DEP ARTMENT OF ELECTRICAL & ELECTRONIC ENGINEERINGBANGLADESH UNIVERSITY OF ENGINEERING & TECHNOLOGY

COURSE NO.: EEE 208EXPT. NO. 07

Name of the Experiment: Linear Application of Operational Amplifier

Objec'tive

To investigate the use of operational amplifier as inverting multiplier, invertingsummer, inverting integrator, inverting differentiator and differential amplifier.

Prelab Calculations and Simulation Using Spice (Home Work)

Students must petjorm the following SPICE simulations at home before attending thelab

I. For the different configurations shown in Fig (2), using SPICE, simulatethe different configurations and sketch the simulated output waveforms.

Circuit Diagrams

N.S. The Op Amp ).lA74 I has eight terminals to be connected as shown in Fig. (I).

+15V

_____ 4

-15V

6OIP

liP

2

).lA74 I-15V

(a)

7

III-+111+15V

(b) DC Supply Connection

Fig. (I)Page I of I

CircuitsRr

VolYo

2. Inverting SummerRr=1 k, 10k, lOOk

VI=2Y pp, lkHz(reet or sin)V2=ly DC

o erations

1. Inverting MultiplierRi =lk, Rr=lk, 10k, lOOk

Vi~2Y p-p sin, I kHz

3. Inverting IntegratorVi= 2Y pp, 1kHz or suitable

4. Inverting DifferentiatorVi= 2Y pp, 1kHz or suitable

5.Differential Amplifier(as Subtraetor)

If R2/R)= R4/R3 •••••••• (A)

then Vo= R2/R)(V2 - VI)VI =2y pp, 1kHz (sin or reet, to RI)

V2 =1 y DC, (to R3)

Satisfy (A) & set R2/R)=1 or 2

YolDifferential

Amplifier

Fig. 2

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Procedures

I. Construct the circuits shown in Fig. (2). Apply the appropriate voltage ineach case with frequencies of the order of 1 KHz. Itl each case measure theoutput waveform by the oscilloscope and sketch it in the Table 2

Table2: Linear Application Outputs.

o erations

1. Inverting MultiplierRj=lk, Rr=lk,IOk,IOOkVi=2v p-p sin, 1kHz

Out uts

Rr=lk Rr=IOk Rr=IOOk

2. Inverting SummerRr= Ik, 10k, lOOkV)=2v pp, IkHz(rect or sin)V2=lv DC

3. Inverting IntegratorVi= 2v pp, 1kHz or suitable

4. Inverting DifferentiatorVi= 2v pp, 1kHz or suitable

Rr=lk Rr=IOk

Out ut For Vj=Vsin Out ut For Vj=Vrec

Out ut For Vj=Vsin Out ut For Vj=Vrec

Rr= I_OO_k__

1

OuI put For V;~I V I

5.Differential Amplifier(as Subtractor)IfR2/R)= ~/R3 then (A)Vo= R2/R)(Vr VI)V I =2v pp, 1kHz (sin or rect, to R1)

V2 =1 v DC, (to R3)

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I

(A) & set R2/R)=1 or 2• ?>\ 8£ lee!t:- ...

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CY(, ~ ~- ...,':: U;dated by: Ycasir Arafat on 7'11 February, 2006

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