ECAD2 Complete Manual

7/28/2019 ECAD2 Complete Manual

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S.E. SEM IV (BIOMEDICAL): ELECTRONIC CIRCUIT ANALYSIS AND

DESIGN-II

Date:

EXPERIMENT NO: - 1

_______________________________________________________________________

AIM: TO MEASURE THE Op-AmprPARAMETERS

a. Slew rateb. Input Bias currentc. Input and output offset Voltage

APPARATUS: Bread Board, CRO, Function Generator, Dual power supply, multimeter.

COMPONENTS: Connecting wires, resistors: 1K,1M, 10K, 100K, LM741

,IC,0.1F,0.001 F

PROCEDURE:

a. Slew rate Connect the circuit as shown in figure. Give square wave input having amplitude 1Vp-p and vary the frequency from

low to high. At each frequency observe the output square wave and note down amplitude

and frequency of output waveform . Calculate the slew rate.

PIN DIAGRAM OF OP-AMPrIC-:


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DESIGN-II

CIRCUIT DIAGRAM:

OBSERVATION TABLE:

Sr. No Frequency(Hz) X-axis Vt Y- axis Vd

1. 2k2. 30k3. 100k

b. Input Bias current Connect the circuit to measure input current of inverting input Noninverting terminal is grounded The inverting terminal is at virtual ground and we have ( IoRf)=VO Measure VO and calculate IB Now connect the circuit to measure input current of non-inverting input inverting terminal is grounded through Rf VO = ( IB.Rf ) So measure VO and IB

c. Input Offset Current:

Iio=IB1-IB2


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DESIGN-II

CIRCUIT DIAGRAM:

0.01F

+Vcc

-VEE

1M

V0

-

+

+Vcc

-VEE

V0

-

+

1M 0.1F

CALCULATION:IB1= VO1/ Rf

IB2= VO2/ RfIB= IB1+ IB2

d. Input and Output Offset Voltage

Connect the circuit as shown in figure Ground the both input terminal Measure the output offset voltage using voltmeter Calculate input offset voltage


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DESIGN-II

CIRCUIT DIAGRAM:

+Vcc

-VEE

V0

-

+

100K

1K

CALCULATION:

Output Offset Voltage VO =_________________mVInput Offset Voltage Vio = VO .R2/ (R1+R2)

CONCLUSION:

Questions:

Q1. Define slew rate and explain the significance of large value of slew rate.


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DESIGN-II

Q.2 A 10Khz square wave is to be amplified by an op-amp to have an output

voltage swing 10V. Two op-amps are1) uA741 having S.R. of 0.5V/us &

2) TL081 with a S.R. of 13V/us. Determine the stability of op-amp.

Q.3 List out the ideal and practical specification for op-amp 741?

References:

Date: Grade: Faculty Incharge:


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DESIGN-II

EXPERIMENT NO: - 2

_______________________________________________________________________

AIM: TO STUDY DIFFERENT APPLICATION OF OPERATIONAL AMPLIFIER

a. Inverting Amplifierb. Non-Inverting Amplifierc. Voltage follower

APPARATUS REQUIRED: function generator, CRO, Dual power supply

COMPONENTS: Connecting wires, resistors: 1K, 10K, LM741 IC, Bread Board,

Connecting wires and probes.

a .Inverting Amplifier

THEORY:

The input signal Vi is applied to the inverting input terminal through R1 and the non-

inverting input terminal of the op-amp is grounded. The output voltage Vo is fed back to

the inverting input terminal through the Rf- R1network, where Rf is the feedback resistor.

The output voltage is given as,

Vo = - ACL Vi

Here the negative sign indicates that the output voltage is 1800

out of phase with the input

signal.

PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. By adjusting the amplitude and frequency knobs of the function generator,

appropriate input voltage is applied to the inverting input terminal of the Op-Amp.


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DESIGN-II

The output voltage is obtained in the CRO and the input and output voltagewaveforms are plotted in a graph sheet.

CIRCUIT DIAGRAM OF INVERTING AMPLIFIER: Rin=1k, Rf=10k

+Vcc

-VEE

V0

-

+

!

20mV

1KHZ

10K

1K

AC

OBSERVATIONS:

A=-Rf/R1

Sr. No Vin Vo(theoretical) Vo(practical)

1.2.3.4.5.

b. NON - INVERTING AMPLIFIER:

THEORY:


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DESIGN-II

The input signal Vi is applied to the non - inverting input terminal of the op-amp. This

circuit amplifies the signal without inverting the input signal. It is also called negativefeedback system since the output is feedback to the inverting input terminals. The

differential voltage Vd at the inverting input terminal of the op-amp is zero ideally and the

output voltage is given as,

Vo = ACL Vi

Here the output voltage is in phase with the input signal.

PROCEDURE:

Connections are given as per the circuit diagram.

+ Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. By adjusting the amplitude and frequency knobs of the function generator,

appropriate input voltage is applied to the non - inverting input terminal of the Op-

Amp.


CIRCUIT DIAGRAM OF NON INVERITNG AMPLIFIER: R1=1k, R2=10k


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DESIGN-II

+Vcc

-VEE

V0

-

+

!

20mV

1KHZ

10K

1K

OBSERVATIONS:A=1+(R2/R1)


1.2.3.4.5.

c. VOLTAGE FOLLOWER:PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. By adjusting the amplitude and frequency knobs of the function generator,

appropriate input voltage is applied to the non - inverting input terminal of the Op-

Amp.


CIRCUIT DIAGRAM:


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DESIGN-II

OBSERVATIONS:

A=1


1.2.3.4.5.CONCLUSION:

Questions:

Q.1 Explain applications of voltage follower.

Q.2 Differentiate NINV & INV configuration.


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DESIGN-II

Q.3 For the circuit shown below, find the range of input voltages that are to be

applied at input terminals, such that output remains undistorted.

(Vcc = 15V)

Q.4 Enlist the features of voltage follower & draw its equivalent circuit.


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DESIGN-II

Q 5. Explain transfer characteristics of op-amp.

References:



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DESIGN-II

EXPERIMENT NO: - 3

_______________________________________________________________________

AIM: Design the op-amp circuits for the following equations

a. AdderVo= -3V1+4V2

b. SubstractorVo= Vb-Va

APPARATUS REQUIRED: Multimeter, Dual power supply

COMPONENTS: Connecting wires, resistors: 1K, 10K, LM741 IC, Bread Board,

Connecting wires.

a. ADDER:i. Inverting adder

CIRCUIT DIAGRAM:

THEORY: Vo= -Rf{(Va/Ra)+(Vb/Rb)}

1) Summing Amplifier :- If Ra=Rb=Rf,Vo= -(Va+Vb)

2) Scaling or weighted amplifier: Each i/p volt is amplified by a different factorsi.e weighted differentially at the o/p , the circuit is then a scaling or weighted

amplifier

Vo= -{(Rf/Ra).Va+(Rf/Rb).Vb}

+Vcc

-VEE

V0

-

+

Vb

Rf=100K

Va

Ra

Rb


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DESIGN-II

3) Averaging Circuit: Ra=Rb=R & Rf/R=1/n where n=no. of i/psEg. n=2 then Vo=Va=Vb/2

PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC.

Give appropriate input voltage to the non - inverting input terminal of the Op-Amp.

The output voltage is obtained on the multimeter.OBSERVATIONS:


1. Va=Vb=

2. Va=Vb=

3. Va=Vb=

4. Va=Vb=

5. Va=Vb=


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DESIGN-II

b. SUBTRACTOR

CIRCUIT DIAGRAM:

+Vcc

-VEE

V0

-

+

10K

10K

10K

10K

Va

Vb

THEORY: Ra=Rb=Rf

Hence gain of differential amplifier is 1.

Vo= -Rf/R(Va-Vb)

Vo=Va-VbHence o/p voltage is equal to the voltage applied to the non inverting terminal

minus voltage applied to the inverting terminal. Hence the circuit is called as

Subtractor.

PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. Give the appropriate input voltage is applied to the non - inverting input terminal of

the Op-Amp.

The output voltage is obtained on the multimeter.

OBSERVATIONS:

Sr.No Vin Vo(theoretical) Vo(practical)

1. Va=Vb=

2. Va=


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DESIGN-II

Vb=

3. Va=Vb=

4. Va=

Vb=

5. Va=Vb=

CONCLUSION:

Questions:

Q.1 How adder can be used as scalar & averaging amplifier? (Explain for NINV)

Q.2 Supply voltage=+15V, Va=2v, Vb=-3v, Vc=4v,R=R1=1k ,Rf=2k. Determine V1 at

the NINV terminal & Vo for NINV adder.


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DESIGN-II

Q.3 When the resistances used in subtractor are not equal, how Vo can be calculated?

References:

Date: Grade: Faculty Incharge


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DESIGN-II

EXPERIMENT NO: - 4

_______________________________________________________________________

AIM: To design an Integrator circuit for the given specifications using

Op-Amp IC 741 & study its frequency response.

APPARATUS REQUIRED: Function generator, CRO, Dual power supply

COMPONENTS: Connecting wires, resistors: 1.5K, 15K, 0.1F, LM741 IC, BreadBoard, Connecting wires and probes

THEORY:

A circuit in which the output voltage waveform is the integral of the input voltagewaveform is the integrator. Such a circuit is obtained by using a basic inverting amplifier

configuration if the feedback resistor Rf is replaced by a capacitor Cf . The expression for

the output voltage is given as,

Vo = - (1/RfC1) Vi dt

Here the negative sign indicates that the output voltage is 1800

out of phase with the inputsignal. Normally between fa and fb the circuit acts as an integrator. Generally, the value of

fa < fb . The input signal will be integrated properly if the Time period T of the signal islarger than or equal to RfCf. That is,

T RfCf

The integrator is most commonly used in analog computers and ADC and signal-waveshaping circuits.

Design:

Let fa = 100Hz , fb= 1KHz

fb= 1/2R1CFLet CF= 0.1FThus, R1= 1.5KNow, Fa= 1/2RFCFThus, RF= 15K


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DESIGN-II

CIRCUIT DIAGRAM OF INTEGRATOR:

PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. Give 20mV with 1KHz input signal to obtain sinusoidal waveform The output voltage is obtained in the CRO and the input and output voltage

waveforms are plotted in a graph sheet.
http://2.bp.blogspot.com/-CC9rQIyxn3I/TZpYuhvkVgI/AAAAAAAAAO8/hhah0KBnAVQ/s1600/2.JPG


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DESIGN-II

OBSERVATIONS:

S.No. Amplitude

( No. of div x Volts per div )

Time period

( No. of div x Time per div )

Input

Output

CONCLUSION:

Questions:

Q.1 Draw basic integrator circuit & draw its frequency response curve. Mark

operating frequencies.

Q.2Write the problems associated with basic integrator. How to overcome it?


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DESIGN-II

Q.3 Enlist the application of integrator.

References:


Date:


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DESIGN-II

Date:

EXPERIMENT NO: - 5

_______________________________________________________________________

AIM: To design a Differentiator circuit for the given specifications using

Op- Amp IC 741 & study its frequency response.


COMPONENTS: Connecting wires, resistors: 1.5K, 15K, 0.1F, LM741 IC, Bread

Board, Connecting wires and probes

THEORY: To design differentiator, we follow the steps as given:

1) Fa= 1KHz = 1/(2 Rf C1)Let C1 = 0.1uF; then Rf = 1/(2)(103)(10-7) = 1.59KLet Rf be 1.5 K

2) Fb = 20KHz = 1/ (2 R1 C1)Hence, R1= 79.5Let R1 = 82. Since R1C1 = RfCfCf = 0.0055uF

CIRCUIT DIAGRAM OF DIFFERENTIATOR:
http://3.bp.blogspot.com/-wXDU6F9Kg0o/TZpYVel_bxI/AAAAAAAAAO0/5XcvFx_nr4w/s1600/2.JPG


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DESIGN-II

PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. Give 20mV with 1KHz input signal to obtain sinusoidal waveform The output voltage is obtained in the CRO and the input and output voltage

waveforms are plotted in a graph sheet.

OBSERVATIONS:

Input - Sine wave

S. No. Amplitude


Time period


Input

Output

InputSquare wave

S. No. Amplitude


Time period


InputOutput

Questions:

Q.1 If a sine wave of 1 V peak at 1000hz is applied to the designed differentiator, find

its output voltage.
http://1.bp.blogspot.com/-25cHj0cJFfc/TZpYQmzuW6I/AAAAAAAAAOw/JitQbBtdom4/s1600/3.JPG


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DESIGN-II

Q.2 Draw basic differentiator circuit and draw its frequency response curve. Mark

operating frequencies.

Q.3 Write the problems associated with basic differentiator. How to overcome it?


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DESIGN-II

Q.4 Enlist the application of differentiator.

References:

Date: Grade: Faculty Incharge


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DESIGN-II

EXPERIMENT NO: - 6

_______________________________________________________________________

AI M: To design a wein bridge oscillator using 741 to generate a frequency of 965 Hz.


COMPONENTS: Connecting wires, resistors: R=3.3k, R1=12k, Rf=50k, C= 0.05F,

LM741 IC, Bread Board, Connecting wires and probes.

DESIGNING PROCEDURE:

Assume output frequency=965Hz

Step 1:The frequency oscillations is,

F0=1/2RCRC=1/2fo=1/2*965Hz=secAssume C=0.05 F, R=3.3K

Step 2: The condition of oscillation is

Av=3Gain of op-amplifier in non inverting mode is Av=1+Rf/R1,

3=1+Rf/R1

Rf= 2 R1

Thus, Rf= 2*12k = 24k


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DESIGN-II

CIRCUIT DIAGRAM OF WEIN BRIDGE OSCILATOR:

R1 RF

+VCC

2 76

3 4VO

C

-VEE

RR C

PROCEDURE:

Connections are given as per the circuit diagram. + Vcc and - Vcc supply is given to the power supply terminal of the Op-Amp IC. The output voltage is obtained in the CRO

OBSERVATIONS:

S.No. Amplitude Time period

( No. of div x Time per

div )

Frequency

Output

CONCLUSION:


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DESIGN-II

Questions

Q.1 What is oscillator.

Q.2 Draw the block diagram of oscillator.

Q.3 Write the barkhausen criterion.

References



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DESIGN-II

EXPERIMENT NO: - 7

___________________________________________________________________

AIM: To study effect of feedback using CE amplifier

APPARATUS:Transistor BC-547

Regulated power Supply (0-30V, 1A)

Function Generator

CROResistors [33K, 3.3K, 330, 1.5K

1K, 2.2K, 4.7K]Capacitors- 10F -2NoBread Board

Connecting Wires

THEORY:1) Current-Series feedback:

i. GMs = Transconductance including Rsii. GMl = Io/Vi = Transconductance including load (without feedback)

iii. GMf= Io/Vs = Transconductance with feedbackiv. = Vf/Io = Gain of feedback circuit

2) Voltage-Series feedback:i. AVs = Voltage gain including Rs

ii. AVl = Vo/Vi = Voltage gain without feedback but with RLiii. AVf= Vo/Vs = Voltage gain with feedbackiv. = Vf/Vo = Gain of feedback network


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DESIGN-II

CIRCUIT DIAGRAM:

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 Generator3. 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 expressionAv=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.10. The band width of the amplifier is calculated from the graph using the express,Bandwidth, BW=f2-f1

Where f1 lower cut-off frequency of CE amplifier, and

Where f2 upper cut-off frequency of CE amplifier11. The bandwidth product of the amplifier is calculated using expression

Gain Bandwidth product=3-dBmidband gain X Bandwidth


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DESIGN-II

OBSERVATIONS:

Input voltage Vi =20mV

FREQUENCY (Hz) OUTPUT

VOLTAGE (V0)

GAIN

AV=(V0/Vi)

GAIN IN dB

Av=20log10(V0/Vi)

FREQUENCY RESPONSE:

Vi =20mv

Sr. no FREQUENCY(Hz) OUTPUT

VOLTAGE (V0)

GAIN IN dB

Av=20 log10 (V0/Vi)

1. 102. 503. 1004. 2005. 3006. 4007. 5008. 6009. 70010. 80011. 90012. 1K13. 2K14. 10K15. 20K16. 80k17. 90K18. 100k19. 120K20. 150k21. 160K22. 180K23. 220K24. 300K25. 400K26. 500K

Sr. no FREQUENCY(Hz) OUTPUT GAIN IN dB


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DESIGN-II

VOLTAGE (V0) Av=20 log10 (V0/Vi)

27. 600K

27. 700K28. 800K29. 900K30. 1M

CALCULATIONS:1. Bandwidth:

2. fH, fL:

MODELWAVE FORMS:

INPUT WAVE FORM:

OUTPUT WAVE FORM
http://4.bp.blogspot.com/-2kjjgBsTBlI/ToxtKA6QKGI/AAAAAAAAAnY/jQHhoPuQcFg/s1600/2.JPG


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DESIGN-II

FREQUENCY RESPONSE

Questions:

Q.1 Differentiate features of CE amplifier with feedback & without feedback.
http://3.bp.blogspot.com/-JzFSngOf5FE/ToxtScy6hGI/AAAAAAAAAng/eB94liJSH5o/s1600/4.JPGhttp://3.bp.blogspot.com/-PQGmUKEruDg/ToxtOXlVbvI/AAAAAAAAAnc/sP5m2yxVnqM/s1600/3.JPGhttp://3.bp.blogspot.com/-JzFSngOf5FE/ToxtScy6hGI/AAAAAAAAAng/eB94liJSH5o/s1600/4.JPGhttp://3.bp.blogspot.com/-PQGmUKEruDg/ToxtOXlVbvI/AAAAAAAAAnc/sP5m2yxVnqM/s1600/3.JPG


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DESIGN-II

Q.2 Derive relations of Rif, Rof, gain of the amplifier.

References:



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DESIGNING-II

CONCLUSION:

Questions:

Q.1 What is the function of capacitor in regulated power supply circuit?

Q.2 How 78XX can be used as adjustable current source?

Q.3 Write the specifications of 78XX


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S E SEM IV (BIOMEDICAL) ELECTRONIC CIRCUIT ANALYSIS AND

Q.4 Write the specifications of 79XX

References:


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ECAD2 Complete Manual