Operational Amplifier: Principles and Applications

This paper presentation, authored by Dr. G.S. Virdi, Ex-Chief Scientist, CSIR–Central Electronics Engineering Research Institute, Pilani, provides an in-depth study of Operational Amplifiers (Op-Amps) and their wide-ranging applications in analog electronic systems. The paper begins with the fundamentals of op-amp construction, pin configuration, and ideal versus non-ideal operational characteristics.
It systematically explains various amplifier configurations such as inverting, non-inverting, summing, and differential amplifiers, supported by analytical derivations and practical examples. The paper further extends to instrumentation amplifiers, active filters (low-pass, high-pass, and band-pass), and controlled source applications like voltage- and current-controlled circuits.
Designed for engineering and science students, researchers, and professionals in microelectronics and analog circuit design, this presentation bridges theoretical concepts with practical implementation, offering valuable insights into one of the most versatile components in electronics.
This lecture paper will be uploaded on SlideShare.

1. OPERATIONALAMPLIFIER(OP-AMP)
Dr.G.S.Virdi
Ex.Chief Scientist
CSIR-Central Electronics Engineering Research Institute
Pilani—33303 1,India

2. INTRODUCTION
⚫Anopampisanactivecircuitelementdesignedtoperform
mathematicaloperationsofaddition,subtraction,multiplication,
division,differentiationandintegration.
⚫Typicaluses:providevoltageamplitudechanges(amplitudeand
polarity),oscillators,filtercircuitandinstrumentationcircuit.
⚫Operationalamplifier(opamp)arelinearintegratedcircuitand
widelyusedinanalogueelectronicsystem.
⚫Theinternalconstructionsincludesanumberoftransistors,diodes,
resistorsandcapacitorsinasingletinychipofsemiconductor material.
⚫Itispackagedinasinglecasetoformafunctionalcircuit.
G.S.VIRDI

3. ⚫Atypicalopamphaseight-pinpackagewherepin/terminal8is
unusedandpin1&5areoflittleconcerntous.
⚫Theother5are:
⚫Pin2–invertinginput
⚫Pin3–noninvertinginput
⚫Pin6–output
⚫Pin7–+vepowersupply
⚫Pin4--vepowersupply
G.S.VIRDI

4. ⚫Thecircuitsymbolisthetriangleandithastwoinputsandone
output.
⚫Theinputsaremarkedwithminus(-)andplus(+)tospecifyinverting and
noninvertinginputs,respectively.
⚫Aninputappliedtothenoninvertingpinwillappearwiththesame polarity
attheoutput.
⚫Aninputappliedtotheinvertingpinwillappearinvertedatthe
output.

5. TYPEOFINPUTMODEOPERATION
Single-EndedInput
-Inputsignalisconnectedtooneop-ampinputandanotherop-amp
inputisconnectedtoground.
Inputsignalthatappliedto+veinput
Outputhassamepolarityasinput
signal
Inputsignalthatappliedto–veinput
Outputisoppositetotheinput
signal
G.S.VIRDI

6. TYPEOFINPUTMODEOPERATION
BothinputappliedwithVdamplified
outputinphasewithinputapplied
between+veand–veinputs.
Wheninputisappliedwithseparate
signalsthedifferencesignalVi1-Vi2
Double-Ended(Differential)Input
-Applysignalstoeachinputs
G.S.VIRDI

7. OP-AMP:TYPEOFINPUTMODE
OPERATION
CommonModeOperation
Sameinputsignalsareappliedtobothinputsof
op-amp
0Voutputbecausethetwosignalsequally
amplifiedbutwithoppositepolaritysignals.
G.S.VIRDI

8. ⚫Here,thebasicoperationwillbedividedintotwosection:
⚫Nonidealoperation
⚫Idealoperation
⚫However,theidealoperationwillbeusedthroughoutthe
analysis.
BASICOPERATION
G.S.VIRDI

9. NONIDEALOPERATION
⚫Thenonidealopamphasthefollowingcharacteristics:
⚫Highopen-loopgain,A10
6
⚫Highinputresistance,R
i2M
⚫Lowoutputresistance,R
075
V
disthedifferentialinputvoltage
givenby:
v
d=v
2–v
1
wherev
1isthevoltagebetweenthe
invertingterminalandgroundandv
2
isthevoltagebetweenthe
noninvertingterminalandground.
G.S.VIRDI

10. IDEALOPERATION
⚫Thecharacteristicsofanidealopampisdefinedasfollows:
⚫Infiniteopen-loopgain,A
⚫Infiniteinput resistance,R
i
⚫Zerooutputresistance,R
00
G.S.VIRDI

11. ⚫Fortheidealopamp,thefollowingisconsidered:
⚫Thecurrentintobothinputpinarezero:
i
1=0, i
2=0
⚫Thevoltageacrosstheinputterminalsarezero:
v
d=v
2–v
1=0
thus
v
1=v
2
G.S.VIRDI

12. EXAMPLE:
Calculatetheclosed-loopgainv
o/v
sandfindi
owhenv
s=1V.
Solution:
Rememberthat,foridealopamp
i
1=0, i
2=0, v
1=v
2
Sincei
1=0,the40kand5kresistorsareinseries;thesamecurrent
flowthroughthem.Andnoticethat
v
2=v
s
G.S.VIRDI

13. Hence,usingthevoltagedivisionprinciple,
Sincev
1=v
2,thus
AtnodeO,
Whenv
s=1V,v
o=9V.Substitutingforv
o=9Vintheaboveequation
gives
i
o=0.2+0.45=0.65mA
1
9
o
5k
v v
v
o
5k40k
s
s
9v
v
v
o
v
o
9
0
iv
o

v
o
5k40k20k
G.S.VIRDI

14. ⚫Figurebelowshowstheconnectionofinvertingamplifier.
⚫Thenoninvertinginputisgrounded,v
iisconnectedtotheinverting
inputthroughR
1andthefeedbackresistorR
fisconnectedbetween
theinvertinginputandoutput.
INVERTINGAMPLIFIER
G.S.VIRDI

15. ⚫Ourgoalistoobtaintherelationshipbetweenthe inputvoltagev
iand
theoutputvoltagev
0.
⚫ApplyingKCLatnode1,
⚫Butforanidealopamp,v
1=v
2=0,sincethenoninvertingpinis
grounded.Hence,
f
R
1 R
i
1i
2
v
iv
1

v
1v
o
1
o i
f
R
1R R
v
i

v
o
v
R
f
v
G.S.VIRDI

16. ⚫Thevoltagegainis
⚫Thedesignationofthecircuitasaninverterarisesfromthenegative
sign.
1
v
i
v R
A
v
o

R
f
G.S.VIRDI

17. EXAMPLE:
Ifv
i=0.5V,calculate:a)theoutputvoltagev
oandb)the currentinthe
10kresistor.
G.S.VIRDI

18. Solution:
a)
v
o
v
i R
110

R
f

25
2.5
v
o2.5v
i2.50.51.25V
b)Thecurrentthroughthe10kresistoris
i
v
i
0

0.50
50A
R
1 10k
G.S.VIRDI

19. ⚫Figurebelowshowstheconnectionofnoninvertingamplifier.
⚫Theinputvoltagev
iisapplieddirectlyatthenoninvertinginputpin
andresistor R
1isconnectedbetweenthegroundandtheinvertingpin.
NONINVERTING AMPLIFIER
G.S.VIRDI

20. ⚫ApplyingKCLattheinvertingpingives
NONINVERTING AMPLIFIER
1 f
R R
i
1i
2
0v
1

v
iv
o
butv
1=v
2=v
i,theaboveequationbecomes
v
i

v
iv
o
R
1 R
f
or
f
o
i
R 
vv1
R

 1
G.S.VIRDI

21. ⚫Thevoltagegainis
⚫Theoutputhasthesamepolarityastheinput,thusitisnamedas
noninvertingamplifier.
NONINVERTING AMPLIFIER
1
v
i
v R
A
v
o
1
R
f
G.S.VIRDI

22. thecircuitwillbecomeasshownfigurewhichiscalledavoltage
follower(or unitygainamplifier)becausetheoutputfollowstheinput.
Thus,
NONINVERTING AMPLIFIER
VOLTAGEFOLLOWER/BUFFER
⚫IfweletR
f=0andR
1=
i
v
o
v
v
ov
i 1
G.S.VIRDI

23. EXAMPLE:
Calculatetheoutputvoltagev
o
Solution:
ApplyingKCLatnodea,
NONINVERTING AMPLIFIER
4k 10k
6v
a

v
av
o
G.S.VIRDI

24. Butv
a=v
b=4,andalso
NONINVERTING AMPLIFIER
o
64

4v
o
4k10k
v1V
G.S.VIRDI

25. ⚫Figurebelowshowstheconnectionofsummingamplifier.Alsocalleda
summer.
⚫Asummingamplifierisanopampcircuitthatcombinesseveralinputs
andproducesanoutput thatistheweightedsumoftheinputs.
SUMMINGAMPLIFIER
G.S.VIRDI

26. notethat v
a=0,thus
SUMMINGAMPLIFIER
⚫Itisaninvertingamplifierwithmultipleinputs.Itcanhavemorethan
threeinputs.
⚫ApplyingKCLatnodeagives
i=i
1+i
2+i
3
but
1 2 3
1 2 3 f
R R R R
v
a
,
v
o
i
v
1v
a
,i
v
2 i
v
3v
a
,i
v
a
1 2
2
o
R R

vv
R
f
v
R
f
v
R
f

R
3
1 3
G.S.VIRDI

27. EXAMPLE:
Calculatev
oandi
oofthefollowingopampcircuit
Solution:
Therearetwoinputs.
SUMMINGAMPLIFIER
1o
R
v

v
R
f
v
R
f

R
2
1 2
G.S.VIRDI

28. Thecurrenti
oisthesumofthecurrentsthroughthe10kand2k
resistors.Bothoftheseresistorshavevoltagev
o=-8Vacrossthem,since
v
a=v
b=0.Hence,
SUMMINGAMPLIFIER
o
v
10
2
10
1

448V

5k 2.5k

 
o
10k 2k
i
v
o
0

v
o0
0.8m4m4.8mA
G.S.VIRDI

29. ⚫Thistypeofampisusedtoamplifythedifferencebetweentwoinput
signals.
DIFFERENCEAMPLIFIER
G.S.VIRDI

30. ⚫ApplyingKCLatnodeagives
or
or
DIFFERENCEAMPLIFIER
R
1 R
2
v
1v
a

v
av
o
1
1
R



R
v
R
2
1v
R
2
v
o a
1
R
3 R
4
⚫ApplyingKCLatnodebgives
v
2v
b

v
b0
2b
R
4

vv


RR

3 4
G.S.VIRDI

31. or
DIFFERENCEAMPLIFIER
sincev
a=v
b
2 1
1
o
R
4
v
R
2
1

R
v
2
v
R

RR

R
13 4
2 1
1 34 1
o
R
v
R
21R
1R
2
v
R
2
v
R1RR
G.S.VIRDI

32. EXAMPLE:
Obtaini
ointheinstrumentationamplifiercircuitbelow
Answer:2A
DIFFERENCEAMPLIFIER
G.S.VIRDI

33. Example1:
Calculatetheoutputvoltageofthecircuitforresistor
componentofvalueRF=470kΩ,R1=4.3kΩ,R2=33
kΩ,r3=33kΩforaninputof80µV
G.S.VIRDI

34. Example2:
Determinetheoutputvoltageofthecircuitinfigurebelow.
Answer:
Vo1=0.587mV
Vo2=-18.13mV
Vo3=412mV
G.S.VIRDI

35. Exercise:
Basedonthemultiplestagesoperationalamplifier(op-amp)
showninFigure1:
(a)Namethetypeoftheamplifierofcircuit1,2and3.
(b)Determinetheoutputvoltageoftheamplifier,V
O1,V
O2andV
O3.
G.S.VIRDI

36. Figure1
G.S.VIRDI

37. Op-AmpApplications: ControlSources
37
⚫Op-Ampcanbeusedtoformavarioustypesofcontrolled
sources:
1.Voltage-ControlledVoltageSource
2.Voltage-ControlledCurrentSource
3.Current-ControlledVoltageSource
4.Current-ControlledCurrentSource
⚫Voltage/Currentcanbeusedtocontrolanoutputvoltageor
current
G.S.VIRDI

38. Op-AmpApplications:
Voltage-ControlledVoltageSource
38
⚫VoiscontrolledbyV1
⚫Theoutputvoltageseetobedependent ontheinputvoltage
(scalefactork)
⚫Invertinginput
⚫Non-invertinginput
G.S.VIRDI

39. Op-AmpApplications:
Voltage-ControlledCurrentSource
39
⚫Provideoutputcurrentcontrolledbyinputvoltage.
G.S.VIRDI

40. Op-AmpApplications:
Current-ControlledVoltageSource
40
⚫Theoutputvoltageisdependentontheinputcurrent
G.S.VIRDI

41. Op-AmpApplications:
Current-ControlledCurrentSource
41
⚫Outputcurrentisdependentontheinputcurrent
G.S.VIRDI

42. Op-AmpApplications:
ControlSources
42
⚫Example:
⚫CalculateIL.
G.S.VIRDI

43. Op-AmpApplications:
ControlSources
43
⚫Example:
⚫CalculateVo.
G.S.VIRDI

44. Op-AmpApplications: InstrumentAmplifier
44
⚫ProvidesanoutputbasedonthedifferencebetweenTWO
inputs.
⚫Potentiometer=providethescalefactoradjustment.
G.S.VIRDI

45. Op-AmpApplications: InstrumentAmplifier
45
⚫Example:
⚫Calculatetheoutputvoltageexpressionofthecircuit:
G.S.VIRDI

46. Op-AmpApplications: ActiveFilters
46
⚫Toprovidevoltageamplificationandsignalisolationor
buffering.
⚫Low-PassFilter–providesaconstantoutputfromDCupto
cut-offfrequency,fOHandthenpassesnosignalabovethatfrequency.
⚫High-PassFilter–providesorpassessignalsaboveacut-offfrequency,FOL.
⚫Band-PassFilter–filterpassessignalsthataboveonecut-offfrequencyand
belowanothercut-offfrequency.
G.S.VIRDI

47. Op-AmpApplications: Low-PassFilter
47
⚫First-OrderLow-PassFilter
⚫UsingsingleRandC
⚫Haspracticalslopeof-20dBperdecade
G.S.VIRDI

48. Op-AmpApplications: Low-PassFilter
48
⚫Second-OrderLow-PassFilter
⚫2sectionsoffilter
G.S.VIRDI

49. Op-AmpApplications: Low-PassFilter
49
⚫Example:
⚫Calculatethecut-offfrequencyofafirst-orderlowpassfilter
forR1=1.2kΩandC1=0.02µF
⚫Answer:
G.S.VIRDI

50. Op-AmpApplications: High-PassFilter
50
⚫First-Order
⚫Second-Order
G.S.VIRDI

51. Op-AmpApplications: High-PassFilter
⚫Example:
⚫Calculatethegainandcut-offfrequencyofasecond-orderhigh-
passfilterforR1=R2=2.1kΩ,C1=C2=0.05µFandRG
•=10kΩ,RF=50kΩ
⚫Answer:
51G.S.VIRDI

52. Op-AmpApplications: Band-PassFilter
52
⚫Combinehigh-passfilterandlow-passfilter
G.S.VIRDI

53. Op-AmpApplications: Band-PassFilter
53
⚫Example:
⚫Calculatethecut-offfrequenciesoftheband-passfilterwithR1
=R2=10kΩ,C1=0.1µFandC2=0.002 µF
G.S.VIRDI

54. Thank You…
G.S.VIRDI