Upload
others
View
6
Download
0
Embed Size (px)
Citation preview
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
Page 2 of2
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)
·-1,
..-J
I
(A) & set R2/R)=1 or 2• ?>\ 8£ lee!t:- ...
~•......c r-~~ 0".(fOv r '\" 0i;j { B.D tJ ,~
••..•( "h'~~? ,10. Reference: Op-amps & Linear ICs - Coughlino VI" .-
~ ( n ng\ades'n ) ~ .~ (DC ) 11><ll \. 'J :'? \. ) ~'/.".~\.. .) ',I'
Q:. '- J ",'
CY(, ~ ~- ...,':: U;dated by: Ycasir Arafat on 7'11 February, 2006
Page 3 of3