Chapter 1 Notes

Basic electronics Module 1

Vinay H S,Dept.of ECE, CEC. 1

MODULE -1

ANALOG ELECTRONICS

.................

Semiconductor diode & Applications : p-n junction,

Characteristics & parameters. Diode approximations, DC load line

analysis, Half wave rectifier. Two diode full wave rectifier, Bridge

rectifier, Capacitor filter circuits. Zener diode voltage regulators

(with & without load).Series &shunt diode clipping circuits,

Clamping circuits, Numerical examples as applicable.

Bipolar Junction Transistors : BJT operation, BJT voltages &

currents, BJT amplification, common base, common emitter,

common collector characteristics, Numerical examples as

applicable.

Text book :

David A Bell, Electronic Devices and Circuits: Oxford University Press,

5th Edition , 2008.

Chapter 2: 2.1,2.2,2.3&2.4.

Chapter 3: 3.1,3.2,3.3,3.7,3.8,3.9 & 3.10

Chapter 4 : 4.1,4.2,4.3,4.5,4.6 &4.7

..

PN JUNCTION DIODE:

When a p-type semiconductor material is suitably joined to

n-type semiconductor, the contact surface is called a p-n

junction. The p-n junction is also called as semiconductor

diode

The left side material is a p-type semiconductor having ve

acceptor ions and +vely charged holes. The right side

material is n-type semiconductor having +ve donor ions and

free electrons

Suppose the two pieces are suitably treated to form pn

junction, then there is a tendency for the free electrons from

n-type to diffuse over to the p-side and holes from p-type to

the n-side . This process is called diffusion.

Note : p type region: Holes are majority charge carrier

n type region: Electrons are majority charge carriers



Diode circuit symbol:

Current flows in the arrowhead direction when the diode is forward-

biased: positive (+) on the anode and negative (-) on the cathode.

Biasing: Connecting a p-n junction to an external d.c. voltage

source is called biasing.

Types of biasing:

1.Forward biasing

2. Reverse biasing

1. Forward biasing

When external voltage applied to the junction is in such a

direction that it cancels the potential barrier, thus permitting

current flow is called forward biasing.

To apply forward bias, connect +ve terminal of the battery

to p-type and ve terminal to n-type as shown in fig. below.

2. Reverse biasing

When the external voltage applied to the junction is in such

a direction the potential barrier is increased it is called

reverse biasing To apply reverse bias, connect ve terminal

of the battery to p-type and +ve terminal to n-type as shown

in figure below



CHARACTERISTICS AND PARAMETERS

Ge & Si VI Characteristics :

.

There are two operating regions and three possible biasing conditions for the standard Junction Diode and these are:

1. Zero Bias No external voltage potential is applied to the PN junction diode.

2. Reverse Bias The voltage potential is connected negative, (-ve) to the P-type material and positive, (+ve) to

the N-type material across the diode which has the effect of

Increasing the PN junction diodes width.

3. Forward Bias The voltage potential is connected positive, (+ve) to the P-type material and negative, (-ve) to

the N-type material across the diode which has the effect of

Decreasing the PN junction diodess width.

Diode Parameters :

The diode parameters of greatest interest are

VF forward voltage drop

IR reverse saturation current

VBR reverse breakdown voltage

rd dynamic resistance

IF(max) maximum forward current

Note :

Determination of the dynamic resistance (rd)of a diode from

the forward characteristic.

The dynamic resistance, also known as the incremental

resistance or acresistance, is the reciprocal of the slope of

the forward characteristics beyond the knee.



Determination of diode forward and reverse resistance.

Example :

Calculate the forward and reverse resistances offered by a silicon

diode with the characteristics, at IF= 100 mA and at VR= 50 V.

Determine the dynamic resistance at a forward current of 70 rnA for

the Silicondiode characteristics estimate the diode dynamic

resistance (IF & VF Refer characteristics graph ).

Solution

Diode approximation:

Ideal diode characteristics

We know that a diode is one way device, offering low resistance

when forward biased and a high resistance when reverse biased. On

the other hand an ideal diode (a perfect diode) would, zero forward

drop and infinite reverse resistance and thus behave electrically

open circuit. Figure below shows the characteristics of ideal diode.



Although an ideal diode does not exist, some situations demand

such assumptions where diodes can be assumed to be near ideal

devices. In situations, for example, when supply voltages much

larger than the diode forward drop VF is used then the diode

forward can be ignored without introducing any serious error.

Also, the diode reverse current is normally so much smaller than the

forward current that the reverse current can be ignored. These

assumptions lead to the near-ideal, or approximate characteristics

for Si and Ge diodes as shown in figure (b) and (c) .

Piecewise Linear Characteristic

When the forward characteristic of a diode is not available. A

straight-line approximation, called the piecewise linear

characteristic, may be employed. To construct the piecewise linear

characteristic, VF is first. marked on the horizontal axis, as shown in

Fig. Then, starting at VF, a straight line is drawn with a slope equal

to the diode dynamic resistance. Ex. demonstrates the process.

Example:

Construct the piecewise linear characteristic for a silicon diode

which has a 0.25 Ohm dynamic resistance and a 200 mA maximum

forward current.



Example:

Calculate IF for the diode circuit in Fig. a assuming that the diode

has VF= 0.7V and rd= 0.Then recalculate the current taking rd=

0.25

.

Diode equation :

General characteristics of a semiconductor diode can be

defined by the following equation, referred to as shockleys

equation, for the forward and reverse bias regions:

1T

FnV

V

SF eII (A)

Where

IS reverse saturation current

VF applied forward bias voltage across the diode

IF diode forward current

Ideality factor (1 for Ge& 2 for Si diode)

VT thermal voltage or voltage equivalent

VT = kT (V)

K =Boltzmans constant = 8.62*10-5eV/oK

T Absolute temperature in Kelvins = 273 + the temperature in oC



DC Load line : It is a graphical analysis of a diode circuit, giving precise levels of

diode current and voltage. It is a straight line that illustrates all dc

conditions that could exists within the diode circuit.

Figure : a. Diode circuit and fig.b. Plotting the dc load line on the

diode characteristics

Explanation of a DC load line:

Consider the diode circuit shown in figure below.

Applying the KVL we get,

E= IF R1+ VF (1)

When IF=0, in eqn 1 becomes E= VF

When VF =0 in eqn.1 becomes V= IFR1 or IF =V/R

Plotting these two conditions as shown in fig, that is identifying

point F equal to V/R and point E equal to VF and drawing line EF

which represents the dc load line and represents all dc conditions

that could exist within the circuit.

The Q point: It is the point of intersection of the diode forward

characteristics with the load line

The dc load line is explained in the figure above. There is only one

point on the dc load line where the diode voltage and current are

compatible with the circuit conditions.

DC load line analysis

The DC load line figure shows graphical representations of dc load

line drawn on the diode forward characteristics. This is a straight line

that illustrates all dc conditions that could exist within the circuit.

The analysis can best be made by taking a practical example.

Example : Draw the dc load line for the circuit in Fig. (a). The

diode forward characteristic is given in Fig. (b).



Example 2.

Using the device characteristics in Fig. b, determine the required

load resistance for the circuit in Fig. a to give IF= 30 mA.



3) Determine a new supply voltage for the circuit in Fig.a to give a

50 mAdiode forward current when Rl = 100 .

Point A may now be plotted (on Fig.) at IF= 0 and E = 6.1 V,and the

new dc load line may be drawn through points A and Q.

Fig .Determination of the required supply voltage for a diode-

resistor circuit witha given resistor and a specified load current.



RECTIFIER

A rectifier is a device which converts a.c. voltage to pulsating d.c.

voltage, using one or more p-n junction diodes.

Types of rectifiers

Half wave rectifier

Full wave rectifier (Center tap & Bridge wave)

Half Wave Rectifier :

The half wave rectifier conducts only positive or negative half

cycles of input a.c.supply.

This rectifier circuit consists of resistive load, rectifying element,

i.e. p-n junction diode, and the source of a.c. voltage, all connected

in series. The circuit diagram is shown in the fig

To obtain the desired d.c. voltage across the load, the a.c. voltage is

applied to rectifier circuit using suitable step-up or step-down

transformer, mostly a step-down one, with necessary turns ratio.

The input voltage to the half-wave rectifier circuit shown in the Fig.

is a sinusoidal a.c. voltage, having a frequency which is the supply

frequency, 50 Hz given by,

The transformer decides the peak value of the secondary voltage. If

Nlare the primary number of turns and N2 are the secondary number

of turns and Epm is the peak value of the primary voltage then,

Where Esm=Peak value of the secondary a.c. voltage.

Operation of the Circuit: During the positive half cycle of input

a.c voltage, terminal (A) becomes positive with respect to terminal

(B). The diode is forward biased and the current flows in the circuit

in the clockwise direction, as shown in the Fig. (a). This current is

also flowing through the load resistance RL hence denoted as iL

(load current). During negative half cycle when terminal (A) is

negative with respect to terminal (B), diode becomes reverse biased.

Hence no current flows in the circuit as shown in the Fig. (b).

Thus the circuit current, which is also the load current, is in the

form of half sinusoidal pulses.

The load voltage, being the product of load current and load

resistance, will also be in the form of half sinusoidal pulses. The

different waveforms are illustrated in Fig.



The peak value of the load current is given by,

If Rs is not given it should be neglected while calculating Im'

Fig. Load current and load voltage waveforms for half wave

rectifier

The average or d.c. value of the load current (Idc):

It is obtained by integration.

The average d.c. load voltage (Edc) :

It is the product of average D.C. load current & the load resistance

RL

The winding resistance Rs and forward diode resistance Rf are

practically very small compared to RL hence neglecting them,

Note: When Rfand Rs are finite, calculate Im, then IDC and from that

calculate Edc as Idc RL



The R.M.S. value of the load current:

The d.c. power output is:

Rectifier efficiency:

The rectifier efficiency is defined as the ratio of output d.c. power

to input a.c. power.

Ripple Factor

It is seen that the output of half wave rectifier is not pure d.c. but a

pulsating d.c. The output contains pulsating components called

ripples. Ideally there should not be any ripples in the rectifier

output.

The measure of ripples present in the output is with the help

of a factor called ripple factor denoted by . It tells how

smooth is the output.

Smaller the ripple factor closer is the output to a pure d.c.

Mathematically ripple factor is defined as the ratio of

RM.S. value of the a.c. component in the output to the

average or d.c. component present in the output.



This is the general expression for ripple factor and can be used for

any rectifier circuit.

This indicates that the ripple contents in the output are 1.21 times

the d.c. component i.e. 121.1 % of d.c. component.

. The ripple factor for half wave is very high which indicates that

the half wave circuit is a poor converter of a.c. to d.c.

Peak Inverse Voltage (PIV)

The Peak Inverse Voltage is the peak voltage across the

diode in the reverse direction i.e. when the diode is reverse

biased. This is called PIV rating of a diode.

In half wave rectifier, the load current is ideally zero when

the diode is reverse biased and hence the maximum value of

the voltage that can exist across the diode is nothing but Esm'

Advantages and Disadvantages

The advantages of half wave rectifier are

1. Only one diode is sufficient.

2. The circuit is easy to design.

3. No centre tap on the transformer is necessary.

The disadvantages of half wave rectifier are

1. The ripple factor of half wave rectifier circuit is 1.21, which is

quite high.

2. The maximum theoretical rectification efficiency is found to be

40% which is very low.

3. The d.c. current is flowing through the secondary winding of the

transformer which may cause dc saturation of the core of the

transformer. To minimize the saturation, transformer size have to be

increased accordingly. This increases the cost.

4. The TUF(Transformer Utilization factor) is very low showing

that the transformer is not fully utilized.



Full wave rectifier: The full wave rectifier conducts during both positive and

negative half cycles of input a.c.supply.

In order to rectify both the half cycles of a.c. input, two diodes are used in this circuit.

The diodes feed a common load RL with the help of centre tap transformer

The a.c. voltage is applied through a suitable power transformer with proper turns ratio.

The full wave rectifier circuit is shown in the Fig

Operation of the Circuit:

Consider the positive half cycle of ac input voltage in which

terminal (A) is positive and terminal (B) negative due to center tap

transformer.

The diode D1 will be forward biased and hence will conduct; while

diode D2will be reverse biased and will act as an open circuit

and will not conduct. The diode Dl supplies the load current, i.e. iL

= id1. In the next half cycle of ac voltage, polarity reversesand

terminal (A) becomesnegative and (B) positive. The diode D2

conducts, being forward biased, while D1 does not, being reverse

biased. The diode D2supplies the load current, i.e. iL = id2 .

The load current flows in both the half cycles of ac voltage and in

the same direction through the load resistance. Hence we get

rectified output across the load.

The load current is sum of individual diode currents flowing in

corresponding half cycles. It is also noted that the two diodes do not

conduct simultaneously but in alternate half cycles.

The individual diode currents and the load current are shown in the

Fig.



The output load current is still pulsating d.c. and not pure d.c.

Average D C Current (Idc) :



Peak Inverse Voltage (PIV) :

Note that Esm = maximum value of a.c. voltage across half the

secondary of transformer.

Advantages and Disadvantages of Full Wave Rectifier

The advantages of full wave rectifier are

1. The d.c. load voltage and current are more than half wave.

2. No d.c. current through transformer windings hence no

possibility of saturation.

3. T.U.F. is better as transformer losses are less. 4. The efficiency is

higher.

5. The large d.c. power output. 6. The ripple factor is less.



The disadvantages of full wave rectifier are

1. The PIV rating of diode is higher.

2. Higher PIV diodes are larger in size and costlier.

3. The cost of centre tap transformer is higher



Bridge Rectifier: The basic bridge rectifier circuit is shown in Fig.

The bridge rectifier circuit is essentially a full-wave rectifier circuit,

using four diodes, forming the four arms of an electrical bridge.

To one diagonal of the bridge, the ac voltage is applied through a

transformer ifnecessary, and the rectified dc voltage is taken from

the other diagonal of the bridge.

The main advantage of this circuit is that it does not require a center

tap on the secondary winding of the transformer.

Operation of the Circuit: Consider the positive half of ac input

voltage. The point A of secondary becomes positive. The diodes Dl

and D2 will be forward biased, while D3 and D4 reverse biased.

The two diodes Dl and D2 conduct in series with the load and the

current flows as shown in Fig.

In the next half cycle, when the polarity of ac voltage reverses

hence point B becomes positive diodes D3 and D4 are forward

biased, while Dl and D2 reverse biased. Now the diodes D3 and D4

conduct is series with the load and the current flows as shown in

Fig.

It is seen that in both cycles of ac, the load current is flowing in the

same direction hence, we get a full-wave rectified output.

Advantages

1) The current in both the primary and secondary of the power

transformer flows for the entire cycle and hence for a given power

output, power transformer of a small size and less cost may be used.

2) No center tap is required in the transformer secondary. ~

3) The currents in the secondary of the transformer are in opposite

directions in two half cycles. Hence net d.c. component flowing is

zero which reduces the losses and danger of saturation

4) As two diodes conduct in series in each half cycle, inverse

voltage appearing across diodes get shared. Hence the circuit can be

used for high voltage applications.

5) The transformer gets utilized effectively.

Peak inverse voltage(PIV) :

PIV=Esm



The waveforms of load current and voltage are shown in the Fig.

NOTE :Bridge full wave rectifier & Center tap full wave rectifiers

Efficiency, Ripple factor, Idc,Edc,Pac,Pdc,Im, equations & Derivations

are same Except PIV.



Comparison of HW, FW & BW Rectifiers:



Filters:

We know that the output of the rectifier is pulsating d.c.ie the

output obtained by the rectifier is not pure d.c. but it contains some

ac components along with the dc o/p. These ac components are

called as Ripples, which are undesirable or unwanted. To minimize

the ripples in the rectifier output filter circuits are used. These

circuits are normally connected between the rectifier and load as

shown below

Filter is a circuit which converts pulsating dc output from a rectifier

to a steady dc output In other words, filters are used to reduce the

amplitudes of the unwanted ac components in the rectifier

Note: A capacitor passes ac signal readily but blocks dc.

Capacitor filter Circuits :

To convert to direct voltage (dc voltage), a smoothing circuit or

filter must be employed. Figure a shows a half-wave rectifier circuit

with a single capacitor filter (C1) and a load resistor (RL), and Fig.b

shows the output waveform. The capacitor, termed a reservoir

capacitor, is charged almost to the peak level of the circuit input

voltage when the diode is forward-biased. This occurs at Vpi, as

illustrated in Fig. c, giving a peak capacitor voltage:

Figure A reservoir capacitor smooths the output from a rectifier circuit by charging to the peak output voltage and retaining most of

its charge between peaks.

Ripple Amplitude and Capacitance

The ripple amplitude can be calculated from the capacitor value, the

load current, and the capacitor discharge time. Consider the circuit

output voltage waveform illustrated in Fig.a. The waveform

quantities are as follows:



Figure a. The capacitance value for a reservoir capacitor can be

calculated from the load current, ripple voltage, and input

frequency. Figure b shows that, because the input wave is

sinusoidal.

Where

f is the frequency of the ac input waveform.

The input waveform goes through a 3600 phase angle during time T,

which gives the time per degree as

Approximation calculations



Zener Diode:

Zener Diode is a Two terminal semiconductor device

A conventional solid-state diode allows significant current if it is reverse-biased above its reverse breakdown voltage Circuit symbol



ZENER DIODE VOLTAGE REGULATORS

Regulator Circuit with No Load:

The circuit in Fig.is usually employed as a voltage reference source

that supplies only a very low current (much lower than Iz) to the

output. Resistor Rl in Fig. limits the Zener diode current to the

desired level. Izis calculated as follows:

The Zener current may be just greater than the diode knee current

(IZK).However, for the most stable reference voltage, Iz should be

selected as IZT(the specified test current). Example demonstrates the

circuit design procedure.

Example

A 9.1 V reference source is to use a series-connected Zener diode

and resistor connected to a 30 V supply (see Fig.). Select suitable

components, and calculate the circuit current when the supply

voltage drops to 27 V.(if IZT = 20 mA.)

Solution:



Loaded Regulator :

When a Zener diode regulator has to supply a load current (IL), as

shown in Fig.the total supply current (flowing through resistor R1)

is the sum of ILandIz.

Fig: Zener diode voltage regulator circuit supplying a load current

(IL).The diode must be able to pass a maximum current of (IL+IZ).

In some cases, the load current in the type of circuit shown in Fig.

may be reduced to zero. Because the voltage drop across Rl remains

constant, the supply current remains constant: IR1=IZ+IL

Example

Design a 6 V dc reference source to operate from a 16V supply (see

Fig.)The circuit is to use a low-power Zener diode and is to produce

the maximum possible load current. Calculate the maximum load

current that can be taken from the circuit.(Vz= 6.2 V and PD = 400

mW.)

Solution

IZM =PD/VZ=400mW/6.2 v

=64.5mW

IL(max) +IZ(max) =IZ=64.5mA



CLIPPING CIRCUITS

The function of a clipper (or limiter) is to clip off an unwanted

portion of a waveform.

Types of clippers

1 Series clipper

2 Shunt (parallel) clipper

SERIES CLIPPING CIRCUITS (positive &negative )

A half-wave rectifier can be described as a clipper because it passes

only the positive (or negative) half-cycle of an alternating

waveform and clips off the other half-cycle. In fact, a diode series

clipper is simply a half-wave rectifier circuit.

Figure a shows a negative series clipper circuit with a square wave

input symmetrical above and below ground level. While the input is

positive, D1is forward-biased and the positive half-cycle is passed

to the output.

During the negative half-cycle of the input, the diode is reverse-

biased.

Consequently, the output remains at zero and the negative half-

cycle is effectively clipped off. The' zero level output from a series

clipper circuit is not exactly zero. The reverse saturation current (IR)

of the diode produces a voltage

drop across resistor Rl:

This voltage drop is almost always so small that it can be ignored.



SHUNT(parallel) CLIPPING CIRCUITS

Here the diode is connected in shunt (or parallel) with the output

terminals. When the input is negative, the diode is reverse-biased

and there is only a small voltage drop across R1' due to load current

IL. This means that the circuit output voltage (V0) is approximately

equal to the negative input peak ( -E). When the input is + E, D1 is

forward-biased and the output voltage equals the diode voltage drop

(+ Vp). Thus, the positive half of the waveform is clipped off. As

illustrated, the upper and lower levels of the output of a positive

shunt clipper are approximately + VF and - E.

A negative shunt clipper circuit is exactly the same as a positive

shunt clipper with the diode polarity reversed (see Fig.b). The

negative half-cycle of the waveform is clipped off.

The load current on a shunt clipper produces a voltage drop (ILR1)

across the resistor; this might be insignificant where the load

current is very low:

As in the case of series clipping circuits, shunt clippers may be used

with square, sinusoidal, or other input waveforms.



Note:

Zener Diode Shunt Clipper

A Zener diode shunt clipper produces the same kind of result as a

biased shunt clipper without the need for bias voltages. The clipper



circuit in Fig. has two back-to-back series-connected Zener

diodes. When the input voltage is positive and has sufficient

amplitude, D1 is forward-biased and D2 is biased into reverse

breakdown. At this time, the output voltage is limited to (VF + Vz2).

A negative input voltage produces a maximum negative output of

-(VF+ VZ1) With equal-voltage Zener diodes, the maximum output

voltage is

VO= (VF+VZ)

The resistor voltage is (E V0), and the resistor current is (IL + Iz). A minimum level of Iz (greater than the device knee current) is

selected, and the resistor value is calculated as

Example

A Zener diode shunt clipper, as in Fig, is to be connected between a

20V square wave signal and a circuit that cannot accept inputs

greater than 5V. Select suitable Zener diodes, and determine

R1.The clipper output current is to be 1mA.(VZ=4.3V IZ(min)=5mA)



CLAMPING CIRCUITS

A clamping circuit, also known as a dc restorer, changes the dc

voltage level of a waveform but does not affect its shape.

.

Negative and Positive Voltage Clamping Circuits

Consider the clamping circuit shown in Fig. When the square wave

input is positive, diode Dl is forward biased and the output voltage

equals the diode forward voltage drop VF.

During the positive half-cycle of the input, the voltage on the right

side of the capacitor is +VF, while that on the left side is + E. Thus,

C1is charged with the polarity shown to a voltage:

The peak-to-peak output voltage (V0(pp))is the difference between

the positive output peak (VF) and negative output peak -(2E - VF):

A positive voltage clamping circuit passes the complete input

waveform to the output but clamps the negative peak of the output

close to ground level Shown in below fig

Output Slope

The output voltage age from a clamping circuit has a slope

(Vc)produced by capacitor discharge. The capacitance value is determined from the acceptable slope.



Example

The diode clamping circuit in Fig. has a 10V,1 kHz square wave

input. Calculate the tilt on the output waveform.

TRANSISTORS

BIPOLAR JUNCTION TRANSISTORS:

The transistor is a three terminal semiconductor device.

The term bipolar reflects the fact that holes and electrons

participate in the conduction process.

It is constructed with 3 doped semiconductor regions, either

two n and one p-type layers of material or two p and one n-

type layers of material.

Transistors are classified into two types

1. npn transistor

npn transistor is obtained when a p-type layer of

silicon is sandwiched between two n-type silicon

materials.

2.pnp transistor

pnp transistor is obtained when a n-type layer of

silicon is sandwiched between two p-type silicon

material.

Below fig shows the schematic representations of a transistor which

is equivalent of two diodes connected back to back



The terminals have been indicated by the capital letters E for

emitter, C for collector, and B for base.

Two pn junctions exist in each transistor: The pn junction

joining the base region and the emitter region is called the

base-emitter junction. The pn junction joining the base

region and the collector region is called the base-collector

junction, as indicated in above Figure .

BJT graphic(Circuit) symbols

The symbols used for bipolar junction transistors in circuit

diagrams are given in Fig below.

The symbol shown at fig (b), in which the emitter arrow

head is directed towards the base (p to n), is used for a pnp

transistor and in the symbol shown at fig (a), in which the

emitter arrow head is directed away from the base (p to n), is

used for an npn transistor.

The arrow head in the emitter terminal always indicate the

conventional current direction, not the electron flow

direction.

npn transistor Operations :

For normal BJT operation, the base-emitter junction is forward

biased and the collector-base junction is reverse-biased. Charge

carriers (electrons in an npn device) are emitted into the base region

from the emitter region, and most of them are collected by the

collector region. Very few charge carriers flow through the base

terminal.

pnp transistors Operations :

In a pnp BJT, the forward bias on the base-emitter junction causes

charge carriers (holes) to be emitted into the base region. Most of

the charge carriers are collected by being drawn across the reverse-

biased collector-base junction.



BJT voltages and currents:

Terminal voltages for npn transistor:

The terminal voltage polarities are shown in fig (a) and the

connection of voltage sources is shown in fig (b).

The arrow head indicates the conventional current direction

and arrow head points from base to emitter.

Terminal voltages for pnp transistor:

The terminal voltage polarities are shown in fig (a) and the

connection of voltage sources is shown in fig (b).

The arrow head indicates the conventional current direction

and arrow head points from emitter to base.



Note: Normally in the transistor, the base-emitter (BE)

junction should be forward biased and the collector-base junction

(CB) should be reverse biased. But in the case of switching

transistors collector-base junction will become forward biased by

0.5V and base-emitter junction will become reverse biased.

Transistor currents:

The current flowing into the emitter terminal is called as

emitter current. In the case of pnp transistor this current is

due to holes.

The emitter current is denoted as IE, base current is denoted

as IB, and the collector current is denoted as IC as shown in

fig .

Both IC and IB flow out of the transistor while IE flows into

the transistor. Therefore by KCL

BCE III (1)

This is shown in fig

Almost all of IE diffuses to the collector (around 96% to

99.5%) and only a small portion flows out of the base terminal.

Hence we can write IC as a percentage of IE,

EdcC II (2)

Where dc is the emitter to collector current gain, or the ratio

of collector current to emitter current i.e.,

E

C

dcI

I (3)



Relationship between &

Example 1:

Calculate IC and IE for a transistor that has dc = 0.98 and IB =

100A. Determine the value of dc for the transistor.

Solution:

mAAI

Idc

Bdc

C 9.498.01

10098.0

)1(

mAmAI

Idc

C

E 598.0

9.4

4998.01

98.0

)1(

dc

dc

dc

Example 2:

Calculate dc and dc for the transistor Q1 in fig, if IC is measured as

1mA and IB is 25A. Determine the new base current to give IC =

5mA.



Solution: 4025

1

A

mA

I

I

B

C

dc

976.0025.1

1

mA

mA

I

I

E

C

dc

AmAI

Idc

C

B

12540

5

BJT AMPLIFICATION:

Current amplification:

Let us assume the common emitter configuration (discussed

later).

IB is the input current and IC is the output current.

From the transistor operation we found that IB is very small

and IC is very large i.e., BdcC II (output current is dc

times the input current).

So that a small change in base current (IB) produces a large

change in collector current (IC). From this we say that a

small input current is amplified to a large output current

(shown in fig ).

mAAmAIII BCE 025.1251



In terms of current level changes base to collector current

gain can be written as

Change in IC is denoted as CI =Ic (changing quantities are

called as ac quantities)

Change in IB is denoted as bB II

Now

b

cac

I

I

Alternate symbol for ac is hfe.

Voltage amplification:

The transistor in fig has dc =50, forward bias voltage

VB=0.7V, ac signal source in series with VB is vi = 20mV,

VCC =20V.

Q1 has the IBVs VBE characteristic shown in fig 3-10.

The corresponding IB for the 0.7V level is 20A.

mAAII BdcC 12050

DC collector voltage is calculated by using KVL,

VKmAVRIVV CCCC 8)121(20)( 1

If the ac input voltage vi is zero, then CV will remains at 8V

itself. But vi causes a base voltage variation of 20mV, so

B

C

acI

I



that the base current changes by 5A (from fig 3-10). Now

the change in collector current is

AAII BdcC 250)5(50

Because of this change in collector current, the change in

collector voltage is given by

VAKIRV CC 3250121

From this we can conclude that a 20mV change in input

base voltage produces 3V change in the output collector

voltage; hence called voltage amplification. The parameter

to define voltage amplification is called as voltage gain (Av).

15020

3

mV

V

V

VA

B

C

v

The equation for ac voltage gain

i

vv

vA 0

TRANSISTOR CONFIGURATIONS:

Depending upon which terminal is made as common between the

input and output, there are 3 configurations:

1. Common Base configuration (CB)

2. Common Emitter configuration (CE)

3. Common collector configuration (CC)

COMMON BASE CHARACTERISTICS:

Common base circuit:

Fig shows a npn&pnp transistor with its base terminal

common to both input and output terminals.

Input is applied between emitter and base terminals.

Output is taken from the collector and base terminals.



Input characteristics

It is the curve between input current IE and input voltage

VEB at constant collector base voltage VCB as shown in fig .

After the cut-in voltage IE increases rapidly with small

increase in VEB i.e., the characteristics is similar to that of

the forward biased PN junction diode.

It is observed from the graph that there is a slight increase in

emitter current IE with increase in VCB.

This is due to, the larger collector base voltage cause the

depletion region at the collector base junction to penetrate

deeper in to the base region, thus shortening the distance and

reducing the resistance between the emitter-base and

collector-base depletion regions.

Common base output characteristics

It is the curve between collector current IC and VCB at

constant emitter current IE.

VCB is adjusted in convenient steps and corresponding

values of IC are noted and the characteristic is plotted in Fig

When the emitter base junction is forward biased and the

collector base junction is reversed biased the transistor is in

the active region. Under normal condition the transistor will

be operated in the active region. This is shown in the output

characteristics.



Common emitter circuit:

Fig shows a npn & pnp transistor with its emitter terminal


Input is applied between base and emitter terminals.

Output is taken from the collector and emitter terminals.

breakdown




It is the curve between input current IB and input voltage

VBE at constant collector emitter voltage VCE as shown in fig

The characteristics are similar to the forward biased pn

junction, but the current IB is only a small portion of the

total current IE.

It is observed from the graph that there is a slight reduction

in base current IB for a particular VBE with increase in VCE

levels.

This is because higher VCE will increase the width of the CB

depletion region, so that it will penetrate more into the base

and reducing the distance between the CB and EB depletion

region, increasing collector current and reducing base

current.

Common emitter output characteristics:

It is the curve between collector current IC and VCE at

constant base current IB.

VCE is adjusted in convenient steps and corresponding


When we increase the level of VCE the width of the CB

depletion region will increase, so that it will penetrate more

into the base and reducing the distance between the CB and

EB depletion region, increasing collector current.

This increase in IC will give some slope in the output

characteristics. The slope in the output characteristics is

termed as early effect and the transistor is in the active

region.

breakdown



When we extend the characteristics to the left of the current

axis they will meet at a point on the horizontal axis and the

voltage at that point is known as early voltage.

For a given level of IC the ac resistance at the collector

terminal is given by

rc= VA/IC

In the output characteristics, VCE = VCB+VBE (shown in fig ),

and at the knee of the characteristics VCE = VBE, VCB = 0V

and we will get more current at this point.

Further reduction in VCE will forward bias the CB junction

and IC will reduce to zero when VCE is zero. At this point

both the BE and CB junctions are forward biased and the

device is said to be in saturation region.

When both the junctions are reverse biased input and output

currents are zero and the transistor is said to be in cut-off

region.

Current gain characteristics:

This characteristics is drawn between output current IC and

input current IB with constant VCE.



Common collector circuit:

Fig shows a npn & pnp transistor with its collector terminal


Input is applied between base and collector terminals.

Output is taken from the emitter and collector terminals.


It is the curve between input current IB and input voltage

VBC at constant emitter collector voltage VEC as shown in fig

Apply KVL around the transistor

-VBC + VEC - VEB = 0

VEB = VEC-VBC --------------- (1)

From eqn (1) we can say that, when VBC increases for a

constant VEC, VEB will reduce and hence IB reduces shown

in fig .

When VEC increases there is a slight increase in IB.



Output and current gain characteristics:

It is the curve between emitter current IE and VEC at constant

base current IB.

VCE is adjusted in convenient steps and corresponding


Since IE IC, the characteristic is now between IC and

VEC.This is nothing but CE output characteristics. (Only

difference is sign of VEC).

Applications:

Active region: amplifier

Cut-off region-Saturation region: switch.

Biasing conditions:

Region of

operation

Emitter base

junction

Collector base

junction

Cut-off Reverse biased Reverse biased

Active Forward biased Reverse biased

Saturation Forward biased Forward biased



.END..................

Chapter 1 Notes

Documents

pn junction

ntype material

ptype material

ntype semiconductor

ptype semiconductor

p type region

standard junction diode

forward biasing