EC 2257- ELECTRONIC CIRCUITS II AND SIMULATION LAB

Dr.NNCE ECE / IV SEM EC II & S LAB - LM

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EC 2257- ELECTRONIC CIRCUITS II AND SIMULATION LAB

LABORATORY MANUAL

FOR FOURTH SEMESTER B.E (ECE)

(PRIVATE CIRCULATION ONLY)

ACADEMIC YEAR 2013-2014

ANNA UNIVERSITY,CHENNAI-600025

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

DR.NAVALAR NEDUNCHEZHIYAN COLLEGE OF ENGINEERING,

THOLUDUR 606 303, CUDDALORE DIST.


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UNIVERSITY PRACTICAL EXAMINATION

Allotment of Marks

Internal assessment = 20 marks

Practical examination = 80 marks

----------------------

Total = 100 marks

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INTERNAL ASSESSMENT (20 MARKS)

Staff should maintain the assessment register and the Head of the Department should

monitor it.

Split up of internal marks

Record = 10 marks

Model exam = 5 marks

Attendance = 5 marks

----------------------

Total = 20 marks

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UNIVERSITY EXAMINATION

The exam will be conducted for 100 marks. Then the marks will be calculated to 80

marks

Split up of practical examination marks

Aim and Procedure = 25 marks

Program = 30 marks

Execution = 30 marks

Result = 05 marks

Viva voce = 10 marks

----------------------

Total = 100 marks

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GENERAL INSTRUCTIONS FOR LABORATORY CLASSES

o Enter the Lab with CLOSED FOOTWEAR.

o Boys should “TUCK IN” the shirts.

o Students should wear uniform only.

o LONG HAIR should be protected, let it not be loose especially near ROTATING

MACHINERY.

o Any other machines / equipments should not be operated other than the

prescribed one for that day.

o POWER SUPPLY to your test table should be obtained only through the LAB

TECHNICIAN.

o Do not LEAN and do not be CLOSE to the rotating components.

o TOOLS, APPARATUS and GUAGE sets are to be returned before leaving the

lab.

o HEADINGS and DETAILS should be neatly written

i. Aim of the experiment

ii. Apparatus / Tools / Instruments required

iii. Procedure / Theory / Algorithm / Program

iv. Model Calculations

v. Neat Diagram / Flow charts

vi. Specifications / Designs Details

vii. Tabulations

viii. Graph

ix. Result / discussions .

o Before doing the experiment, the student should get the Circuit / Program

approval by the FACULTY - IN - CHARGE.

o Experiment date should be written in the appropriate place.

o After completing the experiment, the answer to the viva-voce questions should be

neatly written in the workbook.

o Be PATIENT, STEADY, SYSTEMATIC AND REGULAR.


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LIST OF EXPERIMENTS

DESIGN OF THE FOLLOWING CIRCUITS

1. Series and shunt feedback amplifiers:

Frequency response, input and output impedance calculation

2. RC phase shift oscillator, Wien bridge oscillator

3. Hartley oscillator, Colpitts oscillator

4. Tuned class C amplifier

5. Integrators, Differentiators, Clippers and Clampers

6. Astable, Monostable and Bistable multivibrators

SIMULATION USING PSPICE

1. Differential amplifier

2. Active filters : Butterworth 2 nd order LPF, HPF (Magnitude and Phase response)

3. Astable, Monostable and Bistable multivibrator - Transistor bias

4. D/A and A/D converters (Successive approximation)

5. Analog multiplier

6. CMOS inverter, NAND and NOR

Total: 45


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CONTENTS

S.NO LIST OF EXPERIMENTS

Pa

ge

No

01 RC PHASE SHIFT OSCILLATOR 6

02 WEIN BRIDGE OSCILLATOR 9

03 COLPITTS OSCILLATOR 12

04 HARTLEY OSCILLATOR 15

05 ASTABLE MULTIVIBRATOR 18

06 MONOSTABLE MULTIVIBRATOR 21

07 BISTABLEMULTIVIBRATOR 24

08 CLASS C SINGLE TUNED AMPLIFIER 27

09 CURRENT SERIES FEEDBACK AMPLIFIER 30

10 VOLTAGE SERIES FEEDBACK AMPLIFIER 33

11 SECOND ORDER LOW PASS BUTTERWORTH

FILTER 36

12 DIFFERENTIATOR SINE WAVE INPUT 39

13 DIFFERENTIATOR SQUARE WAVE INPUT 42

14 ASTABLE MULTIVIBRATOR –

SYMMETRICAL 45

15 ASTABLE MULTIVIBRATOR –

ASYMMETRICAL 48

16 MONOSTABLE MULTIVIBRATOR 51

17 BISTABLE MULTIVIBRATOR 54

18 CMOS INVERTER 57

19 CMOS NOR 60

20 CMOS NAND 63

21 DIGITAL TO ANALOG CONVERTER 66

BEYOND THE SYLLABUS

22 Inverting, Non-Inverting , Differential Amplifiers 67

23 Differential Amplifier Using BJT

24 Class B Complementary Symmetry Power

Amplifier

25 Cascade Amplifier


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Exercise Number : 1

Title of the Exercise : RC PHASE SHIFT OSCILLATOR

Date of the Exercise :

OBJECTIVE (AIM) OF THE EXPERIMENT

To design a RC phase shift oscillator and to observe its output waveform.

FACILITIES REQUIRED AND PROCEDURE a) FACILITIES REQUIRED TO DO THE EXPERIMENT:

S.NO APPARATUS SPECIFICATION QUANTITY

1. OPAMP IC74 1

2. Resistors 1.2KΩ,

13KΩ,377Ω,12.5KΩ

3,Each1

3. Capacitors 0.1µF 3

4. RPS ±12V 1

5 CRO 1MHz 1

6. Connecting wires - Req.

b) DESIGN PROCEDURE:

Frequency of oscillator F=1//2πRC.

Assume C and find R to prevent loading of the amplifier by RC networkR1≤10R.

c) THEORY: The amplifier stage is self biased with a capacitor by passed source resistor (Rs) and

drain bias resistor (Rd). the expression for voltage gain of the amplifier is given by Av =gm. rl.

The feedback network consists of three identical RC sections. Each section produces a phase

shift of 60. Therefore the net phase shift of the feedback network is 180 degree. Since the

amplifier stage also introduces a phase shift of 180°, therefore total phase shift is 360° or 0°.

For the variable frequency oscillators, the three capacitors are ganged and varied

simultaneously. When the circuit is energized by switching on the supply, the circuit starts

oscillating. The oscillations may start due to the minor variation in dc supply or inherent

noise.

d) PROCEDURE:

Hook up the circuit as shown in the circuit diagram.

Switch on the power supply.

Observe the output waveform in CRO.


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e) CIRCUIT DIAGRAM:

f) MODEL GRAPH:


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g) TABULATION:

Amplitude(Volts) Time(ms) Frequency (KHz)

RESULT:

Thus the RC phase shift oscillator was designed and its output waveform was verified.

VIVA QUESTIONS:

1. What is amplifier?

Amplifier is a device which is used to amplification purpose.

2. What is amplification?

A low strength signal is converted into strengthen signal ie) boost up process

3. List the disadvantages of Rc phase shift Oscillator. (or)

What are the merits of Rc phase shift Oscillator.

i. It is ideal for frequency adjustment over a wide range.

ii. It requires a high β transistor to overcome losses in the network.

4. What is the difference between amplifier and oscillator/

Amplifier is working in the negative feedback while oscillator working in the positive

feedback.


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Exercise Number : 2

Title of the Exercise : WEIN BRIDGE OSCILLATOR



To design a wein bridge oscillator and to observe its output waveform.



1. OPAMP IC741 1

2. Resistors 1.5KΩ,

1KΩ,500Ω,1.5KΩ

Each1

3. Capacitors 0.1µF 1

4. RPS ±12V 1

5 CRO 1MHz 1


b) THEORY:

Wein Bridge Oscillator uses a non inverting amplifier and hence does not produce any phase

shift during amplifier stage as total phase shift req. is 0. In wein bridge oscillator type no

phase shift is necessary through Feedback. Thus the total phase shift around a loop is 0.

c) DESIGN PROCEDURE:

Select approximate transistor and note down its specification such as

Vce,Vcc(max),hoe(min) and hfe(max) and Vbe(sat).

Vcc=VCEQ+ICQ(RC+RE)

Assuming appropriate stability factor and hence I2 flowing through the biasing resistor and

differentiator. Determine R1 and R2.

Using the condition for sustained oscillation R3>2R4, compute C for designed frequency for

the frequency of oscillator.

F=1/2πRC

d) PROCEDURE:





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e) CIRCUIT DIAGRAM:

f) MODEL GRAPH:


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g) TABULATION:


h) RESULT:

Thus the Wein bridge oscillator was designed and its output waveform was verified.

VIVA QUESTIONS:

1. What is an Oscillator?

An Oscillator is a Circuit, which generates an alternating voltage of any desired

frequency. It can generate an a.c output signal without requiring any externally applied input

signal.

2. What is a beat frequency oscillator?

Beat frequency Oscillator (BFO) is an Oscillator in which a deserved signal frequency

such as the beat frequency produced by combining the different signal frequencies such as on

different radio frequencies.

3. What is sustained Oscillation?

The electrical oscillations in which amplitude does not change with time are called as

sustained oscillations. It is also called as Undamped Oscillation.

4. What is meant by resonant Circuit Oscillators?

LC Oscillators are known as resonant circuit oscillator because the frequency

of operation of LC Oscillator is nothing but a resonant frequency of tank circuit or LC tank

circuit produces sustained Oscillation at the resonant circuit oscillator.


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Exercise Number : 3

Title of the Exercise : COLPITTS OSCILLATOR



To design a Colpitts oscillator and to observe its output waveform.



1. Transistor BC 107 1

2. Resistors 11.64 KΩ,

552.2Ω,10.02KΩ1.67kΩ

Each 1

3. Capacitors 53.5nF,80µF, 100mF 2,1,1

4. Inductor 0.78mH 1

5. RPS ±12V 1

6. CRO 1MHz 1



Select a appropriate transistor and note down its specification such as VCE,IC(MAX), hfe(min) and

Vbe(sat).

VCC= VCEQ

R2=S* RE

VCC[R2/( R1+ R2)= VBE+VBE(SAT)

VR1+VR2=VCC

hfe ≥ C1* C2/( C1+ C1)

XCE ≤ RE/10

PROCEDURE:



Slight modification in value of C1 and C2 can be made to get perfect sine wave output.



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c) CIRCUIT DIAGRAM: COLPITTS OSCILLATOR

d) MODEL GRAPH:


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e) TABULATION:


f) RESULT:

Thus the Colpitts oscillator was designed and its output waveform was verified.

VIVA QUESTIONS:

1. What is piezo electric effect?

The piezo electric Crystals exhibit a property that if a mechanical stress is

applied across one face the electric potential is developed across opposite face. The inverse

is also live. This phenomenon is called piezo electric effect.

2. List the disadvantages of crystal Oscillator.

It is suitable for only low power circuits

Large amplitude of vibrations may crack the crystal.

It large in frequency is only possible replacing the crystal with another one by

different frequency.

3. What are parasitic Oscillators?

In a practical amplifier circuit due to stray capacitances and lead inductances,

oscillations result, since the circuit conditions satisfy the Barkhavsen’s criterion. These

Oscillators are called as unwanted or parasitic Oscillations

4. What is damped Oscillation?

The electrical Oscillations in which the amplitude decreases with time are called as

damped Oscillation.


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Exercise Number : 4

Title of the Exercise : HARTLEY OSCILLATOR



To design a Hartley oscillator and to observe its output waveform.



1. Transistor BC 107 1

2. Resistors 2.74 KΩ,

1.76KΩ,10.58KΩ

1,2,1

3. Capacitors 0.1µF, 0.1µF Each 2

4. Inductor 0.1mH,0.33mH Each 1

5. RPS ±12V 1

6. CRO 1MHz 1



Select a appropriate transistor and note down its specification such as VCE,IC(MAX), hfe(max) and

Vbe(sat).

VCC= VCEQ+ ICQ(RC+RE)

R2=S* RE

VCC[R2/( R1+ R2)= VBE+VBE(SAT)

VR1+VR2=VCC

c) PROCEDURE:



Slight modification in value of L1 and L2 can be made to get perfect sinewave output.



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d) CIRCUIT DIAGRAM: HARTLEY OSCILLATOR

e) MODEL GRAPH:


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f) TABULATION:


g) RESULT:

Thus the Hartley oscillator was designed and its output waveform was verified.

VIVA QUESTIONS:

1. What are the types of sinusoidal oscillator? Mention the different types of

sinusoidal oscillator?

RC phase shift Oscillator.

Wein bridge Oscillator.

Hartley Oscillator

Colpitts Oscillator

Crystal Oscillator

2. What is Barkhausan criterion?

The conditions for oscillator to produce oscillation is given by Barkhausan

criterion. They are :

(i). The total phase shift produced by the circuit should be 360o or 0o

(ii).The Magnitude of loop gain must be greater than or equal to 1

i.e. . A׀β1≤ ׀.

3. Name two high frequency Oscillators.

i. Hartley Oscillator

ii. Colpitts Oscillator

iii.Crystal Oscillator

4. What are the essential parts of an Oscillator?

i. Tank circuit (or) Oscillatory circuit.

ii. Amplifier (Transistor amplifier)

iii. Feedback Circuit.


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Exercise Number : 5

Title of the Exercise : ASTABLE MULTIVIBRATOR



To test astable multivibrator and to observe the two quasi-stable states.


S. NO. COMPONENTS RANGE QUANTITY

1 Breadboard - 1

2 Resistors 100k, 2k 2each

3 Capacitors 0.01μF 2

4 Transistor BC107 1

5 CRO 0-30MHz 1

6 RPS 0-30V 1

b) THEORY:

Astable multivibrator consists of two common emitter amplifying stages.When the d.

c. power supply is switched ON (t=0) one of the transistor will start conducting more than the

other due to some imbalance in circuit. Then because of the other due to some positive

feedback, the transistor Q1 will be driven into saturation and Q2 to cut off. Thus at t>0 Q1 is

ON and Q2 is OFF. During t>0, capacitor C1 is charging and VB2 increases. As it increases

above cut-in-voltage, Q2 starts conducting at t>t1. As the transistor Q2 goes into saturation,

VC falls to VCE(sat). Thus at t>t1, Q1 is OFF and Q2 is ON. During t>t1, VB1 rises

exponentially with time constant. At T=t2, VB1 reaches cutin level and a reverse transition

take place. It is used to generate square waveform.

c) PROCEDURE: 1. The components are connected as per the circuit diagram.

2. The DC supply is switched ON.

3. The output voltage and the time period is measured across transistors T1 and T2 at

both the base and the collector.

4. The output wave is plotted in the graph.


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d) CIRCUIT DIAGRAM: ASTABLE MULTIVIBRATOR

e) MODEL GRAPH:


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f) TABULATION:


g) RESULT:

Thus the Astable multivibrator was designed and its output waveform was verified.

VIVA QUESTIONS:

1. What is a linear waveform-shaping circuit? The process by which the shape of a nonsinusoidal signal is changed by passing the

signal through the network consisting of linear elements is called Linear Wave Shaping.

2. What is meant by multivibrator?

Multivibrators are two stage switching circuits in which the output of the first stage is

fed to the input of the second state and vice-versa. The outputs of two stages are

complementary.

3. Define Astable multivibrator. Astable multivibrator is a multivibrator in which neither state is stable. There are two

temporary states. The circuit changes state continuously from one quasi stables state to

another at regular intervals without any triggering. This generates continuous square

waveform without any external signal.


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Exercise Number : 6

Title of the Exercise : MONOSTABLE MULTIVIBRATOR



To test a monostable multivibrator and to observe the two quasi-stable states.



1 Breadboard - 1

2 Resistors 3k, 1k, 280k, 22k,

1.75k

1 each

3 Capacitors 1nF 2


5 Diode 1N4007 1

6 CRO 0-30MHz 1

7 RPS 0-30V 1

b) THEORY:

Monostble multivibrator is also called as one shot or univibrator and can be used to

generate a gating pulse, whose width can be controlled. The monostable multivibrator

provides a single pulse of desired duration in response to an external trigger. The external

trigger causes the circuit to go to the quasi stable state. After a certain interval of time, the

circuit returns to its original state. It consists of two NPN transistors. In this case, when a

pulse is applied to the input circuit, the circuit state is changed abruptly to unstable state for a

determined time after which the circuit returned to its original state automatically. The two

outputs are the complement of each other i.e when one of the output is at VCC level, the

other is at VCE(sat) level. The monostable multivibrators are used for the generation of well

defined pulses, the logic design of pulse delay, variable pulse width, etc. The width or

duration of the pulse obtained at the collector or output of either transistor of the monostable

multivibrator is given by the expression t=0.69 RC.

c) PROCEDURE:

1. The components are connected as per the circuit diagram.

2. The output is measured at the collector terminals of the two transistors.

3. The required output voltages and the total time period is noted.



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d) CIRCUIT DIAGRAM: MONOSTABLE MULTIVIBRATOR

e) TRIGGER GENERATION:


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f) TABULATION:


g) RESULT:

Thus the Monostable multivibrator was designed and its output waveform was

verified.

VIVA QUESTIONS:

1. Define monostable multivibrator. When a trigger pulse is applied to the input circuit, the circuit state is changed

abruptly to unstable state for a predetermined time after which the circuit returned to its

original stable state automatically.

2. Define integrator. Integrator is a circuit that passes low frequencies of the input and attenuates high

frequencies. Integrator implies that the output voltage is an integral of the input voltage.

3. What is the use of commutating capacitors?

The Commutating capacitors can be used to reduce the transition time in a low to

high level and vice versa.

4. Define transition time. The time interval during which the conduction transfer from one transistor to another

transistor is defined as transition time.


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Exercise Number : 7

Title of the Exercise : BISTABLE MULTIVIBRATOR



To test a Bistable multivibrator and to observe the two quasi-stable states.



1 Breadboard - 1

2 Resistors 3k, 1k, 280k, 22k,

1.75k

1 each

3 Capacitors 1nF 2


5 Diode 1N4007 1

6 CRO 0-30MHz 1

7 RPS 0-30V 1

b) THEORY:

Bistble multivibrator is also called as one shot or univibrator and can be used to

generate a gating pulse, whose width can be controlled. The bistable multivibrator provides a

single pulse of desired duration in response to an external trigger. The external trigger causes

the circuit to go to the quasi stable state. After a certain interval of time, the circuit returns to

its original state. It consists of two NPN transistors. In this case, when a pulse is applied to

the input circuit, the circuit state is changed abruptly to unstable state for a determined time

after which the circuit returned to its original state automatically. The two outputs are the

complement of each other i.e when one of the output is at VCC level, the other is at VCE(sat)

level. The bistable multivibrators are used for the generation of well defined pulses, the logic

design of pulse delay, variable pulse width, etc. The width or duration of the pulse obtained at

the collector or output of either transistor of the bistable multivibrator is given by the

expression t=0.69 RC.

c) PROCEDURE:

1. The components are connected as per the circuit diagram.

2. The output is measured at the collector terminals of the two transistors.

3. The required output voltages and the total time period is noted.



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d) CIRCUIT DIAGRAM: BISTABLEMULTIVIBRATOR

e) MODEL GRAPH:

Output Vb1

V

Time (s)

Output Vb2 Vbe(sat)

Output Vc2

Vce(sat)

Vbe(sat)

Vce(sat)

Output Vc1


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f) RESULT:

Thus the Bistable multivibrator was designed and its output waveform was verified.

VIVA QUESTIONS:

1. Define the Bistable multivibrator. Bistable multivibrator signifies a circuit which can exist indefinitely in either of two

stable states and which can be induced to make an abrupt transition from one state to other by

applying an external triggering signal.

2. Define resolving time. It is the minimum time interval between two consecutive trigger pulses and equals to

transition time plus the settling time.

3. What is meant by linear wave shaping circuit? The action of a linear network in producing a waveform at its outputs different from

its output is known as linear wave shaping circuit.

4. What are different types of triggering of bistable multivibrator? Asymmetrical triggering.

Symmetrical triggering.


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Exercise Number : 8

Title of the Exercise : CLASS C SINGLE TUNED AMPLIFIER



To design and find the frequency response of class c single tuned amplifier.



1 Breadboard - 1

2 Resistors 1k,4.7k,10k 1 each

3 Capacitors 1nF,0.01mF,0.1 mF 2


5 CRO 0-30MHz 1

6 RPS 0-30V 1

b) THEORY:

In class C amplifier, the output current flows only for one half of the cycle of the

input signal. The transistor dissipates no power with zero input signal. The average current

drawn by the circuit in class C is smaller than that in class A. Complementary symmetry

amplifier requireds neither an input nor an output transformer. This arrangement uses

transistors having complementary symmetry in the emitter follower configuration. The term

‘complementary’ means that it uses two identical transistors one NPN and the other PNP. The

term ‘symmetry’ means that biasing resistors are equal. This amplifier circuit has a unity gain

because of the emitter follower configuration. Moreover there is no phase inversion of the

output signal. The split supply used in the circuit gives us an advantage that the dc

component of the output voltage can be made zero. Thus the only ac component of the power

is available across the load resistor.

c) PROCEDURE:

1. The components are connected as shown in the circuit diagram.

2. The input voltage is given using function generator.

3. The load resistance is varied using the decade resistance box and output voltage is

measured and readings are tabulated


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d) CIRCUIT DIAGRAM: CLASS C SINGLE TUNED AMPLIFIER

e) MODEL GRAPH:


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f) TABULATION:

Sl.

No.

Input

frequency

(Hz)

Input

Voltage

(mV)

Output

Voltage

(V)

Voltage gain

20log ogain

i

VV db

V

g) RESULT:

Thus the Class C single tuned amplifier was designed and its frequency response was

observed

VIVA QUESTIONS:

1. What is meant by tuned amplifiers?

Tuned amplifiers are amplifiers that are designed to reject a certain range of

frequencies below a lower cut off frequency ωL and above a upper cut off frequency ωH and

allows only a narrow band of frequencies.

2. Classify tuned amplifiers.

1.Single tuned amplifier.

2.Double tuned amplifier.

3.Synchronously tuned amplifier.

4.Stagger tuned amplifier.

3. What are the advantages of double tuned amplifier?

In double tuned amplifiers, the tuning is done both at the primary and

secondary.

The double tuned amplifiers provide a wider bandwidth, flatter pass band and

a greater selectivity.

4. What is the other name for tuned amplifier?

Tuned amplifiers used for amplifying narrow band of frequencies hence it is also

known as “ narrow band amplifier” or “Band pass amplifier”.


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Exercise Number : 9

Title of the Exercise : CURRENT SERIES FEEDBACK AMPLIFIER



To design and find the frequency response of current series feedback amplifier.



1 Breadboard - 1

2 Resistors 10k, 11k, 200Ω,

283 Ω

1each

3 Capacitors 1μF

10μF

2

1

4 Transistor BC107

5 CRO 0-30MHz 1

6 RPS 0-10V 1

7 Digital multimeter 1

b) THEORY:

In the current series amplifier, the output current flows only for one half of the cycle

of the input signal. The transistor dissipates no power with zero input signal.. Complementary

symmetry amplifier requires neither an input nor an output transformer. This arrangement

uses transistors having complementary symmetry in the emitter follower configuration. This

amplifier circuit has a unity gain because of the emitter follower configuration. Moreover

there is no phase inversion of the output signal. The split supply used in the circuit gives us

an advantage that the dc component of the output voltage can be made zero. Thus the only ac

component of the power is available across the load resistor.

c) PROCEDURE:




measured and readings are tabulated.


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d) CIRCUIT DIAGRAM: CURRENT SERIES FEEDBACK AMPLIFIER

e) MODEL GRAPH:


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f) TABULATION:

Sl. No. Input

frequency

(Hz)

Input

Voltage (mV)

Output

Voltage (V)

Voltage gain

20log ogain

i

VV db

V

g) RESULT:

Thus the Current series feedback amplifier was designed and its frequency response was

observed.

VIVA QUESTIONS:

1. What is feedback amplifier?

The part of the output is given to the input of the circuit called as feedback amplifier.

2. Classify the feedback amplifiers.

1) Voltage series feedback amplifier

2) Current series feedback amplifier

3) Voltage shunt feedback amplifier

4) Current shunt feedback amplifier

3. What is meant by neutralization?

It is the process by which feedback can be cancelled by introducing a current that is

equal in magnitude but 180o out of phase with the feedback signal at the input of the active

device. The two signals will cancel and the effect of feedback will be eliminated. This

technique is termed as neutralization.

4. What is the application of tuned amplifiers? The application of tuned amplifiers to obtain a desired frequwnct and rejecting all other

frequency in

(i). Radio and T .V broadcasting as tunning circuit.

(ii). Wireless communication system.


33

Exercise Number : 10

Title of the Exercise : VOLTAGE SERIES FEEDBACK AMPLIFIER



To design and find the frequency response of voltage series feedback amplifier.



1 Breadboard - 1

2 Resistors 10k, 11k, 200Ω,

283 Ω

1each

3 Capacitors 1μF

10μF

2

1

4 Transistor BC107

5 CRO 0-30MHz 1

6 RPS 0-10V 1

7 Digital multimeter 1

b) THEORY:

The feedback voltage is connected in series with the input circuit means called as

voltage series amplifier. If the feedback voltage is equal to the output voltage then it is called

as voltage series feedback amplifier.

c) PROCEDURE:




measured and readings are tabulated.


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d) CIRCUIT DIAGRAM: VOLTAGE SERIES FEEDBACK AMPLIFIER

e) MODEL GRAPH:


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f) TABULATION:

Sl. No. Input

frequency

(Hz)

Input

Voltage

(mV)

Output

Voltage (V)

Voltage gain

20log ogain

i

VV db

V

g) RESULT:

Thus the Current series feedback amplifier was designed and its frequency response was

observed.

VIVA QUESTIONS:

1.List out the types of feedback circuits.

Positive feedback

Negative feedback

2.Write the advantages of negative feedback.

Higher input and lower output impedance.

Improved higher sensitivity

Reduced noise

Increased bandwidth

Reduced distortion

3.Write the disadvantages of negative feedback

reduced overall circuit gain

reduced distortion


36


Title of the Exercise : SECOND ORDER LOW PASS BUTTERWORTH FILTER



To simulate and obtain the frequency response of second order low pass filter by using

ORCAD capture.


ORCAD capture

b) PROCEDURE:

1. Start the program

2. Select the ORCAD release 9 capture CIS

3. Go to new and select project

4. Create the title of the project

5. Drag the elements as per the circuit diagram requirement.

6. Make connections as per the circuit diagram using wire icon.

7. Create the new simulation

8. Set the output level setting.

9. Placed the voltage markers in input and output mode.

10. Run the circuit diagram and print the output.


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c) CIRCUIT DIAGRAM: SECOND ORDER LOW PASS BUTTERWORTH

FILTER

d) MODEL GRAPH:


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e) RESULT:

Thus the second order low pass filter circuit was simulated using ORCAD capture and

its frequency response was obtained.

VIVA QUESTIONS:

1. Write the equation for finding the bandwidth.

Bandwidth= f2-f1

Where f1= lower cut off frequency and f2= higher cut off frequency

2. What is blocking oscillator? The circuit which uses a regerative feedback, producing a single pulse or pulse

train is called a blocking oscillator.

3. Which are the two important elements of a blocking oscillator?

1. Active element like transistor.

2. A pulse transformer.

4. What is pulse transformer?

A pulse transformer is basically a transformer which couples a source of pulses of

electrical energy to the load, keeping the shape and other properties of pulses unchanged. The

voltage level of the pulse can be raised or lowered by designing the proper turns ratio for the

pulse transformer.


39


Title of the Exercise : DIFFERENTIATOR SINE WAVE INPUT



To simulate and obtain the output waveform of differentiator circuit by using ORCAD

capture.


ORCAD capture

b) PROCEDURE:












40

c) CIRCUIT DIAGRAM: DIFFERENTIATOR SINE WAVE INPUT

d) MODEL GRAPH:

:

Amplitude

in volts

0

Time in ms

0

Time in ms

0

0

0 0

AD741 3

2

7

4

6

1

5 +

-

V+

V-

OUT

OS1

OS2

1.59k

15.9k

1.44k

0.01u

0.1u

12v

12v

C2

R2

R1 C1

R3

OUTPUT

INPUT

Frequency=100

AC=1

DC=0

Voff=0

Vamp=1V

Maximum Step Size= 0.2us

Analysis Type : Time Domain ( Transient) Run to Time =40ms


41

e) RESULT:

Thus the Differentiated Sine wave input circuit was simulated using ORCAD capture

and its frequency response was obtained.

VIVA QUESTIONS:

1. Define differentiator. Differentiator is a circuit that passes high frequencies of the input and attenuates low

frequencies. It implies that the output voltage is the differential of the input.

2. What is meant by clippers?

The circuit with which the waveform is shaped by removing a portion of the input

signal without distorting the removing part of the alternating waveform is called a clipper.

3. What is meant by clampers? Clamping network shifts (clamp) a signal to a different d.c level, i.e., it introduces a

d.c level to an a.c signal.Hence,the clamping network is known as d.c restorer.


42


Title of the Exercise : DIFFERENTIATOR SQUARE WAVE INPUT



To simulate and obtain the output waveform of differentiator circuit by using ORCAD

capture.


ORCAD capture

b) PROCEDURE:












43

CIRCUIT DIAGRAM: DIFFERENTIATOR - SQUARE WAVE INPUT

c) MODEL GRAPH:

0

0

0 0

1.59k AD741 3

2

7

4

6

1

5 +

-

V+

V-

OUT

OS1

OS2

12v

12v

0.1u

0.01u

1.44k

15.9k

C2

R2

R1 C1

R3

OUTPUT

INPUT

V1= -1V V2= 1V AC=10 DC=0 TR=0ms TF=0ms TD=0ms PW=5ms PER=0.01s

Analysis Type : Time Domain ( Transiant) Run to Time =20ms

Maximum Step Size= 0.2us


44

d) RESULT:

Thus the Differentiated Square wave input circuit was simulated using ORCAD

capture and its frequency response was obtained.

VIVA QUESTIONS:

1. What is delay time?

The time required for the current to rise to 10% of its maximum (saturation) value Ics

is called the delay time td.

2. What is the total turn on time? The total turn on time is ton is the sum of the delay time and rise time,

ton = td + tr

Where,

td = Delay time.

tr = Rise time.

3. What is storage time? The interval that elapses between the transition of the input waveform and the time

when the collector current has dropped to 90 % of total output is called the storage time ts.

4. Define transition time. The time interval during which the conduction transfer from one transistor to another

transistor is defined as transition time.


45


Title of the Exercise : ASTABLE MULTIVIBRATOR –SYMMETRICAL



To simulate and obtain the output waveform of astable multivibrator(symmetrical) by using

ORCAD capture.


ORCAD capture

b) PROCEDURE:












46

c) CIRCUIT DIAGRAM: ASTABLE MULTIVIBRATOR –SYMMETRICAL

d) MODEL GRAPH:

Vce(sat) V

Vce(sat)

Vbe(sat)

Vbe(sat)

Time(s)

Output Vc2

Output Vc1

Output Vb2

Output Vb1

0 0

0

C1

0.01U

1k 36K 1k

SMBT2222A/SIE SMBT2222A/SIE

C2

0.01U

36K

8V

Analysis type: Time domain transient Run to time: 5ms Maximum step size: 0.2us

V1

Rc1 R2 R1 Rc2

Vc1 Vc2

Vb2 Vb1 Q23 Q22


47

e) RESULT:

Thus the astable multivibrator(symmetrical) circuit was simulated using ORCAD

capture and its output waveform was obtained.

VIVA QUESTIONS:

1. What is Leading edge response?

At start there is an overshoot and then the pulse settles down. The response till it settles

down after the overshoot is called leading edge response.

2. What is trailing edge response?

The response generally extends below the zero amplitude after the end of pulse width

is called back swing. The portion of response from backswing till it settes down is trailing

edge response.

3. What is flat top response? The portion of the response between the trailing edge and the leading edge is called

flat top response.

4. Define rise time of a pulse.

The rise time is an important parameter related to this part of the response.It is defined

by the time required by the pulse to rise from 10 % of its amplitude to 90 % of its amplitude.


48


Title of the Exercise : ASTABLE MULTIVIBRATOR – ASYMMETRICAL



To simulate and obtain the output waveform of astable multivibrator(Asymmetrical) by using

ORCAD capture.


ORCAD capture

b) PROCEDURE:












49

c) CIRCUIT DIAGRAM: ASTABLE MULTIVIBRATOR- ASYMMETRICAL

d) MODEL GRAPH:

Vce(sat)

V

Vce(sat)

Vbe(sat)

Vbe(sat)

Time(s)

Output Vc2

Output Vc1

Output Vb2

Output Vb1

0 0

0

43K 1k

C1

0.01U

C2

0.01U

28K

8V

SMBT2222A/SIE SMBT2222A/SIE

1k

Analysis type: Time domain transient

V1

Rc1 R2 R1 Rc2

Vc1 Vc2

Vb2 Vb1 Q23 Q22

Run to time: 5ms Maximum step size: 0.2us


50

e) RESULT:

Thus the astable multivibrator(asymmetrical) circuit was simulated using ORCAD

capture and its output waveform was obtained.

VIVA QUESTIONS:

1. What is current time base generator?

The circuit which produces current which linearly increases with time is called current

time base generator.

2. What are the application of the blocking oscillator? The blocking oscillator can be used as low impedance switch used to discharge a capacitor

very quickly. To produce large peak power pulses, both the types of oscillators cab be used.

The output of the blocking oscillator can be used to produce gating waveform with very low

mark space ratio. It may be used as frequency divider or counter in digital circuits.

3. List varies sweep circuits Exponential charging circuit

Constant-current charging circuit.

Miller circuit

poot strap circuit

Inductor circuit.

4. What do you mean by voltage time base generators?

Circuits used to generate a linear variation of voltage with time are called voltage time

base generators.


51


Title of the Exercise : MONOSTABLE MULTIVIBRATOR



To simulate and obtain the output waveform of Monostable multivibrator by using ORCAD

capture.


ORCAD capture

b) PROCEDURE:












52

c) CIRCUIT DIAGRAM: MONOSTABLE MULTIVIBRATOR

d) MODEL GRAPH:

Output Vc2

Output Vc1

Output Vb1

Output Vb2

Time (s)

V

Vce(sat)

Vce(sat)

Vbe(sat)

Vbe(sat)


53

e) RESULT:

Thus the monostable multivibrator circuit was simulated using ORCAD capture and

its output waveform was obtained.

VIVA QUESTIONS:

1. Define resonance.

The reactance of the capacitor equals that of the inductor reactance.

i.e ωC. = 1 / ωL.

2. What is Quality factor?

The ratio of inductive reactance of the coil at resonance to its resistance is known

as quality factor.

Q = XL / R

3. Define gain bandwidth product of a tuned amplifier.

The gain bandwidth(GBW) product is a figure of merit defined in terms of mid band

gain and upper 3-db frequency fh as GBW = | Aim fh | = gm / 2πc

4. What is the other name for tuned amplifier?

Tuned amplifiers used for amplifying narrow band of frequencies hence it is also

known as “ narrow band amplifier” or “Band pass amplifier”.


54


Title of the Exercise : BISTABLE MULTIVIBRATOR



To simulate and obtain the output waveform of bistable multivibrator by using ORCAD

capture.


ORCAD capture

b) PROCEDURE:












55

c) CIRCUIT DIAGRAM: BISTABLE MULTIVIBRATOR

d) MODEL GRAPH:

Output Vb1

V

Time (s)

Output Vb2 Vbe(sat)

Output Vc2

Vce(sat)

Vbe(sat)

Vce(sat)

Output Vc1


56

e) RESULT:

Thus the Bistable multivibrator circuit was simulated using ORCAD capture and its

output waveform was obtained.

VIVA QUESTIONS:

1. What is unilateralisation?

It is the phenomenon by which a signal can be transmitted from the input to the output

alone and not viceversa. In an unilateralised amplifier both resistive and reactive effects are

cancelled.

2. What is stagger tuned amplifier?

In this configuration one or more tuned amplifiers are cascaded each amplifier stage is

tuned to different frequencies. This results in decreased gain and increased bandwidth.

3. What is the effect of ‘Q’ on stability?

Higher the value of Q,provides better selectivity, but smaller bandwidth and larger

gain. Hence it provides less stability.

4. What is meant by unloaded and loaded Q of tank circuit.[ APR – 2003 ]

Unloaded Q is the ratio of stored energy to dissipated energy in a reactor or resonator.

The loaded Q (or) QL of a resonator is determined by how tightly the resonator is coupled to

its terminations.


57


Title of the Exercise : CMOS INVERTER



To simulate and obtain the output waveform of cmos inverter by using ORCAD capture.


ORCAD capture

b) PROCEDURE:












58

c) CIRCUIT DIAGRAM: CMOS INVERTER

e) RESULT:

Thus the CMOS Inverter circuit was simulated using ORCAD capture and its output

waveform was obtained.

0

0

V1 5v

M1

MbreakP

CLK DSTM1

M2

MbreakN

INPUT

OUTPUT

Analysis type:Time domain (transient)

Run to time=0.01ms

Maximum step size=0.2ms

Off time=0.5us On time=0.5us

1 1 1 1

1 1 1 1

0 0 0 0

0 0 0 0

Time(s)

V

Input

Output

d) MODEL

GRAPH


59

VIVA QUESTIONS:

1. Mention the applications of class ‘c’ tuned amplifier. One of the most common applications for mixer is in radio receivers. The mixer is

used to convert incoming signal to a lower frequency where it is easier to obtain the high gain

and selectivity required.

Mixer circuits are used to translate signal frequency to some lower frequency or to

some higher frequency. When it is used to translate signal to lower frequency it is called

down converter. When it is used to translate signal to higher frequency, it is called up

converter.

2. Mention the need for stagger-tuned amplifier.

The double tuned amplifier gives greater 3 db bandwidth having steeper sides and flat

top. But alignment of double tuned amplifier is difficult. To overcome this problem two

single tuned amplifiers are cascaded.

3. What is principle of Hazel tine neutralization?

Hazel tine introduced a circuit in which the troublesome effect of the collector to base

capacitance of the transistor was neutralized by introducing a signal which cancels the signal

coupled through the collector to base capacitance.

4. List the performance measure of a tuned amplifier.

Selection of a desired radio frequency signal.

Effective quality factor.

Gain

Bandwidth.


60


Title of the Exercise : CMOS NOR



To simulate and obtain the output waveform of CMOS NOR by using ORCAD capture.


ORCAD capture

b) PROCEDURE:












61

c) CIRCUIT DIAGRAM: CMOS NOR

d) MODEL GRAPH:

1 1 1 1

1 1 1 1 1

1 1 1

0 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0 0 0

Input A

Input B

Output

Time(s)

V

0

0

0

V1 5V

M4

MbreakN

M2

MbreakP

M1

MbreakP

CLK DSTM1

CLK DSTM2

M3

MbreakN

Input A

Input B Output

Off time=1us On time=1us



Run to time=0.01ms

Maximum step size=0.2us


62

e) RESULT:

Thus the CMOS NOR circuit was simulated using ORCAD capture and its output


VIVA QUESTIONS:

1. What are the characteristics of an ideal tuned amplifier?

Selects a single radio frequency and amplifiers the same by rejecting all other

frequencies.

Bandwidth is zero.

Harmonic distortion is zero.

2. Write down the relationship between bandwidth and effective Q of a tuned

amplifier? Bandwidth = ωo / Q effective.

3. What are the different methods of coupling? (or) Point out different methods of

coupling the load to a tuned amplifier.

The different methods of coupling the load to a tuned amplifier are:

Capacitive coupling.

Inductive coupling.

4. Why tuned amplifier cannot be used at low frequency?

For low frequencies the size L and C are large. So the circuit will be bulky and

expensive, hence the tuned amplifiers cannot be used at low frequency.


63


Title of the Exercise : CMOS NAND



To simulate and obtain the output waveform of CMOS NAND by using ORCAD capture.


ORCAD capture

b) PROCEDURE:












64

c) CIRCUIT DIAGRAM: CMOS NAND

d) MODEL GRAPH:

Input B

V

Output 0 0

Input A

1

1

1 0

1

0

0

0

0

0 0

0

0 0

0 0

0

0

1

1 1

1

0

1

Time(s)

1

1

0 0

0

0

M4

MbreakP

M5

MbreakP

M6

MbreakN

M7

MbreakN

V2

5V

CLK DSTM3

CLK DSTM4

Input A

Input B

Output

Off time=1us On time=1us


Run to time=0.01ms

Maximum step size=0.2us



65

e) RESULT:

Thus the CMOS NAND circuit was simulated using ORCAD capture and its output


VIVA QUESTIONS:

1. What is the need for differential amplifiers?

Differential amplidiers are small signal direct coupled amplifiers used to amplify the

difference between two signals. The need for differential amplifier arises in physical

measurements, instrumentation amplifiers and medical instrumentation.

2. What are the advantages of Differential Amplifiers? * High voltage gain.

* High input impedence

* High Bandwidth

* Good bias stability.

3. Define CMRR.

Common Mode Rejection Ratio(CMRR) is the ability of the differential amplifiers to

reject the common mode signals. It is defined as the ratio of difference mode gain Ad to

common mode gain Ac.

4. Why Differential amplifiers are widely used in Integrated Circuits?

It has good bias stability and good voltage gain without the use of large bypass

capacitors. Hence it is used in ICs.


66


Title of the Exercise : DIGITAL TO ANALOG CONVERTER



To simulate and obtain the output waveform of digital to analog converter by using ORCAD

capture.


ORCAD capture

b) PROCEDURE:












67

c) CIRCUIT DIAGRAM:

d) MODEL GRAPH:

Time(s)

Input B0

Input B1

Input B2

Input B3

Output

V

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1


68

RESULT:

Thus the digital to analog converter circuit was simulated using ORCAD capture and

its output waveform was obtained.

VIVA QUESTIONS:

1. What are the basic elements of power supply ?

(i) Transformer

(ii) Rectifier.

(iii) Filter.

2. What is ripple factor(Υ)?

Ripple factor (γ) may be defined as the ratio of the root mean square (rms) value of the ripple

voltage to the absolute value of the dc component of the output voltage, usually expressed as

a percentage. However, ripple voltage is also commonly expressed as the peak-to-peak value.

This is largely because peak-to-peak is both easier to measure on an oscilloscope and is

simpler to calculate theoretically. Filter circuits intended for the reduction of ripple are

usually called smoothing circuits.

3. What is a rectifier?

A rectifier is an electrical device that converts alternating current (AC), which periodically

reverses direction, to direct current (DC), which flows in only one direction. The process is

known as rectification.

4. Define SMPS .

A switched-mode power supply (switching-mode power supply, SMPS, or simply switcher)

is an electronic power supply that incorporates a switching regulator in order to be highly

efficient in the conversion of electrical power.

An SMPS is usually employed to efficiently provide a regulated output voltage, typically at a

level different from the input voltage.

http://en.wikipedia.org/wiki/Root_mean_square

http://en.wikipedia.org/wiki/Voltage

http://en.wikipedia.org/wiki/Absolute_value

http://en.wikipedia.org/wiki/Peak-to-peak

http://en.wikipedia.org/wiki/Oscilloscope

http://en.wikipedia.org/wiki/Rectifier#Rectifier_output_smoothing

http://en.wikipedia.org/wiki/Alternating_current

http://en.wikipedia.org/wiki/Direct_current

http://en.wikipedia.org/wiki/Power_supply


69

Exercise/Experiment Number: 22

Title of the exercise/experiment : Inverting, Non-inverting and Differential amplifiers

Date of the experiment :

AIM: To construct and test the performance of an Inverting, Non-inverting amplifier

and Differential amplifier using IC µA 741

APPARATUS REQUIRED:

DESIGN

:

INVERTING AMPLIFIER:

Let A = -5; R1 = 1KΩ

A = − Rf / R1

Rf = 5 KΩ

Rcomp = R1 Rf / R1 + Rf

= 833 Ω

NON-INVERTING AMPLIFIER:

Let A = 6; R1 = 1KΩ

A = 1 + (Rf / R1)

Rf = 5 KΩ

Rcomp = R1 Rf / R1 + Rf

= 833 Ω

NON-INVERTING AMPLIFIER:

Let A = 100; R1 = 1KΩ

A = R2 / R1

R2 = 100 KΩ

S. No. Name Range Quantity

1 Dual Power Supply (0-30)V 1

2 Resistors 1KΩ;5 KΩ;100 KΩ Each 2

3 Regulated Power Supply (0-30)V 1

4 IC µA 741 - 1

5 Voltmeter (0-50)V 1

6 Connecting Wires - -


70

CIRCUIT DIAGRAM:

NON-INVERTING AMPLIFIER: MODEL GRAPH:

DIFFERENTIAL AMPLIFIER:


71

THEORY:


The fundamental component of any analog computer is the

operational amplifier or op-amp and the frequency configuration in which it is used as an

inverting amplifier. An input voltage Vin is applied to the input voltage. It receives and

inverts its polarity producing an output voltage. this same output voltage is also applied to a

feedback resistor Rf, which is connected to the amplifier input analog with R1. The amplifier

itself has a very high voltage gain.

If Rf = R1 then Vo=Vi

NON- INVERTING AMPLIFIER:

Although the standard op-amp configuration is as an inverting

amplifier, there are some applications where such inversion is not wanted. However, we

cannot just switch the inverting and non inverting inputs to the amplifier itself. We will still

need negative feedback to control the working gain of the circuit .Therefore, we will need to

leave the resistor structure around the op-amp intact and swap the input and ground

connections to the overall circuit.

VO/VI = (Rf / Ri) +1

From the calculations, we can see that the effective voltage

gain of the non-inverting amplifier is set by the resistance ratio. Thus, if the two resistors are

equal value, then the gain will be 2 rather than 1.


A circuit that amplifies the difference between two signals is

called as a differential amplifier. This type of amplifiers is very useful in instrumentation

circuits. From the experimental setup of a differential amplifier, the voltage at the output of

the operational amplifier is zero. The inverting and non-inverting terminals are at the same

potential. Such a circuit is very useful in detecting very small differences in signals. Since the

gain can be chosen to be very large. For example, if R2=100R1, then a small difference V1-

V2 is amplified 100 times.


72

TABULATION:


S.No Input Voltage (in volts) Output Voltage (in volts)

1

1

−9.93

NON- INVERTING AMPLIFIER:


1

1

11.2



1

2

V1 V2

9.74

9.80

3 2

2 3

PROCEDURE:

Connections are made as per the EXPERIMENTAL SETUP.

The supply is switched ON.

Output is connected to anyone channel of CRO.

The V1 and V2 voltages are fixed and measured from the other channel of CRO and

the corresponding output voltages are also noted from the CRO.

The above step is repeated for various values of V1 and V2.V1 and V2 may be AC or

DC voltages from function generator or DC power supply.


73

Readings are tabulated and gain was calculated and composed with designed values.

RESULT Thus the Inverting, Non-inverting and Differential amplifier using op-amp was designed and tested.

VIVA QUESTIONS:

1. Define an operational amplifier.

An operational amplifier is a direct-coupled, high gain amplifier consisting of one or more

differential amplifier. By properly selecting the external components, it can be used to

perform a variety of mathematical operations.

2. Mention the characteristics of an ideal op-amp.

* Open loop voltage gain is infinity. *Input impedance is infinity. *Output

impedance is zero. *Bandwidth is infinity. *Zero offset.

3. Define input offset voltage.

A small voltage applied to the input terminals to make the output voltage as zero when the

two input terminals are grounded is called input offset voltage.

4. Define input offset current.

The difference between the bias currents at the input terminals of the op-amp is called as

input offset current.


74


Title of the exercise/experiment : DIFFERENTIAL AMPLIFIER USINGBJT


AIM:

To construct a differential amplifier using BJT and to determine the dc collector current of

individual transistors and also to calculate the CMRR.

APPARATUS REQUIRED:

S.No. Name Range Quantity

1. Transistor BC107 2

2. Resistor 4.7kΩ, 10kΩ 2,1

3. Regulated power supply (0-30)V 1

4. Function Generator (0-3) MHz 2

5. CRO 30 MHz 1

6. Bread Board 1

OBSERVATION

VIN = V1 – V2

V0 =

Ad = V0/ VIN

For the differential mode operation the input is taken from two different sources and the

common mode operation the applied signals are taken from the same source

Common Mode Rejection Ratio (CMRR) is an important parameter of the differential

amplifier. CMRR is defined as the ratio of the differential mode gain, Ad to the common

mode gain, Ac.

CMRR = Ad / Ac

In ideal cases, the value of CMRR is very high.

OBSERVATION

VIN =VO =AC = VO / VIN

FORMULA:

Common mode Gain (Ac) = VO / VIN

Differential mode Gain (Ad) = V0 / VIN

Where VIN = V1 – V2


75

CIRCUIT DIAGRAM


76

Common Mode Rejection Ratio (CMRR) = Ad/Ac

Where, Ad is the differential mode gain

Ac is the common mode gain.

THEORY:

The differential amplifier is a basic stage of an integrated operational amplifier.

It is used to amplify the difference between 2 signals.

It has excellent stability, high versatility and immunity to noise

. In a practical differential amplifier, the the common mode signal.

Transistor Q1 and Q2 have matched characteristics. The values of RC1 and RC2 are

equal. Re1 and Re2 are also equal and this differential amplifier is called emitter coupled

differential amplifier. The output is taken between the two output terminals

PROCEDURE:

1. Connections are given as per the circuit diagram.

2. To determine the common mode gain, we set input signal with voltage Vin=2V

and determine Vo at the collector terminals. Calculate common mode gain, Ac=Vo/Vin. 3. To determine the differential mode gain, we set input signals with voltages V1 and

V2. Compute Vin=V1-V2 and find Vo at the collector terminals. Calculate differential mode

gain, Ad=Vo/Vin.

4. Calculate the CMRR=Ad/Ac. 5. Measure the dc collector current for the individual transistors.

RESULT:

Thus, the Differential amplifier was constructed and dc collector current for the

individual transistors is determined. The CMRR is calculated as

Viva questions 1.What is an amplifier?

An amplifier is a device which produces a large electrical output of similarcharacteristics to

that of the input parameters.

2.. How are amplifiers classified according to the input?

1. Small – signal amplifier 2. Large – signal amplifier

3.How are amplifiers classified according to the transistor configuration?

Common emitter amplifier 2. Common base amplifier 3. Common collector

amplifier

4. What is the different analysis available to analyze a transistor?

1. AC analysis 2. DC analysis


77


Title of the exercise/experiment :CLASS B COMPLEMENTARY SYMMETRY

POWER AMPLIFIER


AIM:

To construct a Class B complementary symmetry power amplifier and observe the

waveforms with and without cross-over distortion and to compute maximum output power

and efficiency.

APPARATUS REQUIRED:


1. Transistor CL100, BC558 1,1

2. Resistor 4.7kΩ,15kΩ 2,1

3. Capacitor 100µF 2

4. Diode IN4007 2

5. Signal Generator (0-3)MHz 1

6. CRO 30MHz 1


8. Bread Board 1

FORMULA:

Input power, Pin=2VccIm/П

Output power, Pout=VmIm/2

Power Gain or efficiency, η=л/4(Vm/Vcc) 100

. THEORY: A power amplifier is said to be Class B amplifier if the Q-point and the input signal

are selected such that the output signal is obtained only for one half cycle for a full input

cycle. The Q-point is selected on the X-axis. Hence, the transistor remains in the active

region only for the positive half of the input signal.

There are two types of Class B power amplifiers: Push Pull amplifier and complementary

symmetry amplifier. In the complementary symmetry amplifier, one n-p-n and another p-n-p

transistor is used. The matched pair of transistor are used in the common collector

configuration. In the positive half cycle of the input signal, the n-p-n transistor is driven into

active region and starts conducting and in negative half cycle, the p-n-p transistor is driven


78

into conduction. However there is a period between the crossing of the half cycles of the

input signals, for which none of the transistor is active and output, is zero.

MODEL GRAPH

f 1 f2 f (Hz)


79

TABULATION:

FREQUENCY RESPONSE OF CASCODE AMPLIFIER

Keep the input voltage constant (Vin) =

Frequency (in Hz) Output Voltage (in volts) Gain = 20 log (Vo / Vin) (in dB)

RESULT:

Thus, the Class B amplifier was constructed and the gain was determined.

VIVA QUESTION:

1. What is feed back?

It is the process of injecting some energy from the output and then returns it back

tothe input.

2. What is the disadvantage of negative feed back?

Reduces amplifier gain.

3. Define sensitivity.

It is the ratio of percentage change in voltage gain with feedback to the percentage

change in voltage gain without feed back.

4. Define Desensitivity.

It is the ratio of percentage change in voltage gain without feedback to thepercentage

change in voltage


80


Title of the exercise/experiment :CASCADE AMPLIFIER


AIM:

To construct a cascode amplifier circuit and to plot the frequency response

characteristics.

APPARATUS REQUIRED:


1. Transistor BC107 2

2. Resistor

22kΩ,6 kΩ,700 Ω,470Ω

16 kΩ,6.2 kΩ,3.3 kΩ

1.1 kΩ

1,1,1,1,

1,1,1,

1


4. Signal Generator (0-3)MHz 1

5. CRO 30 MHz 1

6. Bread Board 1

7. Capacitor 0.01µF 3

THEORY:

A cascode amplifier consists of a common emitter amplifier stage in series with a common

base amplifier stage. It it one approach to solve the low impedance problem of a common

base circuit. Transistor Q1 and its associated components operate as a common emitter

amplifier, while the circuit of Q2 functions as a common base output stage. The cascade

amplifier gives the high input impedance of a common emitter amplifier, as well as the good

voltage gain and frequency performance of a common base circuit.

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. The waveforms at the input and output are observed for cascade operations varying the

input frequency.

3. The biasing resistances needed to locate the Q-point are determined.


81

4. Set the input voltage as 1V and by varying the frequency, note the output voltage.

5. Calculate gain=20 log (Vo / Vin.)

6. A graph is plotted between frequency and gain.

CIRCUIT DIAGRAM:


82

TABULATION:

Keep the input voltage constant, Vin =

Frequency (in Hz) Output Voltage (in volts) Gain= 20 log(Vo/Vin) (in dB)

RESULT:

Thus, the Cascade amplifier was constructed and the gain was determined.

VIVA QUESTIONS:

1.What is biasing?

To use the transistor in any application it is necessary to provide sufficient voltage

and current to operate the transistor. This is called biasing. 2. What is the necessary of the coupling capacitor?

It is used to block the c signal to the transistor amplifier. It allows a c &blocks the d c.

3. What is reverse saturation current?

The current due to the minority carriers is called the reverse saturation current.


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LIST OF QUESTION SET

1. a) Design a differential amplifier circuit and calculate its CMRR.

b) Design and simulate a Monostable Multivibrator with frequency 2 KHz

using ORCAD.

2. a) Design a Wein bridge oscillator with frequency of oscillation 1 KHz.

b) Simulate a CMOS NAND using ORCAD.

3. a) Design a Colpitts oscillator with frequency of oscillation 12 KHz.

Specifications:

VCC=12V, S=6, ICQ=1.7mA, VRE=3V.

b) Simulate a CMOS Inverter using ORCAD.

4. a) Design a RC Phase shift oscillator with frequency of oscillation 1 KHz.

b) Design and simulate a Symmetrical Astable Multivibrator with frequency

2 KHz using ORCAD.

5. a) Design a differential amplifier circuit and calculate its CMRR.

b) Design and simulate a Asymmetrical Astable Multivibrator with frequency

2 KHz using ORCAD.

6. a) Design a single tuned Class C amplifier with resonant frequency 9KHz.

Specifications:

QL=1, RL= 1KΩ, L= 1mH.

b) Design and simulate a second order Butterworth low pass filter with

frequency 1 KHz using ORCAD.

7. a) Design a Class B amplifier with resonant frequency 9KHz.

b) Simulate a differentiator with sine wave input using ORCAD.

8. a) Design a Hartley oscillator with frequency of oscillation 12 KHz.

Specifications:


b) Simulate a CMOS NOR using ORCAD.

9. a) Design a Colpitts oscillator with frequency of oscillation 12 KHz.

Specifications:


b) Design and simulate a Bistable Multivibrator with frequency 2 KHz using

ORCAD.


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10. a) Design a Wein bridge oscillator with frequency of oscillation 1 KHz.

b) Simulate a differentiator with square wave input using ORCAD.

11. a) Design a RC Phase shift oscillator with frequency of oscillation 1 KHz.

b) Simulate a Digital to Analog converter using ORCAD.

12. a) Design a single tuned Class C amplifier with resonant frequency 9KHz.

Specifications:

QL=1, RL= 1KΩ, L= 1mH.

b) Simulate a CMOS Inverter using ORCAD.

13. a) Design a Hartley oscillator with frequency of oscillation 12 KHz.

Specifications:


b) Simulate a CMOS NAND using ORCAD.

14. a) Design a Class B amplifier with resonant frequency 9KHz.

b) Design and simulate a second order Butterworth low pass filter with frequency 1

KHz using ORCAD.

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EC 2257- ELECTRONIC CIRCUITS II AND SIMULATION LAB

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