Electronic Circuit Analysis Laboratory Mannual

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SIR VISHVESHWARAIAH INSITUE OF SCIENCE & TECHNOLOGY

MADANAPALLE –

517325, CHITTOOR DIST. ANDHRA PRADESH.(APPROVED BY AICTE – NEW DELHI , AFFILIATED TO JNTU ANANTAPUR, ANANTAPUR)

CONTACTS: 08571 – 280888. email: [email protected]. url: www.vist.ac.in

Department of ELECTRONICS & COMMUNICATION ENGINEERING

ELECTRONICS CIRCUIT ANALYSIS

LABORATORY MANNUAL

For II B.Tech, II Semister Students

Compiled by: K. MOHAN

Assistant Professor

Department of ECE.



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ELECTRONIC CIRCUIT ANALYSIS LABORATORY

INTRODUCTION TO PSPICE

Electronic circuit design requires accurate methods for evaluating circuit performance.

Because of the enormous complexity of modern integrated circuits, computer – aided circuit

analysis is essential and can provide information about circuit performance that is almost

impossible to obtain with laboratory prototype measurements. Computer – aided analysis

permits.

1. Evaluating the effects of variations in elements, such as resistors, transistors,

transformers, and so on.

2. The assessment of performance improvements or degradations.

3. Evaluating the effects of noise and signal distortion without the need of expensive

measuring instruments.

4. Sensitivity analysis to determine the permissible bonds due to tolerances on each and

every element value of parameter of active elements.

5. Fourier analysis without expensive wave analyzers

6. Evaluating the effects of nonlinear elements on the circuit performance

7. Optimizing the design of electronic circuits in terms of circuit parameters.

SPICE is a general – purpose circuit program that simulates electronic circuits. SPICE can

perform various analyses of electronic circuits: the operating (or the quiescent) points of

transistors, a time-domain response, a small-signal frequency response, and so on. SPICE

contains models for common circuit elements, active as well as passive, and is capable of

simulating most electronic circuits. It is a versatile program and is widely used both in industries

and universities. The acronym SPICE stands for Simulation Program with Integrated Circuit

Emphasis.

Unit recently, SPICE was available only on mainframe computers. In addition to the initial

cost of the computer system, such a machine can be expensive and inconvenient for classroom

use. In 1984, MicroSim introduced the PSpice simulator, which is similar to the Berkeley SPICE

and runs on an IBM-PC or compatible. It is available at no cost to students for classroom use.

PSpice, therefore, widens the scope for the integration of computer-aided circuit analysis into



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electronic circuits courses at the undergraduate level. Other versions of PSpice that will run on

computers such as the Macintosh II, VAX, SUN, and NEC are also available.

DESCRIPTION OF SPICE:

The development of SPICE spans a period of about 30 years. During the mid-

1960s, the program ECAP was developed at IBM [1]. Later ECAP served as the starting point

for the development of program CANCER at the University of California (UC), Berkeley in early

1970s. SPICE2, which is an improved version of SPICE, was developed during the mid-1970s

at UC – Berkeley.The algorithms of SPICE2 are general in nature but are robust and powerful

for simulating electrical and electronics circuits, and SPICE2 has become a standard tool in the

industry for circuit simulations. The development of SPICE2 was supported by public funds at

UC-Berkeley, and the program is in the public domain. SPICE3, which is a variation of SPICE2,

is designed especially to support the computer – aided design (CAD) research program at UC –

Berkeley.

SPICE2 has become an industry standard and is now referred to simply as SPICE.

The input syntax for SPICE is a free-format style; it does not require that data be entered in

fixed column locations. SPICE assumes reasonable default values fro unspecified circuit

parameters. In addition, it performs a considerable amount of error checking to ensure that a

circuit has been entered correctly.

PSpice, which uses the same algorithms as SPICE2 and is a member of the

SPICE family, is equally useful for simulating all types of circuits in a wide range of applications.

A circuit is described by statements that are stored in a file called the circuit file. The circuit file

is read by the SPICE simulator. Each statement is self-contained and independent; the

statements do not interact with each other. SPICE (or PSpice) statements are easy to learn and

use.

TYPES OF SPICE:

The commercially supported versions of SPICE2 can be divided into two types: mainframe

versions and PC-based versions. Their methods of computation may differ, but their featuresare almost identical to those of SPICE2. However, some may include such additions as a pre-

processor or shell program to manage input and provide interactive control, as well as a post-

processor for refining the normal SPICE output. A person who is familiar with one SPICE

version (e.g., PSpice) should be able to work with other versions.



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1.The mainframe versions are:

HSPICE (Meta-Software), which is designed for integrated circuit design with special device

models RAD-SPICE (Meta-Software), which simulates circuits subjected to ionizing radiation.

IG-SPICE (A.B. Associates)

I-SPICE (NCSS Time Sharing). IG-SPICE and I-SPICE are designed for interactive circuit

simulation with graphic output.

Precise (Electronic Engineering Software)

PSpice (Mentor Graphics), Cadence –SPICE (Cadence Design), SPICE –Plus (Valid Logic)

2. The PC-versions are:

All Spice (Acotech), IS-SPICE (Intusoft), Z-SPICE(Z-Tech), SPICE-Plus (Analog Design Tools)

DSPICE (Daisy Systems),PSpice (MicroSim)

TYPES OF ANALYSIS :

PSpice allows various types of analysis. Each analysis is invoked by including its command

statement. For example, a statement beginning with the .DC command invokes the DC sweep.

The types of analysis and their corresponding .(dot) commands are described below.

Dc Analysis is used for circuits with time-invariant sources (e.g., steady state dc

sources). It calculates all node voltages and branch currents over a range of values, and their

quiescent (dc) values are the outputs.

Dc sweep of and input voltage/current source, a model parameter, or temperature over a

range of values (.DC)

Determination of the linearized model parameters of nonlinear devices (.OP)

Dc operating point to obtain all node voltages (.OP)

Small – signal transfer function with small – signal gain, input resistance, and output

resistance (Thevenin’s equivalent) (.TF)

Dc small – signal sensitivities (.SENS)

Transient Analysis is used for circuits with time-variant sources (e.g., ac sources and switched

dc sources). It calculates all node voltages and branch currents over a time interval, and their

instantaneous values are the outputs.



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Circuit behavior in response to time varying sources (.TRAN)

Dc and Fourier components of the transient analysis results (.FOUR)

Ac Analysis is used for small-signal analysis of circuits with sources of variable

frequencies. It calculates all node voltages and branch currents over a range of frequencies,

and their magnitudes and phase angle are the outputs.

Circuit response over a range of source frequencies (.AC)

Noise generation at an output node for every frequency (.NOISE)

LIMITATIONS OF PSpice :

As a circuit simulator, PSpice has the following limitations:

1. The student version of PSpice is restricted to circuits with 10 transistors only. However,

the professional DOS (or production) version can simulate a circuit with up to 200 bipolar

transistors (or 150 MOSFETs).

2. The program is not interactive; that is, the circuit cannot be analyzed for various

component values without editing the program statements.

3. PSpice does not support an interactive method of solution. If the elements of a circuits

are specified, the output can be predicted. On the other hand, if the output is specified,

PSpice cannot be used to synthesize the circuit elements.

4. The input impedance cannot be determined directly without running the graphic post –

processor, probe. The student version does not require a floating-point co-processor for

running Probe, but the professional version does require such a co-processor.

5. The PC version needs 512 kilobytes of memory (RAM) to run.

6. Distortion analysis is not available in PSpice . SPICE2 allows distortion analysis, but it

gives wrong answers.

7. The output impedance of a circuit cannot be printed or plotted directly.

8. The student version will run with or without the floating – point co-processor (8087,

80287, or 80387). If the co –processor is present, the program will run at full speed;otherwise it will run 5 to 15 times slower. The professional version requires a co-

processor; it is not optional.

Note: The component values in circuits may change depending on specifications. So the

experimental procedure should be given more weightage.



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TABLE OF CONTENTS

1. COMMON EMITTER AMPLIFIER

2. COMMON SOURCE AMPLIFIER

3. VOLTAGE SERIES FEEDBACK AMPLIFIER and CURRENT SERIES FEED BACK AMPLIFIER

4. RC PHASE SHIFT OSCILLATOR

5. CLASS B COMPLEMENTARY SYMMETRY AMPLIFIER

6. CASCADE AMPLIFIER

7. TWO STAGE RC COUPLED AMPLIFIER CLASS C POWER AMPLIFIER

8. CLASS A POWER AMPLIFIER

9. HARTLEY AND COLPITTS OSCILLATOR

10. SINGLE TUNED VOLTAGE AMPLIFIER

11. WIEN BRIDGE OSCILLATOR USING TRANSISTOR

12. CLASS C POWER AMPLIFIER



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1. COMMON EMITTER AMPLIFIER

AIM: i)To analyze the performance of given CE amplifier circuit with respect to gain, band width,

i/p resistance & o/p resistance.

ii) To design and simulate common emitter amplifier as per given specifications.

APPARATUS:

S.NO NAME OF EQUIPMENT RANGE QUANTITY REMARKS

1

2

3

4

5

6

7

8

Transistor BC-107

Regulated power Supply

Function Generator

CRO

Resistors

Capacitors-Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM :



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THEORY:

The CE amplifier provides high gain and wide frequency response. The emitter lead is

common to both the input and output circuits are grounded. The emitter base junction is at

forward biased .The collector current is controlled by the base current rather than the emitter

current. The input signal is applied to the base terminal of the transistor and amplified output

taken across collector terminal. A very small change in base current produces a much larger

change in collector current. When the positive is fed to input circuit it opposes forward bias of

the circuit which cause the collector current to decrease, it decreases the more negative. Thus

when input cycle varies through a negative half cycle, increases the forward bias of the circuit,

which causes the collector current increases .Thus the output signal in CE is out of phase with

the input signal.

PROCEDURE:

1. Connect the circuit as shown in circuit diagram

2. Apply the input of 20mV peak-to-peak and 1 KHz frequency using Function Generator

3. Measure the Output Voltage Vo (p-p) for various load resistors

4. Tabulate the readings in the tabular form.

5. The voltage gain can be calculated by using the expression

Av= (V0 /Vi)

6. For plotting the frequency response the input voltage is kept Constant at 20mV peak-to-

peak and the frequency is varied from 100Hz to 1MHz Using function generator

7. Note down the value of output voltage for each frequency.

8. All the readings are tabulated and voltage gain in dB is calculated by Using The

expression Av=20 log10 (V0 /Vi)

9. A graph is drawn by taking frequency on x-axis and gain in dB on y-axis on Semi-log graph.

The band width of the amplifier is calculated from the graph. Using the expression,

Bandwidth, BW=f2-f1

Where f1 lower cut-off frequency and f2 upper cut-off frequency of CE amplifier

The bandwidth product of the amplifier is calculated using the Expression

Gain Bandwidth product=3-dBmidband gain X Bandwidth



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OBSERVATIONS:

Input voltage Vi=20mV

LOAD

RESISTANCE(KΩ)

OUTPUT

VOLTAGE (V0)

GAIN

AV=(V0 /Vi)

GAIN IN dB

Av=20log10 (V0 /Vi)

FREQUENCY RESPONSE: Vi=20mv

FREQUENCY(Hz) OUTPUTVOLTAGE (V0)

GAIN IN dBAv=20 log10 (V0 /Vi)

MODELWAVE FORMS:

INPUT WAVE FORM:



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OUTPUT WAVE FORM

FREQUENCY RESPONSE

RESULT: The voltage gain and frequency response of the CE amplifier are obtained. Also

gain bandwidth product of the amplifier is calculated.

VIVA QUESTIONS:

1. What is phase difference between input and output waveforms of CE amplifier?

2. What type of biasing is used in the given circuit?

3. If the given transistor is replaced by a p-n-p, can we get output or not?

4. What is effect of emitter-bypass capacitor on frequency response?

5. What is the effect of coupling capacitor?

6. What is region of the transistor so that it is operated as an amplifier?

7. How does transistor acts as an amplifier?



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2.COMMON SOURCE AMPLIFIER

AIM : i)To analyse the performance of given CS amplifier circuit with respect to gain,band width,

i/p resistance & o/p resistance.

ii) To design and simulate common emitter amplifier as per given specifications.

APPARATUS:


1

2

3

4

56

7

8

FET-Transistor 2N5047


Function Generator

CRO

ResistorsCapacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:



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THEORY:

A weak signal is applied between gate and source and output is obtained at drain.

For the proper operation of FET, gate must be reverse biased. A small change in reverse bias

on the gate produces a large drain current. This fact makes FET capable of raising the strength

of a weak signal. The gain of the common source FET amplifier is very high which is greater

than unity.

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. For calculating the voltage gain the input voltage of 0.2V(p-p) amplitude and 1KHz

frequency is applied, then the circuit is simulated and output voltage is noted.

3. The voltage gain is calculated by using the expression Av = Vo / Vi

4. For plotting frequency response the input voltage is kept constant at 0.2V(p-p) andfrequency is varied.

5. Note down the output voltage for each frequency.

6. All readings are tabulated and Av in dB is calculated using 20 Log Vo / Vi.

7. A graph is drawn by taking frequency on X-axis and gain in dB on Y-axis on a Semi-log

graph sheet.

OBSERVATIONS:

S.NO INPUTVOLTAGE(Vi)

OUTPUTVOLTAGE(V0)

VOLTAGE GAIN

]Av= (V0/Vi)



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THEORITICAL CALCULATIONS :

RD*RL

rL = ----------

RD + RL

IDSS = 10mA , VGS = 4v

2 IDSS

gmo = -------------

-VGS(off)

VGSgm = gmo 1 - ------------

(VGS off)

Av = gm * rL

Vin = Vpp

Vout = Av * vin

PRACTICAL CALCULATIONS :

Vin = Vpp

Vout =

Vout

Av = ---------------- =

Vin



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MODEL GRAPH:

RESULTS: The frequency response of the common source FET Amplifier and Bandwidth is

obtained.

VIVA QUESTIONS:

1. How does FET acts as an amplifier? 2. What are the parameters of a FET?

3. What is an amplification factor?

4. Draw the h-parameter model of the FET.

5. What are the advantages of FET over BJT?

6. What is the region of FET so that it acts as an amplifier?

7. What are the differences between JFET and MOSFET?

8. What type of biasing is used in the given circuit?





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THEORY:

When any increase in the output signal results into the input in such a way as to

cause the decrease in the output signal, the amplifier is said to have negative feedback.

The advantages of providing negative feedback are that the transfer gain of the amplifier with

feedback can be stabilized against variations in the hybrid parameters of the transistor or the

parameters of the other active devices used in the circuit. The most advantage of the negative

feedback is that by prepare use of this , there is significant improvement in the frequency

response and in the linearity of the operation of the amplifier. This disadvantage of the

negative feedback is that the voltage gain is decreased.

In Voltage-Series feedback, the input impedance of the amplifier is decreased and

the output impedance is increased. Noise and distortions are reduced considerably.

PROCEDURE:

1. Connections are made as per circuit diagram.

2. Keep the input voltage constant at 20mV peak-peak and 1kHz frequency. For different

values of load resistance, note down the output voltage and calculate the gain by using

the expression

Av = 20log(V0 / Vi ) dB

3. Add the emitter bypass capacitor and repeat STEP 2.And observe the effect of

Feedback on the gain of the amplifier

4. For plotting the frequency the input voltage is kept constant at 20mV peak-peak and the

frequency is varied from 100Hz to 1MHz.

5. Note down the value of output voltage for each frequency. All the readings are tabulated

and the voltage gain in dB is calculated by using expression


6. A graph is drawn by taking frequency on X-axis and gain on Y-axis on semi log graph

sheet

7. The Bandwidth of the amplifier is calculated from the graph using the expression

Bandwidth B.W = f2 – f1.

Where f1 is lower cut off frequency and f 2 is upper cut off frequency of CE amplifier

The gain-bandwidth product of the amplifier is calculated by using the expression

Gain-Bandwidth Product = 3-dB mid band gain X Bandwidth.



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OBSERVATIONS:

Voltage Gain:

S.NO Output Voltage(Vo) with feedback

Output Voltage (Vo)without feedback

Gain(dB) withfeedback

Gain(dB)withoutfeedback

Frequency Response: Vi = 20mV

S.NO Frequency (Hz) Output Voltage (Vo) Gain A = Vo /Vi Gain in dB

20log(Vo /Vi)



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MODEL WAVEFORMS:

PRECAUTIONS :

1. While taking the observations for the frequency response , the input voltage must be

maintained constant at 20mV.

2. The frequency should be slowly increased in steps.

3. The three terminals of the transistor should be carefully identified.



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RESULT:

The effect of negative feedback (Voltage -Series Feedback ) on the amplifier is

observed. The voltage gain and frequency response of the amplifier are obtained. Also gain-

bandwidth product of the amplifier is calculated.

VIVA QUESTIONS

1. What is meant by Feedback?

2. What are the types of feedback amplifiers? Explain?

3. Draw the circuit for voltage series feedback?

4. What are the differences between positive and negative feedback?

5. What is the effect of negative feedback on gain of an amplifier?

6. What is the formula for voltage gain with negative feedback?

7. What are the other names for positive and negative feedback circuits?

8. What is the formula for input resistance of a voltage series feedback?

9. What is the formula for output resistance of a voltage series feedback?



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3.b. CURRENT-SERIES FEEDBACK AMPLIFIER

AIM: To measure the voltage gain of current - series feed back amplifier.

APPARATUS:


1

2

3

4

5

6

7

8

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:



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THEORY:

When any increase in the output signal results into the input in such a way as to

cause the decrease in the output signal, the amplifier is said to have negative feedback.

The advantages of providing negative feedback are that the transfer gain of the amplifier withfeedback can be stabilized against variations in the hybrid parameters of the transistor or the

parameters of the other active devices used in the circuit. The most advantage of the negative

feedback is that by prepare use of this, there is significant improvement in the frequency

response and in the linearity of the operation of the amplifier. This disadvantage of the

negative feedback is that the voltage gain is decreased.

In Current-Series Feedback, the input impedance and the output impedance are

increased. Noise and distortions are reduced considerably.

PROCEDURE:

1. Connections are made as per circuit diagram.

2. Keep the input voltage constant at 20mV peak-peak and 1kHz frequency. For differentvalues of load resistance, note down the output voltage and calculate the gain by using theexpression


3. Remove the emitter bypass capacitor and repeat STEP 2.And observe the effect offeedback on the gain of the amplifier.

4. For plotting the frequency the input voltage is kept constant at 20mV peak-peak and thefrequency is varied from 100Hz to 1MHz.

5. Note down the value of output voltage for each frequency. All the readings are tabulatedand the voltage gain in dB is calculated by using expression Av =20log (V0 / Vi ) dB

6. A graph is drawn by taking frequency on X-axis and gain on Y-axis on semi log graphsheet

7. The Bandwidth of the amplifier is calculated from the graph using the expression

Bandwidth B.W = f2 – f1.

Where f1 is lower cutt off frequency of CE amplifier

f 2 is upper cutt off frequency of CE amplifier

8. The gain-bandwidth product of the amplifier is calculated by using the expression

Gain-Bandwidth Product = 3-dB midband gain X Bandwidth.



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OBSERVATIONS:

Voltage Gain: Vi = 20 mV

S.NO Output Voltage(Vo) with feedback

Output Voltage (Vo)without feedback

Gain(dB) withfeedback

Gain(dB)withoutfeedback

Frequency Response:

S.NO Frequency (Hz) Output Voltage (Vo) Gain A = Vo /Vi Gain in dB 20log(Vo /Vi)



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MODEL WAVEFORM:

Frequency response



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PRECAUTIONS:

1. While taking the observations for the frequency response , the input voltage must be

maintained constant at 20mV.

2. The frequency should be slowly increased in steps.

3. The three terminals of the transistor should be carefully identified.

4. All the connections should be correct.

RESULT:

The effect of negative feedback (Current-Series Feedback) on the amplifier is

observed. The voltage gain and frequency response of the amplifier are obtained. Also gain-bandwidth product of the amplifier is calculated.

VIVA QUESTIONS

1. What is the effect of Current-Series Feedback amplifier on the input inmpedance of the

amplifier?

2. What is the effect of negative feedback on the Bandwidth of an amplifier?3. State the reason for the usage of negative feedback in an amplifier?

4. What are the fundamental assumptions that are made in studying feedback amplifiers?

5. What are the advantages of providing negative feedback amplifier?

6. What are the ideal characteristics of a voltage amplifier?\

7. Draw the circuit for the current series feedback?

8. What is the other name for current series feedback amplifier?

9. What is the formula for input resistance of a current series feedback?

10. What is the formula for output resistance of a current series feedback?



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4. RC PHASE SHIFT OSCILLATOR

AIM: To construct the RC phase shift oscillator to give output signal of specified frequency.

APPARATUS:


1

2

3

4

5

6

7

8

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:



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THEORY: RC-Phase shift Oscillator has a CE amplifier followed by three

sections of RC phase shift feed back networks. The out put of the last stage is

return to the input of the amplifier. The values of R and C are chosen such that

the phase shift of each RC section is 60º. Thus the RC ladder network producesa total phase shift of 180º between its input and output voltage for the given

frequencies. Since CE Amplifier produces 180 º phases shift the total phase shift

from the base of the transistor around the circuit and back to the base will be

exactly 360º or 0º. This satisfies the Barkhausen condition for sustaining

oscillations and total loop gain of this circuit is greater than or equal to 1, this

condition used to generate the sinusoidal oscillations.

The frequency of oscillations of RC-Phase Shift Oscillator is,

1

f = -----------

2∏RC* √6

PROCEDURE:

1. Make the connection as per the circuit diagram as shown above.

2. Observe the output signal and note down the output amplitude and time period (Td).

3. Calculate the frequency of oscillations theoretically and verify it practically (f=1/Td).

4. Calculate the phase shift at each RC section by measuring the time shifts (Tp) between

the final waveform and the waveform at that section by using the below formula.

MODEL WAVE FORMS:

OUT PUT WAVE FORM :



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OUT PUT WAVE FORM : θ = 600



OBSERVATIONS:

THEORITICAL CALCULATIONS: R = 10000, C = 0.001 μf

1f = -----------------

2∏RC* √6

PRACTICAL CALCULATIONS:



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Td =165*10-6

1f = -----------------

Td

Tp(1) θ 1= --------* 3600 =

Td

Tp(2) θ 2 = --------- * 3600

=

Td

Tp(3) θ 3= ---------- * 3600 =

Td

RESULT:

VIVA QUESTIONS:

1. Mention the conditions for oscillations in RC phase shift oscillator?

2. Give the formula for frequency of oscillations in RC phase shift oscillator?

3. The phase produced by a single RC network is RC phase shift oscillator?

4. RC phase shift oscillator uses positive feedback or negative feedback?

5. The phase produced by basic amplifier circuit in RC phase shift oscillator is?

6. What is the difference between damped oscillations undamped oscillations?

7. What are the applications of RC oscillations?

8. How many resistors and capacitors are used in RC phase shift feed back network.

9. How the Barkhausen criterion is satisfied in RC phase shift oscillator



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5. CLASS B COMPLEMENTARY SYMMETRY PUSH PULL AMPLIFIER

AIM: i)To simulate and verify the efficiency of class B complementary symmetry push pull

amplifier.

II) To study the phenomenon of cross over distortion in a Class B amplifier

APPARATUS:


1

2

3

4

5

67

8

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:



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THEORY:

Complementary means the circuit uses two identical transistors but one is NPN and

other is PNP. The symmetry means the biasing resistors connected in both transistors are

equal. As a result of this, emitter base junction of each transistor is biased with the same

voltage.

During the positive half cycle of ac input the base emitter voltage of both transistors

becomes positive. Under this condition only NPN transistor conducts, while PNP transistor is

cutoff. During this process positive half cycle current flows through load resistor R5.

During negative half cycle of ac input only PNP transistor conducts and NPN transistor is

cutoff and the negative half cycle current flows through R5. We get a complete amplified wave

form of input signal. This amplifier circuit has a unity gain because of the emitter follower

configuration is used

PROCEDURE:

1. Select different components and place them in the grid. 2. Apply the input ac signal voltage of 0.8V (p-p) and simulate the circuit. 3. Observe the output wave form on CRO and measure the output voltage Vo.

4. Now connect the ammeters at collector terminals of NPN and PNP transistors.

5. Disconnect the ac signal from input and measure the collector currents Ic1 and Ic2 in

ammeters. 6. Calculate the efficiency by using practical calculations 7. Compare it with theoretically calculated efficiency

OBSERVATIONS :


VCC/RL

ICQ = --------------2 π

Pin(d.c) = VCC*ICQ

VCC*VCC (VCC) 2

Pin(d.c) = ------------- = ------------

2πRL 2 πRL



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( Vmax –Vmin)*(Imax – Imin)

Po(a.c) = -----------------------------------------

8

VCC

(Imax – Imin) = ------

RL

( Vmax –Vmin) = VCC

VCC*VCC (VCC ) 2

Po(a.c) = ------------------- = -----------

8RL 8RL

Po(a.c) ( VCC ) 2 /8RL

% of efficiency = ----------- *100 = ------------------- *100

Pin(d.c) ( VCC ) 2 /2πRL

π

= -----------*100 = 78.5%

4


IC1 =

IC2 =

IC1+IC2IC = -------------- =

2IC

ICQ = ---------- =2 π

VCC = , Vo(p-p) =

Vo(p-p)2

Po(a.c) = ------------

8RL

Pin(d.c) = VCC*ICQ

Po(a.c)% of efficiency = ----------- *100

Pin(d.c)



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MODEL WAVE FORMS:

RESULT: The frequency response of the Amplifier and Bandwidth is obtained.

VIVA QUESTIONS:

1. Explain complementary and symmetry concept? 2. What is the conduction angle in class B operation? 3. What is the efficiency of class B power amplifier?

4. what will be change in the above circuit if the two transistors are interchanged? 5. what is the formula for out put power in class B power amplifier?



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6. TWO STAGE RC COUPLED AMPLIFIER

AIM: To calculate voltage gain, to observe frequency response.

APPARATUS:


1

2

3

4

5

6

78

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:



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THEORY:

This is most popular type of coupling as it provides excellent audio fidelity.

A coupling capacitor is used to connect output of first stage to input of second stage.

Resistances R1, R2,Re form biasing and stabilization network. Emitter bypass capacitor offers

low reactance paths to signal coupling Capacitor transmits ac signal, blocks DC. Cascade

stages amplify signal and overall gain is increased total gain is less than product of gains of

individual stages. Thus for more gain coupling is done and overall gain of two stages equals to

A=A1*A2

A1=voltage gain of first stage

A2=voltage gain of second stage.

When ac signal is applied to the base of the transistor, its amplified output appears across

the collector resistor Rc.It is given to the second stage for further amplification and signal

appears with more strength. Frequency response curve is obtained by plotting a graph

between frequency and gain in db .The gain is constant in mid frequency range and gain

decreases on both sides of the mid frequency range. The gain decreases in the low frequency

range due to coupling capacitor Cc and at high frequencies due to junction capacitance Cbe.

PROCEDURE: 1. Apply input by using function generator to the circuit.

2. Apply input by using function generator to the circuit and simulate the circuit. 3. Observe the output waveform on CRO. 4. Measure the voltage at

a. Output of the first stage

b. Output of the second stage

5. From the readings, calculate voltage gain of first stage, second stage and overall gain.

Disconnect second stage and then measure output voltage of first stage and

calculate voltage gain. 6. Compare it with the voltage gain obtained when second stage was connected. 7. For plotting the frequency response, the input voltage is kept constant at 2mv (p-p) and

the frequency is varied from 100Hz to 1MHz.

8. Note down the value of output voltage for each frequency. 9. All the readings are tabulated and voltage gain in dB is calculated by using the

expression Av =20 Log 10 (Vo/Vi)



36

10. A graph is drawn by taking frequency on X-axis and gain in dB on Y-axis on a Semi log

graph sheet. 11. The bandwidth of the amplifier is calculated from the graph using the expression

Bandwidth = f2 – f1.

Where f1 = Lower cutoff frequency of CE amplifier.

f2 = Upper cutoff frequency of CE amplifier


DC Analysis :

Calculation of RC& RE :

Given Ic = 1mA, Vce= 5v , VE =2v for the operating point to be approximately for the centre

point with this data.

Ic

IB = ------ =

β

IE = IB + Ic

VE

Choose VE =2v then RE = ------ =

IE

Then Vc =VCC – (VE + Vce ) =

Vc 2v

RC = ------ = -------- = 2kΩ

Ic 1mA

It is recommended that RE must be less than RC selected

RE = 1kΩ and RC = 2kΩ

Calculation of R3 & R4 :

The Thevenins equivalent voltage base

R4

VB = --------- VCC and is equal to sum of VBE & VE.

R3 + R4

VBE + VE =

R4 VB

--------- = ------

R3 + R4 VCC



37

R3 = 2.33* R4

Choose the current flowing through R4 is I4

IC 1mA

I4 = ------ = -------- = 100 μA

10 10

VBE 2.7

R4 = -------- = -------- = 27kΩ

I4 100 μA

R4 = 2.33kΩ, R3 = 27kΩ

Select R3 = 2.2k Ω, R4 = 27k Ω

AC Analysis:

The voltage gain of an amplifier can be taken as

RL’

Av = -------- =

Re

20mv

Where Re = ------- =

IE

RL’

Av = -------- = 10 => RL’ = 250Ω

Re

RL’ = RC//RL = 2.2kΩ// RL => RL = 282Ω

There is resistance offered between collector and emitter choose RL = 300 Ω for ac analysisselect

Ce = 100μf, Cc = 1 μf, Rs =2kΩ



38


Vi1 = 1.5mv, Vo1 =

Vo1

Av1 = -------- =

Vi1

Vi2 = Vo1, Vo2 =

Vo2

Av2 = --------- =

Vi2

Av =Av1*Av2 =

Vo2

Av = ----------- =

Vi1

OBSERVATIONS: -

APPLIED FREQUENCY O/P VOLTAGE

(Vo)

VOLTAGE GAIN

in dB (20 log10Vo/Vi)



39

MODELGRAPH:-

INPUT WAVE FORM:

FIRST STAGE OUTPUT:

SECOND STAGE OUTPUT:

FREQUENCY RESPONSE:



40

PRECAUTIONS:

1) All connections should be tight.

2) Transistor terminals must be identifying properly.

3) Reading should be taken with out any parallax error.

RESULT: Thus voltage gain is calculated and frequency response is observed along with

loading affect.

VIVA QUESTIONS:

1) What is the necessity of cascading?

2) What is 3dB bandwidth?

3) Why RC coupling is preferred in audio range?

4) Which type of coupling is preferred and why?

5) Explain various types of Capacitors?

6) What is loading effect?

7) Why it is known as RC coupling?

8) What is the purpose of emitter bypass capacitor?

9) Which type of biasing is used in RC coupled amplifier?



41

7. CLASS A POWER AMPLIFIER

AIM: To calculate the efficiency of class A power amplifier by simulation

APPARATUS:


1

2

3

4

5

6

7

8

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:





43

VCC

(Imax – Imin) = ---------

RL

(Vmax –Vmin) = VCC

VCC*VCC ( VCC ) 2

Po(a.c) = ------------------- = -----------

8RL 8RL

Po(a.c) ( VCC ) 2 /8RL

% of efficiency = --------- *100 = ------------------- *100 = 25%

Pin(d.c) ( VCC ) 2 /2RL


IC =

Pin(d.c) = VCC*ICQ =

Vo2

Po(a.c) = -------- =

8RL

Po(a.c)

% of efficiency = ------------- *100 =

Pin(d.c)



44

MODEL WAVEFORM:

RESULT: The efficiency of class A Power amplifier is _______

VIVA QUESTIONS:

1. Explain class A operation?

2. What is phase shift of input and out put signals in class A operation?

3. What is the efficiency of class A power amplifier?

4. Distinguish class A and class B operations

5. What is the formula for the in put and out put power in class A power amplifier?



45

8.a.. HARTLEY OSCILLATOR

AIM: To study and calculate frequency of oscillations of Hartley oscillator. Compare the

frequency of oscillations, theoretically and practically.

APPARATUS:


1

2

3

4

5

6

7

8

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM:



46

THEORY:

Hartley oscillator is very popular and is commonly used as a local oscillator in

radio receivers. It has two main advantages viz... Adaptability to wide range of frequencies and

easy to tune. The tank circuit is made up of L1, L2, and C1. The coil L1 is inductively coupled

to coil L2, the combination functions as auto transformer. The resistances R2 and R3 provide

the necessary biasing. The capacitance C2 blocks the d.c component. The frequency of

oscillations is determined by the values of L1, L2 and C1 and is given by,

F=1/(2(C1(√L1+L2)))

The energy supplied to the tank circuit is of correct phase. The auto transformer provides 180˚

out of phase. Also another 180˚ is produced

By the transistor. In this way, energy feedback to the tank circuit is in phase with the generated

oscillations.

PROCEDURE:


2. Connect CRO at output terminals and observe wave form.

3. Calculate practically the frequency of oscillations by using the expression.

F=1/T, Where T= Time period of the waveform

4. Repeat the above steps 2, 3 for different values of L1 and note down practical values of

oscillations of colpitts oscillator.

5. Compare the values of frequency of oscillations both theoretically and Practically.

OBSERVATIONS:

CAPACITANCE(μF) Theoritical frequency(KHZ)

Practical frequency(KHZ)



47

MODEL GRAPH:

RESULT: Frequency of oscillations is calculated and compared with theoretical values.

VIVA QUESTIONS:

1. What are the applications of LC oscillations?

2. What type of feedback is used in oscillators?

3. What the expression for frequency of oscillations?

4. Whether an oscillator is dc to ac converter?

5. What is the loop gain of an oscillator?

6. What is the difference between amplifier and oscillator?

7. What is the condition for oscillations?

8. How many inductors and capacitors are used in Hartley Oscillator?



48

8.b. COLPITT’S OSCILLATOR

AIM: To study and calculate frequency of oscillations of colpitt’s oscillator.

APPARATUS:


1

2

3

4

5

6

7

8

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUITDIAGRAM:



49

THEORY:

The tank circuit is made up of L1,C4 and C5 .The resistance R2 and R3 provides the

necessary biasing. The capacitance C2 blocks the D.C component. The frequency of

oscillations is determined by the values of L1,C4 and C5, and is given by

f = 1 / (2 (CTL1)1/2) Where CT = C1C2 / ( C1 + C2)

The energy supplied to the tank circuit is of correct phase. The tank circuit provides 1800 out of

phase. Also the transistor provides another 1800 . In this way, energy feedback to the tank

circuit is in phase with the generated oscillations.

PROCEDURE:

1. connections are made as per circuit diagram.

2. Connect CRO output terminals and observe the waveform.

3. Calculate practically the frequency of oscillations by using the expression

f = 1 / T ( T= Time period of the waveform)

4. Repeat the above steps 2,3 for different values of L, and note down the practically

values of oscillations of the collpitt’s oscillator.5. Compare the values of oscillations both theoritically and practically.

OBSERVATIONS:

Inductance ( mH ) Theoretical Frequency( Hz ) Practical Frequency( Hz )



50

MODELWAVEFORM:

PRECAUTIONS:

1. The connections should be correct.

2. Transistor terminals should be identified properly.

3. Readings should be taken without parallalox error.

RESULT: Frequency of oscillations of colpitts oscillator is measured practically and compared

with theoretical values .

VIVA QUESTIONS:

1. What are the applications of LC oscillators?

2. What type of feedback is used in oscillators?

3. What is the expression for the frequency of oscillations of colpitt’s oscillator?

4. Is an oscillator DC to AC converter?

5. What is the loop gain and loop phase shift of an oscillator?

6. How does colpitts differ from Hartley?

7. Which pair in circuit forms stabilizing circuit?

8. What is the function of input and output capacitor?

9. What is the condition for sustained oscillations in this oscillator?

10. Output capacitor acts as a?





52

THEORY:

The signal to be amplified is applied between the terminals base and emitter. The tank

circuit is tuned (i.e L or C may be varied) in such a way that the resonant frequency becomes

equal to the frequency of the input signal. At resonance the tuned circuit offers very high

impedance and thus, the given input signal is amplified by the amplifier and appears with large

value across it and other frequencies will be rejected. So the tuned circuit selects the derived

frequency and rejects all other frequencies.

PROCEDURE:


2. Set Vi = 50mV using the signal generator

3. Keeping the input voltage constant, vary the frequency from 10Hz to 1MHz in regular

steps and note down the corresponding output voltage.

4. Plot the graph of gain in dB vs Frequency (Hz)

5. Calculate the bandwidth from the graph.

OBSERVATIONS:

THEORITICAL CALCULATIONS: L = 100mH, C = 0.1 μf

1 1

f = ----------------- = ----------------------------- = 1.6 kHZ

2∏√LC 2∏√100*10-3*0.1* 10-6


Td =

1

f = ----------------- =

Td

PRECAUTIONS:

Transistor terminals must be identified properly



53

RESULT:

VIVA Questions:

What is the purpose of tuned amplifier?

What is Quality factor?

Why should we prefer parallel resonant circuit in tuned amplifier.

What typef of tuning we need to increase gain and bandwidth.?

What are the limitations of single tuned amplifier?

What is meant by Stagger tuning?

What is the conduction angle of an tuned amplifier if it is operated in class B mode?

What are the applications of tuned amplifier?



54

10. WIEN BRIDGE OSCILLATOR

AIM : To construct and simulate the Wien Bridge oscillator and then to verify the frequency of

oscillation theoretically and by simulation.

APPARATUS:


1

2

3

4

5

6

78

Transistor BC 107


Function Generator

CRO

Resistors

Capacitors-

Bread Board

Connecting Wires

(0-30V, 1A)

CIRCUIT DIAGRAM :



55

THEORY:

The Wien Bridge oscillator consists of two RC coupled Amplifiers which provide a phase

of 360° or 0°. A balanced bridge is used as the feedback network which has no need to provide

additional phase shift to satisfy the Barkhausen criteria for sustain the oscillations. The loop gain

of the circuit is greater than equal to one, this condition generates the sinusoidal oscillations.

The feed back network consists of Lead Lag network and a Voltage divider. The Lead Lag

network provides the positive feedback to the input of the first stage and the voltage divider

provides a negative feed back to the emitter of the first stage.The frequency of oscillations is

1

f = ----------

2πRC

PROCEDURE:

1. Select different components and place them in the grid and simulate the circuit.

2. Observe the output signal and note down the output amplitude and time period (T). 3. Calculate the practical frequency of oscillation (f=1/T) and compare with the theoretical

value.

OBSERVATIONS :

Theoretical calculation :

1 1

f = ---------- = ------------- = 16 kHz

2πRc 6.283*10-5

Practical calculation :

1

f = ------- =

T

MODEL WAVE FORM:



RESULT : The frequency of the Wein bridge oscillator is calculated and is verified

VIVA QUESTIONS:

1. Give the formula for frequency of oscillations?

2. What is the condition for wien bridge oscillator to generate oscillations?

3. What is the total phase shift provided by the oscillator?

4. What is the function of lead-lag network in Wein bridge oscillator?

5. which type of feedback is used in Wein bridge oscillator

6. What is gain of Wein bridge oscillator?

7. what are the application of Wein bridge oscillator

8. What is the condition for oscillations?

9. What is the difference between damped oscillations un damped Oscillations?

10.Wein bridge oscillator is either LC or RC oscillator.

Documents

Electronic Circuit Analysis Laboratory Mannual