waves.

In FM transmitter, we make use of reactance modulator which changesthe amplitude variations of an audio signal into frequency variations.

FM transmitter is much more efficient than an equivalent AM transmitter.

Indirect FM

Phase modulation may be used to achieve frequency modulation by indirectmethod. It is only necessary to integrate the modulating signal prior toapplying it to the phase modulator.

This transmitter is widely used in VHF and UHF radio telephoneequipment.

Radio Receiver

A radio receiver is a device that picks up the desired signal from the numeroussignals propagating at that time through the atmosphere, amplifies the desired

NOTES



signal to the requisite level, recovers from it theoriginal modulating signal and eventuallydisplays it in the desired manner.

Basic functions of a receiver are shown inFig. 5.6.

Requirements of Radio Receivers

Sensitivity. It is the capability of a radioreceiver to detect weak signals. It is defined asthe amount of r-f input voltage needed to producea specified amount of audio output power.

Selectivity. It is the ability of a receiver toreject unwanted signals and amplify the desiredone.

In addition to the above requirements, everyradio receiver should have :

— stability in frequency— constant bandwidth of operation.

Types of Receivers

Following two types of receivers have practical or commercial significance :

(i) Tuned Radio Frequency (TRF) receiver.

(ii) Superheterodyne receiver.

(i) Tuned Radio Frequency Receiver :

— It consists of several stages of R-Famplification with tuned LC tankcircuit at the input of each stage, anAM detector, one or more audiofrequency amplifier stages drivingloud-speaker or headphones asshown in Fig. 5.7.

— The provision of r-f stage at theinput of the receiver provides thefollowing advantages :(a) Greater gain.

(b) Prevention of re-radiation of thelocal oscillator (in case ofsuperheterodyne receiver).

(c) Improved rejection of adjacentunwanted signals.

(d) Better coupling of antenna withthe receiver.

A tuned radio frequency receiver possesses the following good features :1. High sensitivity 2. Simplicity.

Selectiontuned circuit

Amplification(R.F.)

Antenna

Detection

Amplification(A.F.)

Reproduction

Fig. 5.6. Basic functions of areceiver.

1st R.F.amplifier

Antenna

A.M.detector

A.F.amplifier

Loudspeaker

2nd R.F.amplifier

Poweramplifier

Gauged

Fig. 5.7. Tuned radio frequencyreceiver.


NOTES


Engineering

It has the following shortcomings :

1. Bandwidth variations 2. Instability

3. Insufficient adjacent frequency rejection.

(ii) Superheterodyne Receiver :

These days, practically, all receivers are superheterodyne.

— The superhet has the same essential components as that of a TRF receiverin addition to the mixer, local oscillator and interfrequency amplifier(Fig. 5.8).

RFamplifier

Antenna

IFamplifier

Detector

Mixer

Audioamplifier

Localoscillator

Tuni

ngG

auge

d

AG

C

Fig. 5.8. Superheterodyne receiver.

— The RF amplifier is tuned to the required incoming frequency.— The output of RFA is combined with the local oscillator voltage and

normally converted into a signal of a lower fixed frequency calledintermediate frequency. The signal at intermediate frequency (IF) containsthe same modulation as the original carrier.

— IF signal is then amplified and detected to obtain the original information.— A fixed frequency difference is maintained between the local oscillator

and RF frequency with the help of capacitance tuning, in which all thecapacitors are ganged together and operated in unison.

— IF stage consists of a number of transformers, each consisting of a pair ofmutually coupled tuned circuits, providing a large gain. The characteristicsof the IF amplifier are kept independent of the frequency to which thereceiver is tuned, so that sensitivity and selectivity of supperhet usuallyremain fairly uniform throughout its tuning range.

— The radio frequency section provides coupling from the antenna inputterminals of the receiver to the first stage of RF amplifier so as to amplifythe incoming signal before its frequency is changed.

— The RF section carries out the following main functions :(i) To provide discrimination or selectivity against image and interacted

frequency signals.

NOTES



(ii) To provide an efficient coupling between the antenna and first stageof RF amplifier.

(iii) To reduce the noise figure of the receiver.

5.5 TELEVISION SYSTEMS

Television means seeing at a distance. To be successful, a television system may be required to reproduce faithfully

the following :(i) The shape of each object, or structural content.

(ii) The relative brightness of each object, or tonal content.(iii) Motion, or kinematic content.(iv) Sound.(v) Colour, or chromatic content.

(vi) Perspective, or stereoscopic content. Television system deals with the transmission and reception of visual ‘live’

scene by means of radio broadcasting. Along with the signals, the soundsignals associated with the scene are also transmitted at the same time toprovide a complete sight and reproduction at the receiver of the televisedprogramme.Fig. 5.9 shows a block diagram of a rudimentary television system,indicating basically how the requirements of monochrome TV transmissionand reception may be met.

Principle of Colour Television

The basic principle of colour television is same as for black and whitetelevision with the main difference that colour TV system requires thetransmission and reception of specific colour information along withmonochrome (black and white) signals.

Crystaloscillator

RFamplifiers

Poweramplifier

Combiningnetwork

Soundtransmitter

FMmodulating

amplifier

Audioamplifiers

Scanning andsynchronizing

circuits

AMmodulating

amplifier

Videoamplifiers

Microphone

Cameratube

( ) Transmittera


NOTES


Engineering

Audioamplifiers

Sounddemodulator

SoundIF

amplifiers

TunerCommon

IFamplifiers

Videodetector

Videoamplifiers

Picturetube

Scanning andsynchronizing

circuits

Sync

Loudspeaker( ) Receiverb

Fig. 5.9. Basic monochrome television system.

Colour television must have two way compatibility with monochrometelevision. That is, a colour transmission can be reproduced in black-and-white shades by a monochrome receiver and monochrome transmission isreproduced in black-and-white transmission by a colour receiver.

Table 5.1. Selected Standards of Major Television Systems

Standard American System EuropeanSystem (India)

Number of frames per second 30 25

Number of lines per frame 525 625

Field frequency, Hz 60 50

Line frequency, Hz 15750 15625

Channel width, MHz 6 7

Video bandwidth, MHz 4.2 5.5

Colour subcarrier, MHz 3.58 4.43

Sound system FM FM

Maximum sound deviation, kHz 25 50

Intercarrier frequency 4.5 5.5

5.6 MICROWAVES

The waves of high frequencies usually above 300 MHz are termed asmicrowaves.

The wavelength of such waves is less than 1 metre. The microwaves frequencies span the following three major bands at the

highest end of RF spectrum :(i) Ultra High Frequency (UHF) ... 0.3 to 3 GHz

(ii) Super High Frequency (SHF) ... 3 to 30 GHz

(iii) Extra High Frequency (EHF) ... 30 to 300 GHz.

At such high frequencies, the components depend on the varyingelectromagnetic field rather than on the current in a wire conductor or the voltage

NOTES



across two points. Instead of resonant LC circuits and conventional wire conductors,therefore, resonant cavities and waveguides are often used at microwave frequencies.

Applications of Microwaves

Microwaves are widely used in the following fields :1. Telecommunications 2. Radar3. Televisions for relaying signals 4. Satellite communications5. Industrial heating 6. Research7. Cooking.

Waveguides

— Waveguides are hollow metal tubes used to propagate microwave energyin the form of electric and magnetic fields.

— The purpose of waveguide is to guide the wavefrom source end to load end.The fundamental difference is that the propagation in the waveguides isin the form of electric and magnetic fields, whereas propagation intransmission line or free space is in the form of voltage and currents.

— These are usually made of brass or aluminium. The inner surface is usuallysilver plated to minimise the losses at higher frequencies. As the frequencyincreases, the size of waveguide reduces.

Advantages

The waveguides entail the following advantages :

1. These are simpler to manufacture than co-axial lines.

2. Improved power handling ability.

3. Power losses lower in comparison to transmission lines.

4. These have mechanical simplicity and a much higher maximum operatingfrequency as compared to co-axial lines.

Applications

1. It is observed that waveguides have dimensions that are convenient in the3 to 100 GHz range and somewhat inconvenient much outside the range. Withinthis range, waveguides are generally superior to co-axial transmission lines for awhole spectrum of microwave application, for either power or low level signals.

2. Waveguides as well as transmission lines can pass several signalssimultaneously, but in waveguides, it is sufficient for them to be propagated indifferent modes to be separated. They do not have to be of different frequencies.

Guided Wavelength

The wavelength in the waveguide is different from the wavelength in air orfree space. It can be shown that the relationship among guide wavelength, cut offwavelength, and free space wavelength is :

1 1 12 2 2λ λ λg c

= −


NOTES


Engineeringλ λ

λλ

g

c

=

−FHGIKJ

12

where, λg = Guide wavelength,

λ = Free space wavelength, and

λc = Cut off wavelength

The above equation is true for any mode in waveguide of any cross-sectionprovided the value of λc corresponds to the mode and cross-section.

Modes

The two kinds of modes in the waveguides are :

1. Transverse electric mode (TE) ... Electric field is always transverseto the direction of propagation.

2. Transverse magnetic mode (TM) ... Magnetic filed is always transverseto the direction of propagation.

Cavity Resonators

Cavity resonators are used as tuned circuits at microwave frequencies.This results in lower losses and Q of a resonate frequency as high as 20000.

The cavity resonator can be used to select one frequency and reject thefrequencies above and below a function similar to that of a band passfilter.

Mainly a resonant cavity is used to control the frequency of oscillations whenit is used with an active device to form a microwave oscillator.

Microstrip

A microstrip is a co-axial transmission line made as a printed circuit forapplication in an amplifier or oscillator in which microwave diodes and transistorsare used.

The characteristic impedance for microstrip is determined by :

(i) Thickness of the substrate ;

(ii) Dielectric constant of the substrate ;

(iii) Width of the conductor strip.

Microwave Vacuum Tubes

The mechanism of operation of various microwave tubes differ in details, butall of them involve transfer of power from a source of D.C. voltage to a source ofA.C. voltage by means of a current density modulated electron beam. This is achievedby accelerating electrons in a static electric field and retarding them in an A.C.field. The density modulation of the electron beam allows more electrons to beretarded by the A.C. field than accelerated by A.C. field, and therefore, makespossible a net energy to be delivered to the A.C. electric field.

NOTES



1. Klystron A klystron has a thermionic cathode which releases a stream of electrons.

The stream passes through two cavities called a buncher and a catcher. It is used as microwave amplifier.

2. Magnetron

It consists of an anode and a heated cathode. The anode is a metal blockwith machined cavities.

The cavity dimensions determine the frequency of oscillations. The pathof electrons is cycloidal.

The process of accelerating or decelerating the electrons is called velocitymodulation.

Used in radar and linear particle accelerators, current efficiencies are oforder of 50 percent.

3. Travelling Wave Tubes (TWT)

It is a wide band device compared with the klystron and magnetron. Travelling wave tubes are employed when larger bandwidth (both klystrons

and magnetrons have very high Q resulting in narrow bandwidth) isrequired.

These are used as linear amplifiers of microwave signals in satellitecommunications.

The characteristics of TWT are :

(i) Power gain : 60 db

(ii) Power output : 10 kW average

(iii) Efficiency : 20 % to 40%

(iv) Bandwidth : about 0.8 GHz

(v) Frequency range : greater than 3 GHz.

4. Crossed-field Amplifier (CFA)

It is a microwave power amplifier, based on the magnetron. It is a cross between the TWT and the megnetron in its operation. Pulsed CFAs are available for the frequency ranges from 1 to 50 GHz. The maximum powers available are well over 10 MW in UHF range,

1 MW at 10 GHz and 400 kW in the S-band.Power gain ... upto 20 dbEfficiency ... upto 70%Bandwidth ... 25% of centre frequencyApplications. Used widely for radar and electronic counter-measures.

5. Tunnel Diode

A tunnel diode is a thin junction diode. It exhibits negative resistance under low forward bias conditions. It is highly suitable for microwave frequencies, because of thin junction

and short transit time.


NOTES


Engineering

It is a low noise device. The tunnel diode amplifier (TDA) is a broadband, high gain microwave

amplifier.

Applications

(i) Suitable for space work (because TDAs are immune to the ambientradiation encountered in inter-planetary space).

(ii) Used as self-excited mixers.

(iii) Used for high speed switching and logic operations.

(iv) Due to their simplicity, frequency stability and immunity to radiation theseare used as low power oscillators upto 100 GHz.

6. Gunn Diodes

In bulk semiconductor materials the Gunn effect is instrumental in thegeneration of microwave oscillators.

Gunn diodes are grown epitaxially out of GaAs or Inp doped with silicon. Gunn diodes have an efficiency of the order of 2.5 to 5 per cent.

7. Impatt Diodes

IMPATT diode exhibits negative resistance and delivers high pulsed powerat the lower microwave frequencies.

A typical IMPATT diode works at about 50 GHz. These diodes are used at higher frequencies and for the highest output

powers.

8. Trapatt Diodes

These diodes also exhibit negative resistance and deliver high pulsedpowers at lower microwave frequencies.

These have an efficiency of 30 per cent. Most TRAPATT oscillators and amplifiers are still in laboratory stage. These diodes may find application in airborne and marine radars.

9. Laser

It is an oscillator. Optical fibre communication using laser operates at wavelength of about

0.85 µm.Non-communication application of lasers include the following :(i) Industrial welding ;

(ii) Distance and speed measuring equipment ;(iii) Formation of three-dimensional holograms ;(iv) Optical and other surgery etc.10. Parametric Amplifiers :

The name of the amplifier stems from the fact that capacitance is aparameter of a tuned circuit.

It is a low-noise microwave amplifier.

NOTES



Parametric amplifiers find application in the following :(i) Radio-telescopes.

(ii) Space probe tracking and communications.(iii) Tropospheric scatter receivers.

Monolithic Microwave Integrated Circuits (MMICs)

These circuits claim the following advantages over discrete circuits :1. Small size 2. Light-weight3. Improved performance 4. Improved reproducibility

5. Low cost 6. High reliability.

Applications. Monolithic microwave integrated circuits are suitable for spaceand military applications since they meet the requirements for shock, temperatureconditions, and severe vibrations.

Microwave Antenna

An antenna is a structure capable of radiating or receiving electromagneticwaves ; their function is to couple the transmitter or receiver to space. In case ofmicrowaves the transmitting and receiving antennas should be highly directive.Some of the essential requirements are given below :

(i) Omnidirectional antennas are not required because no broadcasting isdone at these frequencies.

(ii) The signals at the input terminals of the receiver should be as large aspossible.

(iii) Directional application is often required.

(iv) Need not be physically large in size.

(v) Power gain should be high.

Horn Antenna

— An horn antenna is just an open section of a rectangular guide. It is flaredout to match the characteristic impedance Z0.

— The flare angle of the horn is usually between 40° and 66° (changing overto Horn from open ended waveguide increases the directivity and alsodiffraction is reduced).

This antenna has the following features :

(i) Fairly good directivity.

(ii) Adequate bandwidth.

(iii) Simple mechanical construction.

Antenna with Parabolloidal Reflectors

— A parabola is a very suitable reflector for light rays or microwaves.— In these antennas all the waves are in phase as a result, radiation is very

strong and concentrated.


NOTES


Engineering

— The gain is influenced by the aperture ratio and the uniformity ofillumination.

Lens Antenna

— A lens antenna is used as a collimator at frequencies greater than 3 GHzand works just like a glass lens in optics.

— These antennas are oftenly made of polystyrene.— It is used to correct the curved wave front from a horn that has to be made

a plane wave front.

Advantages

(i) Greater design tolerances.

(ii) No obstruction of radiation.

Disadvantages

(i) Greater bulk.

(ii) Costly.

(iii) Design difficulties.

Helical Antenna

— It consists of a loosely wound helix backed up by a ground plane which issimply a screen made of “chicken wire”.

— This is a broad band VHF and UHF antenna which is used to providecircular polarization characteristics.

— It is used either singly or in an array for transmission and reception ofVHF signals through the ionosphere.

— It is employed for satellite and probe communications, particularly forradio telemetry.

Discone Antenna

— This is low gain antenna, but it is omnidirectional.— It is a combination of disc and cone and has enormous bandwidth for both

input impedance and variation pattern.— It is used for VHF and UHF signals, specially at airports where

communication must be maintained with aircraft coming from anydirection.

Log Periodic Antenna

— Directive gains are low to moderate.— Radiation pattern may be unidirectional or bidirectional.— Used in HF range.

5.7 SATELLITE COMMUNICATION

The need for satellite communication was realised when the other conventionalsystems became overcrowded. A number of channels can simultaneously be operated

NOTES



using a satellite. The process of sending information around the world has beenrevolutionised by the advent of satellites.

— The real beginning in the field of satellite was made in October 1957 whenfirst man-made satellite Sputnik I was launched.

— Sputnik I was followed by a rally man-made satellites. SatellitesTELESTAR I and II of AT and T launched in July 1962 and May 1963formed the basis of communication satellites. TELESTAR I was the firstsatellite to demonstrate a television link between USA and Europe. Thiswas followed by a series of communication satellites by various compa-nies.

Communication Satellite—General Aspects

A communication satellite is basically a microwave-link repeater. Itreceives the energy beamed up from an earth station, amplifies and returnsit to the earth at a frequency of a couple of GHz away to prevent interferencebetween the uplink and the downlink.

Satellite is an artificial body that is projected from earth, which revolvesaround the earth in a particular orbit. It keeps on revolving in the orbitdue to balance of force between inertia of the revolving satellite andgravitational pull of the oriented body.

The satellites remain in geostationary orbit, i.e, they have the same angularvelocity as the earth and hence they appear to be stationed over one spoton the globe.

A satellite 35800 km from the earth will complete a revolution in 24 hours. These stationary satellites present no tracking problems, but require large

antennas, high power and high sensitivity. Satellite communication come under the Transionospheric wave

propagation. Satellites must be launched into the space with a certain velocity so that

it can overcome the gravitational pull of the earth and reach the orbit.That velocity with which the satellite is launched is called Escapevelocity. The escape velocity of earth is 25000 miles/hour.

The satellites revolve in particular orbits allotted to them ; they may becircular orbits, elliptical orbits, polar orbits, equatorial orbits or anycombination of them.— If the direction of the satellite’s revolution is the same as the earth

rotation, then the orbit is said to a Posigrade orbit.— If the direction of the satellite’s revolution in an orbit is in opposite

direction to that of the earth’s rotation, then the orbit is said to be aRetrograde orbit.

— When the satellite is in an elliptical orbit, the highest point from theearth of that orbit is called Apogee.

— In an elliptical orbit, the lowest point of that orbit from the earth iscalled Perigee.


NOTES


Engineering

The rocket that carries the satellite to the required height and puts it intothe orbit is called Satellite Launch Vehicle [SLV].

The place on the earth where the transmitting and receiving equipmentused to communicate with satellite is installed, is called Earth station.

Transponder. It is a combination of transmitter and receiver. It receivessignal, amplifies and retransmits the signal with different frequency. Theyare used in “Active satellites”.

The orbit in which Geostationary satellite is kept is called Geostationaryorbit which is most widely used.

The circular orbits are used for navigational purposes for ships and for airtraffic control.

The inclined orbit is not widely used. It is mainly used to cover the polarregions.

The position of the satellite may be changed due to magnetic field of earth,sun and moon. The changed position is corrected by the earth station ;this process of correcting the satellite position is called Station keeping.

The satellite in position with transmitter, power supply and control circuitis called On-Board equipment.

The power supply to the satellite is generally provided through solar cells(main source). In case of an availability of solar energy they are poweredfrom rechargeable Nickel-Cadmium (Ni-Cd) cells.

Types of Communication Satellites

The communication satellites can be of two basic types :

1. Passive SatellitesA passive satellite acts merely as a reflector of signals. These satellites need

very powerful ground systems. These satellites do not have electronic parts (incomparison to active satellites) which may fail and render the satellite useless.

Passive communication satellites can be of two types :

(i) Discrete structure satellites ; or

(ii) Where a large volume of space is filled with a large number of tiny passivesatellites (dipoles).

The large structure usually takes the form of a sphere. When a radio signal isbeamed at a sphere much large than the signals wavelength, it gets reflected in alldirections. This kind of sphere is called an Isotropic reflector, because the power inthe signal is equally distributed in all directions. The sphere, a balloon, can bemade of a thin plastic material ; the reflector surface can be made of a thin coatingof metal. The sphere is folded into the nose cone of the booster rocked and laterinflated in orbit.

2. Active SatellitesAn active satellite receives the signal, amplifies it and transmits it back to

other points on the earth.

Active satellites can be of the following two types :

(i) Store and Forward Type : It is a rather primitive concept in satellitetechnology ; the message is stored in tape-recorder in the satellite, and

NOTES



then played back when the satellite has moved over the appropriate re-ceiving point on the earth.

(ii) Line of Sight Repeater Type : It is the type being used now-a-days.

Like the microwave repeaters which must keep each other in view, thesatellites, placed at a high altitude, must be visible to both transmitting andreceiving stations.

Synchronous or Geostationary Satellite

— If a satellite can be lifted into orbit 35680 km above the earth in circularorbit lying in the plane of the equator, it is termed as synchronous satellite.

— In this satellite the satellite period of revolution is the same as the earth’speriod for one complete rotation about its axis. This means, satelliteremains fixed relative to a point on the earth, thus to a ground observer itappears stationary.

Synchronous systems claim the following advantages over lower altitudesystems :

1. As few as three properly positioned satellites can entirely cover the earthwhere as 50 lower altitude satellites are required to serve the same purpose.

2. A single synchronous satellite can provide uninterrupted communicationservice 24 hours daily. Lower altitude systems require the ground station trackand follow the satellites as the spacecraft crosses the sky, and the stations switchfrom satellite to satellite as one dips below the horizon and another comes withinthe range.

Problems encountered with synchronous communication satellites :(i) Time delay concerning telephone.

(ii) Difficulty in establishing synchronous orbit. The space communication field, according to the requirements, can be

subdivided into three parts :1. Communication with and tracking of fast moving satellite orbit of about

145 km radius.

2. Communication via geostationary satellites.

3. Communication and tracking connected with interplanetary probes.

Orbiting period of satellite, T = 2π r

kM

2

where, r = Radius of circular orbit,

k = Gravitational constant, and

M = Mass of earth.

The height (h) of orbit above the earth in which the satellite moves is givenby :

h = (rkM – 6372) km

— The operating frequency for the satellite should be such that these signalsare able to pass the ionosphere without getting reflected by its layer andshould also not get absorbed or attenuated by the troposphere.


NOTES


Engineering

— At lower limit frequencies used are 10 MHz to 100 MHz and the upperlimit is 100 GHz.

Satellite Description

A satellite is an extremely complex and sophisticated electronic and mechanicaldevice. A satellite typically consists of many subsystems functions, all carefullyintegrated into a single system.

The “subsystems” include :

1. Electrical power subsystem.

2. Telemetry tracking and control subsystem.

3. Main and auxiliary propulsion subsystem.

4. Communication channel subsystem.

5. Antennas.

General Structure of a Satellite Communication System

Fig. 5.10 shows the general structure of a satellite communication system. Asatellite in space links many earth stations.

— The user is connected to the earth station through terrestrial network.This network may assume various configurations including a telephoneswitch or a dedicated link to the earth station.

— Signal generated by the user is processed and transmitted from the earthstation to the satellite. The satellite receives the modulated RF carriersat the predetermined uplink (earth-to-satellite) frequencies from all theearth stations in the network, amplifies these carriers and then re-transmits them back to earth at downlink (satellite to earth) frequencies.The downlink frequencies are kept different from the uplink frequenciesin order to avoid interference.

Satellite

Earthstation

Earthstation

UserUser

Terrestrialnetwork

Terrestrialnetwork

Fig. 5.10. General structure of a satellite communication system.

NOTES



— The modulated carrier received at receiving earth station is processed toget back the original base band signal. This signal is then sent to the userthrough a terrestrial network.

Satellite Spacecraft Systems

A complete satellite consists of several subsystems, but most important ofthese are :

1. Power Supply System

(i) Solar array (ii) Battery

(iii) Power supply control circuit.

2. Altitude and Orbit Control

(i) Sensor (ii) Propulsion system

(iii) Altitude control (iv) Antenna control

(v) Orbit control.

3. Telemetry and Telecontrol

4. Communications.

Launching, Power Supplies and Control of Satellites

Launching. The launcher places the satellite first into a low altitude circularinduced orbit and then into an elliptical transfer orbit with apogee at the altitudeof geostationary orbit. At an approximate apogee, the apogee boost motor is firedto circularise the orbit and remove most of the remaining orbit inclinations.

Power Supplies. Silicon solar cells are the accepted source of primary powerexcept during eclipse when power is maintained by Nickel-Cadmium cell batteries(for about 70 minutes).

Control :

— Body-stabilized designs generally employ an internal momentum wheelswith axis perpendicular to the orbit plane.

— Control about the pitch axis is through the wheel’s drive motor electrodeswhile control about the yaw and roll axes may be by gimballing the wheelor by use of hydrazine mono-propellant thrusters correct the axis direction.

— The orbit of satellite is controlled by ground command of hydrazinethrusters.

International Regulation of Satellite Service

The regulations concerning frequency allocation for different purposes,standard of communication etc., are laid down by InternationalTelecommunication Union (ITU), a body of United Nation Organization.

World Administrative Radio Conference (WARC) held in 1979, has allocatedfrequency band for satellite communication servicewise under 17 categoriesnamely: (1) fixed, (2) intersatellite, (3) mobile, (4) land mobile, (5) maritimemobile, (6) acronautical mobile, (7) broadcasting, (8) earth exploration,


NOTES


Engineering

(9) space research, (10) meterological, (11) space operation, (12) amateur,(13) radio transmission, (14) radio navigation, (15) aeronautical radionavigation, (16) maritime radio navigation, (17) standard frequency andtime signal.

WARC further divided the globe into the following three geographic regionsfor frequency allocation :

Region I : Europe, Africa, USSR and Mongolia.

Region II : North America, South America, Greenland.

Region III : Asia (except USSR and Mongolia), Australia and South WestPacific.

5.8 FIBER OPTIC COMMUNICATIONS

Owing to rapidly increasing demands for telephone communicationsthroughout the world, multiconductor copper cables have become not only veryexpensive but also an inefficient way to meet these information requirements. Thefrequency limitations inherent in copper conductor system (approximately 1 MHz)make a conducting medium for high-speed communications necessary. The opticalfiber, with its low weight and high frequency characteristics (approximately 40GHz) and its imperviousness to interference from electromagnetic radiation, hasbecome the choice for all heavy-demand long-line telephone communications systems.

Advantages and Limitations of Fiber Optic System

Advantages. The advantages of fiber optic system over conventional co-axialcable or microwave systems are as follows :

1. Light weight.

2. Larger information carrying capacity (A single fiber can handle as manyvoice channels as 1500-pair cable can).

3. Less space and easy installation.

4. Information tapping from a fiber cable is virtually impossible.

5. Fiber cables are cheaper.

6. No capacitance and inductance formation.

7. The fiber cables can operate over larger temperature and are less affectedby corrosive liquids and gases.

8. Immunity to interference (from lightning, cross talk and electromagneticradiation).

Limitations

1. Higher initial cost.

2. Switching and routing of fiber optic signals is difficult.

3. Difficult to splice optical fibers to make them longer or to repair breaks.

4. Connectors for fiber optics are more complex to attach to cable and requireprecise physical alignment.

NOTES



Optical Fiber Communication System

Fig. 5.11 shows the simplified blockdiagram for an optical fiber communicationsystem. This system converts electrical signalinto light, transmits the light through fibercable and converts light back to electricalsignal.

— The information source provides theelectrical signal to a transmitterconsisting of an electrical stage whichdrives an optical source to givemodulation of the light wave carrier.The amount of light emitted by theoptical source is proportional to thedrive current.

— The transmission medium consists ofan optical fiber cable with plasticor glass core, a cladding and protective jacket.

— The receiver consists of an optical detector which drives a further electricalstage and hence provides demodulation of the optical carrier.

Optical Fiber Construction

Refer to Fig. 5.12.

An optical fiber consists of a centralcore and an outer cladding. Thematerial of core has higher index ofrefraction as compared to cladding.The optical fibers are available eitheras a single fiber [Fig. 5.12 (a)] or asseveral fibers bundled together[Fig. 5.12 (b)].

Materials used for optical fibers areglass (fused silica) and plastic.

— Glass fibers are more costly ascompared to plastic fibers. Theypropagate light more efficiently than plastic. These are used for high speedapplication and also used for long transmission paths.

— Plastic fibers are more flexible and rugged than glass. These fibers areless expensive.

Information source

Electrical transmitter

Optical source

Optical fiber cable

Optical detector

Electrical receiver

Destination

Fig. 5.11. Optical fibercommunication system.

Cladding

Singlecore

( ) Single fibera

( ) Bundled fiberb

Jacket

Bundledfibers

Fig. 5.12. Optical fiber construction.


NOTES


EngineeringSUMMARY

1. The term communication refers to the sending, processing and reception ofinformation by electrical means.

2. Modulation is the process of combining the low-frequency signal with a veryhigh-frequency radio wave called carrier wave (CW).

3. The process of extracting the low-frequency modulating signal from themodulated wave is known as demodulation or detection.

4. Television means seeing at a distance.

5. The waves of high frequencies usually above 300 MHz are termed asmicrowaves.

6. A communication satellite is basically a microwave-link repeater.

GLOSSARY

• Transmitter: It processes the information and makes it fit for transmission.• Amplitude modulation (AM): The process by which the amplitude of a

carrier wave is varied in accordance with the modulating signal.• Frequency modulation (FM): The process by which the frequency of a carrier

wave is varied in accordance with the modulating signal.• Radio transmitter: It is a device that transmits information by means of

radio waves.• Waveguides: It is a hollow metal tube used to propagate microwave energy

in the form of electric and magnetic fields.

REVIEW QUESTIONS

1. What do you mean by the term ‘Communication’ ?

2. Explain briefly the following :

(i) Analog signals ; (ii) Digital signals.

3. Define the term ‘Modulation’ ?

4. What is the need of modulation ?

5. What are the various methods of modulation ?


(i) Amplitude modulation (AM)

(ii) Frequency modulation (FM).

7. What are the limitations of amplitude modulation ?

8. What is frequency modulation ? What are its advantages and limitations ?

9. What is ‘Pulse modulation’ ? Explain.

10. Name the various types of digital modulation techniques.

NOTES



11. What is a radio transmitter ?

12. Give the block diagram of a basic radio transmission system.


(i) AM transmitters

(ii) FM transmitters.

14. What is a radio receiver ? What are its basic functions ?

15. Explain briefly the following receivers :

(i) Tuned Radio Frequency (TRF) receiver.

(ii) Superheterodyne receiver.

16. Write a short note on ‘Television system’ ?

17. What are ‘Microwaves’ ?

18. What are applications of microwaves ?

19. What are ‘waveguides’ ? What are its advantages and applications ?

20. What is a communication satellite ?

21. Explain briefly the following communication satellites :

(i) Passive satellites ; (ii) Active satellites.

22. What is a ‘Synchronous satellite’ ? Explain.

23. Give the general structure of a satellite communication system.

24. What are the advantages and limitations of fiber optic system ?

25. Explain briefly ‘optical fiber communication system’, with the help of a blockdiagram.

26. Write a short note on optical fiber construction.

FURTHER READINGS

• Mehta V.K., “Principles of Electronics”, S. Chand & Company Ltd. (1994).• R.K. Rajput, “Basic Electrical and Electronics Engineering”, Laxmi

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