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SAlT IABALPURDEPARTMENT OF ELECTRONICS &COMMUNICATION ENGG.EC-404 ELECTRONICS CIRCUITS LAB

LIST OF EXPERIMENT

1) TO MEASURE THE INPUT OFFSET VOLTAGE AND INPUT BIASCURRENT OF OP-AMP USING IC-741.2) TO MEASURE THE CMRR OF OP-AMP USING IC-741.3) OP-AMP AS INVERTING AND NON-INVERTING AMPLIFIERUSING IC-741.4) OP-AMP AS ADDER AND SUSTRACTOR USING IC-741.5) OP-AMP AS AN INTEGRATOR AND DIFFERENTIATOR USINGIC-741.6) DESIGN &CONSTRUCT A SHUNT & SERIES REGULATOR ANDFINE LINE AND LOAD REGULATION.7) PERFORMANCE EVALUATION OF FEEDBACKAMPLIFIERS(OSCILLATOR).8) TO MEASURE THE SLEWRATE OF OP-AMP USING IC-741.

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S.NO. NAME OF PAGE DATE GRADEl REMARK SIGNEXPERIMENT NO. TOTAL1.

2.

3.

4.

5.6.7.

8.

9.

10.

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EXPERIMENT NO: 1oBIECTIVE: -Tomeasure the Input offset voltage (Vio)& Input Biascurrent of op-amp.EQUIPMENT: -1.OPT Trainer Kit2. Power supply.3. Patch cords.4.MultimeterTHEORY:-1) Input offset voltage (Vio):-An ideal op-amp will give olp of 0 v if both its input are shorted together. A realworld op-amp will have non-zero voltage, even if its inputs are shorted together.This is the result of liP offset voltage which is slight voltage present at its inputsbrought about its non-zero input offset voltage which is slight voltage present atits inputs brought about its non-zero input offset current. In essence, the inputoffset voltage is also the input voltage that's needs to be applied across the inputsof an OP-Amp to make the output voltage zero.2) Input offset current (Iio):-It is defined as the difference b/w the separator currents entering the inputterminal of a balanced amplifier.lio= (IBl) - (IB2)3) Input bias current (IB):-

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It is defined as the average of the currents that flow into the inverting and non-inverting input terminals of the OP-AMP.IB= (IBl) - (IB2)

2PROCEDURE:-MEASUREMENTS OF I / P OFFSET VOLTAGE (Vio)

RF=100K!!

Rl=100k!! +12V7

1)Make the connections as shown in the diagrams.2) Apply +VCCand --VEEas + 12V.3)Measure the output voltage on DMM.4) Find the input offset voltage using formula.Vio = vo*Rlj (Rl+Rf)

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MEASUREMENT OF INPUT OFFSET CURRENT (lio) AND INPUTBIAS CURRENT (Ib)

C=O.lf.lf

1-----------,R=lM!!

Vo

+12VIBI 2

>--------L...---O

IB2 3A -12V

R=lM!! C=O.lf.l

B

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Measurement of IB1:-

C =0.1,...

R=1M!!

Measurement of IB2:-

R=1M!!

I B l>-__ ---L..- __ -{ V0"

-12V

+12V

IB2

Yo'

-12V

T C=O.l

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Measurement of input offset current (lio) and input bias current (IB):-1) Make the connections as shown in the diagrams.2) Apply +VCC and -VEE as + 12V.3) Measure the output voltage on DMM.4) To measure IBl, short A & B, measure Vo as Vo". Calculate IBI using formula,IBl= Vo" jR.5) To measure IB2, short C & D, measure Vo as Vo". Calculate IB2 using formula,IB2= Vo' jR.6) Calculate input offset current lio = (IBl) - (IB2).7) Calculate input bias current, using IB = (IBl) - (IB2)j2.DESIGNS:-1) Input offset voltage (Vio).-Vio = Vo*Rlj (Rl +Rf)2) Input bias current = IB = (IBl) - (IB2)

2OBSERVATION TABLE:-IC-741:-1.ljp offset voltage (Vio) =Output voltage reading Vo = -1.225 V.2. Vo" = 29 mV. Therefore IBI = 29 nA.3. Vo' = -33 mV. Therefore IB2 = 33 nA

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Serial no. OpAmp Ideal Typical Practicalparameter Value Value Value

1. if p offset voltage 0 2mV 1.25mV2. if p offset current 0 20nA 4nA(Iio)3. ifp bias 500nA BOnA 31 nA

current(IB)

CONCLUSION:-In this experiment, we studied different parameters like Vio, lio & IBfor the IC741.Weobserved that the IC 741as its parameter values is verified with the idealvalues.

VIVA QUESTIONSQ.l. What is Input offset voltage?

Q2.What is Input Bias current?

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Q3. What is the I/P and OjP impedance of an ideal OP-AMP?---------------------------------

EXPERIMENT NO: 2OBJECTIVE: -To measure the OP-Amp parameter for IC-741.Common mode rejection ration (CMRR)EQUIPMENT:-1. OPT- 01 Trainer kit.2. Power supply.3. Patch chords.4.DSO5. Multi-meter.THEORY:-COMMON MODE REJECTION RATIO (CMRR):-It is defined as the ratio of differential voltage gain to common mode voltagegaIn.

CMRR= Ad/AcAd =Vo/Vd

And common mode voltage gain can be determined byACM=VocM

VicMVicM is the common mode l /P voltage.ACM is common mode voltage gain. The CMRR is very large hence CMRR ismost often expressed in decibels.PROCEDURE:-

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MEASUREMENT OF CMRR:-

RF=100Kn

Rl=100n VCC=+12

Rl=100n >-------~----~Jo/P

Ivp-pR2=100Knin

1.Make the connection as shown in fig.2. Apply 1Vpp sine wave at input.3.Measure ojp and calculate CMRRas.

CMRR = 1+ (R2jR1)*(VsjVo)DESIGNS:-

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EXPERIMENT NO: 3

oBIECTIVE: -To measure the OP-Amp parameter for IC-741.Slew Rate (SR)EQUIPMENT:-1. OPT- 01 Trainer kit.2. Power supply.3. Patch chords.4.DSO5. Multi-meter.THEORY:-SLEWRATE:-This is the minimum rate of change of o/p Voltage/unit time.

SR =? VO/? tIt is expressed in V/ J . l second.The SR is measured in the unity gain voltage follower ckt as shown. The SR ismeasured in the amplifier driven by high frequency wave of sufficientmagnitude. The SR slope of the transition between o/p levels frequently +ve and-ve swings will have different SR and both must be examined. In such case lowerSr is commonly specified.

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MEASUREMENT OF SLEW RATE:-SR = ! : : : : . V0 -VEE

l : : . t

-vcc

6Vp-p, 25KHZSquare wave

SLEWRATE:-1.Make the connections as shown in fig.2. Apply 6VPP, 25 khz square wave input at non-inverting terminal.3.Measure? Vow.r.t.?t and calculate SR asS R = ! : : : : . V 0

l : : . tDESIGNS:-Slew rate (SR)

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SR =f:. Vof : : : : . t

OBSERVATION TABLE:-IC-741:-

Sr. No. PARAMETER IDEAL VALUE PRACTICAL VALUE

1 SR 0.5 V/Ilsec .73V/ usee

CONCLUSION:-In this experiment we studied different parameter like SLEW RATE for the IC-741.WE observedthat the IC-741 as its parameters values is verified with the ideal values.

VIVA QUESTIONSQl.What is Slew Rate?

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EXPERIMENT NO: 4oBIECTIVE: -1. To study Ie 741 as Inverting Amplifier.2. To observe & note effect of change in RF& R i value on OIP Vo,3. To study Ie 741 as Non-Inverting Amplifier.4. To observe & note effect of change in RF& R i value on OIP Vo,EQUIPMENT:-Signal generator, eRO - dual channel, patch chord, DMM.COMPONENT VALUE:-Rn = lK, R12= 10K Rl3 = 4k7 pot, RFl = 10K ,RF2= lOOK,RF3= 33K

THEORY:-An op-amp can be used for number of application likeSubtractors, Rectifier, Multivibrators, Analog computer etc.Inverting & Non-inverting amplifier.

amplifier, Adder'sIe 741 is used as a

1. IC 741 AS INVERTING AMPLIFIER.When the input is applied to the inverting input terminal (negative terminal) ofop-Amp, then Op-Amp is said to be operated in inverting amplifier mode. Theamplified as well as the inverted output signal is obtained from output pin 6 ofOp-Amp. This output signal is applied to inverting input via feedback resistorRF.It forms a negative feedback because any increase in the output signal resultsin a feedback signal into inverting causing a decrease in the output signal.

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Note that non-inverting terminal is grounded. Feedback circuit usesanyone resistor RF, However an extra resistor R i is connected in series with theinput signal source Yin. As shown in fig. 1. It is also known as voltage shuntfeedback amplifier.

Fig. lie 741 As Inverting Amplifier.Closed Loop Voltage Gain (AP):The closed loop voltage gain - AF of the amplifier can be obtained by applyingKCL at node V2.

il = IB2 + iF - - - - - - - - - - - - - - - - - ( 1 )Since input impedance Ri of Op -Amp is very large, so 1 B 2 ::::;0.Hence neglecting 1 B 2we get, h = iF----------------(2)

h =From fig. 1

From equation (2) we get iF =V 2- Vo---

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= - - - - - - - - - - - - - - - - ( 3 )

We know, Va= A (Vid):.As Vl=O

: .Vo = -A V 2-V a:.V2= --ASubstituting the value of V2 in equation (3)

We get, Yin +VoLA = ::Y. . .o~o -------------(4)

Va - -- -- -- -- -- -- -- -- -- - ( 5)

As gain A is very large (ideally infinity).We get,

A F = _ A R FARlVa =Yin

.'. V a = -R F-- Vi n ------------------- (6)R l

The negative sign in the above equation indicates the phase difference of 1800between input and output. By changing the value of R i & RF the gain can bechanges.

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PROCEDURE:-For AC I/P:1. Study the circuit provided on front panel of the kit.2. Make the circuit as shown in above figure.3. Switch on the power supply.3. Select desired R i &RF.5. Connect dual trace CRO at input and output side to observe Yin & Vorespectively.6. Apply 100Hz, sine wave input Yin from signal generator. Adjust it's

amplitude so that op-amp should not enter in saturation.7. Observe & note input Yin amplitude & output Vo amplitude on CRO

calculate it's gain using AF = Vo/Vin8. Calculate theoretical gain of Op-amp using selected R i &RFvalues.9. Compare the above results.10.Vary the input frequency observe change in output.11.Draw the waveforms on graph.12.Repeat the above procedure for different combination of R i &RF.

OBSERVATION TABLE:-Rn= _ R12= _ R13= _

RFl= _ RF2= _ RF3= _

Sr. R l RF Vin Vo Theoretical gain Practical gainNo. A F = - RF /R l A F = VeVVin1 Rn RFI

RF2RF3

2 R12 RFIRF2

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RF3

3 R13 RFIRF2

RF3

RESULT:The input signal is amplified & inverted at the output of inverting amplifier.

2. rc 741 AS NON-INVERTING AMPLIFIER.When the input is applied to the non-inverting input terminal (positive terminal)of op-Amp, then Op-Amp is said to be operated in non-inverting amplifiermode. The amplified in phase output signal is obtained from output pin 6 of Op-Amp. This output signal is applied to inverting input via feedback resistor RF.Itforms a negative feedback because any increase in the output signal results in afeedback signal into inverting causing a decrease in the output signal. Notethat inverting terminal is connected with OIP VOthrough Feedback resistor RF,with R i is grounded form inverting terminal. However an extra resistor R i isconnected in series with the input signal source Vin. As shown in fig. 1. It is alsoknown as voltage series feedback amplifier.

alP v;Vid

-Vee

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Fig. 1: IC 741 As Non-Inverting Amplifier.

Closed Loop Voltage Gain (AF)

AF= Va-. - -----------------( 1)V inWe know, closed loop gain as defined as

Also, Vo = A ( VI - V2)- - - - - - - - - - - - - - - - - - - (2)Referring to Fig. 1: We get VI = Vin

Since RiRl

. . v ; = A ( Vin

Finally we get, Va = A(Rl+RF) VinRl+RF+ARl

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Va__ =1+AF = V inRF) -----------( ~,~ - - - - - - - - - - - - - - - - - - - - - ( ~ )

- - - - - - - - - - - - - - - - - - - - - ( 5 )

Bychanging the value of R i &RF,the closed loop gain can be changed. The phasedifference between I/P &O/P is zero, so it is called as Non-inverting amplifier.

PROCEDURE:-For AC I/P:1) Study the circuit provided on front panel of the kit.2) Make the circuit as shown in above figure.3) Switch on the power supply.4) Selectdesired R i &RF.5) Connect dual trace CRO at input and output side to observe Vin & Vo

respectively.6) Apply 100Hz, sine wave input Vin from signal generator. Adjust it's

amplitude so that op-amp should not enter in saturation.7) Observe &note input Vin amplitude &output Voamplitude on CRO calculate

it's gain using AF= Vo/Vin8) Calculate theoretical gain of Op-amp using selected R i &RFvalues.9) Compare the above results.10)Vary the input frequency observe change in output.11)Draw the waveforms on graph.

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12)Repeat the above procedure for different combination of Ri & RF.

OBSERVATION TABLE:-R n = _ R 1 2 = _ R 1 3 = _R F l = _ R F 2 = _ R F 3 = _

Sr. R l R F Vin Vo Theoretical Practical gainNo. gain A F = VeVVin

A F = 1 R F / R l1 R n R F I

R F 2R F 3

2 R 1 2 R F IR F 2R F 3

3 R 1 3 R F IR F 2R F 3

RESULT:-The Ie 741 as a Non-inverting amplifier provided amplified and in phase O/Pwhen input is applied to non-inverting l /P, By changing Ri & RF gain can bechanged.

VIVA QUESTIONS

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Q1.What is OP-AMP?

Q2.What is inverting & non-inverting terminal of OP-AMP and why it is called so?

Q3.What is the name of the IC used in OP-AMP?

Q4.IC 741is ----------pin LC,Q5.What is IC?

Q6.What are the type of IC's?

Q7.What is the gain of inverting amplifier?

Q8.What is the gain on non-inverting amplifier?

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EXPERIMENT NO: 5oBIECTIVE: -1. To study IC 741 as an Adder circuit.2. To study IC 741 as a Subtractor circuit.EQUIPMENT:-

Digital Multimeter, Patch Cords.COMPONENT USED:-Rl = 10K, R = 10K, RL= 10K, RF= 10K, 20K, IC 741, +12V.THEORY:An op-amp can be used for number of application like amplifier, Adder'sSubtractors, Rectifier, Multivibrators, Analog computer etc.

1. IC 741 AS ADDER &AVERGER CIRCUIT.If the different l /P voltage sources (Va, Vb, Vc) & resistors are connected to thenon-inverting terminal as shown in fig. 1, then it forms Non-inverting adder,averager circuit, depending on the selection of resistor Ri and RF

R R FV2

Vi

vaVa R

+ 0--------"~Rl

+ Vb R0--------"+V C R

0--------" Fig. 1:- Non-inverting adder averager circuit.

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As the input Resistance Rif of the Non-inverting amplifier is very large.Therefore, using the superposition theorem, the voltage Vi at the Non-invertingterminal is

RI2 RI2 RI2V1 = Va + Vb +

R+RI2 R+RI2 R+RI2

Va Vb Vc Va + Vb + VcV1 = . . . . . .

3 3 3 3

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The gain of the Non-inverting amplifier is given by

RFVO = (1 + ) V1

R1

RF Va +V b+VcVO = (1 + ) (1 + J ----------------(1)

R1 3The equation (1) shows the scaling O/P.

If in equation (1)RF

(1 + )=3 1.e.

thenVo = Va + Vb + Vc

Hence the circuit will be adder/summing amplifier circuit. Where as if Inequation (1).

1+ =1 i.e. RF = 0 OR Rl = 00

Then from equation (1),Va + Vb + Vc

VO =3

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The circuit will be average amplifier.

2. IC 741 AS SUBTRACTOR CIRCUITA basic differential amplifier can be used as a subtractor as shown in fig.l.In thisfigure, input signal can be scaled to the desired values by selecting appropriatevalues from the external resistors.

R

10 KVb R

a lP10 K

10 K R,

~RO KFig.1:- Subtractor Ckt. ~10 K

When this is done, the circuit is referred as scaling amplifier. However if all theexternal resistors are equal in value, then the gain of the amplifier is equal to 1.

From fig.l, the 0 jP voltage of the differential amplifier with a gain of 1 isR

Va = - - - -R

:. Va =

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Vo = V t. - V a

As the O/P voltage is difference of the liP voltages applied. Hence, the circuit iscalled as subtractor.

PROCEDURE:-(1) Study the circuit provided on the front panel of the kit.(2) Switch ON the power supply and Note the Voltages formed by Network.

Select Adder operation by Patch Cord shown in circuit.(3) Connect liPs of OP-AMP Va, Val & Va2to desired voltages v,V2,V3or

V4,note its value, measured O/P voltage Vo,Compare with theoretical Vo = Va + Vb +Vc

(4) Repeat step (3) for different combinations of l/Ps.(5) Now select averager operation by using Patch Cord.Then repeat step (3 ) . Compare the result with theoretical

Va + Vb + Vc

3(6 ) For Subtraction then Connect OP-AMP l/Ps Va & Vb to the desired

voltages of potential divider network (VI to V4) & note it's values.Measure the O/P voltage Vo & compare it with theoretical Vo = - (Va - Vb)

(7) Repeat step (4)for different combinations of Va and Vb.OBSERV ATIONS:For Adder Operations:-

I j P s Theoretical Observed

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Va Vb Vc VQ= VoVa+Vb+Vc

VI V2 V3V2 V3 V4

For Subtractor Operations:-

VI = - - - - - - - - - - V., V2 = - - - - - - - - - - v . , V3 = - - - - - - - - - - V, V4 = - - - - - - - - - - V.Theoretical ObservedlIPsVo=Vb-Va VO-

Va (volts) Vb (volts)Va = - - - - - - Vb = - - - - - -Va = - - - - - - Vb = - - - - - -

RESULT:-By selecting the proper values of resistors circuit can be operates as an adder, orsubtractor circuit.

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VIVA QUESTIONS

Ql. What is the OjP voltage of non-inverting adder circuit?

Q2. What is the OjP voltage of inverting adder circuit?

Q3. What is the OjP voltage of non-inverting subtractor circuit?

Q4. What is the OjP voltage of inverting amplifier subtractor circuit?

Q5. Why IC741 is called 741 IC?

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EXPERIMENT NO: 6OBIECTIVE:-

1) Study of Differentiator Circuit using Op-Amp IC 741.2) Study of Integrator Circuit using Op-Amp IC741.

INSTRUMENTS REQUIRED: - eRO (dual channel) function generator,

THEORY:The output waveform of differentiator is derivative of input signal.

By interchanging the capacitor & resistor in integrator circuit differentiator circuit canbe formed. The basic circuit of differentiator is shown in fig.

C F Ieo 1 1 - - . . . . . . . . . . . . .--1Vin

Va a dVin

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ByKCL

As Ri = 00, Ib ~ 0

:. Ie= I f

d:. CI dt (Vin-V2) =

As VI = 0, so by virtual ground V2~ o.d: . V O =-R p -Yindt

Rl &CFare added in practical differentiator to compensate problem of stability & highfrequency noise.The input signal will be differentiated properly if the time period of input signalapplied (T)is larger than RlCr.

Waveforms for Differentiator Circuit

Input Vi\ A I......,..L-. ...

Vin a1 time-Viva1 VOutput aWavefor time-V

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PROCEDURE:-1) Study the circuit given on the panel kit.2) Connect function generators output to the circuit as input source and CRO to

observe output waveform.3) Apply input signal (T> RF.Cl)and observe the output waveform as CRO.4) Now replace C1by C2 and repeat step (3).5) Vary IfP frequency and observe the change in OIP waveform.OBSERVATION TABLE:-

( 1 ) Input signal Output signalSquare wave - - - - - - - - - - - - - - - - - - - - - -Sinewave - - - - - - - - - - - - - - - - - - - - - -

RESULT:-The input to the circuit of differentiator is integrated and differentiated. From

observation it is seen that the circuit gives proper response and this is depend on l/Psignal frequency.

OP-AMP IC741 AS AN INTEGRATORINSTRUMENTS REQUIRED:-eRO (dual channel) function generator, Patch chords.THEORY:-

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Integrator circuit are used in analog computer, A to D convertors and signalwave shaping. The main reason for using active components (741) withintegrator is to give response for low level input signals.DEFINATION:-The circuit in which the output is a integral of input voltage / signal is known asIntegrator Circuit. Integrator acts as a low pass filter. The basic circuit ofintegrator is shown in fig below.

VinBy Kirchoff's current law.h = Ib+ If

Since input impedance of Op-Amp is very high, Ib~ o .:. h = If ---------------------(1)

Current through capacitor is given by

If = IedVcIf = Cr-+dt

Where Vc = voltage across capacitor.dVin - V2 ( V 2 - V o ) - - - - - - - - - - - - - - - ( 2 )

Rl rlt

As VI = 0, by virtual ground V 2 = 0From equation (2) Vin

=C,Rld-(-Va)dt

Now integrating both sides we get

t Vin tddt

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fa

:. finally we get tf Vin dt + Co - 1 a

Where C is constant.

(-Vo)dt

The proper response of integrator iswith a particular range of input frequencywh 1 . in by formula as given bellowfb: 2I1R1CfWhere tb IS the frequency at which gain is zero dB.fa= 1

Where fa = gain limiting frequency.Input signal will be integrated properly if the time period of input signal is largerthan product of Rr Cr, i.e. T > Rr CrWhere T, is time period of input signal.Waveforms for Integrator Circuit

Input Vi'XT"up-tnr a

-V i

v aOutput

Vi

v a

timeI Il+-t-+l

time

time

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In practical circuit feedback resistor Rs used to reduce the error in outputvoltage

PROCEDURE:-1) Study the circuit given on the front panel of the kit.2) Switch on the power supply.3) Apply input from function generator i.e. sine or square wave (The input

should be such that T >RC).4) Connect Ci in parallel with Rby patch chords.5) Connect output of integrator to CRO by probe and observe the waveform.6) Vary input frequency and observe the change in output waveform.7) Now replace Cl by C2 and repeat step (6)

Graph:InputSquar +V

-V----II I:+"-t--,

I I

Output V 1 0 R ?1 ~"ViI II I~ I I t f\

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Draw the graph with C r as C1 or C2Now increase the frequency of input signal (T

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Q1. Draw the circuit diagram of Integrator circuit?

Q2.What is the OjP voltage equation of Integrator?

Q3.What is the voltage across capacitor?

Q4.What is the current across capacitor?

Q5. Draw the circuit diagram of Differentiator circuit?

Q6.What is the OjP voltage equation of Differentiator?

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EXPERIMENT NO: 7OBIECTIVES:-To Study operation of Zener Diode as a Shunt Voltage Regulator.INSTRUMENT:-Voltmeter (0-15V), milliammeter ( 0- 25mA) optional., Digital Multimeter.Components used:-R = 3800, RL= 10k pot, Zener Diode Vz = 6.2V & 6.8VTHEORY:-A Zener diode shunt regulator is shown in Fig 1.Since the Zener is connected in parallel(or shunt) with the load, the circuit is said to be a shunt regulator. A resistance (RS)isconnected in a series to the zener Diode which limit the current in the circuit. Hence itis also known as series current limiting resistor. The output voltage VL is taken acrossthe load resistance RL. Which is also the voltage across Zener diode. To observe theproper operation, the input voltage (Vi)must be greaterthan the Zener Voltage (Vz). Thus ensures that Zener diode operates in the Reversebreakdown Region. The Input Current (I)through Limiting resistor R is given by therelation.I =Vi -Vz (1)

Rs

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Where Vi = Unregulated DCl/P Voltage to the circuit.

_[----+~ 1\1'['1+ V iT Unregulated

+I z I L Reg. a lP Voltage

RL

Fig 1:- Zener Diode Shunt Voltage RegulatorVz = Zener VoltageThe ideal Zener diode acts as a constant voltage source of voltage (Vz). But a practicalzener diode has a finite value of resistance called zener resistance (Rz). Because of thezener resistance, there is a voltage drop across it which is equal to IzRz. Therefore thevoltage across the zener diode is given by expression.

VL =Vz = IzRz -------------------------2)Let the zener resistance is negligible, and then the load voltage isVL =Vz ------------------( 3 )And the current through the load resistance is given by the relation.IL =VLIRL -----------------------(4 )From the circuit it is clear that1= Iz + IL& Iz = I - IL-----------------( 5 )

OPERATION:-The operation of the circuit can be discussed under following two heads.1). Line Regulation with Varying input Voltage:-The Load RL is constant & if l/P voltage Vi varies within the limits then the circuitprovides constant O/P voltage. As RL is constant. So IL is constant, hence fromequation (5),Iz = 1. As the l/P voltage increase, the current I increases so Iz increaseswithin the limits Izmin - Izmax. So zener diode remains in Reverse breakdown state.But the voltage drop across series Resistance R increases therefore keeping the O/Pvoltage constant at Vzvalue &ViceVersa. SoAs the l/P voltage changes within limitsoI P remains constant.

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2). Load Regulation with varying load Resistance (RL)Consider the operation with Load RL Changes & input voltage Vi constant. As l/Pvoltage Vi constant so from equation (3) I = Iz + IL, current I is constant so as Izdecreases I increases then IL increases I decreases respectively. If Load Resistance RLdecreases then current IL increases, current IZ decreases within limits IZmin - Izmax .SoZener diodes remains in breakdown thereby keeping a constant alP voltage equalsto Vz value. As the circuit provides a constant alP voltage nearly equal to Vz if inputvoltage &Load RL changes so it is called as Zener diode shunt voltage regulator.

PROCEDURE:-1) Study the circuit provided on the front panel of the kit.2) Connect Saner diode ZDl of Vz = 6.2Vin the circuit.3) Connect all the voltmeters at the input Vi&alP Voside.4) Connect all the milliammeter at their respective places ( Optional or short the

terminals by patch cords)5) Switch ON the Power Supply.6) Keep Load Resistance RL constant vary the input voltage Vi from 0 to 12V in

steps & note down the corresponding alP voltage Vofor each step.7) Now, keep IfP voltages vi constant at the voltage greater than Vz value say Vi =

10Vconstant, vary the Load RL & note alP voltage VOin steps.8) Keep input Vi = 10v,note RL&Rvalue using digital multimeter.9) Find current I, Ir.& Iz from equation (1),(4)& (5)respectively.10) Repeat the above steps for the other Diode.

OBSERVATION TABLE:-

R = ---------0, VZl = ---------V, VZ2 = ---------V.

1). Line Regulation with RL constant:-

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RL= KQ

Unregulated I / P(Vi)Regulated (Vo)

2). Load Regulation with Input Vi constant:-Vi= V.

Load Resistance Regulated OjPRL VoMin.

Med.

Max.

Repeat the above for the other Zener Diode

RESULT:-The Circuit provides a constant OIP voltage for Zener DiodeZD1, Vo = V. Ifthe l /P voltage Vi is changes from V to V & if RLchanges from __ to___ KQ.

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EXPERIMENT NO: 8OBIECTIVE:-Study to design IC741 as aWein Bridge Oscillator.

EQUIPMENTS:-1.OPT-1Trainer Kit2. Power Supply3. Patch cords4. DSO and Function Generator.5.Multimeter.THEORY:-Wein bridge oscillator in which the wein bridge circuit is connected between theamplifiers input terminals and output terminal. The bridge has a series RC network inone arm and a parallel RC network in the adjoining arm. I the remaining two arms ofthe bridge resistors R1 and Rf are connected. See the practical figure1.The phase angle criterion for oscillation is that the total phase shift around the circuitmust be aO.Thiscondition occurs only when the bridge is balanced.i.e.At resonance. Thefrequency of oscillation fo is exactly the resonant frequency of the balanced Wien bridgeand is given by

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fo = 12nx Rx C

fo = O.159jRCAssuming that the resistors are equal in value and capacitors are equal in value in thereactive leg of the Wien bridge. At this frequency the gain required for sustainedoscillation is given by

Av =ljB=31 =RfjRl=3

OrRF=2Rl

Wienbridge oscillator ismostly used as audio frequency oscillator.

PROCEDURE:-

Rl RF

Va

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R

R c

Figure 1:Wien Bridge Oscillator Using OP-AMP1. Make the connections on OPT-01board as shown in practical figure 1 as

per your design.2. Apply +ve 15Vto pin 7 and -ve 15Vto pin 4 of op-amp.3. Switch on power supply.4. Observe the waveform on oscilloscope.5. Adjust the RF to required value to get the proper output.6. Measure the frequency and amplitude output signal.DESIGN: - ConsiderR=lKO resistorC=O.OlJ-lfapacitorR1=10KOresistorRF =must be at least 20K.(here put RF =lB.03K)Do the calculations using formulas.OBSERVATION TABLE:-

Sr. No. Descriptions Calculated Observed (from

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waveform 1)1 Frequency 15.9KHZ 11.65KHZ

WORK ASSIGNMENT:-For Fo = 500hz find R=? When C=O.OlJ-lfcapacitor, R1=10KQ resistor. Observethe output of circuit.

DESIGN FORMULAS1. Oscillator frequency is given by

fo = 12nx Rx C

fo = O.159jRC2. Gain requirement: - RF=2R1 minimum.

CONCLUSION:-We studied the design of Wien bridge oscillator circuit using op-amp.Calculated and observed readings are almost same.

VIVA QUESTIONSQl. What is Oscillator?

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Q2.Which type of feedback is provided in Oscillators?

Q3.What is balance condition?

Q4. What is lead -lag compensation circuit?

Q5.What is frequency of oscillation for Wien bridge oscillator?