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]
V0 =[
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
A
AD741
2
-
V-
D1
4
1k
I
OS1
D1N4001
0
B
+
OS2
6
VO
5
U3
7
V4
VOFF =
VAMPL =
FREQ =
3
V+
OUT
I
1
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
AD741
2
-
V-
ROM
4
15. Draw the Schmitt trigger and give its application?[ AUC May2009]
OS1
1k
OUT
3
+
V+
V1
VOFF =
VAMPL =
FREQ =
OS2
1
vo
6
5
7
U1
0
R1
R3
1k
1k
R2
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]
Rf
1k
2k
4
5
Ri
2
-
Vin
AD741
3
6
+
7
1
U2
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
R4
vf
1k
I0
AD741
2
-
VD
OS1
OUT
3
+
V+
0
-vee
4
ib=0
V-
I0
OS2
1k
1
6
5
I0
7
U1
RL
V2
vin
vee
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
3
+
V+
OUT
OS2
1
R4
5
OS1
OUT
6
3
1k
+
V+
OS1
V-
4
AD741
2
-
V-
C1
4
1n
AD741
2
-
OS2
1
V02
6
5
U3
7
7
U2
0
R7
R5
1k
1k
R6
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
2. a)Draw the circuit diagram of an op-amp differentiator and derive an expression for
the output in terms of the input. [AUC Nov,04]
Differentiator
A differentiator circuit produces differentiated version of the input voltage applied to it
Circuit diagram
R2
19.30uV
-15.00V
15Vdc
V2 = 0v
V10
0.01u
V
TD = 0.5ms
+
OS2
1
6
5
V3
0
V
U1
7
TR = 1ms
TF = 1ms
PW = 0.5ms
PER = 0.5ms
OS1
OUT
3
V+
V1 = 5v
AD741
2
-
-60.47uV
V-
C1
5.000V
4
1k
0
15.00V
-79.72uV
V2
15Vdc
R3
0
1k
0
Waveforms
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 11
Analysis
Ic = IB +IF
IB ~ 0
IC = IF
w.k.t IC =C DV/DT
In this circuit C is replaced by C1
IC=C1 DVC/DT
Voltage across c is given by
Vc =C1 D/DT(VIN-V2)
But IF=V2- VO / RF
But IC and IF are equal because IB =0
V2-VO /RF =C1 D/DT(VIN-V2)
V1=V2=0
-VO/RF=C1D/DT(VIN)
VO=-RFC1 D/DT(VIN)
Thus the output is –rfc1 times the time derivative of the input voltage
Applications :
1. P-I-D Controllers
2. High pass filter
3. Wave shaping circuit to generate narrow pulses
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 12
b)Design a schmitt trigger for UTP=0.5v and LTP=-0.5v
Vut =+0.5V , Vlt =-0.5V , Supply +/- 15V
=
Op-amp 741 with Ib (max) =500nA
Let I2 be much higher than Ib(max)
I2 = 100 Ib(max) = 50
R2 = Vut /I2 = 0.5 / 50
=10 KΩ
R1=V0-Vut / I2 =13.5 – 0.5 /50
= 260 KΩ =270 KΩ (practically)
3. a)State the advantages and limitations of active filters.
Advantages :
Gain and Frequency adjustment flexibility:
Since the op-amp is capable of providing a gain, the i/p signal is not attenuated as it
is in a passive filter. [Active filter is easier to tune or adjust].
No loading problem:
Because of the high input resistance and low o/p resistance of the op-amp, the active
filter does not cause loading of the source or load.
Cost:
Active filters are more economical than passive filter. This is because of the variety of
cheaper op-amps and the absence of inductors.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 13
Limitations :
Finite bandwidth limits high frequency operation.
Sensitivity : Active filters are sensitive to the temperature and environmental
changes.
Requirement of dc power supply.
b)A second order low pass filter at cut off frequency 1KHz [AUC Dec 2006]
fc =
Assume :R1=R2=R , C1=C2=C
f=
Given f=1 Khz ; Af=
; Af =3-1.1414 ; Af=1.586
Choose C=0.1µF
R=
=
=1.59 KΩ
Af=1+
0.586=
Rf =0.586 *10 =5.86 KΩ
Let Ri=10KΩ
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 14
4. a)Explain the working of an instrumentation amplifier. [NOV 2013]
Instrumentation amplifiers
An instrumentation amplifier is basically a differential amplifier which meets certain
requirements.
In industries the measurement and control of quantities such as temperature , pressure is
required
A transducer is used to convert these quantities into a proportional electrical signal.
Output of transducer is then applied to an amplifier called instrumentation amplifier
This amplifier amplifies the low level output signal of the transducer to such a level it can drive
the indicator or display.
Requirements
Precise
Low level signal amplification
High CMRR
Low noise
High input resistance
Low power dissipation
High slew rate
Circuit diagram
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 15
Analysis
Voltages at nodes A and B are given as
Va=V2 and Vb=V1
Expression for current is
I = Va-vb/ r2 = v2-v1 / r2
Output voltage of op-amp 2 is given by
V02 = Va + ir1
=V2+V2-V1/r2 * r1
=v2r2 +(v2-v1)r1 / r2 = v2(r2+r1) – v1 r1 / r2
Similarly output of op-amp1 is given by
V01=(r1+r2) v1 – r1v2 / v2
Output of first stage is given by
V02 – V01 = v2(r2+r1) – v1 r1 / r2 - (r1+r2) v1 – r1v2 / v2
=(2r1+r2)(v2-v1) /r2
Vo2-v01 = [1+ 2r1/r2][v2-v1]
Av1=vo2-v01/ v2-v1 = 1+ 2r1/r2
Similarly Av2= r4/r3
Overall gain of instrumentation amplifier is given by
Av=Av1 * Av2 =(1+2r1/r2) * r4/r3
Output voltage is then given by
Vo=av(v2-v1)
Applications
Temperature indicator
Temperature controller
Light intensity meter
Medical equipments
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 16
5. a)Draw the circuit of a second order butterworth low pass filter and derive its
transfer function.
The above diagram is transferred to S domain
A second order LPF having a gain 40dB/decade in stop band. A First order LPF can be
converted into a II order type simply by using an additional RC network.
The gain of the II order filter is set by R1 and RF, while the high cut off frequency fH is
determined by R2,C2,R3 and C3.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 17
Writing Kirchoff’s current law at node VA (S) .
I1 = I2 + I3 ------------(1)
V in - V A / R2 = V A-V 0 / 1/SC2 + V A-V1/R3 ------------------2
using voltage divider rule,
V 1 = [1/SC3] / [R3 + 1/SC2] VA
V A =V1/[R3C3 s + 1]
V1 =V A [ R3 C3 S + 1]
Substituting the value of VA in eqn 2 and solving for V1 , we get,
V in-V A [1+R3 C3 S] / R2=V A[1+s R3 C3]-V0 / [1/sc2] + V A[1+s R3 C3]-VA/ R3
]
]
]
For an op-amp in non inverting configuration
Vo=Af VA
Af=
VA= The voltage at the non inverting terminal
Vo=Af
]
= V0[1-
]
= V0
The denominator quadratic in the gain (V0/Vin) eqn must have two real and equal roots.
This means that
w 2 H = 1/R2 R3 C2 C3
wH = 1/R2 R3 C2 C3 q
(2 πf H)2 = 1/ R2 R3 C2 C3
Fh =1/ 2π√R2 R3 C2 C3
For a second-order LP Butterworth response, the volt gain magnitude eqn is,
V 0 /Vin =AF / √1+(f/fh)4
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 18
b)Draw the circuit of an astable multivibrator using op-amp and derive an
expression for its frequency of oscillation[AUC June 2006]
R2
1k
V18
C1
-15Vdc
0.01u
0
4
0
V-
AD741
2
-
vd
OS1
V+
OUT
3
+
OS2
1
6
V0V
5
7
U5
R3
R
1k
1k
V19
15Vdc
0
0
Expression for frequency of oscillation
Similarly
Total time period is given by T= t1+t2=
=
[a]
=
=
[b]
Subtracting 1 on both sides in equation[a]
=
=
-1
[c]
Dividing equation [b] and [c] we get
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 19
T=T1+T2 =
f0 =
6. Explain briefly about log amplifiers[AUC Dec 2007]
Log amplifier :
These circuits are used in analog computers.
Circuit diagram
D1
R1
AD741
2
-
V-
I1
4
D1N4001
OS1
1k
VIRTUAL GROUND
3
+
V+
OUT
OS2
1
VO
6
5
V01
U3
V4
7
VOFF =
VAMPL =
FREQ =
0
0
PN junction diode is connected in the feedback path. Output voltage is nothing but the
voltage across the diode.
V0= - Vf-------1
Due to infinite input impedance, the current going into the inverting terminal is zero.
i1 = if= Vin/R-------2
Generally,
Vf = ή vt[log (if) –log(i0)]-----3
Substituting 3 in 1 we get
V0=- ή vt[log (if) –log(i0)]
= ή vt[log (Vin/r) –log(i0)]
Vo= -ή vt log [ vin / r *i0]
Vo= -ή vt log [ vin / vref]
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 20
Where vref = r.Io
Vo is directly proportional to the logarithm of the input voltage.
7. With diagram explain the working principle of ICL 8038 function generator. [AUC
Dec 2007]
Functional Diagram
Output Waveform :
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 21
It consists of two current sources, two comparators, two buffers, one FF and a sine wave
converter.
Pin description:
Pin 1 & Pin 12: Sine wave adjusts:
The distortion in the sine wave output can be reduced by adjusting the 100KΩ pots connected
between pin12 & pin11 and between pin 1 & 6.
Pin 2 Sine Wave Output:
Sine wave output is available at this pin. The amplitude of this sine wave is 0.22 Vcc. Where
5V ≤ Vcc ≤ 15 V.
Pin 3 Triangular Wave output:
Triangular wave is available at this pin. The amplitude of the triangular wave is 0.33Vcc. Where
5V ≤ Vcc ≤ 15 V.
Pin 4 & Pin 5 Duty cycle / Frequency adjust:
The symmetry of all the output wave forms & 50% duty cycle for the square wave output is
adjusted by the external resistors connected from Vcc to pin 4. These external resistors &
capacitors at pin 10 will decide the frequency of the output wave forms.
Pin 6 + Vcc:
Positive supply voltage the value of which is between 10 & 30V is applied to this pin.
Pin 7 : FM Bias:
This pin along with pin no8 is used to TEST the IC 8038.
Pin9 : Square Wave Output:
A square wave output is available at this pin. It is an open collector output so that this pin can be
connected through the load to different power supply voltages. This arrangement is very useful
in making the square wave output.
Pin 10 : Timing Capacitors:
The external capacitor C connected to this pin will decide the output frequency along with the
resistors connected to pin 4 & 5.
Pin 11 : -VEE or Ground:
If a single polarity supply is to be used then this pin is connected to supply ground & if ( )
supply voltages are to be used then (-) supply is connected to this pin.
Pin 13 & Pin 14: NC (No Connection)
Important features of IC 8038:
1. All the outputs are simultaneously available.
2. Frequency range : 0.001Hz to 500kHz
3. Low distortion in the output wave forms.
4. Low frequency drift due to change in temperature.
5. Easy to use.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 22
Parameters:
(i) Frequency of the output wave form:
The output frequency dependent on the values of resistors R1 & R2 along with the external
capacitor C connected at pin 10.
If RA= RB = R & if RC is adjusted for 50% duty cycle then fo = RC 0.3 ; RA = R1, RB = R3, RC
= R2
(ii) Duty cycle / Frequency Adjust : (Pin 4 & 5):
Duty cycle as well as the frequency of the output wave form can be adjusted by controlling the
values of external resistors at pin 4 & 5.
The values of resistors RA & RB connected between Vcc * pin 4 & 5 respectively along with the
capacitor connected at pin 10 decide the frequency of the wave form.The values of RA & RB
should be in the range of 1kΩ to 1MΩ.
(iii) FM Bias:
· The FM Bias input (pin7) corresponds to the junction of resistors R1 & R2.
· The voltage Vin is the voltage between Vcc & pin8 and it decides the output frequency.
· The output frequency is proportional to Vin as given by the following expression
For RA = RB (50% duty cycle).
fo = CRAVcc
1.5Vin ; where C is the timing capacitor
· With pin 7 & 8 connected to each other the output frequency is given by fo = RC0.3
where R = RA = RB for 50% duty cycle.
(iv) FM Sweep input (pin 8):
· This input should be connected to pin 7, if we want a constant output frequency.
· But if the output frequency is supposed to vary, then a variable dc voltage should be applied to
this pin.
· The voltage between Vcc & pin 8 is called Vin and it decides the output frequency as,
1.5 Vin
fo = C RA Vcc
A potentiometer can be connected to this pin to obtain the required variable voltage
required to change the output frequency.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 23
8. Design a differentiator to differentiate an input signal that varies in frequency from
10hz to about 1khz.[ AUC April 2008]
Steps to be followed to design a differentiator
Select fa equal to the highest frequency of the input signal.
Assume a value C1 < 1 µf and calculate Rf.
Select fb =20fa and calculate the values of R1 and Cf so that R1C1 =RfCf
Step 1 : select f and calculate Rf
fa= 1 KHz as the highest input frequency is 1 KHz.
Let C1 =0.1 µf
But fa =
Rf =
Step 2 : Calculate R1 and Cf
fb = 20 fa =20 KHz
fb =
R1 =
As R1 C1 = Rf cf
Cf =
Rcomp =Rf =
Designed circuit
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 24
9. Explain with neat sketch about the working of an instrumentation amplifier and
show with derivation that gain of this amplifier can be varied by using a variable
resistance R.[ AUC June 2009]
Instrumentation amplifiers
An instrumentation amplifier is basically a differential amplifier which meets certain
requirements.
In industries the measurement and control of quantities such as temperature , pressure is
required
A transducer is used to convert these quantities into a proportional electrical signal.
Output of transducer is then applied to an amplifier called instrumentation amplifier
This amplifier amplifies the low level output signal of the transducer to such a level it can drive
the indicator or display.
Requirements
Precise
Low level signal amplification
High CMRR
Low noise
High input resistance
Low power dissipation
High slew rate
Circuit diagram
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 25
Analysis
Voltages at nodes A and B are given as
Va=V2 and Vb=V1
Expression for current is
I = Va-vb/ r2 = v2-v1 / r2
Output voltage of op-amp 2 is given by
V02 = Va + ir1
=V2+V2-V1/r2 * r1
=v2r2 +(v2-v1)r1 / r2 = v2(r2+r1) – v1 r1 / r2
Similarly output of op-amp1 is given by
V01=(r1+r2) v1 – r1v2 / v2
Output of first stage is given by
V02 – V01 = v2(r2+r1) – v1 r1 / r2 - (r1+r2) v1 – r1v2 / v2
=(2r1+r2)(v2-v1) /r2
Vo2-v01 = [1+ 2r1/r2][v2-v1]
Av1=vo2-v01/ v2-v1 = 1+ 2r1/r2
Similarly Av2= r4/r3
Overall gain of instrumentation amplifier is given by
Av=Av1 * Av2 =(1+2r1/r2) * r4/r3
Output voltage is then given by
Vo=av(v2-v1)
Applications
Temperature indicator
Temperature controller
Light intensity meter
Medical equipments
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 26
10. Show with help of circuit diagram an op-amp used as i) Scale changer ii) Phase
Shifter iii)inverting adder iv)Non-inverting adder.
Scale changer
Applying KCL
Phase shifter
The circuit that introduces phase shift as signal transmits from output to input is called
phase shifter.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 27
Applying KCL at Node B
Applying KCL at Node A
Since VA =VB
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 28
Inverting adder
The summing amplifier is same as the inverting amplifier.
It has several inputs.
The feedback applied through Rf from the output to the input terminal is negative.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 29
Applying KCL
Due to virtual ground node B and node A is at ground potential
Output voltage is equal to
Non inverting Summer
Circuit diagram
Applying KCL
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In non inverting amplifier output voltage must be equal to
The output voltage is equal to
Draw an op-amp circuit whose output is v1+v2-v3-v4.[ AUC June 2009,nov 09]
11. Draw the circuit of a fourth order butterworth low pass filter having an upper cutoff
frequency of 1 Khz.[ AUC May2009]
Solution :
The general equation of butterworth filter denomination contains
i
ii
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
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In the fourth order filter n=4
Consider equation i
The second order filter general equation is
Let Ri1 =Ri2 =10KΩ
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12. Explain voltage follower, operation of op-amp in detail[AUC Nov 2009]
Voltage Follower: [Non-Inverting Buffer]
A circuit in which the output voltage follows the input voltage is called voltage follower.
Va=Vb =Vin
Due to virtual ground node b is at potential vin and node a is also at the potential Vin.
Vo=Va
V0=Vin
The voltage gain is unity.
13. Draw the circuit diagram and explain the working of precision rectifier. [AUC Nov
2009,2013]
Precision Rectifier:
The signal processing applications with very low voltage, current and power levels require
rectifier circuits. The ordinary diodes cannot rectify voltages below the cut-in-voltage of the
diode.
A circuit which can act as an ideal diode or precision signal – processing rectifier circuit for
rectifying voltages which are below the level of cut-in voltage of the diode can be designed
by placing the diode in the feedback loop of an op-amp.
Precision diodes:
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Half – wave Rectifier:
A non-saturating half wave precision rectifier circuit is shown in figure. When Vi > 0V ,
the voltage at the inverting input becomes positive, forcing the output VOA to go
negative.
This results in forward biasing the diode D1 and the op-amp output drops only by ≈ 0.7V
below the inverting input voltage.
Diode D2 becomes reverse biased. The output voltage V0 is zero when the input is
positive. When Vi > 0, the op-amp output VOA becomes positive, forward biasing the
diode D2 and reverse biasing the diode D1 .
The circuit then acts like an inverting amplifier circuit with a nonlinear diode in the
forward path. The gain of the circuit is unity when Rf = Ri .
Figure shows the arrangement of a precision diode. It is a single diode arrangement
and functions as a non-inverting precision half – wave rectifier circuit.
If V1 in the circuit of figure is positive,the op-amp output VOA also becomes positive.
Then the closed loop condition is achieved for the op-amp and the output voltage V0
= Vi .
when Vi < 0, the voltage V0A becomes negative and the diode is reverse biased. The
loop is then broken and the output V0
= 0.
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Full wave Rectifier:
The Full wave Rectifier circuit commonly used an absolute value circuit is shown in figure. The
first part of the total circuit is a half wave rectifier circuit considered earlier in figure. The second
part of the circuit is an inverting.
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For positive input voltage Vi > 0V and assuming that RF =Ri = R, the output voltage VOA = Vi .
The voltage V0 appears as (-) input to the summing op-amp circuit formed by A2 ,
The gain for the input V’0 is R/(R/2), as shown in figure.
The input Vi also appears as an input to the summing amplifier. Then, the net output is V0 = -Vi 2V’0 = -Vi -2(-Vi ) = Vi
Since Vi > 0V, V’0 will be positive, with its input output characteristics in first quadrant.
For negative input Vi < 0V, the output V’0 of the first part of rectifier circuit is zero. Thus, one input
of the summing circuit has a value of zero. However, Vi is also applied as an input to the summer
circuit formed by the op-amp A2 .
The gain for this input id (-R/R) = -1, and hence the output is V0 = -Vi . Since Vi is negative, v0
will be inverted and will thus be positive.
This corresponds to the second quadrant of the circuit.
To summarize the operation of the circuit,
V0 = Vi when Vi < 0V and V0 = Vi for Vi > 0V, and hence V0 = |Vi |
It can be observed that this circuit is of non-saturating form.
14. Draw and explain the commonly used three op-amp instrumentation amplifiers and
derive expression for its gain [AUC MAY 2010]
An instrumentation amplifier is basically a differential amplifier which meets certain
requirements.
In industries the measurement and control of quantities such as temperature , pressure is
required
A transducer is used to convert these quantities into a proportional electrical signal.
Output of transducer is then applied to an amplifier called instrumentation amplifier
This amplifier amplifies the low level output signal of the transducer to such a level it can drive
the indicator or display.
Requirements
Precise
Low level signal amplification
High CMRR
Low noise
High input resistance
Low power dissipation
High slew rate
Circuit diagram
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 36
Analysis
Voltages at nodes A and B are given as
Va=V2 and Vb=V1
Expression for current is
I = Va-vb/ r2 = v2-v1 / r2
Output voltage of op-amp 2 is given by
V02 = Va + ir1
=V2+V2-V1/r2 * r1
=v2r2 +(v2-v1)r1 / r2 = v2(r2+r1) – v1 r1 / r2
Similarly output of op-amp1 is given by
V01=(r1+r2) v1 – r1v2 / v2
Output of first stage is given by
V02 – V01 = v2(r2+r1) – v1 r1 / r2 - (r1+r2) v1 – r1v2 / v2
=(2r1+r2)(v2-v1) /r2
Vo2-v01 = [1+ 2r1/r2][v2-v1]
Av1=vo2-v01/ v2-v1 = 1+ 2r1/r2
Similarly Av2= r4/r3
Overall gain of instrumentation amplifier is given by
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
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Av=Av1 * Av2 =(1+2r1/r2) * r4/r3
Output voltage is then given by
Vo=av(v2-v1)
Applications
Temperature indicator
Temperature controller
Light intensity meter
Medical equipments
15. Explain the working principles of regenerative comparator with necessary
diagrams[AUC MAY 2010,2011]
Schmitt trigger
Comparator which use positive feedback is known as Schmitt trigger.
AD741
2
-
V-
ROM
4
Circuit diagram
OS1
1k
OUT
3
+
V+
V1
VOFF =
VAMPL =
FREQ =
OS2
1
vo
6
5
7
U1
0
R1
R3
1k
1k
R2
1k
0
0
This circuit converts an irregular shaped waveform to a square wave or pulse. The
circuit is known as Schmitt Trigger or squaring circuit.
The input voltage Vin triggers (changes the state of) the o/p V0 every time it exceeds
certain voltage levels called the upper threshold Vut and lower threshold voltage. These
threshold voltages are obtained by using the voltage divider R1 –R2, where the voltage
across R1 is feedback to the (+) input.
The voltage across R1 is variable reference threshold voltage that depends on the value
of the output voltage.
When V0 = +Vsat, the voltage across R1 is called “upper threshold” voltage Vut.
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The input voltage Vin must be more positive than Vut in order to cause the output V0 to
switch from +Vsat to –Vsat. As long as Vin < Vut , V0 is at +Vsat, using voltage divider
rule,
V ut =(R1 / R1 + R2 ) +V sat
Similarly, when V0 = -Vsat, the voltage across R1 is called lower threshold voltage Vlt .
the vin must be more negative than Vlt in order to cause V0 to switch from –Vsat to
+Vsat.
In other words, for Vin > Vlt , V0 is at –Vsat. Vlt is given by the following eqn. V lt =
(R1/R1 + R2) -V sat
Thus, if the threshold voltages Vut and Vlt are made larger than the input noise voltages,
the positive feedback will eliminate the false o/p transitions. Also the positive feedback,
because of its regenerative action, will make V0 switch faster between +Vsat and –Vsat.
Resistance Rcomp tR1 || R2 is used to minimize the offset problems.
The comparator with positive feedback is said to exhibit hysteresis, a dead band
condition. (i.e) when the input of the comparator exceeds Vut its output switches from
+Vsat to –Vsat and reverts to its original state, +Vsat
when the input goes below Vlt. The hysteresis voltage is equal to the difference between
Vut and Vlt.
Therefore
Vref = Vut – Vlt
Vref = R1 /(R1 + R2) [+Vsat -(-Vsat)]
Operation :
Ref voltage v1 is developed across r2
V1=r2/r1+r2 * v0
Two different triggering voltages are defined for Schmitt trigger
Upper threshold voltage (Vut)
Lower threshold voltage (Vlt)
Vut forces transition from + Vsat to – Vsat
Vlt forces transition from –vsat to + Vsat
Vut = r2/r1+r2 * Vsat
Vlt = r2/r1+r2 * -vsat
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Waveforms
Hysteresis
The center of the hysteresis loop may be shifted by choosing a centre voltage which is the
average of Vut and Vlt
Vcenter = [Vut+Vlt] /2
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Applications
To convert sine wave to square wave.
Over voltage and over current protection circuit.
On /off temperature controllers.
16. What are the limitations of an ideal op-amp differentiator ?How are these
limitations overcome in practical differentiator ? Explain with necessary diagrams.
[AUC MAY 2011]
The ideal differentiator suffers from stability and noise problems which are overcome by
means of practical differentiator
Analysis
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PAGE 42
Applications :
Wave shaping circuits
Used as detector in FM demodulators
17. Explain the working of logarithmic and antilogarithmic amplifiers using op-amp
with necessary diagrams . [AUC MAY 2011]
Log amplifier :
These circuits are used in analog computers.
Circuit diagram
D1
R1
AD741
2
-
V-
I1
4
D1N4001
OS1
1k
VIRTUAL GROUND
3
+
V+
OUT
OS2
1
VO
6
5
V01
U3
V4
7
VOFF =
VAMPL =
FREQ =
0
0
PN junction diode is connected in the feedback path. Output voltage is nothing but the
voltage across the diode.
V0= - Vf-------1
Due to infinite input impedance, the current going into the inverting terminal is zero.
i1 = if= Vin/R-------2
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Generally,
Vf = ή vt[log (if) –log(i0)]-----3
Substituting 3 in 1 we get
V0=- ή vt[log (if) –log(i0)]
= ή vt[log (Vin/r) –log(i0)]
Vo= -ή vt log [ vin / r *i0]
Vo= -ή vt log [ vin / vref]
Where vref = r.Io
Vo is directly proportional to the logarithm of the input voltage.
Antilog amplifier
Antilog amplifiers are used to produce the antilog or exponential function.
Circuit diagram
RF
A
AD741
2
-
V-
D1
4
1k
I
OS1
D1N4001
VOFF =
VAMPL =
FREQ =
0
B
+
OS2
6
VO
5
U3
7
V4
3
V+
OUT
I
1
0
Analysis
Node B is connected to ground, A is also at ground potential (virtual ground)
Since ,A is virtual ground , Vin=Vf ,If =I0 . e vin / ήVt
Assume input current of the op-amp will be zero, the current “I” flowing through the feedback
resistor Rf is given by
I=If = -Vo/ rf
- Vo / rf = I0 . e vin / ήVt
Vo=- I0 rf . e vin / ήVt
V0 is directly proportional to the exponential function of vin
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 44
(ii) Draw and explain the circuit of a voltage to current converter if the load is (1)
Floating (2) Grounded. [AUC MAY 2012]
Voltage to Current Converter with floating loads (V/I):
Voltage to current converter in which load resistor RL is floating (not connected to
ground).
Vin is applied to the non inverting input terminal, and the feedback voltage across R1
devices the inverting input terminal.
This circuit is also called as a current – series negative feedback amplifier.
Because the feedback voltage across R1 (applied Non-inverting terminal) depends on
the output current i0 and is in series with the input difference voltage Vid .
Writing KVL for the input loop,
Vin =V id +V f
V id t 0v , since A is very largeA
Vin =V f
Vin = R1 i0 or
i0 = Vin/R1
From the fig input voltage Vin is converted into output current of Vin/R1 [Vin -> i0 ] .
In other words, input volt appears across R1. If R1 is a precision resistor, the output current (i0 =
Vin/R1 ) will be precisely fixed.
Applications:
1. Low voltage ac and dc voltmeters
2. Diode match finders
3. LED
4. Zener diode testers.
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PAGE 45
Voltage – to current converter with Grounded load:
This is the other type V – I converter, in which one terminal of the load is connected to ground.
Analysis of the circuit:
The analysis of the circuit can be done by following 2 steps.
1. To determine the voltage V1 at the non-inverting (+) terminals and
2. To establish relationship between V1 and the load current IL .
Applying KCL at node V1 we can write that,
I L = I1 + I 2 -------------1
But I1 = V in -V 1/R
and I 2 = V 0-V1/R
Sub these values into eqn 1
I L =V in -V1/R +V 0-V 1/R
I L =(Vin R -V1 R +V 0 R-V1 R)R2
=V inR +V 0 R - 2V1 R/R2
=R V in +V 0 -2V1 R /R2
I L =V in +V 0-2V1/R
RI L =V in +V 0 -2V1
V 1 =V in +V 0-I L R/2
the op-amp is connected in the non-inverting node A hence ,gain of the circuit is
0-V2 /R =V2-V0/R , 2V2=V0
The output voltage is given by
V 0 = 2V1
V 0 = 2[ V in +V 0 -I L R/2]
V 0 =V in +V 0 -I L R
V in = I L R
I L =V in /R
The load current I Lis dependent on the input voltageV in and Resistor R
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PAGE 46
18. Explain the working of Integrator [AUC NOV 2013]
19. INTEGRATOR
The circuit in which the output voltage waveform is the integration of the input voltage
waveform is called as integrator.
CIRCUIT DIAGRAM
0.01u
C2
5.000V
V2 = 0v
V10
5.000V
OS1
OUT
3
+
OS2
1
6
5
V3
0
V
U1
7
TD = 0.5ms
TR = 1ms
TF = 1ms
PW = 0.5ms
PER = 0.5ms
4
V
V+
V1 = 5v
1k
AD741
2
-
V-
R4
-14.80V
-15.00V
15Vdc
0
15.00V
0V
V2
15Vdc
0
0
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE
PAGE 47
Analysis
Applying KCL at node V2 we get
I1= Ib + if
Due to high input Z , Ib<< If
I1~If
Relation between current and voltage across a capacitor is given by
Ic = c.dvc / dt
In case of an integrator if = icf
I1=cf dvc/dt-----------------1
But i1=vin-v2/r1 and vc=v2-vo
Equating i1 in 1
Vin-v2 /r1 =n cf dvc/dt =cf d/dt(v2-vo)
V1-v2=0
Vin/r1=cf d/dt(-vo)
Vo=-1/r1cf ∫vin dt +c
Applications
Triangular wave or ramp generators.
A/D converters.
Integral type controllers.
In analog computers to solve differential equations.
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PAGE 48
20. Design an OP-AMP based second order active low pass filter with cut off
Frequency 1kHz. (8) [AUC MAY 2012]
Choose C=0.01µf
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