ec-ii

ELECTRONIC CIRCUITS II AND SIMULATION LAB

(FOR ECE)

1

EX NO:1a

DATE :DESIGN AND ANALYSIS OF CURRENT SERIES FEEDBACK

AMPLIFIERAim:

To design and test the current-series feedback amplifier and to calculate the

following parameters with and without feedback.

1. Mid band gain.

2. Bandwidth and cut-off frequencies.

3. Input and output impedance.

Components & Equipment required:

S.NO APPARATUS RANGE QUANTITY

1. Power supply (0-30)V 1

2. Function generator (0-20M)Hz 1

3. CRO 1

4. Transistor BC107 1

5. Resistors

6. Capacitors

7. Connecting wires

2

Circuit diagram:(i) Without feedback:

(ii) With feedback:

3

Theory:

The current series feedback amplifier is characterized by having shunt sampling and series

mixing. In amplifiers, there is a sampling network, which samples the output and gives to the

feedback network. The feedback signal is mixed with input signal by either shunt or

series mixing technique. Due to shunt sampling the output resistance increases by a factor of

‘D’ and the input resistance is also increased by the same factor due to series mixing. This is

basically transconductance amplifier. Its input is voltage which is amplified as current.

Design:

(i) Without feedback:

VCC = 12V;

IC = 1mA;

fL = 50Hz;

S = 2;

RL = 4.7KΩ;

hfe = re = 26mV / IC = 26Ω;

hie = hfe re = VCE= Vcc/2 (transistor Active) = VE = IERE = Vcc/10

Applying KVL to output loop, we get

VCC = ICRC + VCE + IERE

RC = ?

Since IB is very small when compare with IC,

IC ≈ IE

RE = VE / IE = ?

S = 1+ RB / RE = 2

RB = ?

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = ? R2 =?

XCi = Zi / 10 = (hie || RB) / 10 = ?

Ci = 1 / (2πf XCi) = ?

Xco = (RC || RL)/10 =?

4

Co = 1 / (2πf XCo) = ?

XCE = RE/10 = ?

CE = 1 / (2πf XCE) ?

(ii) With feedback (Remove the Emitter Capacitor, CE):

Feedback factor, β = -RE =

Gm = -hfe / (hie + RE) =

Desensitivity factor, D = 1 + β Gm =

Transconductance with feedback, Gmf = Gm / D =

Input impedance with feedback, Zif = Zi D

Output impedance with feedback, Z0f = Z0 D

Procedure:

1. Connect the circuit as per the circuit diagram.

2. Keeping the input voltage constant, vary the frequency from 50Hz to 3MHz in regular steps

and note down the corresponding output voltage.

3. Plot the graph: Gain (dB) Vs Frequency

4. Calculate the bandwidth from the graph.

5. Calculate the input and output impedance.

6. Remove Emitter Capacitance, and follow the same procedures (1 to 5).

Tabular column:

(i) Without feedback:

Vi=

S.NoFrequency (Hz)

Output Voltage(V0) (volts) Gain = V0/Vi

Gain = 20 log(V0/Vi) (dB)

5

(iii) With feedback:Vi=

S.NoFrequency (Hz)

Output Voltage(V0) (volts)

Gain = V0/ViGain = 20

log(V0/Vi) (dB)

Theoretical Practical

With feedbackWithoutfeedback

With feedback Withoutfeedback

Inputimpedance

Output impedance

Gain(midband)

Bandwidth

Model graph: (frequency response)

7

VIVA Questions:

1. Define feedback.

2. Define positive feedback.

3. Define negative feedback.

4. Define sensitivity.

5. What are the types of feedback?

Result:

STAFF SIGNATURE

8

EX NO:1b

DATE :DESIGN AND ANALYSIS OF VOLTAGE SHUNT FEEDBACK

AMPLIFIER

Aim:

To design and test the voltage-shunt feedback amplifier and to calculate the following

Parameters with and without feedback.

1. Mid band gain.

2. Bandwidth and cut-off frequencies.

3. Input and output impedance.




2.Function generator (0-20M)Hz

1

3.CRO

1

4.Transistor BC107

1

5. Resistors

6.Capacitors

7. Connecting wires

9

Circuit Diagram:

(i) Without Feedback:

(ii) With Feedback:

10

Theory:

In voltage shunt feedback amplifier, the feedback signal voltage is given to the base

of the transistor in shunt through the base resistor RB. This shunt connection tends to decrease

the input resistance and the voltage feedback tends to decrease the output resistance. In the

circuit RB appears directly across the input base terminal and output collector terminal. A part

of output is feedback to input through RB and increase in IC decreases IB. Thus negative

feedback exists in the circuit. So this circuit is also called voltage feedback bias circuit. This

feedback amplifier is known a trans-resistance amplifier. It amplifies the input current to

required voltage levels. The feedback path consists of a resistor and a capacitor.

11

Design


VCC = 12V;

IC = 1mA;

AV = 30;

Rf = 2.5KΩ; S = 2;

hfe = ;

β=1/ Rf = 0.0004

re = 26mV / IC = 26Ω;

hie = hfe re =

VCE= Vcc/2 (transistor Active) =

VE = IERE = Vcc/10 =

Applying KVL to output loop, we get VCC = ICRC + VCE + IERE

RC =

Since IB is very small when compare with IC, IC ≈IE

RE = VE / IE =

S = 1+ RB / RE

RB =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =

(ii) With feedback:

RO = RC || Rf =

Ri = (RB || hie ) Rf =

Rm = -(hfe (RB || Rf) (RC || Rf)) / ((RB || Rf) + hie) =

Desensitivity factor, D = 1 + β Rm

Rif = Ri / D =

Rof = Ro / D =

12

Rmf = Rm / D =

XCi = Rif /10 =

Ci = 1 / (2πf XCi) =

Xco = Rof /10 =

Co = 1 / (2πf XCo) =

RE’ = RE || ((RB + hie) / (1+hfe))

XCE = RE’/10 =

CE = 1 / (2πf XCE) =

XCf = Rf/10

Cf = 1 / (2πf XCf) =

Procedure:


2. Keeping the input voltage constant, vary the frequency from 50Hz to 3MHz in regular

steps and note down the corresponding output voltage.


4. Calculate the bandwidth from the graph.

5. Calculate the input and output impedance.

6. Remove Emitter Capacitance, and follow the same procedures (1 to 5).

13

Tabular Column:


Vi=10mV

Frequency (Hz)Vo (Volts)


log(V0/Vi) (dB)

(ii) With Feedback:

Vi=10mV



log(V0/Vi) (dB)

Model graph: (frequency response)

15

VIVA Questions:

1. Give an example for voltage-series feedback.

2. Give the properties of negative feedback.

3. Define voltage shunt feedback.

4.Define voltage series feedback.

5.Define current shunt feedback.

Result:

STAFF SIGNATURE

16

EX NO:2a

DATE :DESIGN AND ANALYSIS OF RC PHASE SHIFT OSCILLATOR

Aim:

To design and construct a RC phase shift oscillator for the given frequency (f0).



1.Power supply (0-30)V

1


1

3.CRO

1

4.Transistor BC107

1

5. Resistors

6.Capacitors

7.Connecting wires

Theory:

In the RC phase shift oscillator, the required phase shift of 180˚ in the feedback

loop from the output to input is obtained by using R and C components, instead of tank

circuit. Here a common emitter amplifier is used in forward path followed by three

sections of RC phase network in the reverse path with the output of the last section being

returned to the input of the amplifier. The phase shift Ф is given by each RC section Ф=tan¯1

(1/ωrc). In practice R- value is adjusted such that Ф becomes 60˚. If the value of R and C

are chosen such that the given frequency for the phase shift of each RC section is 60˚.

Therefore at a specific frequency the total phase shift from base to transistor’s around circuit

and back to base is exactly 360˚ or 0˚. Thus the Barkhausen criterion for oscillation is

satisfied

17

Circuit Diagram:

Design:

VCC = 12V; IC = 1mA; C = 0.01μF; fo = ; S = 2; hfe =

re = 26mV / IC = 26√Ω;

hie = hfe re =

VCE= Vcc/2 (transistor Active) =

VE = IERE = Vcc/10

Applying KVL to output loop, we get

VCC = ICRC + VCE + IERE

RC =

Since IB is very small when compare with IC,

18

IC ≈ IE

RE = VE / IE =

S = 1+ RB / RE = 2

RB = VB = VBE + VE =

VB = VCC R2 / (R1 + R2)

RB = R1 || R2

R1 = R2 =

Gain formula is given by,

Effective load resistance is given by, Rleff = Rc ||

RL RL =

XCi = [hie+(1+hfe)RE] || RB/10 =

Ci = 1 / (2πf XCi) =

Xco = Rleff /10 =

Co = 1 / (2πf XCo) =

CE = RE/10 =

CE = 1 / (2πf XCE) =

19

Feedback Network:

f0 = ;

C = 0.01μf;

R =

Procedure:

1. Connections are made as per the circuit diagram.

2. Switch on the power supply and observe the output on the CRO (sine wave).

3. Note down the practical frequency and compare with its theoretical frequency.

Model Graph:

21

Tabular Column:

AMPLITUDE(V) TIME(ms) FREQUENCY(HZ)

VIVA Questions:

1. What is an Oscillator circuit?

2.What are the different types of oscillators?

3. What are the conditions for oscillations?

4.Define frequency oscillation.

5. What is the application of RC phase shift oscillator?

Result:

STAFF SIGNATURE

22

EX NO:2b

DATE :DESIGN AND ANALYSIS OF WEIN BRIDGE OSCILLATOR

Aim:

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


S.no Name of the component Range Quantity1 Op-amp IC741 12 CRO - 13 Capacitor 2.88uF,0.01uF,0.08uF 54 Resistor 15k,8.8k,12k,1.18k 75 Power supply - 16 Bread Board - 17 Connecting wires - As

required

Theory:

The wein bridge oscillator is a standard circuit for generating low frequencies in the range of 10

Hz to about 1MHz.The method used for getting +ve feedback in wein bridge oscillator is to

use two stages of an RC-coupled amplifier. Since one stage of the RC-coupled amplifier

introduces a phase shift of 180 deg, two stages will introduces a phase shift o f 360 deg. At the

frequency of oscillations f the +ve feedback network shown in fig makes the input & output in

the phase. The frequency of oscillations is given as f =1/2π√R1C1R2C2 In addition to the

positive feedback

Design:

Select appropriate transistor and note down its specification such as

Vce(max),IcQ(max),hfe(min)) and hfe(max) and VBE(SAT). Here the transistor is allowed

to work in the active region.Assume Vcc,VceQ,IcQ,Vcc. where

Vcc=VcEQ+IcQ(Rc+RE),determine Rc.Assuming appropriate stability factor and hence I2

current flowing through the biasing resistor R2 and determine R1 and R2.

23

R2=SXRE

Vcc[(R2/R1)+R2]=VRE +VBE(sat)

VR1+VR2=Vcc

Using the condition for sustained oscillation hfe>4k+23+29/k, where k=Rc/R. compute C

for designed desired frequency f1 using formula for frequency of oscillation.

F=1/2 1 ߨ 2ܥ1ܥ2

Compute Cin, Xcin<=Zin/10, where Xcin is the impedance offered by the coupling

capacitor for the frequency of interest and Zin is the input resistance at the transistor.

Compute CE, the impedance XCE<=RE/10.

Procedure:

1. Connections are made as per the circuit diagram

2. Feed the output of the oscillator to a C.R.O by making adjustments in the Potentiometer connected in the +ve feedback loop, try to obtain a stable sine Wave.

3. Measure the time period of the waveform obtained on CRO. & calculate the Frequency of oscillations.

4. Repeat the procedure for different values of capacitance

Circuit Diagram:

24

Design:

F=1/2πRC

R=1/2πFC

Given F=1KHZ

Assume, C=0.1μF,

R=1.5kΩ.

R3/R4=2.

Assume

R3=1kΩ,R4= 500Ω

R=R1=R2=1.5kΩ

C=C1=C2=0.1μF

Tabulation:


Model Graph:

26

VIVA Questions:

1. What is an Oscillator?

2. What is a tuned amplifier?

3. What happens to the Oscillator circuit when operated above and below resonance?

4. What are the different coil losses?

5. What is Q factor?

Result:

STAFF SIGNATURE

27

EX NO:3a

DATE :

DESIGN AND ANALYSIS OF HARTLEY OSCILLATOR

Aim: To design and construct the given oscillator for the given frequency

(fO). Components & Equipment required:




1

3. CRO 1


5. Resistors

6. Capacitors

7.DIB

8.DCB

9. Connecting wires

Theory:

Hartley oscillator is a type of sine wave generator. The oscillator derives its initial output

from the noise signals present in the circuit. After considerable time, it gains st rength

and thereby producing sustained oscillations. Hartley Oscillator have two major parts namely

– amplifier part and feedback part. The amplifier part has a typically CE amplifier with

voltage divider bias. In the feedback path, there is a LCL network. The feedback network

generally provides a fraction of output as feedback.

28

Circuit Diagram:

Design:

Given Specifications,

Vcc=12V, S=6, VRE=3V, hfe=300, fc=12kHz, VBE(sat)=0.7, IcQ=1.6mA.

Let L1=100uH, L1/L2=hfe, L2=?

i)VcEQ=Vcc/2=6V

ii)VRE=ICQ.RE=1.6x10 -3.RE=3

29

RE=1.857kΩ.

Vcc=VCEQ+ICQ (RC+RE)

12= 6+1.6x10 -3 (RC+1.875x10 3)

RC=1.875kΩ

iii)R2=SxRE=11.25kΩ

iv)Vcc(R2/(R1+R2)=VRE+VBE(sat)

R1=25.29kΩ

L1/L2=hfe=300

Assume, L1=100μH,

L2=0.33 μH.

f=1/2π ( 2ܮ + 1ܮ )

C=0.1753μF.

CE=1/(2πfcXcE)=0.0707 μF.

Zin=R1||R2||hie=2.997kΩ

Xcin=Zin/10

Cin=1/(2ߨfcXcin)=0.0442 μF.

30

Procedure:


2. Switch on the power supply and observe the output on the CRO (sine wave).3. Note down the practical frequency and compare with its theoretical frequency.

Model Graph:

Tabulation:


32

VIVA Questions:

1. What is dissipation factor?

2. Give the classification of tuned amplifiers.

3. What is a single tuned amplifier?

4. What are the advantages of tuned amplifiers?

5. List out the limitations of tuned amplifiers?

Result:

STAFF SIGNATURE



1


1

3. CRO 1


5. Resistors

6. Capacitors

7.DIB

8.DCB

7.Connecting wires

33

EX NO:3b

DATE :

DESIGN AND ANALYSIS OF COLPITTS OSCILLATOR

Aim: To design and construct the given oscillator at the given operating frequency.


Theory:

A Colpitts oscillator is the electrical dual of a Hartley oscillator. In the Colpitts

circuit, two capacitors and one inductor determine the frequency of oscillation. The oscillator

derives its initial output from the noise signals present in the circuit. After considerable

time, it gains strength and thereby producing sustained oscillations. It has two major parts

namely – amplifier part and feedback part. The amplifier part has a typically CE amplifier with

voltage divider bias. In the feedback path, there is a CLC network. The feedback network

generally provides a fraction of output as feedback.

34

Circuit Diagram:

Design:

Vcc=8V, S=6, VRE=3V, hfe=300, fc=12kHz, VBE(sat)=0.7, IcQ=1.8mA.

Let L1=100uH,L1/L2=hfe,L2=?

i)VcEQ=Vcc/2

ii)Vcc=VcEQ+IcQ.RE,VRE=ICQ.RE=IE.RE

RE=VRE/VCQ iii)R2=S.RE

iv)Vcc(R2/(R1+R2)=VRE+VBE(sat)

re’=(26x10-3

)/ICQ

R1=Vcc.R2/(R1+R2) -R2

35

v)Cin=1/(2ߨfcXcin);Xcin=Zin/10

Zin=R1||R2||hie

vi)CE=1/(2πfcXcE)

XcE=RE/10

vii)fc=1/2π ( ݍܮܥ ,Ceq=C1||C2

viii)Vcc=VcEQ+ICQ(Rc+RE)

Procedure:

1. Connections are given as per the circuit diagrams (both oscillators).

2. Switch on the power supply and observe the output on the CRO (sine wave).

3. Note down the practical frequency and compare with its theoretical frequency.

Model Graph:

Tabulation:

AMPLITUDE(V) TIME(ms) FREQUENCY(HZ

37

VIVA Questions:

1. What is neutralization?

2. What are double tuned amplifiers?

3. What is a stagger tuned amplifier?

4. What are the advantages of stagger tuned amplifier?

5. What are the different types of neutralization?

Result:

STAFF SIGNATURE

38

EX NO: 4

DATE :DESIGN AND ANALYSIS OF FREQUENCY RESPONSE OF CLASS C

SINGLE TUNED AMPLIFIER

Aim:

To design and construct a single tuned amplifier and to plot the frequency response.




1


1

3.CRO

1

4.Transistor BC107

1

5. Resistors

6.Capacitors

7.DIB

8.DCB

7.Connecting wires

Theory:

The amplifier is said to be class c amplifier if the Q Point and the input signal are

selected such that the output signal is obtained for less than a half cycle, for a full input cycle

Due to such a selection of the Q point, transistor remains active for less than a half cycle

.Hence only that Part is reproduced at the output for remaining cycle of the input cycle

the transistor remains cut off and no signal is produced at the output. The total Angle during

which current flows is less than 180.This angle is called the conduction angle, Qc.

39

Circuit Diagram:

Design:VCC = 12V; IC = 1mA; fo = ; S = 2; hfe = Q = 5; L = 1Mhre = 26mV / IC = 26Ω;hie = hfe re =

VCE= Vcc/2 (transistor Active) = VE = IERE = Vcc/10Applying KVL to output loop, we get VCC = ICRC + VCE + IERERC =Since IB is very small when compare with IC, IC ≈ IE RE = VE / IE =

S = 1+ RB / RE = 2RB =

VB = VBE + VE =

40

VB = VCC R2 / (R1 + R2) RB = R1 || R2R1 = R2 = RL = XCi = [hie+(1+hfe)RE] || RB/10 =Ci = 1 / (2πf XCi) = Xco = (RC||RL) /10 = Co = 1 / (2πf XCo) = XCE = RE/10 =CE = 1 / (2πf XCE) = Q = RL /ωLRL =

C =

Procedure:


2. Set Vi = 50 mV (say), using the signal generator.

3. Keeping the input voltage constant, vary the frequency from 0Hz to3MHz in regular steps

and note down the corresponding output voltage.


Tabular Column:

Vi = 50 mV


Gain = 20 log(V0/Vi) (dB)

42

Model Graph: (Frequency Response)

VIVA Questions:

1. What is rice neutralization?

2. What is unloaded Q?

3. What are the applications of mixer circuit?

4. What is up converter?

5. What is an amplifier?

Result:

STAFF SIGNATURE

43

EX NO: 5

DATE :DESIGN AND ANALYSIS OF WAVE SHAPING CIRCUITS

(Differentiator, Integrator, Clipper and Clamper)

Aim:

To design and implement different wave shaping circuits (Differentiator, Integrator,Clipper and Clamper).

Components / Equipments Required:


1.Function / Pulse generator (0 – 3M)Hz

1

2.CRO (0-20M)Hz

1

4. Resistors 1KΩ / 100KΩ1

5.Capacitors 0.1μF

1

6.Connecting wires

Theory:

(i) Differentiator: The high pass RC network acts as a differentiator whose output

voltage depends upon the differential of input voltage. Its output voltage of the differe

ntiator can be expressed as,

Vout = d/dt .Vin

(ii) Integrator: The low pass RC network acts as an integrator whose output voltage

depends upon the integration of input voltage. Its output voltage of the integrator can be

expressed as,

Vout = ∫ Vin dt

44

(iii) Clipper: This circuit is basically a rectifier circuit, which clips the input waveform

according to the required specification. The diode acts as a clipper. There are several

clippers like positive clipper, negative clipper, etc. Depending upon the connection of diode it

can be classified as series and shunt.

(iv) Clamper: The clamper circuit is a type of wave shaping circuit in which the DC level of the input signal is altered. The DC voltage is varied accordingly and it is classified as positive clamper or negative clamper accordingly.

Circuit Diagram:

(i) Differentiator:

(ii) Integrator:

45

(iii) Clipper:

(a) Series Positive Clipper:

(b) Shunt Positive Clipper:

(c) Series Negative Clipper:

46

(d) Shunt Negative Clipper:

(e) Positive Biased Series Positive Clipper:

(f) Positive Biased Shunt Positive Clipper:

47

(g) Positive Biased Series Negative Clipper:

(h) Positive Biased Shunt Negative Clipper:

(i) Negative Biased Series Positive Clipper:

48

(j) Negative Biased Shunt Positive Clipper:

(k) Negative Biased Series Negative Clipper:

(l) Negative Biased Shunt Negative Clipper:

\

49

(m) Combinational Clipper

(iv) Clamper:

(a) Positive Clamper:

(b) Negative Clamper:

50

Design:

(i) Differentiator:

f = 1KHz

т = RC = 1ms If

C = 0.1μF Then

R = 10KΩ

For T << т, Choose R = 1KΩ and

For T >> т, Choose R = 100KΩ

(ii) Integrator:

f = 1KHz

т = RC = 1ms If

C = 0.1μF Then

R = 10KΩ

For T << т, Choose R = 1KΩ and

For T >> т, Choose R = 100KΩ

Procedure:


2. Set Vin = 5V and f = 1KHz.

3. Observe the Output waveform and plot the graph.

51

Model Graph:

(i) Differentiator

(ii) Integrator

52

(iii) Clipper:

54

(iv) Clamper:

60

VIVA Questions:

1.Define Integrator.

2.Define differentiator.

3.Define Clipper.

4.Define Clampers.

5.Define Comparator.

RESU LT:

STAFF SIGNATURE

61

EX NO:6a

DATE : DESIGN AND ANALYSIS OF ASTABLE MULTIVIBRATOR

Aim: To design and construct an astable multivibrator using transistor and to plot the output waveform.




2. CRO (0-20M)Hz 1

3.Transistor BC107

2

4. Resistors 4.9KΩ, 1.6MΩ2 each

5.Capacitors 0.45nF

2

6.Connecting wires

Circuit Diagram:

62

Theory:

Astable multivibrator is also known as free running multivibrator. It is rectangular wave

shaping circuit having non-stable states. This circuit does not need an external trigger to change

state. It consists of two similar NPN transistors. They are capacitor coupled. It has 2 quasi-stable

states. It switches between the two states without any applications of input trigger pulses. Thus it

produces a square wave output without any input trigger. The time period of the output

square wave is given by, T = 1.38RC.

Design Procedure:

VCC = 10V; IC = 2mA;VCE (sat) = 0.2V; f = 1KHz; hfe =

RC =( VCC - VCE (sat) )/ IC =(12 – 0.2 )/ 0.002 = 5.9kΩ.

R ≤hfe RC = 315 * 5.9 * 103 = 1.85MΩ

R = 1.5MΩ.

T = 1.38RC

C = T / (1.38R) = (1 * 10-3) / (1.38 * 1.5 * 106)= 0.48nF

Procedure:


2. Switch on the power supply.

3. Note down the output TON, TOFF and output voltage from CRO.

4. Plot the output waveform in the graph.

Tabular Column:

Amplitude (V) TON (ms) TOFF (ms) Frequency(HZ)

VO1

VO2

64

Model Graph:

VIVA Questions:

1.What is a Multivibrator?

2.List the types of Multivibrators.

3. How many stable states a bistable Multivibrator have?

4. When will the circuit change from stable state in bistable Multivibrator ?

5. What are the different names given to a bistable Multivibrator?

RESULT:

STAFF SIGNATURE

65

EX NO:6b

DATE :

DESIGN AND ANALYSIS OF MONOSTABLE MULTIVIBRATOR

Aim:

To design and construct monostable multivibrator using transistor and to plot the output waveform.




2. CRO (0-20M)Hz 1


4. Resistors 4.9KΩ, 1.6MΩ2 each

5.Capacitors 0.45nF

2

6.Connecting wires

Circuit Diagram:

66

Theory:

Monostable multivibrator has two states which are (i) quasi-stable state and (ii)

stable state. When a trigger input is given to the monostable multivibrator, it switches

between two states. It has resistor coupling with one transistor. The other transistor has

capacitive coupling. The capacitor is used to increase the speed of switching. The resistor

R2 is used to provide negative voltage to the base so that Q1 is OFF and Q2 is ON. Thus an

output square wave is obtained from monostable multivibrator.

Design Procedure:

VCC = 12V; VBB = -2V; IC = 2mA; VCE (sat) = 0.2V; f = 1KHz; hfe =

RC =( VCC - VCE (sat) )/ IC =(12 – 0.2 )/ 0.002 = 5.9kΩ.IB2(min) = IC2 / hfe =

Select IB2 > IB2(min)

IB2 =

R=( VCC - VCE (sat) )/ IB2

T = 0.69RC

C = T / 0.69R =

VBBR1 = VCE (sat) R2

R2 =10R1 (since, VBB = 2V and VCE (sat) = 0.2V)

Let R1 = 10KΩ, then R2 = 100KΩ

Choose C1 = 25pF.

67

Procedure:


2. Switch on the power supply

3. Observe the output at collector terminals.

4. Trigger Monostable with pulse and note down the output TON, TOFF and voltage from CRO.

5. Plot the waveform in the graph.

Tabular Column:

Width (ms)Input Output

TON(ms) TOFF(ms) Voltage(V) TON(ms) TOFF(ms) Voltage(V)

Model Graph:

69

VIVA Questions:

1. What are the applications of a Multivibrator?

2. What are the other names of monostable Multivibrator?

3. Why is monostable Multivibrator called gating circuit?

4. Why is monostable Multivibrator called delay circuit?

5. What is the main characteristics of Astable Multivibrator?

Result:

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EX NO:6c

DATE : DESIGN AND ANALYSIS OF BISTABLE MULTIVIBRATOR

Aim:

To design a bistable multivibrator and to plot its output waveform.




1

2.CRO (0-20M)Hz

1


4. Resistors 4.9KΩ, 1.6MΩ 2 each

5. Capacitors 0.45nF 2

6. Connecting wires

Circuit Diagram:

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Theory:

The bistable multivibrator has two stable states. The multivibrator can exist

indefinitely in either of the two stable states. It requires an external trigger pulse to change

from one stable state to another. The circuit remains in one stable state until an external trigger

pulse is applied. The bistable multivibrator is used for the performance of many digital

operations such as counting and storing of binary information. The multivibrator also

finds an application in generation and pulse type waveform.

Design:

VCC =12V; VBB = -12V; IC = 2mA; VCE (sat) = 0.2V; VBE (sat) = 0.7V

R2 = 1.8MΩ

Let R1 = 10KΩ, C1 = C2 = 50pF

Procedure:


2. Set the input trigger using trigger pulse generator.

3. Note the output waveform from CRO and plot the graph.

Tabular Column:

InputVoltage

(V)Width (ms)

Input Output

TON(ms) TOFF(ms) Voltage(V) TON(ms) TOFF(ms) Voltage(V)

Model Graph:

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VIVA Questions:

1. What is the other name of Astable Multivibrator?

2, What are the two types of bistable Multivibrator?

3. Why does one of the transistor starts conducting ahead of other?

4. What are the two stable states of bistable Multivibrator?

5. What finally decides the shape of the waveform for bistable multivibrator?

Result:

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EX NO: 7

DATE : SIMULATION OF ACTIVE FILTERS

Aim:

To simulate the Active filters: Butterworth 2nd order Low Pass and High Pass Filter by using PSICE.

System Required:

PC with SPICE software

Circuit Diagram:

SEC O N D O R D E R L O W P A SS BUT T E RW O R TH F IL T E R :

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SE COND ORDER HIGH PASS BUTTERWORT H FILTER:

Procedure:

EDWin 2000 -> Schematic Editor: The circuit diagram is drawn by loading components from

the library. Wiring and proper net assignment has been made. The values are assigned for

relevant components.

EDWin 2000 -> Mixed Mode Simulator: The circuit is preprocessed. The waveform marker is

placed at the output of the circuit. GND net is set as reference net. The Transient Analysis

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parameters have been set. The Transient Analysis is executed and output waveform is

observed in Waveform Viewer.

EDWin 2000 -> EDSpice Simulator: The circuit is preprocessed. The waveform marker is

placed at the output of the circuit. The Transient Analysis parameters are also set. The

Transient Analysis is executed and output waveform is observed in Waveform Viewer.

Model Graph:

Low Pass Filter:

High Pass Filter:

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OUTPUT:

Low Pass Filter:

High Pass Filter:

Result: -

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EX NO:8a

DATE : SIMULATION OF ASTABLE MULTIVIBRATOR

Aim: To simulate the Astable Multivibrator by using PSICE.

System Required:


Circuit Diagram:

A sta ble M ultiv i bra t o r- S y mm et ri ca l

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A s tab l e M u l tiv i b r a t o r - A s y m m e t r i c a l

Procedure:





placed at the output of the circuit. GND net is set as reference net. The Transient

Analysis parameters have been set. The Transient Analysis is executed and output waveform is


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Model Graph:

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OUTPU T :

Astable Multivibrator-Symmetrical

Astable Multivibrator-Asymmetrical

RE SU LT

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EX NO: 8b

DATE : SIMULATION OF MONOSTABLE MULTIVIBRATOR

Aim:

To simulate the Monostable Multivibrator by using PSICE.

System Required:


Circuit Diagram:

Procedure:




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Model Graph:

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O U TP U T

Result:

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EX NO: 8c

DATE : SIMULATION OF BISTABLE MULTIVIBRATOR

Aim:

To simulate the Bistable Multivibrator by using PSICE.

System Required:


Circuit Diagram:

Procedure:




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Model Graph:

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OU T P U T :

Resu lt : -

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EX NO: 9

DATE : SIMULATION OF ANALOG TO DIGITAL CONVERTER

Aim: To simulate the Analog to Digital Converter by using PSICE.

System Required:


Circuit Diagram:

Procedure:




EDWin 2000 -> Mixed Mode Simulator: The circuit is preprocessed. The waveform marker

is placed at the output of the circuit. GND net is set as reference net. The Transient

Analysis

Am

plitu

de i

n vo

lts

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parameters have been set. The Transient Analysis is executed and output waveform is





Model Graph:

M O D E L G R A P H :

Input (A)

Time in ms

Input (B)

Time in ms

Input (C)

Time in ms

Input (D)

Time in ms

Output Analog Signal

Time in ms

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OUTPUT:

A N A L O G TO D I G I TA L C O N V E R T E R

Result: -

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EX NO: 10

DATE : SIMULATION OF CMOS INVERTER, NAND AND NOR

Aim: To plot the transient characteristics of output voltage for the given CMOS inverter, NAND

and NOR from 0 to 80μs in steps of 1μs.

System Required:


Circuit Diagram:

(i) CMOS Inverter:

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(ii) CMOS NAND:

(iii) CMOS NOR:

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Theory:

(i) Inverter CMOS is widely used in digital IC’s because of their high speed, low

power dissipation and it can be operated at high voltages resulting in improved noise immunity.

The inverter consists of two MOSFETs. The source of p-channel device is connected to

+VDD and that of n-channel device is connected to ground. The gates of two devices are

connected as common input.

(ii) NAND It consists of two p-channel MOSFETs connected in parallel and two n-channel

MOSFETs connected in series. P-channel MOSFET is ON when gate is negative and N-channel

MOSFET is ON when gate is positive. Thus when both input is low and when either of input

is low, the output is high.

(iii) NOR It consists of two p-channel MOSFETs connected in series and two n-channel

MOSFETs connected in parallel. P-channel MOSFET is ON when gate is negative and N-

channel MOSFET is ON when gate is positive. Thus when both inputs are high and when

either of input is high, the output is low. When both the inputs are low, the output is high.

Truth Table:

(i) Inverter

Input Output

(ii) NAND

V1 V2 Output

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(iii) NOR

V1 V2 Output

Model Graph:

(i) Inverter

(ii) NAND

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(iii) NOR

(i) Inverter

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(ii) NAND

(iii) NOR

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VIVA Questions:

1. What are the advantages of monostable Multivibrator?

2. What are the applications of astable Multivibtrator?

3 .What is a complementary Multivibrator

4. Define Blocking Oscillator?

5. What are the two important elements of Blocking Oscillator?

6. What are the applications of blocking Oscillator?

Result: -

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Documents

feedback factor

feedback signal

feedback network

positive feedback

negative feedback

types of feedback

input impedance

output resistance