Eca-labmanual--multisim.pdf 6f581a

SRINIVASA INSTITUTE OF TECHNOLOGY AND SCIENCE,KADAPA

ELECTRONIC CIRCUIT ANALYSIS LABORATORY MANUAL

DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING

1


INDEX S.NO

NAME OF THE EXPERIMENT

PAGE.NO

SOFT WARE 1. COMMON EMITTER AND COMMON SOURCE

1

AMPLIFIER. 2. TWO STAGE RC COUPLED AMPLIFIER.

9

3. CURRENT SHUNT AND FEED BACK AMPLIFIER

16

4. RC PHASE SHIFT OSCILLATOR.

21

5. CLASS A AND CLASS AB POWER AMPLIFIERS

25

6. HIGH FREQUENCY COMMON BASE AMPLIFIER.

32

HARD WARE 7. TWO STAGE RC COUPLED AMPLIFER

38

8. CURRENT SHUNT AND FEED BACK AMPLIFIER

41

9.

44

CLASS A AND CLASS AB POWER AMPLIFIERS.

10. SINGLE TUNED VOLTAGE AMPLIFIER

48

11. SERIES VOLTAGE REGULATOR

50

12. SHUNTVOLTAGE REGULATOR.

52

2


COMMON EMITTER AND COMMON SOURCE AMPLIFIER AIM:

To design CE single stage amplifier with potential divider circuit using

NPN Transistor 2N2923 for the specifications : IC= 3 mA, Vce = 10v,β = 190, & IR1 = 32IB . DC values (Voltage and current) at various nodes using MULTISIM. APPARATUS: - Multisim Soft ware. DESIGN PROCEDURE: Vcc = 25V Select VRE ≤ VCE Select VRE = 5V ∴ RE =

V RE 5 = = 1.66K (select 2.00K ) IC 3m

& VRC=VCC-VCE-VRE=25-10-5=10V ∴ RC =

V RC 10 = = 3.33K 3m IC

I C = 3m = 15A 190 B β ∴ I R1 = 32I B = 32 ×15 = 480 A ∴ I R 2 = I R1 − I B = 480 − 15 = 465A I =

& V B = V BE + V RE = 0.65 + 5 = 5.65V ∴ R2 = R1 =

5.65 VB = = 12.15K I R 2 465

VCC − V B I R1

=

25 − 5.65 = 40.3125K 480

3


CIRCUIT DIAGRAM:-

VCC

25V

VCC R1 40.2kOhm_1%

R4 40.2kOhm_1% 10

V4 0V

9 V3 0V

Q1 7 2N2923

V2 11

4

0V

V1 0V

2 3 R2 12kOhm_5%

R3 2.00kOhm_1%

0

PROCEDURE:- 1. Rig up the circuit using multisim software and the results using DC operating point analysis (simulate ---- analysis ----- DC operating point).

4


2.Rig up the circuit using multisim software and the results using DC transfer characteristic analysis(simulate ---- analysis ----- DC sweep)

Out put voltage VCE variation with V CC (0 to 25V)

RESULT:- The CE single stage amplifier is designed. The D.C voltages and currents at various nodes are observed. The D.C transfer characteristic is plotted.

5


COMMON SOURCE AMPLIFIER AIM: a) To design a single stage FET Common Source amplifier with potential divider circuit using 2n4861 FET-N channel for the following specifications: VDD = 24V,ID = 1ma,VGS=2V,VPMAX =13V,RL=1K. b) To observe dc operating point, frequency response, DC transfer characteristic & C.R.O waveforms. APPARATUS: Multisim soft ware. DESIGN PROCEDURE: VDSmin = VPmax + 1 - VGS = 13 + 1 - 2 =

12V VDD − VDS min

VS = VRD = RD = RS =

VRD ID

=

24 −12 =6 2

2 6 = = 6K(Use s tan dard value6.3K) 1m

VG = VS − VGS = 6 − 2 = 4V = VR 2 Select R2 =1M R1 =

VR1 × R2 V R2

VR1 = VDD − VG = 24 − 4 = 20 20 ×1M = 5M ∴R1 = 4 (Use s tan dard value which is less than or equal to 5M i.e. 4.7M )

6


CIRCUIT DIAGRAM:

PROCEDURE:- 1. Rig up the circuit using multisim software and the results using DC operating point analysis (simulate----analysis ---- DC operating point) 7


2.Rig up the circuit using multisim software and the results using DC transfer characteristic analysis(simulate ---- analysis ----- DC sweep)

3. Rig up the circuit using multisim software and the results using AC analysis (Simulate ---- analysis ----- AC analysis) 8


4.Rig up the circuit using multisim software and the results using Oscilloscope

RESULT: The CS single stage amplifier is designed with the given specifications. The D.C operating point analysis is performed. The frequency response is plotted and the Band width is found.

9


TEST YOUR SELF

1. What is meant by Q- point? 2. What is the need for biasing a transistor? 3. What factors are to be considered for selecting the operating point Q for an amplifier? 4. Distinguish between D.C. and A.C load lines. 5. What are the reasons for keeping the operating point of a transistor as fixed? 6. What is thermal runaway? How can it be avoided? 7. What are the factors which contribute to thermal instability? 8. Define ‘Stability Factor’. Why would it seem more reasonable to call this an instability factor? 9. How will be the output voltage in a CS amplifier? 10.To have good voltage gain and high input resistance which FET amplifier is to be used?

10


TWO STAGE RC COUPLED AMPLIFIER Q1) Design a single stage transistor amplifier with potential divider circuit (using an npn si transistors) with following specifications. IC=1.6ma,VCE=7.6v,RC=2.2k,VCC=12v, I1=10IB and β=54. the DC values (Voltage and current) at various nodes using Multisim software

DESIGN: IB=IC/β = 1.62/54=0.03ma

VCC=IC(RC+RE)+VCE ; 12=1.62(2.2+RE)+7.6 ; RE=0.516k V2=VBE+ICRE ; V2=0.7+1.62*0.516=1.536v V2=I1R2 ; R2=V2/(I1=10IB) ; 1.536/0.3=5.12k I1=VCC/(R1+R2) ; (R1+R2)=12/0.3=38.1k ; R1=38.1-5.12=32.98k

CIRCUIT DIAGRAM: VCC

12V

R3 2.2kohm

R1 32.98kohm

V2 0V

5V1 0V

1 V3

8

BC107BP Q1

4

0V

2R4

R2 5.1kohm

516ohm

0

11


PROCEDURE: Rig up the circuit using multisim software and the

results using DC operating point analysis (simulate

analysis

DC

operating point) DC Operating Point (Results) 1

8.05632v

2

928.05352mv

4

1.58077v

Vv1#branch

315.92573µa

Vv2#branch

1.79258ma

vcc

12.00000v

vccvcc#branch

-2.10851ma

Q2) Design a single stage transistor amplifier with potential divider circuit (using

an

npn

si

transistors)

with

following

specifications.

IC=2.32ma,VCE=5.7v,RC=2.2k,VCC=12v, I1=10IB and β=33. the DC values (Voltage and current) at various nodes using Multisim software DESIGN:

IB=IC/β = 2.32/33=0.07ma VCC=IC(RC+RE)+VCE ; 12=2.32(2.2+RE)+5.7 ; RE=0.51k V2=VBE+ICRE ; V2=0.7+2.32*0.51=1.88v V2=I1R2 ; R2=V2/(I1=10IB) ; 1.88/0.7=2.68k I1=VCC/(R1+R2) ; (R1+R2)=12/0.7=17.14k ; R1=17.14-2.68=14.46k

12


CIRCUIT DIAGRAM: VCC

12V

R3 2.2kohm

R1 14.46kohm V1

V2 0V

12

5 0V V3

BC107BP

3

Q1 0V

2

R2 2.68kohm

R4 510ohm

0

PROCEDURE: Rig up the circuit using multisim software and the

results using DC operating point analysis (simulate operating point)

analysis

DC Operating Point (Results) 1 6.84857v 2 1.19817v 3 1.85869v vv2#branch 2.34156ma vcc 12.00000v vccvcc#branch -3.04290ma vv1#branch 701.33569µa vv3#branch -7.79614µa

13

DC


Q3) Cascade above two stages and find overall gain (choose Cc=4.7f, Ce=470f, hfe=50) find the frequency response, DC operating points and parameter sweep of load resister. ANALYSIS: Stage-2: AI2= -hfe/(1+hoeRL2) ; -50/(1+2/40) = -47.62 Ri2 = hie+hreAI2RL2 ;1.1+2.5e-4*-47.62*2 = 1.076k; Av2= -AI2*RL2/Ri2 ; -47.62*2/1.076 = -88.51 Stage-1: RL1 = 2.2k||14.2||2.5||1.076 = 0.54k AI1 = -50/(1+0.54/40) = -49.3 Ri1 = 1.1+2.5e-4*-49.3*0.54 = 1.106k Av1 = -49.3*0.54/1.106 = -24.07 Overall gain Av = Av1*Av2 = 24.07*88.51= 2130.4 Avs = Av*Ri’/(Ri’+RS) ; Ri’ = 1.106||33||5.1= 0.88k =2130.4*0.88/(0.88+15) =118 CIRCUIT DIAGRAM: VCC 12V

VCC R1

Rc2 2.2kOhm_5% C5

14.0kOhm_1%

Rc1

33kOhm_5%

R12 2.2kOhm_5%

10 9

47uF

C3

Q1 BC107BP

5 Q3 4.7uF BC107BP

C1 Rs 35kOhm_51% 1 4.7uF

R10 22kOhm_5%

2

V1 Re1 R2 15 C2 1mV 5.s1kOhm_5% 510Ohm_5% 0.71mV_rm 1000Hz 0Deg

2.55kOhm_1% R22

470uF

11 Re2

0

XSC1 G T A

14

C4

470Ohm_5%

B

14

470uF


Note: In above two stage CE amplifier, all resistor values are same as trainer kit values. PROCEDURE:1. Rig up the circuit using multisim software and the

results using DC operating point analysis (simulate------- analysis ------DC operating point and AC analysis) DC Operating Point (Results)

Node no

Voltage

2 9 15 5 11 10

1.57966 8.01593 926.65516m 1.83114 1.16896 6.54642

Rig up the circuit using multisim software and the results using Parameter Sweep(Simulate------ analysis ------- Parameter sweep) 2.

15


RESULTS:

Observed the DC voltages/currents for single stage and two stage

amplifiers. It is observed that Two stage amplifier gives a mid band gain of 1220.It is also observed that as load resistance is nearer to RC2, the out put voltage is decreasing, since net load resistance is decreasing.

16


TEST YOUR SELF

1. What do you mean by a multi stage amplifier? Mention its need. 2. What are the objectives of coupling elements? 3. What are the effects of bandwidth in multistage amplifier? 4. What is the expression for the band width of multistage amplifier? 5. Why RC coupling is popular? 6. What are the advantages & disadvantages of RC coupled amplifier? 7. Define the BW, gain BW product and dynamic range of an amplifier? 8. What is the effect of By capacitor in a RC coupled amplifier? 9. What is the use of transformer coupling in the output stage of multistage amplifier? 10.What is a DC amplifier? What are its advantages and drawbacks?

17


CURRENT SHUNT AMPLIFIER AIM:

Design current shunt feed back amplifier with a resistance 5K

using transistor BC 107. Obtain DC operating point and frequency response. APPARATUS: Multisim software.

DESIGN PROCEDURE: β=

RE 470 − IF = = = 0.085 RF + RE 5K + 470 IE

AI =

I R − I C 2 I B 2 I C1 I B1 = × × × IS I B2 I C1 I B1 I S

IC2 I = −hFE = −50, C1 = hFE = 50 I B2 I B1 IC2 − RC1 2.2K = = = 0.67151 I C1 RC1 + RI 2 2.2K +1.076K I B1 4K R = = = 0.8;Where R = RS //(470 + 5K ) = 4K IS R + hie 4K +1.1K AI = −(−50)(0.67151)(50)(0.8) = 1.343K D = 1 + βAI = 1 + (0.085)(1.343K ) = 115.15 AIF = ∴ AVF

AI 1.343K = = 11.66 D 115.15 R V I R 2.2K = 0 = O C 2 = AIF C 2 = 11.66 × = 1.71 RS VS 15K I S RS

18


CIRCUIT DIAGRAM:

VCC 12V

VCC R3 33kohm

R4 2.2kohm

R9 2.2kohm

R7 14kohm

C2

C4 7 4.7uF 4 BC107BP R10 Q1

C1 R1 15kohm 2 4.7uF 1 V1 1mV 0.71mV_rms 1000Hz 0Deg

5kohm C3

3 R2 5.1kohm

5 R5 470uF 510ohm

Q2 47uF BC107BP

R11 22kohm

6 8 R6 R8 2.55kohm 470ohm

0 XSC1 G T A

19

11

B


PROCEDURE:- 1. Rig up the circuit using multisim software and the results using DC operating point analysis (simulate ---- analysis ----- DC operating point)

2. Rig up the circuit using multisim software and the results using Oscilloscope

20


3. Rig up the circuit using multisim software and the results using AC analysis (simulate----- analysis----AC analysis)

4. Rig up the circuit using multisim soft ware and the results using Parameter Sweep(Simulate ---- analysis ----- Parameter Sweep)

RESULT: The current shunt feed back amplifier is designed with feed back resistance of 5K . The DC operating point values are obtained and the frequency response is plotted.

21


TEST YOUR SELF

1. What is in Amplifiers? 2. Explain the feed back factor and open loop gain. 3. What are the types of ? 4. Explain the term negative in amplifiers? 5. What are the disadvantages of negative ? 6. What are the advantages of negative ? 7. When will a negative amplifier circuit be unstable? 8. Compare the negative and Positive . 9. Give the expression for closed loop gain for a negative feed back amplifier? 10.How does negative reduce distortion in an amplifier? 11.How does series differ form shunt ? 12.What is the difference between voltage and current ?

22


RC PHASE SHIFT OSCILLATOR AIM: a) Design RC phaseshift oscillator to have resonant frequency of 6KHz. Assume R1 = 100k, R2 = 22K, RC = 4 K ,RE =1K & VCC = 12V. b) Obtain hfe for the above designed value for AV > - 29, R≥ 2 RC.

APPARATUS: Multisim software.

DESIGN PROCEDURE:

a)

Let R = 10K fr =

1

When K =

2π RC 6 + 4K

RC R

1

∴C =

4K 10K = 0.962nF ≈ 1nF (Select s tan dard ) 2π × 10K × 6K 6 + 4 ×

∴ R = 10K ; C = 1nF

b)

hfe ≥ 23 =

29 + 4K for sustained oscillation K

= 97.1

23


CIRCUIT DIAGRAM:

VCC 12V VCC R1 100kohm

R3 C2

XSC1

10 A

10uF

4

B

Q2

12 T

R2 22kohm

C3 100uF

11 R4 1kohm 0

C6 1.0nF6 9 R7 10kOhm_5%

C5

C4 1.0nF 7

1.0nF

R6 10kOhm_5%

R5 10kOhm_5%

24

G

2N2222A

10uF


PROCEDURE:

Rig up the circuit using multisim software and the

results using Oscilloscope.

RESULT: RC phase shift oscillator with fr =6KHz is designed. The value of hfe for the designed value is computed.

25


TEST YOUR SELF

1. What is an Oscillator circuit? 2. What is the main difference between an amplifier and an oscillator? 3. State Barkhausen criterion for oscillation. 4. State the factors on which oscillators can be classified. 5. Give the expression for the frequency of oscillation and the minimum gain required for sustained oscillations of the RC phase shift oscillator. 6. Why three RC networks are needed for a phase shift oscillator? Can it be two or four? 7. What are the merits and demerits of phase shift oscillator? 8. At low frequency which oscillators are found to be more suitable? 9. What are the two important RC oscillators? 10.What makes Quartz produce stable oscillations? 11.What are the factors which contribute to change in frequency in oscillators?

26


CLASS A,AB,B,C POWER AMPLIFIERS

AIM : To study the operation of Class A, Class AB, Class B, Class C power amplifiers. APPARATUS:

Multisim soft ware.

CIRCUIT DIAGRAM:

V2

12V

R2 1kohm

R5 1kohm 47C uF 2

R1

C1

47uF

R3 30kohm

XSC1

Q1 PN2369A

G T A

B

100ohm V1 50mV 35.36mV_rms 1000Hz 0Deg

R4 100ohm

THEORY: The classification of amplifiers is based on the position of the quiescent point and extent of the characteristics that is being used to determine the method of operation. There are 4 classes of operations.They are 1.Class A

2.Class AB

3.Class B

27

4.Class C


CLASS A:- In class A operation the quiescent point and the input signal are such that the current in the output circuit (at the collector) flows for all times. Class A amplifier operates essentially over a linear portion of its characteristic there by giving rise to minimum of distortion .

CLASS B:- In class B operation , the quiescent point is at an extreme end of the characteristic , so that under quiescent conditions the power drawn from the dc power supply is very small .If the input signal is sinusoidal, amplification takes place for only half cycle. CLASS AB:- A class AB amplifier is the one that operates between the two extremes defined for class A and Class B. Hence the output signal exists for more than 1800 of the input signal. CLASS C :- In class C operation, the quiescent operating point is chosen such that output signal (voltage or current)is zero for more than on half of the input sinusoidal signal cycle.

PROCEDURE: 1. An input sine wave (peak-peak)of 50mV is applied to the circuit. 2. connect the output to the C.R.O.

3. varying R3 value, observe and record the output waveforms for different classes of operation. 4. Also observe the Vi & Vo waveforms using parameter sweep for different classes of operation.

28


OBSERVATIONS: CLASS A:

CLASS AB : 29


30


CLASS B :

31


CLASS C :

RESULT : 1. In class A amplifier output current flows for whole 3600 2. In class AB power amplifier output current flows between 1800 and 3600 3. In class B power amplifier output current flows for 1800 4.

In Class C power amplifier output current flows for less than 1800.

32


TRY YOUR SELF 1. How do you bias class A operation? 2. What is conversion efficiency? 3. Define Class B mode of operation. 4. What are the advantages & disadvantages of Class B mode of operation? 5. Distinguish between voltage and power amplifier? 6. Which power amplifier gives minimum distortion? 7. What are the drawbacks of class C amplifier? 8. In Which class of amplifier, the efficiency is high? And why? 9. Classify power amplifiers on the basis of the mode of operation. 10.Give two drawbacks of Class A power amplifier.

33


HIGH FREQUENCY COMMON BASE AMPLIFIER AIM - Design a common Base high frequency amplifier with a over all gain of 30 and Lower cut off frequency of 130 Hz and Higher cut frequency 10 MHz . Transistor Specifications: hib = 22.6, hfb = -0.98, hrb = 2.9 × 10-4 , hob = 0.49 s, IC = 1.35ma = -IE, VCE = 5.85V, VEB = 0.6V, VCB = 5.25V. the DC values (Voltage and current) at various nodes using Multisim software APPARATUS: Multisim software. DESIGN PROCEDURE: 1. DESIGN OF BIASING CIRCUIT : VBE = 0.6V, VCE = 5.85V, IC = 1.35mA = -IE VCB = 5.25V

34


Find the value of Re : KVL to Input: -2v – Re(1.35ma) + 0.6 = 0 Re =

1.4 =1.03 K 1.35ma

Find the value of RC : KVL to Output : Vcc – ICRC - VCB = 0 12 – 1.35ma RC – 5.25 = 0 1.35mRc = 6.75 RC = 5k 2.DESIGN OF AMPLIFIER CIRCUIT : 1.To find Cb Assume Rs = 100 Calculate the value of Cb fL =

Cb =

1 ; 2π (Rs + Ri)C b

130=

1 6.28(100 + 22.6)C b

1 6.28(100 + 22.6)130

= 9.9 × 10-6 ≈ 10 f

35


2.To Calculate RL AV = -hfb

RL' Ri RL'

= 0.98 × 22.6 Overall gain = AVS = Av

Ri Ri + Rs

For above circuit Ri 1= Ri=hib AVS = -hfb

RL' Ri + Rs

RL1 = RL 11 RC =3.75k

RL = 15k

3. To Calculate Shunt Capacitance Csh fh

=

1 2πRL × C sh

C sh =

1 2π ×10 ×10 6 × 3.75k

= 4.24 pf ≈ 4pf The Internal junction capacitance Cb’c ≈ 3pf C’sh = Cb’c + Csh 4.24 = 2.9pf + Csh Csh = 1.34pf ≈ 2pf

36


CIRCUIT DIAGRAM : XSC1 G T A

Rs 11000ohm V1 10mV 7.07mV_rms 1000Hz 0Deg

Cb 11

10uF 12 Re 1kohm

10uF

BC107BP Q1

Rc 5kohm

8

12V

6

7 VEE 2V

B

Cb1 3 R5 15kohm

Csh 2pF

VCC 0

PROCEDURE: 1.Rig up the circuit using multisim software and the results using DC operating point analysis (Simulate ------Analysis------ DC operating point)

37


2. Rig up the circuit using multisim software and the results using AC analysis (Simulate--- Analysis--- AC analysis)

38


RESULTS: The common base high frequency amplifier is designed. Also DC voltages / currents are observed. The Bandwidth is far greater than the bandwidth of the CE amplifier.

39


TEST YOUR SELF

1. What are the factors to be taken into in the high frequency model of a transistor? 2. What are α and β cut off frequencies? 3. What is unity gain small signal band width? 4. What is the relation between fT & fβ? 5. What is the expression for trans conductance gm in the high frequency model? 6. What is the expression for input conductance gb e in of gm ? 7. What is base spreading resistance? 8. Give the expressions for the hybrid ∏ capacitances.

40


TWO STAGE RC COUPLED AMPLIFIER

AIM: 1. To study the Two-stage RC coupled amplifier. 2. To measure the voltage gain of the amplifier at 1KHz. 3. To obtain the frequency response characteristic and the band width of the amplifier.

EQUIPMENT: Two stage RC coupled amplifier, trainer. 1. Signal Generator. 2. C.R.O 3. Connecting patch cords. CIRCUIT DIAGRAM:

VCC 12V

R1 33kohm

R3

R6 2.2kohm

CC 15kohm

C2

Q1 10uF BC107BP

C1 RG 15kohm

RC 2.2kohm

Q2 10uF BC107BP

10uF C3 OUTPUT

INPUT V 50mV

CE R2 5.1kohm

RE 510ohm

10uF 2.7kohm

41

R4

R5 1kohm

10uF

VO


PROCEDURE: 1.Switch ON the power supply. 2. Connect the signal generator with sine wave output 50mV p-p at the input terminals. 3. Connect the C.R.O at output terminals of the module. 4. Measure the voltage at the second stage of amplifier. 5. Now vary the input frequency from 10Hz to 1MHz in steps,and for every value of input frequency note the output voltage

keeping the input

amplitude at constant value. 6. Calculate the gain magnitude of the amplifier using the formula Gain = Vo/Vi Gain in dB= 20 log (Vo / Vi ) 7. Plot a graph of frequency versus gain (dB) of the amplifier. Sample frequency response graph is as shown in fig. Below. OBSERVATION: Vi = 50mV(p-p) Frequency

Gain =20 log (Vo/Vi)dB

VO

42


FREQUENCY RESPONSE:

0.707 VO/VI

FL

FH

Gain VO/VI

FL

FH

RESULT: The gain of the amplifier at 1 KHz is -----The BW of the amplifier is -------

43

Frequency


CURRENT SHUNT AMPLIFIER

AIM: 1. To study the current shunt amplifier 2. To measure the voltage gain of the amplifier at 1KHz. 3. To obtain the frequency response characteristic and the band width of the amplifier.

EQUIPMENT: Current shunt feed back amplifier trainer. 4. Signal Generator. 5. C.R.O 6. Connecting patch cords. CIRCUIT DIAGRAM:

44


PROCEDURE: 1. Switch ON the power supply. 2. Connect the signal generator with sine wave output 50mV p-p at the input terminals. 3. Connect the C.R.O at output terminals of the module. 4. Measure the voltage at the second stage of amplifier. 5. Now vary the input frequency from 10Hz to 1MHz in steps, and for each value of input frequency note the output voltage keeping the input amplitude at constant value. 6. Calculate the gain magnitude of the amplifier using the formula Gain = Vo/Vi Gain in dB= 20 log (Vo / Vi ) 7. Plot a graph of frequency versus gain (dB) of the amplifier. Sample frequency response graph is as shown in fig. Below.

OBSERVATIONS : Vi = 50mV(p-p) Frequency

VO

Gain =20 log (Vo/Vi)dB

45


46


FREQUENCY RESPONSE:

RESULT: The gain of the amplifier at 1 KHz is -----The BW of the amplifier is -------

47


CLASS A/B/C/AB POWER AMPLIFIER

AIM: To study the operation of Class A, Class B, Class AB and Class C power amplifiers.

EQUIPMENT: 1.Class/A/B/C/AB amplifier trainer 2.Function generator. 3.C.R.O 4. Connecting patch cords. CIRCUIT DIAGRAM:

48


PROCEDURE: 1.Connect the circuit as shown in the circuit diagram, and get the circuit verified by your Instructor. 2. Connect the signal generator with sine wave at 1KHz and keep the amplitude at .5V (peak-to-peak) 3. Connect the C.R.O across the output terminals. 4. Now switch ON the trainer and see that the supply LED glows. 5. Keep the potentiometer at minimum position, observe and record the waveform from the C.R.O. 6. Slowly varying the potentiometer, observe the outputs for the Class A/B/AB/C amplifiers as shown in fig. CLASS A:

49


CLASS B:

CLASS AB:

50


CLASS C :

RESULT: It is observed that for class A amplifier, the transistor conducts for 360deg, for class AB more than 180deg and for Class C less than 180 deg.

51


SINGLE TUNED VOLTAGE AMPLIFIER

AIM: 1.To calculate the resonant frequency of tank circuit. 2. To plot the frequency response of the tuned amplifier. EQUIPMENT: 1. Tuned voltage amplifier trainer. 2. Function generator. 3. C.R.O. 4. Connecting patch cords.

CIRCUIT DIAGRAM:

52


PROCEDURE: 1.Connect the circuit as shown in fig and get the circuit verified by your Instructor. 2. Connect the signal generator with sine wave at the input and keep the amplitude to minimum position, and connect a C.R.O at output terminals of the circuit. 3. Apply the amplitude between 1.6v to 4.4v to get the distortion less output sine wave. 4. Now, vary the input frequency in steps and observe and record The output voltage. 5. Calculate the gain of the tuned RF amplifier using the formula Gain = out put voltage/ input voltage. 6. plot a graph with input frequency versus gain (in dBs) Gain (in dBs) = 20 log (Vo/Vi) Graph :-

Gain

Frequency

RESULT: The tuned amplifier offers maximum gain at resonant frequency of the tank circuit. 53


SERIES VOLTAGE REGULATOR AIM : To study and design a Series voltage regulator and to observe the load regulation feature.

EQUIPMENT : 1. Series voltage regulated power supply trainer. 2. Multimeter. 3. Patch chords. CIRCUIT DIAGRAM: 3055 560E Rs

+ IL

+

IR

Un Regulated Input

+ - IZ

500E

VO 50%

VZ=12V

PROCEDURE: 1. Switch ON the power supply. 2. Observe the Unregulated voltage at the output of rectifier. 3. Connect this voltage to the input of series voltage regulator circuit. 4. Keep the load resistance 1K at constant. 5. Observe the output voltage VO = VZ-VBE

54


6. And also observe the voltage across RS,and values of IR,IL and IZ. 7. Compare the practical values with theoretical values. 8. By changing the load resistance, observe the output voltage and various currents. OBSERVATIONS: RL

VO

IR

IZ

IL

LOAD REGULATION :

VO

RL

RESULT: The regulator maintains a constant output voltage inspite of the changes in load conditions.

55


SHUNT VOLTAGE REGULATOR

AIM : To study and design a Shunt Regulator and to observe the load regulation feature. EQUIPMENT : 1. Shunt regulated power supply trainer. 2. Multimeter. 3. Patch chords.

CIRCUIT DIAGRAM: RS

+

-

+

220E

IL

+ Un Regulated In put

8.2V

IC

1K

-

RL

VO

50%

3055

PROCEDURE: 1. Switch On the main power supply. 2. Observe the unregulated voltage at the output of rectifier. 3. Connect this voltage to the input of shunt Regulator circuit 4 .Keep the load resistance 1K constant. 5. Observe the output voltage across the load resistor V0 =VZ + VBE

56


6. Also observe IL, IS & IC. 7. Compare the practical values with theoretical values. 8. By changing the load resistance, observe the output voltage and various currents. OBSERVATIONS: RL

VO

IS

IC

IL

LOAD REGULATION:

VO

RL RESULT: The regulator maintains a constant output voltage inspite of the changes in load conditions.

57

Eca-labmanual--multisim.pdf 6f581a

Overview 5o1f4z

More details 6z3438

More Documents from "Siva Kumar" j2856

Cs2351 Artificial Intelligence 16 Marks 2x4t68

Gas Compressor - Wikipedia, The Free Encyclopedia 2g1h3q

Eca-labmanual--multisim.pdf 6f581a

Pharmacy Management System 1ko2h

Soundarya-lahari-in-tamil.pdf 4q102g

Design Of A Windmill For Pumping Water.pdf 4vd1r