UNIT – II: APPLICATIONS OF OP-AMP PART -A (2 Marks)

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UNIT – II: APPLICATIONS OF OP-AMP

PART -A (2 Marks)

1. An input of 3v is fed to the non-inverting terminal of an op-amp. The amplifier has
a Ri of 10KΩ and Rf of 10KΩ .Find the output voltage[AUC April 2004]
V =[
0
2. Draw an integrator circuit using op-amp.[ AUC April 2004]

3. Mention two characteristics of instrumentation amplifier.[ AUC May 2005]
high gain accuracy
high CMRR
high gain stability with low temperature coefficient
low output impedance
4. Mention two applications of Schmitt trigger.[ AUC May 2005]
To convert sine wave to square wave.
Over voltage and over current protection circuit.
On /off temperature controllers.
5. Draw an adder circuit using op amp to get the output expression as vo=-
(0.1+v2+5v3)[ AUC May 2006]
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 1
6. Draw the circuit of a voltage follower using op-amp and prove that its gain is
exactly equal to unity.?[ AUC May 2006]

Va = Vb = vin
Node a is directly connected to output
V0= Va
V0=Vin
For the circuit voltage gain is unity.
7. Draw the block diagram of a multiplier using log and antilog amplifiers.[ AUC May
2006]

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 2
8. State the requirements of an instrumentation amplifier. [AUC Nov 2006]
high gain accuracy
high CMRR
high gain stability with low temperature coefficient
low output impedance
9. Name four applications of operational amplifier based comparator.[ AUC Nov
2006,April 2008]
ero crossing detector
Window detector
Marker generator
Phase meter
10. What is an antilog amplifier? Draw the circuit diagram of an antilog amplifier.[ AUC
Nov 2007]
Amplifier that converts logarithmic numbers back to decimal numbers is called antilog
amplifiers.
RF
I 1k
D1 A AD7414
2 - -VOS1 1
D1N4001 6 VO
I OUT
3 + 5
+ VOS2
V4 B U3
VOFF = 7
VAMPL =
FREQ =
0 0

11. What is the principle of regenerative comparator.[ AUC Nov2007]
If positive feedback is added to the comparator circuit, gain can be increased. Hence ,the
transfer curve of comparator becomes more closer to ideal curve. If the loop gain is adjusted
to unity then the gain becomes infinity. This results in an abrupt transition between the
extreme values of the output voltage.

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 3
12. Draw the circuit diagram of a non-inverting amplifier ?[ AUC May 2008]

13. Draw the circuit of an integrator[AUC Nov 2008]

14. What is a V to I converter?[ AUC Nov 2008]
Converter that produces output current which is directly proportional to the input voltage is
called as V to I converter.

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 4
15. Draw the Schmitt trigger and give its application?[ AUC May2009]

ROM AD7414
2 - -VOS1 1
1k 6 vo
OUT
V1
VOFF = 3 + 5
VAMPL = + VOS2
FREQ = U1
0 7
R1

R3
1k
R2 1k
1k 0

0

Applications
o To convert sine wave to square wave.
o Over voltage and over current protection circuit.
o On /off temperature controllers.

16. Draw the input and output waveforms for an integrator for square wave input.[
AUC May 2010]

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 5
17. Mention the applications of an instrumentation amplifier.[ AUC May 2009]
Data acquisition system.
Temperature indicator
Temperature controller.
Light intensity meter.
18. In what way, a precision rectifier using op-amp is superior to a conventional
rectifier. [AUC Nov09 ,MAY 2011]

Conventional rectifier cannot rectify voltages below 0.6V.
Precision rectifiers rectify voltages having amplitude less than 0.7V

19. Draw an op-amp subtractor circuit. . [AUC Nov09]

20. Design an inverter using op-amp [AUC MAY 2010 ,MAY 2011]

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 6
21. Design a peak detector using op-amp [AUC MAY 2010]

22. Distinguish between active and passive filter. [AUC MAY 2011]
ACTIVE FILTER PASSIVE FILTER
Tuning is easy Difficult to tune
Does not cause loading effect Causes loading effect to source or load
More economical Cost is high

23. Why active guard drive is necessary for an instrumentation amplifier?[AUC MAY
2012]
The common ground is shared by variety of circuits.
Due to ground loop interference , additional voltage drop develops and lead to
error in low voltage measurement.
Due to distributed cable capacitances degradation of CMRR occurs.
The active guard drive eliminates all these problems.

24. What is comparator? [AUC MAY 2012]
Comparator is a circuit that compares the voltage applied at one of its input to that
applied at its other input and to produce an output voltage which is either Vh or vl

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 7
25. Draw a non –inverting amplifier with a voltage gain of 3 [AUC NOV 2013]

Ri Rf
1k 2k
45
AD741
2
-
6
Vin 3
+
U2
71

26. Give an application for each of the following circuits : voltage follower ,peak
detector ,Schmitt trigger , clamper

Voltage follower : used in isolator circuit
Peak detector : used for amplitude modulation in communication applications.
Schmitt Trigger : square wave generators , ON –OFF controllers.
Clamper : DC restorer in television receivers.

PART-B (16 marks)

1. a)Explain briefly about the working of voltage to current converter[AUC May 2004]

Voltage to current converter
This circuit converts voltage to current
Circuit diagram
vf
R4
1k I0 RL
I0 ib=0
-vee 1k
AD7414
2 - -VOS1 1
VD 6
OUT

0
3 + 5
+ VOS2 I0
U1
7
V2 vee
vin

0
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 8
Operation
Input voltage is applied to the non inverting terminal .
Load resistance is connected in place of feedback resistor rf.
This circuit is also called as negative feedback amplifier(current series) R1 is directly
proportional to Io
Applying KVL to the input loop
Vin=vd+vf
But open loop gain Av of op-amp is very large
Vd ~ 0
Vin=Vf
But Vin=R1*I0
I0=Vin /r1
Thus vin is converted into proportional output current (I0 = vin / R1)
Applications
Low voltage dc and ac voltmeters
LCD and zener diode testers
b) )Explain briefly about the working of triangular wave generator .
Triangular wave generators
This circuit consist of a square wave generator and an integrator.
The square wave generator generate square wave at its output. This square wave is integrated
by the integrator to generate a triangular waveform.

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 9
Circuit diagram
R8 R9
1k 1k
C2
1n
AD7414
C1 2AD7-414-VOS1 1 R4 2 - -VOS1 61 V02
1n 6 OUT
3 + +VOOUST2 5 1k 3 U+3 +VOS2 5
U2 7
0 R7 7
R5
1k
R6 1k
1k
0
0
Waveform

Triangular wave is generated by alternatively charging and discharging a
capacitor with a constant current source.
Assume V0’ is high at +Vsat .
This forces a constant current (+Vsat / R4) through C.
When V0’ is low at –Vsat .
This forces a constant current (-Vsat / R4) through C.
The frequency of the triangular wave is same as the square wave frequency.
The amplitude of the triangular wave will decrease as the frequency increases.

EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 10