sdc

RCPIT, Shirpur Department of E&TC

Semiconductor Devices & Circuits 1

Semiconductor Devices & Circuits

SE (E&TC) SEM-I

Laboratory Notebook

Department of Electronics & Telecommunication

R C Patel Institute of Technology, Shirpur



The Shirpur Education Societys

R. C. Patel Institute of Technology, Shirpur

CERTIFICATE

This is to certify that Mr / Miss ________________________________Of SecondSecondSecondSecond

YearYearYearYear Electronics & TelecommunicationElectronics & TelecommunicationElectronics & TelecommunicationElectronics & Telecommunication branch, Roll No. ______Has performed practical

work satisfactorily in the Subject Semiconductor Devices & CircuitSemiconductor Devices & CircuitSemiconductor Devices & CircuitSemiconductor Devices & Circuit, in the premises of the

Department of Electronics & Telecommunication Engineering during the academic year

20 -20 .

Date: / / 20 Signature of the Teacher

Place: Shirpur

Principal Head of Department



CONTENTS

Expt No Experiment Title Page No

1 For a half wave rectifier with capacitor filter find line and load regulation and ripple factor.

4

2 For full wave rectifier with capacitor filter find line and load regulation and ripple factor.

13

3 For Bridge wave rectifier with capacitor filter find line and load regulation and ripple factor.

23

4 Square wave testing of an amplifier used to find lower and higher cut off frequency.

31

5 For two cascaded CE-CE stages, find voltage gain and bandwidth.

36

6 For two stage cascode (CE-CB) amplifier determine voltage gain and bandwidth.

43

7 Determine I/P and O/P impedances and voltage gain of a CE stage followed by CC.

49

8 Study the effect of bypass capacitor on frequency response of single stage CE amplifier

55

9 Study frequency response of CSFET

61

10 Determine I/P and O/P impedances and voltage gain and current gain for CSFET.

67

11 Plot characteristics of CSFET. Determine amplification factor, transconductance and dynamic resistance

72

12 Determine H-parameters

78



Laboratory Report

Experiment No.1

Batch Code: __________________

Name of Student: ____________________________________________ Roll No. ____________

Date of Lab: ______________________ Date of Submission: ____________________

Evaluations

1) Lab Attendance [2]

a) Pre lab work done (if any) [1] ___________

b) Records of Lab Notebook [1] ___________

2) Observations and Conclusion [2] ___________

3) Oral [1] ___________

Overall Marks ___________

Sign of practical In-charge



Experiment No.1

Objective: To plot regulation characteristics of Half Wave Rectifier.

Pre-lab: What is a rectifier? Definition of Ripple factor, TUF, Rectification efficiency

Equipments needed:

Sr.No Name of Equipments Specifications Quantity 1 Experiment kit 1 2 Power Supply 0-30V DC 1 3 Auto transformer 1

4 CRO 20 MHz Dual channel

1

5 Digital Multimeter 1

Theory:

Rectifier is defined as an electronic device used for converting AC voltage to DC voltage.

Half Wave Rectifier: It converts an AC voltage to pulsating DC voltage, using only

one half cycle of the applied AC voltage. The rectifying diode conducts only during one half of the AC cycle only.

Let Vi be the voltage of the primary of the transformer and given by equation-

Vi = Vm sinwt Vm>>Vr

Where Vr is the cut-off voltage of the diode.

During positive half cycle of the input signal, the anode of the diode becomes more positive w.r.t. cathode and hence diode conducts. For the ideal diode the forward bias voltage drop is zero and hence the input half appears at the output (as diode acts as a closed switch.)

During negative half cycle of the input signal, the anode becomes negative and cathode is at positive, hence diode becomes reverse bias in which it acts as an open switch. During the negative half cycle no input appears at the o/p. When capacitor is used as filter, then it is connected across the load resistor. The Property of the capacitor is that



it allows to pass only AC components present in o/p and blocks the DC. As the o/p across the load resistor is not constant but it contains Positive half cycles only, during positive half cycle capacitor charges and when the Negative half cycle appears for which no o/p is obtained, due to charged capacitor Voltage across capacitor remains always constant.

Procedure: A) For with capacitor filter: (Short points A & C) 1. Trace and draw the circuit diagram. 2. For load regulation ammeter is connected between points A and B.Voltmeter is Connected between point B and Ground. Vary load resistor RL and note readings Of voltage VL and current IL. 3. For line regulation, Voltmeter is connected between point B and ground. Vary Supply voltage through auto-transformer and note readings of supply voltage Vi And o/p voltage Vo. At that time keep RL constant. 4. For ripple factor, short A and B, connect CRO between B to ground at constant load and measure Vm on CRO. 5. Draw the waveforms on the graph paper.

B) For without capacitor filter: (Open points A & C) 1. Trace and draw the circuit diagram. 2. For load regulation ammeter is connected between points A and B.Voltmeter is connected between point B and Ground. Vary load resistor RL and note readings of voltage VL and current IL. 3. For line regulation, Voltmeter is connected between point B and ground. Vary Supply voltage through auto-transformer and note readings of supply voltage Vi And o/p voltage Vo. At that time keep RL constant. 4. For ripple factor, short A and B, connect CRO between B to ground at

Constant load and measure Vm on CRO. 5. Draw the waveforms on the graph paper.



Observation Table: Line Regulation (Without Capacitor Filter):

Line Regulation (With Capacitor Filter):

Sr.NO. Vi(volts) Vo(volts) 1

2

3

4

5

6

7

8

9

10


2

3

4

5

6

7

8

9

10



Load Regulation (Without Capacitor Filter):

Sr.NO. IL(mA) VL(volts) 1 2 3 4 5 6 7 8 9

10

Load Regulation (With Capacitor Filter):


10



Calculations:

Load regulation=

No load voltage Full load Voltage *100 Full load voltage

Line regulation=

Vo/Vi *100

Load Regulation:

1) (Without Capacitor Filter):

2) (With Capacitor Filter):

Line Regulation:





Result:

Type of Regulation Without capacitor filter With capacitor filter

% Load Regulation

% Line Regulation

Conclusion: _________________________________________________________________________

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Lab Assignment1: What is rectifier .Also give the types of rectifier?

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Lab Assignment2: What is line regulation and load regulation?

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Lab Assignment 3: What is ripple factor of HWR without capacitor filter?

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Conclusion: _________________________________________________________________________

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Laboratory Report

Experiment No. 2

Batch Code: __________________

Name of Student: _________________________________ Roll No. ____________


Evaluations

1) Lab Attendance [2] a) Pre lab work done (if any) [1] ___________ b) Records of Lab Notebook [1] ___________

2) Observations and Conclusion [2] ----------------

3) Oral [1] --------------

Overall Marks ----------------

Sign of Practical In-charge



Experiment No. 2

Objective To plot regulation characteristics of Full Wave Rectifier.

Pre-lab: What is rectification efficiency? Expression for rectification efficiency.

Equipments needed:

Sr.No Name of Equipments Specification Quantity 1 Experiment kit 1 2 Power Supply 0-30V DC 1 3 Auto transformer 1 4 CRO 20 MHz Dual

channel 1

5 Digital Multimeter 1 6 Experiment kit 1

Theory:

Rectifier is defined as an electronic device used for converting AC voltage to DC voltage.

Full Wave Rectifier: It converts an AC voltage to pulsating DC voltage,using both half cycles of the applied AC voltage. The rectifying diode conducts during both half cycles of the AC cycle.

During positive half cycle of the input signal, the anode of the diode D1 becomes more positive w.r.t. cathode and at the same time anode of D2 becomes negative and hence diode D2 does not conduct and o/p current flows through only D1. During negative half cycle of the input signal, the anode of diode D1 becomes negative and anode of D2 becomes positive ,hence diode D1 becomes reverse bias in which it acts as an open switch and only D2 conducts and provides the load current. When capacitor is used as filter, then it is connected across the load resistor. The Property of the capacitor is that it allows to pass only AC components present in o/p

and blocks the DC. As the o/p across the load resistor is not constant but it contains Positive half cycles only which does not give constant DC o/p. During positive half Cycle capacitor charges and when the next positive half cycle appears in between this time, due to charge stored across capacitor o/p across load resistor appears almost



constant.

Procedure: A) For with filter: (Short points A & C) 1. Trace and draw the circuit diagram. 2. For load regulation ammeter is connected between points A and B. Voltmeter is Connected between point B and Ground. Vary load resistor RL and note readings of voltage VL and current IL. 3. For line regulation, Voltmeter is connected between point B and ground. Vary Supply voltage through auto-transformer and note readings of supply voltage Vi And o/p voltage Vo. At that time keep RL constant. 4. For ripple factor, short A and B, connect CRO between B to ground at constant load and measure Vm on CRO. 5. Draw the waveforms on the graph paper.

B) For without filter: (Open points A & C) 1. Trace and draw the circuit diagram. 2. For load regulation ammeter is connected between points A and B. Voltmeter is connected between point B and Ground. Vary load resistor RL and note readings of voltage VL and current IL. 3. For line regulation, Voltmeter is connected between point B and ground. Vary Supply voltage through auto-transformer and note readings of supply voltage Vi And o/p voltage Vo. At that time keep RL constant. 4. For ripple factor, short A and B , connect CRO between B to ground at Constant load and measure Vm on CRO. 5. Draw the waveforms on the graph paper.



Observation Table: Line Regulation: (Without capacitor Filter)-

Line Regulation : (With Capacitor Filter)- Sr.NO. Vi(volts) Vo(volts)

1

2

3

4

5

6

7

8

9

10


2

3

4

5

6

7

8

9

10



Load Regulation :( Without capacitor Filter)-

Sr.NO. IL(mA) VL(volts) 1

2

3

4

5

6

7

8

9

10

Load Regulation (With capacitor Filter)-

Sr.NO. IL(mA) VL(volts) 1

2

3

4

5

6

7

8

9

10



Calculations:

Load regulation=


Line regulation=

Vo/Vi *100

Load Regulation:



Line Regulation:





Result:

Type of Regulation Without capacitor filter With capacitor filter

% Load Regulation

% Line Regulation

Conclusion: ________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________



Lab Assignment1: What is full wave rectifier? Why it is called so? Ans:-----------------------------------------------------------------------------------------

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Lab Assignment2: Derive expression for ripple factor of a full wave rectifier with Capacitor filter?

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Laboratory Report

Experiment No. 3

Batch Code: __________________

Name of Student: _______________________ Roll No. ____________


Evaluations


2) Observations and Conclusion [2] -----------------

3) Oral [1] ---------------


_

Sign of practical In-charge



Experiment No. 3

Objective: - To plot regulation characteristics of Bridge Wave Rectifier.

Pre-lab: What is PIV? What is PIV of BWR?

Equipments needed:-

Sr.No Name of Equipments Specification Quantity 1 Experiment kit 1 2 Power Supply 0-30V DC 1 3 Auto transformer 1 4 CRO 20 MHz Dual

channel 1


Theory:

Rectifier is defined as an electronic circuit used for converting AC voltage to DC voltage.

Bridge Wave Rectifier: It converts an AC voltage to pulsating DC voltage, using

only one half cycle of the applied AC voltage. The rectifying diode conducts only during one half of the AC cycle only. During positive half cycle of the input signal, the point A is more positive w.r.t.

point B. The current path is point A through D1, load, D3 and backs to the point B.During this time diodes D2 and D4 are reverse bias. During negative half cycle of the input signal, the point B is More positive than

point A.The current path is point B through D2, load D4 and back to the point A.During this time D3 and D4 are reverse bias. It is interesting to note that, in both the half cycles the direction of current flow through the load is same. Thus we obtain full wave rectification, the waveform being same. When capacitor is used as filter, then it is connected across the load resistor. The Property of the capacitor is that it allows to pass only AC components

present in o/p and blocks the DC. As the o/p across the load resistor is not constant but it contains Positive half cycles only, during positive half cycle capacitor charges



and when the Negative half cycle appears for which no o/p is obtained, due to charged capacitor Voltage across capacitor remains always constant.

Procedure:

A) For with capacitor filter: (Short points A & C) 1. Trace and draw the circuit diagram. 2. For load regulation ammeter is connected between points A and B.Voltmeter is Connected between point B and Ground. Vary load resistor RL and note readings of voltage VL and current IL. 3. For line regulation, Voltmeter is connected between point B and ground. Vary Supply voltage through auto-transformer and note readings of supply voltage Vi And o/p voltage Vo. At that time keep RL constant. 4. For ripple factor, short A and B, connect CRO between B to ground at constant load and measure Vm on CRO. 5. Draw the waveforms on the graph paper.

B) For without capacitor filter: (Open points A & C) 1. Trace and draw the circuit diagram. 2. For load regulation ammeter is connected between points A and B.Voltmeter is connected between point B and Ground. Vary load resistor RL and note readings of voltage VL and current IL. 3. For line regulation, Voltmeter is connected between point B and ground. Vary Supply voltage through auto-transformer and note readings of supply voltage Vi And o/p voltage Vo. At that time keep RL constant. 4. For ripple factor, short A and B, connect CRO between B to ground at Constant load and measure Vm on CRO.

5. Draw the waveforms on the graph paper.



Observation Table: Line Regulation :( Without capacitor Filter)-

Line Regulation :(With capacitor Filter)- Sr.NO. Vi(volts) Vo(volts)

1

2

3

4

5

6

7

8

9

10


2

3

4

5

6

7

8

9

10



Load Regulation: (Without capacitor Filter)-


10 11

Load Regulation (With capacitor Filter)-


10 11



Calculations:

Load regulation=


Line regulation=

Vo/Vi *100

Load Regulation:



Line Regulation:





Result:

Types of Regulation Without capacitor filter With capacitor filter

% Load Regulation

% Line Regulation

Conclusion: _________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________



Lab Assignment1: Differentiate between FWR and BWR.

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Laboratory Report

Experiment No. 4

Batch Code: __________________

Name of Student: _______________________________________ Roll No. ____________


Evaluations



3) Oral [1] ---------------


Sign of Practical In-Charge



Experiment No. 4

Objective - Square wave testing of an amplifier to find lower and higher cut-off freq.

Pre-lab: Advantage of square wave testing of an amplifier. What is bandwidth?

Equipments needed:-

Sr.No Name of Equipments Specification Quantity 1 Experiment Kit

1 2 Power Supply 0-30 V DC

1 3 Function Generator Sine wave o/p

1 4 CRO 20 MHz Dual

Channel 1

5 Connecting wires

Theory:

It is the method to find out lower and higher cut off freq. of an amplifier without plotting the actual freq. curve is known as square wave testing. The reason for

choosing square wave signal can be cleared from the following. 1) If we examine the fourier series expansion of square wave , we get a series of

sinusoidal components of different magnitude and freq. .So by Square wave testing, we are checking the amplifier response for series of sinusoidal components of the different amplitudes and freq.

2) The use of square wave testing significantly less time consuming than applying a series of sinusoidal signals at diff. freqs. And magnitude to test the freq. response of the amplifier.

3) Square waveform is one which permits small distortions to stand out clearly. For upper 3db freq.:

Fh = 0.35/tr For lower 3 db freq.: FL = p* f / 100 pi



Procedure:

1) Set i/p freq. square wave to 20 mV at 1 KHz. Keep i/p freq. of square wave low

(100 Hz) for FL, observe the waveform on CRO .Calculate P and from P calculate

FL.

2) Keep i/p freq. high (10 KHz / 100 KHz)for FH ,observe waveform on CRO

Measure rise time (tr) .Calculate FH.

3) Set i/p freq. sine wave and observe the response.

Observation Table:

1) Lower cut-off freq:

Sr.NO. Freq. (Hz) V(volts) V (volts) P=V-V/V*100 FL =P*f/100pi 1

2

3

Mean FL = FL1+FL2+FL3 / 3

2) Higher cut-off freq:

Mean FH = FH1+FH2+FH3 / 3

Sr.NO. Freq. (Hz) T1 (s) T2 (s) Tr = T2-T1 FH =(0.35/tr) 1

2

3



Calculations:

Result:

Mean FL =

Mean FH =

Bandwidth = Mean FH - Mean FL

Conclusion:-----------------------------------------------------------------------------------------------

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Lab Assignment 1: What is bandwidth of an amplifier?

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Lab Assignment 2: Explain the need of square wave testing of an amplifier?

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Laboratory Report

Experiment No. 5

Batch Code: __________________

Name of Student: ____________________________________________ Roll No.

Date of Lab: Date of Submission:

Evaluations


2) Observations and Conclusion [2] -----------------

3) Oral [1] ----------------





Experiment No. 5

Objective: - Find voltage gain and bandwidth for CE-CE amplifier. Pre-lab: Compare CE, CC, and CB configuration.

Equipments needed:-

Sr.No Name of Equipments Specification Quantity 1 Experiment Kit 1 2 Power Supply 0-30 V DC 1 3 Function Generator Sine wave o/p 1 4 CRO 20 MHz Dual

Channel 1

Theory:-

If the voltage or power gain obtained from A single stage small signal amplifier is not sufficient for practical, we must have to use more than one stage of amplification to

Achieve necessary voltage and power gain. Such an amplifier is called as a multistage amplifier. A two stage amplifier is shown in fig .

In this amplifier o/p of first stage is fed as i/p to the second stage, such a connection is commonly referred to as cascading. In amplifiers, cascading is also done to achieve i/p and o/p impedances for specific applications.

A multistage amplifier having two or more stages of CE amplifier is called as cascaded amplifier. So overall voltage gain of the amplifier is the product of voltage gain of individual stage .Therefore we get much higher voltage in cascaded amplifier. V1 = i/p voltage of first stage V2 = o/p voltage of first stage

V3 = i/p voltage of second stage V4 = o/p voltage of second stage Therefore Av1 = V2/V1 Av2 = V4/V3

Overall voltage gain = Av = V4/V1 AVS = V4/V3 * V2/V1

Avs = Av2 * Av1



Observation Table: Vi = 20mV

Sr.No. Freq. Output(Vo) Voltage gain (Vo/Vi) 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27



Calculations:

Bandwidth = FH - FL

Result: -----------------------------------------------------------------------------------------------

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28

29

30

31

32

33

34

35

36

37

38

39

40



Conclusion: --------------------------------------------------------------------------------------------

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Lab assignment 1: What is the need of using multistage amplifier?

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Lab assignment 2: Write down expressions for Av, Ri and Ro of CE amplifier

with and without emitter resistance.

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Lab assignment 3: Compare CE, CC, CB configuration on the basis of Av ,Ai ,Ri and Ro

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Laboratory Report

Experiment No. 6

Batch Code: __________________



Evaluations



3) Oral [1] ---------------





Experiment No. 6

Objective: - Find voltage gain and bandwidth for CE-CB Cascode amplifier. Pre-lab: What is difference between Cascade and cascode amplifier?

Equipments needed:-


Channel 1

Theory:

Multistage Amplifier: If the voltage or power gain obtained from A single stage small signal amplifier is not sufficient for practical, we must have to use more than one

stage of amplification to achieve necessary voltage and power gain. Such an amplifier is called as a multistage amplifier. A two stage amplifier is shown in fig .

In this amplifier o/p of first stage is fed as i/p to the second stage, such a Connection is commonly referred to as cascading. In amplifiers, cascading is also done to achieve i/p and o/p impedances for specific applications. Depending upon the type of amplifier, used in individual stages, multistage amplifiers can be classified into several types. A multistage amplifier having two or more stages of CE amplifier is called as Cascaded amplifier. A multistage amplifier with CE as a first stage and CB as a second Stage is called cascade amplifier. It is one approach to solve the low impedance problem of CB circuit. The cascade amplifier gives high i/p impedance stage of a CE amplifier stage as well as good voltage gain and high freq. performance of CB circuit.



Observation Table: Vi = 20mV

Sr. No. Freq. Output(Vo) Voltage gain (Vo/Vi) 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26



Calculations:

Bandwidth = FH - FL

Result: -----------------------------------------------------------------------------------------------

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27

28

29

30

31

32

33

34

35

36

37

38

39

40



Conclusion: -------------------------------------------------------------------------------------------

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Lab assignment 1: What is the need of using multistage amplifier? What is Cascode

amplifier?

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Laboratory Report

Experiment No. 7

Batch Code: __________________

Name of Student: ____________________________________________ Roll No: ____________


Evaluations



3) Oral [1] ---------------





Experiment No. 07

Objective: - Determine i/p , o/p impedance and voltage gain of CE stage followed by CC Stage. Pre-lab: Need of multistage amplifier.

Types of coupling in multistage amplifier

Equipments needed:-

Sr.No Name of Equipments Specification Quantity 1 Experimental Kit

1 2 Power Supply 0-30 V DC 1 3 Function Generator Sine wave o/p 1 4 CRO 20 MHz Dual

Channel 1

Theory:

If the voltage or power gain obtained from A single stage small signal amplifier is not sufficient for practical, we must have to use more than one stage of amplification to

achieve necessary voltage and power gain. Such an amplifier is called as a multistage amplifier. A two stage amplifier is shown in fig .

In this amplifier o/p of first stage is fed as i/p to the second stage, such a connection is commonly referred to as cascading. In amplifiers, cascading is also done to achieve i/p and o/p impedances for specific applications. A multistage amplifier having two or more stages of CE amplifier is called as cascaded amplifier.

Analysis of amplifier: For Second Stage:

Current gain: Current gain of the particular stage is given by-

AI = -hf 1 + ho ZL

And that for CE amplifier with unbypassed RE is given by, For the second stage ZL = RE2 and the current gain of the second stage is,



AI2 = -Ie2 - hfe =

Ib2 1 + hoc RE2

The i/p impedance Ri of the particular stage is given by,

Ri = hi + hr AI ZL

For the second stage,

Ri2 = hic + hrc AI ZL

Thus ,the CC stage has a high i/p impedance.

The voltage gain of a particular stage is

Av = AI ZL Zi

For Second stage

Av2 = Vo = Ai2 Re2 V2

The first stage

RL1 = RC1 II RB

Current gain,

AI1 = -Ic1 = - hfe Ib1 1+ hoe RL1

The i/p impedance of the first stage, which is also the i/p impedance of the cascaded

Amplifier is, Ri1 = hie + hre AI1 RL1

The voltage gain of the first stage is , AV1 = V2 = AI1 RL1 V1 Ri1



The o/p admittance of the first transistor Q1, Yo1 = hoe - hfe *hre hie + Rs

The o/p impedance of the first stage, Ro1 = 1/Yo1

The o/p impedance taking Rc1 into account is Rot1 = Ro1 II Rc1

The o/p admittance of the second stage, Yo2 = hoc hfc * hrc hic + Rot1

o/p impedance, Ro2 = 1/Yo2

The amplifier o/p impedance taking RE2 into account is Ro2II RE2

Hence, Ro2 = Ro2 * RE2 Ro2+Re2

The overall voltage gain taking the source impedance into account,

Avs = - Vo = Av Ri1 Vs Ri1 + Rs

Result:

1) Av =

2) Ri =

3) Ro =

Conclusion:________________________________________________________________________________________________________________________________________

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Lab Assignment: Differentiate CE, CC, CB amplifiers on the basis of Av, Ai, Ri and Ro.

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Laboratory Report

Experiment No. 8

Batch Code: __________________



Evaluations



3) Oral [1] ---------------


Sign Practical In-Charge



Experiment No. 08

Objective: Study the Effect of bypass Capacitor on Frequency Response of Single Stage

CE Amplifier. Pre-lab: What is frequency response? Need of coupling and bypass capacitors

Equipments needed:-

Sr.No Name of Equipments Specification Quantity 1 Experimental Kit

1 2 Power Supply 0-30 V DC 1 3 Function Generator Sine wave o/p 1 4 CRO 20 MHz Dual

Channel 1

Theory:

The bypass capacitor is connected across the emitter resistance RE in BJT amplifier as shown in figure.

1. At medium and high freq. the bypass capacitor CE offers a very low reactance. Therefore it can be replaced by a short circuit as shown fig. Therefore the Impedance ZE which is the parallel combination of RE and XCE will have a Zero .value. Thus RE is bypassed successfully and the bypass capacitor CE has no effect On the freq. response of the amplifier at medium and high freq.

2. But, at low freq. the reactance XCE is not equal to zero,butit has some finite Value. Thus the parallel combination of RE and CE will offer a finite Impedance.So RE is not properly bypassed.

3. Let the impedance of parallel combination be denoted by RE 4. We know that the gain of the CE amplifier with bypassed RE is given by AV= - hfe RL

hie



And that for CE amplifier with unbypassed RE is given by, Av = -hfe RL/hie+(1+hfe)RL So if RE is not properly bypassed then voltage gain will decrease.

Procedure: 1. Trace and draw circuit diagram as shown in Fig. 2. Set supply voltage of +10V. DC input, at input 20mV and 1000Hz. 3. See the output on CRO when XY open i.e. without bypass capacitor effect and

note down it, then find out voltage gain for the same. 4. See the output on CRO when XY shorted i.e. with bypass capacitor effect and

note down it, then find out voltage gain for the same. 5. Then select the position when XY open and vary the freq. from 10 Hz to 1MHz. 6. Repeat the step-5, when XY is shorted.

Observation Table:

When X-Y is Open - Without bypass capacitor When X-Y short - With bypass capacitor

S N Freq. Output voltage Voltage gain X-Y open

(Without bypass) X-Y short

(Withbypass) X-Y open

(Without bypass) X-Y short

(Withbypass) 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18



Calculations:

Result:

__________________________________________________________________________________________

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19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45



Conclusion __________________________________________________________________________________________

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Lab Assignment 1: What is the effect of bypass capacitor on freq.?

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Lab Assignment 2: What is a feedback amplifier?

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Laboratory Report

Experiment No. 9

Batch Code: __________________

Name of Student: ____________________________________________ Roll No: ____________


Evaluations



3) Oral [1] ---------------





Experiment No. 9

Objective: - To plot frequency response of FET amplifier and find bandwidth.

Pre-lab: What is Frequency response and bandwidth of an amplifier?

Equipments needed:-


Channel 1

Theory:

FET Amplifiers: FET amplifiers provide an excellent voltage gain with the added advantage of high input impedance. There are three basic FET circuit configurations as follows:

1. Common Source 2. Common Drain

3. Common Gate.

The common source amplifier which provides good voltage amplification is most frequently used. The common drain amplifier with high i/p impedance and nearly unity voltage gain is used as buffer amplifier and the common gate amplifier is used as high freq. amplifier. The small signal current source model for FET in CS configuration. In Common source amplifier source resistor is used to set the Q-point but bypassed by capacitor for mid-freq. operation.

Procedure: 1) Trace and draw the ckt.dia. 2) Apply 20 V DC using power supply. 3) Set the sinusoidal i/p 50mV at 1 KHz from function generator.



4) Observe output on CRO & measure amplitude of output. 5) Vary i/p freq. from 10Hz to 1 MHz and note down readings for each freq. 6) Draw graph of freq. vs gain & calculate BW.

Observation Table:

Sr.No

Freq. Output(Vo) Voltage gain(Vo/Vi)

Sr.No

Freq. Output(Vo) Voltage gain(Vo/Vi)

1 23

2 24

3 25

4 26

5 27

6 28

7 29

8 30

9 31

10 32

11 33

12 34

13 35

14 36

15 37

16 38

17 39

18 40

19 41

20 42

21 43

22 44



Calculations:

Result: ___________________________________________________________________________

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Conclusion: _________________________________________________________________________________

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Lab Assignment 1. - In which region Q-point of FET is located? Why?

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Lab Assignment 2. - Draw V-I characteristics of CSFET.



Laboratory Report

Experiment No. 10

Batch Code: __________________



Evaluations



3) Oral [1] ---------------


Sign Practical In-Charge



Experiment No. 10

Objective: Determine input, output impedances and voltage gain for common source FET amplifier.

Pre-lab: Compare BJT and FET. Construction of FET.

Equipments needed: Sr.No Name of Equipments Specification Quantity

1 Experiment kit 1 2 Power Supply 0-30V DC 1 3 Auto transformer 1 4 CRO 20 MHz Dual

channel 1


Theory:

FET Amplifiers: FET amplifiers provide an excellent voltage gain with the added a

sdc

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