Download unit – ii applications of operational amplifiers

UNIT – II
APPLICATIONS OF OPERATIONAL AMPLIFIERS
Sign Changer, Scale Changer, Phase Shift Circuits, Voltage Follower, V-to-I
and I-to-V converters, adder, subtractor, Instrumentation amplifier, Integrator,
Differentiator, Logarithmic amplifier, Antilogarithmic amplifier, Comparators,
Schmitt trigger, Precision rectifier, peak detector, clipper and clamper, Low-pass,
high-pass and band-pass Butterworth filters.
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UNIT- II
APPLICATIONS OF OPERATIONAL AMPLIFIERS
ANNA UNIVERSITY SOLVED 2 MARKS AND 16 MARKS QUESTIONS
1. Mention some of the linear applications of op – amps. (DEC 09)
Adder, subtractor, voltage –to- current converter, current –to- voltage converters,
Instrumentation amplifier, analog computation ,power amplifier, etc are some of the linear
opamp circuits.
2. Mention some of the non – linear applications of op-amps:Rectifier, peak detector, clipper, clamper, sample and hold circuit, log amplifier, anti –log
amplifier, multiplier are some of the non – linear op-amp circuits.
3. What are the areas of application of non-linear op- amp circuits?
1. Industrial instrumentation
2. Communication
3. Signal processing
4. What is voltage follower? (MAY 2010)
A circuit in which output folloes the input is called voltage follower.
5. List the features of instrumentation amplifier:
1. High gain accuracy
2. High CMRR
3. High gain stability with low temperature co-efficient
4Low dc offset
5. Low output impedance
6. What are the applications of V-I converter?
1. Low voltage dc and ac voltmeter
2. L E D
3. Zener diode tester
7.Write transfer function of op amp as an integer. (MAY 2010)
The transfer function of the integer is
LAL=1/MR1cf
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8.What do you mean by a precision diode? (DEC 2010)
The major limitation of ordinary diode is that it cannot rectify voltages below the cut – in
voltage of the diode. A circuit designed by placing a diode in the feedback loop of an op – amp is
called the precision diode and it is capable of rectifying input signals of the order of milli volt.
9. Write down the applications of precision diode. (DEC 2010)
1.Half - wave rectifier
2.Full - Wave rectifier
3.Peak – value detector
4.Clipper
5.Clamper
10. Differentiate Schmitt trigger and comparator. (MAY 2010)
comparator.
1. It compares the input signal with references voltage then yields the output voltage
2. It need not consist of feedback
3. comparator output need not to besquare wave
Schmitt trigger
1. It operates between two reference
points namely UTP&LTP.
2. It employs positive feedback
3. Its output is square wave.
.
11. Write down the condition for good differentiation?(MAY 2010)
1.For good differentiation, the time period of the input signal must be greater than or
equal to Rf C1
2.T > R f C1 Where, Rf is the feedback resistance
3.Cf is the input capacitance
12. What is a comparator? (MAY 2010), (MAY 2011)
A comparator is a circuit which compares a signal voltage applied at one input of an opamp
with a known reference voltage at the other input. It is an open loop op - amp with output +
Vsat .
13. What are the applications of comparator?[APRIL/MAY 2008]
1.Zero crossing detector
2.Window detector
3.Time marker generator
4.Phase detector
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14. What is a Schmitt trigger?(DEC 09,MAY 10)
Schmitt trigger is a regenerative comparator. It converts sinusoidal input into a square
wave output. The output of Schmitt trigger swings between upper and lower threshold voltages,
which are the reference voltages of the input waveform.
15. What is a multivibrator?
Multivibrators are a group of regenerative circuits that are used extensively in timing
applications. It is a wave shaping circuit which gives symmetric or asymmetric square output. It
has two states either stable or quasi- stable depending on the type of multivibrator.
16. What are the characteristics of a comparator?
1.Speed of operation
2.Accuracy
3.Compatibility of the output
17. Why integrators are preferred over differentiator in analog computers?
Since the gain of the differentiator increases linearly with frequency and it tends to amplify low
frequency noise, which may result in spurious oscillations, integrators are preferred over
differentiators in analog computer.
18.What are the drawbacks (limitations) of an ideal integrator? (or)
are errors in ideal integrator?
What
The drawbacks of ideal integrator are
 At low frequencies ( d.c), the gain becomes infinite.
 Input offset voltage gets amplified and appears as error voltage, which causes
saturation.
 It is difficult to pull integrator out of saturation hence true integration may not be
possible.
 Limited Bandwidth.
18. What are the important features (characteristics) of an instrumentation
amplifier? (Or)
What are the basic requirements of instrumentation amplifier? [Nov2003][April-2005]
The main characteristics of instrumentation amplifier are
 High gain
 High CMRR.
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



High gain stability
Low output impedance
Low D.C Offset.
Low power loss.
19. What are the disadvantages of passive filters. [Nov 2003].
 Passive filters made of passive components like R, L, C works well for high
frequencies. For audio frequencies inductors become problematic since they are large, heavy
and expensive.
 For low frequency applications, more number of turns of wire must be used, which
degrades inductor performance and lowers Q, which results in high power dissipation.
20.Draw the characteristics of an ideal comparator and that of a commercially
available comparator
V0
V0
+Vsat
10
(Vi-Vref)
0
1
2
(Vi-Vref)
mV
-Vsat
10
(i). Ideal comparator
(i).Practical comparator
21.Mention some characteristics (features) of a comparator.
The important characteristics of a comparator are,
 Speed of operation
 Accuracy
 Compatibility of output.
22.
What is Schmitt Trigger? (Or) What is regenerative comparator?
Schmitt Trigger is an inverting comparator with positive feedback. It converts sinusoidal
input into square wave or pulse output. The o/p of Schmitt trigger swings between Upper
Threshold voltage (VUT) and Lower Threshold voltage (VLT), the reference voltages of i/p
waveform. It is also called as squaring circuit.
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23. What is Hysteresis width? What parameters determine the Hysteresis
width?
The difference between the upper and lower threshold voltage is known as Hysteresis
width.
VH  VUT  VLT 
2.R2 .Vsat
R1  R2
Where,
VUT
Upper threshold voltage
VLT
Lower threshold voltage
Vsat
Saturation voltage
24.What is a precision diode (or) precision rectifier?
When input signal Vi>VD/AOL, then the o/p of Op-amp exceeds V0 and diode D
conducts.
Vi
100mV
-100mV
V0
100mV
D -ON
D-OFF
When Vi < VD/AOL, the diode D is off and no current is delivered to the load RL. This
circuit is called precision diode capable of rectifying input signals less than VD or in order of
millivolts (mv)
25.What are the typical applications of precision diode?
The typical applications of precision diodes are
 Half wave rectifier
 Full wave rectifier
 Peak value rectifier
 Clipper, and clampers.
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26.What are function generators? List out some of the commercial function
generators.
Function generators are circuits used to provide basic waveforms with minimum
number of external components. They are also called as Waveform generators. Some
normally available waveform generators are

ICL 8038 Waveform generator

XR 2206 Function generator
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PART B
1. Explain the working of a Schmitt trigger circuit with necessary diagrams?
[NOV/DEC-2006] , [NOV/DEC-2010] (10m)
Or
Briefly explain the working principle of a Schmitt trigger? [NOV/DEC-2006] (12m)
Or
With neat circuit diagram Explain the working of a Schmitt trigger?
[APRIL/MAY-2008] (16m) & [MAY/JUNE-2007](10m)
Schmitt Trigger is an inverting comparator with positive feedback. It converts sinusoidal
input into square wave or pulse output. The output of Schmitt trigger swings between Upper
Threshold voltage and Lower Threshold voltage, the reference voltages of input waveform. It is
also called as squaring circuit.
 Input is applied to the inverting terminal of the Op-Amp. The inverted output is feedback to
the non-inverting terminal which is in same polarity as that of the output. This ensures
positive feedback.
 When Vin is slightly positive than Vref, V0 becomes – Vsat. When Vin is slightly more negative
than Vref, V0 becomes + Vsat.
 The voltage at which the transition takes place is decided by R1 and R2.
 + Vref is for positive saturation when V0 = +Vsat and is called upper threshold voltage (VUT).
 Vref  VUT  Vsat
R2
R1  R2
--------------- (1)
 -Vref is for negative saturation, when V0 = -Vsat and is called lower threshold voltage (VLT).
 Vref  VLT  Vsat
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R2
R1  R2
--------------- (2)
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 The output voltage remains in a given state until the input exceeds the threshold voltage
either positive or negative.
 Once the output changes its state, it remains there indefinitely until the input voltage (Vi)
crosses any of the threshold voltage levels. This is called Hysteresis of Schmitt Trigger.
The Hysteresis is also called as Dead band or Dead zone.
 The difference between two threshold voltages VUT and VLT is called Hysteresis Width (H).
H  VUT  VLT
---------------- (3)
 Vsat .
H 
  V .R 
R2
  sat 2 
R1  R2  R1  R2 
from (1)&(2)
2Vsat .R2
R1  R2
--------------- (A)
 Schmitt Trigger eliminates the effect of noise voltage present in the input.
 The noise voltages less than the Hysteresis, (H) cannot cause triggering.
 From input and output waveforms, we observe, when Vin is greater than +Vref or VUT, the
output becomes -Vsat.
 When Vin is less than VLT, the output becomes +Vsat. i.e.,
Vin  VLT  V0  Vsat
Vin  VUT  V0  Vsat
VLT  Vin  VUT  V0 (Previous state achieved.)
 Resistor R3 is selected as (R1 R2) to compensate the effect of input bias current.
 For non-inverting Schmitt trigger the Vi and Vref are just interchanged.
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APPLICATION:
The important application of Schmitt Trigger is the conversion of a very
slowly varying input voltage to a square wave output (Sine to Square wave
converter).
2.What are the advantages and limitations of active filters? [NOV/DEC-2006]
(10m)
ADVANTAGES :
(i) Gain & Frequency adjustment flexibility:
Since op-amp provides some gain, input signal is not attenuated.Active filters are easier to time
or adjust.
(ii) No Loading Problem:
Because of high i/p impedance & low o/p impedance of Op-amp, active
cause loading of source or load.
filters doesn’t
(iii) Cost:
Active filters are more economical, because of cheaper op-amps and absence of
inductors.
LIMITATIONS:
 Active RC filters also have some disadvantages:
• limited bandwidth of active devices limits the highest attainable pole frequency
and therefore applications above 100 kHz (passive RLCfilters can be used up to
500 MHz)
• the achievable quality factor is also limited
• require power supplies (unlike passive filters)
increased sensitivity to variations in circuit parameters caused by environmental changes
compared to passive filters
For many applications, particularly in voice and data communications, the economic and
performance advantages of active RC filters far outweigh their disadvantages.
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3.With circuit diagram the explain following applications of op-amp?
[NOV/DEC-2006]
[NOV/DEC-2010]
1) Voltage to current converter
2) Precision rectifier
Voltage to current converter:
 The voltage to current converter is a circuit the output load current is proportional to the input
voltage.
 According to connection of load, there are two types
i. Floating load.
ii. Grounded load.
FLOATING LOAD V-I CONVERTER
R1
R1
+
RL
IL
Ii
R1
IL
V0
V0
Vi
-
Vi
 As the input current to an Op-Amp is zero,
V
I L  Ii  i
R1

I L Vi
 Thus the load current is proportional to input voltage and the circuit acts as a voltage to current
converter.
 If the load is a capacitor, it charges and discharges at a constant rate. Such converters are used to
generate Sawtooth or triangular waveforms.
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 The proportionality constant is generally 1/R1. Hence the circuit is called Transconductance
amplifier. It is also called voltage controlled current source. (VCCS).
 The load current, IL = Vi/R1 is same for all types of loads.Load can be linear (e.g Resistor), or
non-linear (e.g LED) or it can have time dependent.
 No matter what the load is, the Op-amp will draw the current Ii whose magnitude depends only Vi
and R.
GROUNDED LOAD V – I CONVERTER:
‘V2’
R
R
IB = 0
V0 = 2V1
+
I1
+
IB = 0
+
IL
RL
Vi
-
+
I2
-
 When one end of the load is grounded, it is called grounded V-I converter. The circuit is also
called as HOWLAND CURRENT CONVERTER.
 Applying KCL at node, V1
I1  I 2  I L
V  V1 V0  V1
 i

 IL
R
R
 Vi  V0  2V1  I L .R
Vi  V0  I L .R
---------------- (1)
2
The gain of an Op-Amp in non- inverting mode is given as,
R
---------------- (2)
ACL  1  F
R1
For this ciruit, Rf = R1 = R,
R
 ACL  1   2
R
Vi 
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 ACL 
V0
2
V1
 V0  2V1
 V1  V0  I L .R
 Vi  I L R
I LVi
---------------- (3)
from equation (1)
---------------- (A)
---------------- (B)
 Hence the load current depends on input voltage Vi and resistor R.
Precision rectifier:
To achieve precision rectification we need a circuit that keeps V0 equal to Vi for Vi>0. This is
achieved by Op-Amp along with diodes and these circuits are called PRECISION
RECTIFIERS. These circuits are used to rectify voltages having amplitudes less then 0.7V.
INVERTING MODE:
 It consists of two diodes, two resistors and one Op-Amp connected in inverting
configuration.
Case (i) When Vi > 0V
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 From virtual ground concept, VP = VA = OV. For Vi > 0, Vi is positive with respect to Vn,
hence current through R1 flows through D1.
 Diode D1 is forward biased (D1-ON) and D2 is reverse biased (D2-OFF). As current flow
through R2 is zero, output, V0 is zero.
Case (ii) When Vi < 0V
 For Vi < 0, Vi is negative with respect to vn, so current flows from right to left through R1.
 Diode D2 and resistor R2 is the path for the current. Hence diode D1 is OFF and D2 is ON.
 Now the circuit acts as an inverting amplifier and output voltage is given as,
R
----------------- (1)
V0   2 .Vi
R1
when R2 = R1, then
V0 = -Vi
---------------- (2)
NON-INVERTING MODE:
 It consists of diode D1 in the feedback loop of a non-inverting Op-amp.
Case (i) When Vi > 0V
 For closed loop Op-amp Vp = Vn due to virtual ground concept.
 Diode D1 is forward biased and output of Op-amp, V0 = Vn = Vp = vi . as conducting diode
provides closed feedback path.
Case (i) When Vi < 0V
 When Vi < 0 diode D1is reverse biased and Op-amp works is open – loop.
 There is no current through RL. & V0 = 0V
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Since,
Vn = V0 = 0 V
Vp = Vi < 0 V
4. Explain the operation of instrumentation amplifier? [APRIL/MAY-2008]
[NOV/DEC 2010]
 Instrumentation amplifier is a circuit which amplifies the low level output signal of a
transducer to drive the indicator or a display system.
R2
R1 = R3 =1k
R2 = R4 =100k
R1
V1
V0
V2
R3
R4
 The above circuit is a basic instrumentation amplifier. The output voltage, V0 is given by
 R 
 R2
1
V0 
.V2 
.V1 1  2 
R
R1
R1 
1 3 
R4




 R2
R 
1
.V2 

.V1 1  1 
R
R1 
R2  
1 3 


R4


R
R
For 1  3 , we obtain
R2 R4
R2
-------------- (A)
.V1  V2 
R1
The source V1 sees an i/p impedance of R3+R4 (101k) and input impedance seen by source
V2 is only R1 (1k)
This low impedance loads the signal source heavily.
So a high impedance buffer is used preceding each input to avoid the loading effect.
Consider the following where Op-Amps. A1 and A2 have differential input voltage as zero.
V0 




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 Under common mode condition i.e., for V1 = V2, the drop across R is zero.
 No current flows through resistor R and R1. So non-inverting amplifier A1 acts as voltage
follower, hence V2 = V21
 Similarly, A2 acts as voltage follows having output V1  V11
 If V1  V2 , then current flows through R and R1 and V21  V11   V2  V1 
 Thus the circuit has high CMRR and high differential gain.

R

2
 . Using superposition
 The voltage at (+) input terminal of Op-amp A3 is V11 
R

 1 R2 
theorem,
 R2 1  R2   R2 .V11 

V0 
.V2  1  .
R1
 R1   R1  R2 


R2 1
. V1  V21
R1
current passing through R is given by
 V0 
-------------- (1)
V1  V2
-------------- (2)
R
The same current flows upwards through R1 since op-amp i/p current is zero.
I
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R1
V1  V2 
R
-------------- (3)
R1
V1  V2 
V  V2  I .R  V2 
R
-------------- (4)
V11  V1  I .R1  V1 
1
2
1
Substituting equations (3) and (4) in equation (1), we get

R2 R2  2 R1
V1  V2   V1  V2 
V0  V  V


R1 R1  R


1
1
1
2

 2 R1 
R2

V0 
.V1  V2 1 
R1
R 


--------------- (B)
The difference gain of this amplifier can be varied by using a variable resistor R.
Instrumentation Amplifier using Transducer Bridge
+
Vdc
-
Resistive Transducer
( RT  R)
V1
3 op-amp
instrumenta
-tion
amplifier
Indicator
(or)
Display
Device
V2
The above circuit uses a resistive transduces whose resistance changes as a function of
physical quantity to be measured.
 The bridge is initially balanced by a D.C supply Vdc. So that V1 = V2.
 As the physical quantity changes, the resistance RT also changes, causing an unbalance in
the bridge i.e., V1  V2
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 This differential voltage now gets amplified by the three Op-Amp differential
instrumentation amplifier and output is given to display device.
APPLICATION :
 Used in numerous industrial and process control applications like precise measurement and
control of temperature, pressure, humidity, light intensity water flow etc.,
5. Explain the working of log and antilog amplifier [APRIL/MAY-2008] (8m)
 The fundamental log-amplifier circuit with grounded base transistor in feedback path is given
as below
Fig(1) :
 Since collector is at virtual ground and base is also grounded, the V-I relationship of
transistor is (Shockley’s equation)
 qVE

I E  I S  e k .T  1


------------------- (1)
For a grounded-base transistor, IC = IE. Hence from equation (1) we get,
 qVE k .T

IC  I S  e
 1


------------------- (2)
Where,
Is  Reverse saturation current  10 13 A
K  1.38 x 10-23 J/K is Boltzmann Constant.
T  Absolute temperature (in º K).
Q =1.6 x 10-19 C is charge of an electron.
From equation (2)
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I C  qVE k .T

 e
 1
IS 

qVE
IC
I
 1  C  e k .T
IS
IS

Taking natural logarithm on both sides,
I 
q.VE
 ln  C 
kT
 IS 
kT  I C 
 .VE 
ln  
q  I S 
From the circuit diagram,
V
VE  V0 and I C  i
R1
Substituting above results in equation (3) we get,
 kT  Vi 

V0 
ln 
q
R
I
 1. s. 
------------------- (3)
------------------- (4)
Let Vref  R1 .I s
 kT  Vi 
ln
------------------ (A)
V 
q
 ref . 
 Thus the output voltage is proportional to logarithm of the input voltage. To find log10 we
use scaling
------------------ (5)
log  X   0.4343 ln  X 
 The problem in above circuit is Is varies with temperature and from transistor to transistor.
So to obtain stable Vref. Value we go for below circuit.
 V0 
 The input is applied to Op-amp A1 and reference voltage is applied to Op-amp A2. Both
transistors are matched transistors.(IS1 = IS2 = IS )
V1 
 kT  Vi
ln 
q
 R1 I S



-------------------- (6)
V2 
 kT  Vref
ln 
q
 R1 I S



-------------------- (7)
from equations (6) & (7),
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V0  V2  V1 
 V0 
kT
q
  Vi
ln 
  R1 I S
kT  Vi
ln
q  Vref
V

  ln  ref

 R1 I S







-------------------- (B)
 Thus now Vref is set by single external source. Shift V0 is dependent on temperature, T.

R 
 This is compensated by Op-amp A4, which provides non inverting gain 1  2  .
 RTC 
Where,
RTC  Temperature sensitive resistor with positive temperature coefficient.
Antilog amplifier:
 The input Vi for antilog is fed into the temperature compensating voltage divider R 2 and RTC
and then to base of Q2.
 Output, V0 of op-amp A1 is feedback to its inverting input through the resistor Ri
 The base-emitter voltage of transistors Q1 & Q2 are,
VQ1 BE 
kT  V0
ln 
q  R1 I S



----------------- (1)
VQ 2 BE 
kT  Vref
ln 
q  R1 I S



----------------- (2)
 Since base of Q1 is grounded,
V A  VQ1B  E 
 kT  V0
ln 
q
 R1 I S



----------------- (3)
 Base voltage of Q2 is given by, (Using voltage divider rule),
 RTC
VB  Vi 
 R2  RTC



------------------ (4)
 Emitter voltage of Q2 is given as
V A  VQ 2.E
----------------- 4(a)
But
LINEAR INTEGRATED CIRCUITS
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VQ 2.E  VB  VQ 2.E
Subtracting equation (2) from (4)
 RTC
VQ 2 E  Vi 
 R2  RTC
 kT  Vref
 
ln 
q

 Ri I S



---------------- (5)
From equations 3, 4(a) & (5) we get,
 kT  V0
ln 
q
 Ri I S

 RTC
  Vi 

 R2  RTC
 kT  Vref
 
ln 
q

 Ri I S



 RTC
 Vi 
 R2  RTC

kT   V0
  
ln 
q

  Ri .I S

V
  ln  ref

 Ri I S
 RTC
 Vi 
 R2  RTC
  kT 
 
ln
q






V
 ln  0
V
 ref
  q  RTC

.
 kT  R  R
TC
 2

 V0

V
 ref

.Vi

---------------- (6)
 Changing natural logarithm to log 10 we get,
V 
V 
log 10  0   0.4343 ln  0 
V 
V 
 ref 
 ref 
Using equations (6) & (7)
 q  RTC
 0.4343

 kT  R2  RTC
V
 log 10 o
V
 ref
V

.Vi  log 10 o
V

 ref



---------------- (7)





   K 1 .Vi


V0
 10  K Vi
Vref

V0  Vref 10  K .Vi
1

----------------- (A)
Where,
LINEAR INTEGRATED CIRCUITS
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 q  RTC 

.0.4343
kT  R2  RTC 
 Increase in input by one volt, results in decrease of output by one decade.
K1 
 IC755 Log /Antilog amplifier is available as a functional module which requires some
external components also to be connected to it.
6. Draw the circuit diagram of a second order butter worth active low pass filter and
derive an expression for its transfer function?
[MAY/JUNE-2007] (10m) &[Nov/Dec-2010]
Sallen-key filter consists of two RC pairs and have roll-off rate of -40dB/decade.
SALLEN KEY FILTER (GENERAL II ORDER FILTER);
 The circuit is in non-inverting mode, so
 R 
V0  1  F .VB  ACL .VB
R1 

VB  Voltage at node B, and ACL = 1+RF/R1
 KCL at node A gives,
ViY1  V A Y1  Y2  Y3   V0Y3  VB .Y2
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Vi Y1  V A Y1  Y2  Y3   V0Y3 
V0
.Y2
ACL
----------------- (1)
 KCL at node B, gives,
V AY2  VB Y2  Y4  
V0
.Y2  Y4 
ACL
----------------- (2)
 Substituting (2) in (1) and after simplification, voltage gain is given as,
V0
ACL Y1Y2

Vi Y1Y2  Y4 Y1  Y2  Y3   Y2Y3 1  ACL 
 To make a LPF, Y1 = Y2 = 1/R and
----------------- (3)
Y3 = Y4 = sC, then
From equation (3) we get,
H ( s) 
ACL
s R C  sCR(3  ACL )  1
2
2
----------------- (4)
2
 The transfer function of II order LPF is,
H ( s) 
ACL. h
2
s 2  h .s   h
----------------- (A)
2
Where
ACL  Gain of the filter.
h  Upper cut-off frequency in rad/sec.
  Damping coefficient.
 Comparing equations (4) & (A), we get,
1
h 
and   3  ACL 
RC
 put s = j in equation (A), we get,
ACL
H ( j ) 
2
 j /  h   j ( /  h )  1
 The normalized expression for LPF is
A
H ( j )  2 CL
s   .s  1
Where,
Normalized frequency, s = j/h
 Magnitude of Transfer function in dB. is,
LINEAR INTEGRATED CIRCUITS
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ACL
20 log H ( j )  20 log


1  2
  h
2
2
  


  


 h 
---------------- (B)
2
 For heavily damped filter (Bessel, =1.73), the response is stable. But roll-off begins earlier
to pass band. Flattest pass band occurs for  = 1.414.This is called Butterworth filter.
Substitute  = 1.414 in equation (B), we get,
ACL
20 log H ( j )  20 log
  

1  
 h 
 The Roll – off rate is -40dB/decade in stop band.
4
7. Write short notes on inverting amplifier?(OR)
Derive the expression for closed loop gain of an inverting amplifier.
 An amplifier which provides a phase shift of 180o between input and output is
called inverting amplifier.
I1
R1
Vi
Rf
I=0
I1
Vd=0
V0
I=0
RL
 The output voltage is feedback to inverting input terminal through Rf-R1 network,
where Rf is the feedback resistor.
 Input signal (Vi) is applied to the inverting terminal through R1. The noninverting terminal is grounded.
LINEAR INTEGRATED CIRCUITS
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ANALYSIS:
 At node ‘a’ is at ground potential as Vd = 0, so current I1 through R1 is given as
V
--------------- (1)
I1  i
R1
 As input current to op-amp is zero, the same current I1 passes through Rf . So o/p
voltage is I1 through R1 is
V 0   I 1 .R f
V 
  i .R f
 R1 
 Rf 

 V0  Vi 
 R1 
 Hence closed loop gain of inverting amplifier is
ACL 
V0  R f

Vi
R1
Using equation (1)
--------------- (2)
--------------- (A)
 Hence negative sign indicates phase shift of 180o.
 Effective input impedance is R1.
 R1 should be large to avoid loading effect.
 Usually load resistor, RL is connected to prevent the reduction in gain.
 The load current is given as
IL = V0 / RL
 If Rf >R1, the gain is greater than 1.
 If Rf <R1, the gain is less than 1.
 If Rf = R1, the gain is unity.
 If
Rf
 If
Rf
R1
R1
= K (Some constant), the circuit is called Scale Changer.
=1, it is called Phase Inverter.
 Bias current compensation is provided with Rcomp = (R1|| Rf) resistor.
LINEAR INTEGRATED CIRCUITS
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Page 25
Vi
Input wave
t (ms)
V0
Output wave
t (ms)
8. Derive the expression for closed loop gain of non-inverting
(OR) Write short notes on non-inverting amplifier.
amplifier.
 It the signal is applied to non-inverting terminal and feedback is as shown in circuit, the
circuit amplifier without inverting the signal.
 It is a negative feedback system since the output is feedback to inverting terminal.
 Differential input, Vd at i/p terminal of op-amp is zero, so at node ‘a’ the voltage is Vi.
 Now Rf and R1 form a potential divider.
Hence,
V0R1
R1 + Rf
and no current flows into Op-Amp.
Vi =

V0
V1
R1 + Rf
=
R1
Rf
V
 ACL  0  1 
Vi
R1
=
--------------- (A)
 The gain can be adjusted to unity or more by proper selection of Rf and R1.
 The input resistance of non-inverting amplifier is extremely large (Ri) as the op-amp draws
negligible current from signal source.
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9.Explain Voltage follower? [APRIL/MAY-2011]
V0
Vi
 In non-inverting amplifier, if Rf = 0 and Ri =  , we get a modified circuit called
voltage follower.
ACL  1 
 ACL 
Rf
R1
 1
0
1

V0
1
Vi
V0  Vi
 The output voltage is equal to input voltage, both in phase and magnitude.
 Here the output voltage follows the i/p voltage hence the name.
 The main use of this unity gain circuit lies in the fact that its input impedance is very
high (in order of M ) and output impedance is zero.
 The circuit draw negligible current from source.
APPLICATIONS:
Voltage follower is used as buffer for impedance matching, i.e., to connect high
impedance source to a low impedance load.
Possible 2 Mark.
1.
Draw the voltage follower circuit.
LINEAR INTEGRATED CIRCUITS
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10.Write short notes on summing amplifier [APRIL/MAY-2011]
(or)
With the help of ckt. Diagram show that Op-amp can be used as
(i) Inverting adder (ii) non-inverting adder.
[NOTE: If it is asked as just summing amplifier you can write any one either
inverting or non-inverting.]
(i). INVERTING ADDER (or) INVERTING SUMMING AMPLIFIER:
R1
Rf
V1
R2
V2
R3
‘A’
V0
V3
Rcomp = (R1 R2 R3 Rf)
 The above circuit is a inverting adder with three input voltages (V1,V2 & V3), three
input resistors (R1, R 2 & R 3) and feedback resistor Rf.
 Assuming ideal Op-Amp, A0L =  and Ri = 
 Since input bias current is zero there is no voltage drop across Rcomp. Hence the noninverting terminal is at virtual ground.
 So voltage at node ‘A’ is also zero. Nodal equation by KCL at ‘A’ is
V1 V2 V3 V0



0
R1 R2 R3 R f

---------------- (1)
V V
V0
V 
  1  2  3 
Rf
 R1 R2 R3 
Rf
Rf
 Rf

V0  
.V1 
.V2 
.V3 
R2
R3
 R1

---------------- (2)
 Thus the output is an inverted, weighted sum of inputs.
 Considering R1 = R2 = R3 = Rf in equation (2) we get,
LINEAR INTEGRATED CIRCUITS
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V0 = - (V1+ V2+ V3)
---------------- (A)
 The output is inverted sum of the inputs. Hence the circuit is called inverting
summing amplifier.
 When R1= R2 = R3= 3 Rf in (2) we get
V0  
V1  V2  V3 
3
 Thus the output is average of the three inputs. This circuit is called Inverting
Averaging Amplifier.
(ii)
NON – INVERTING ADDER (or) SUMMING AMPLIFIER:
Rf
R
R1
V0
V1
V2
R2
V3
R3
 A summer or adder that produces non-inverted sum is called non-inverting summing
amplifier. Let the voltage at node at be VA.
 Hence voltage at inverting terminal is also VA. since Vd = 0.
 The nodal equation by KCL at ‘A’ given, as,
V1  V A V2  V A V3  V A


0
R1
R2
R3

 1
V1 V2 V3
1
1 



 V A  

R1 R2 R3
 R1 R2 R3 
V A 
V1 / R1  V2 / R2  V3 / R3 
1 / R1  1 / R2  1 / R3 
--------------- (1)
 The Op-amp and two resistors Rf and R form a non-inverting amplifier, with
 Rf 

V0  V A 1 
R


 Substituting equation (1) in (2) we get,
LINEAR INTEGRATED CIRCUITS
--------------- (2)
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 R f  V1 / R1  V2 / R2  V3 / R3 

--------------- (A)
V0  1 
R  1 / R1  1 / R2  1 / R3 

which shows output is a non-weighted sum of inputs.
 Let R1 = R1 = R2 = R3 = R = Rf /2 , then equation (A) becomes

R f  V1 / R  V2 / R  V3 / R 

V0   1 
 R  1 / R  1 / R  1 / R 
f /2 

 V  V2  V3 
 3 1

3



V0  V1  V2  V3
2 Marks:
1. Draw the inverting summing amplifier circuit.
2. Draw the non-inverting adder circuit.
11.Write short notes on Subtractor (or) Difference amplifier.
 A basic differential amplifier can be used as a Subtractor by considering
Rf = R1 = R. Consider all the resistor values as equal value then the o/p
voltage can be derived using superposition principles.
 The output V01 due to input voltage V1 above is found by making V2 = 0.
 Now the circuit becomes a non-inverting amplifier having input voltage V1/2 at
positive terminal and output becomes.
V1  R 
------------- (1)
1    V1
2  R
 Similarly output V02 due to V2 alone (V1 is grounded ) can be given by an inverting
amplifier
V02 = - V2
------------- (2)
 Thus the output voltage due to both inputs is given as
Adding equations (1) and (2),
V0  V01  V02  V1  V2
V01 

V0  V1  V2
2 Marks:
1. Draw the circuit of Subtractor or difference amplifier.
LINEAR INTEGRATED CIRCUITS
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12. Write short notes on current to Voltage (I-V) converter?
Ii
R
A
V0
Ii
B
 The circuit in which the output voltage is proportional to input current is called current to
voltage converter. Consider, input current Ii and output voltage V0, then
V0  A.I i
(1)
Where, A  Gain of the circuit.
 As we measure gain in ohms it is appropriate to denote gain by R. Hence I-V converter is
also called as Transresistance amplifier.
 Node B is grounded. Hence node A is at virtual ground.
VA  0
 Ii 
V A  V0  V0

R
R
 V0  R.I i

V0R.I i
 Here the output voltage (V0) is proportional to input current (Ii), therefore the circuit is a
current to voltage converter.
 The circuit is also called as current controlled voltage source [CCVS].
 It the resistance in the circuit is replaced with impedance Z, the circuit is called
Transimpedance Amplifier.
LINEAR INTEGRATED CIRCUITS
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13.
Write notes on Differentiator. (Or)
Explain the operation of differentiation amplifier with neat circuit
diagram.
[APRIL/MAY 2010]
 Differentiator or Differentiation amplifier is a circuit which performs mathematical operation
of differentiation. i.e., the output waveform is the derivative of the input waveform.
IDEAL DIFFERENTIATOR
RF
IF
C1
IC
Vin
V0
Rcomp Rf
 The node A is at virtual ground i.e., VA= 0
d .Vin
d

 The current through capacitor is I C  C1  viA  V A   C1 .
dt
 dt

The circuit iF though RF is given by
V0  V A V0
------------ (2)

RF
RF
 At node A, by KCL (i.e) equations (1) + (2)
IF 
-------- (1)
since V A  0
IC  I F  0
 C1
C1

d .Vin V0

0
dt
RF
d .Vin
V
 0
dt
RF
V0   RF .C1 .
d .Vin
dt
--------------- (A)
 Thus the outputvoltage,V0 is (-RFC1) times the derivative of input voltage
V in.
The minus indicates phase shift of 180o between i/p signals.
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 Writing equation (A) in frequency domain, we get
V0 (s)   RF C1 .S.VIN (S )
 Now gain ‘A’ of the differentiator is
A
V0 ( s)
  SRF .C1   j.RF .C1  .RF .C1
Vin ( s)
( s = j ,  = 2f)
 A  2f .RF C1
 A
f
fa
where
fa 
1
2RF C1
At f=fa, A  1, i.e 0 dB and gain increases at a rate of +20dB/decade.
DEMERITS:
 The i/p impedance (1/C1) decreases with increase in frequency, thereby making the
circuit sensitive to high frequency noise.
 Also, at high frequencies, the ckt. becomes unstable and enters into oscillation.
Fig(ii) PRACTICAL DIFFERENTIATOR
 Both stability and high frequency noise problems can be corrected by adding two
components, R1 & CF as shown in above circuit. The transfer function of this circuit is
V0 ( s)  Z f
 SRF C1


1  SR1C1 1  SRF C F 
Vin ( s)
Zi
For R1C1 = RFCF,
A
V0 ( s )
 sR F C1
 sR F .C1


2
2
Vin ( s ) 1  sR1C1 

f 
1  j 
fb 

where
fb 
1
1

2R1C1
2R F C F
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FREQUENCY RESPONSE:
 fa is the frequency at which gain is 0 dB.
 fb is the gain limiting frequency
 fC is the unity gain bandwidth of Op-Amp.




From fa to fb gain increases at rate of +20db/decade.
After fb the gain decreases at -20dB/decade.
This change in gain is caused by R1C1 & RFCF combinations.
The value of fb should be selected such that
fa  fb  fc
 For good differentiation, the time period T of input signal is
T  RFC1.
DESIGN STEPS:
Refer Question No. 9 in 2 Marks. (Page No.)
APPLICATIONS:
 In wavaeshaping circuits todetecthigh frequency components in input signal.
 As Rate of change detector in FM Modualtor.
2 Marks :
1. Give applications of Differentiator.
14. Explain the operation of an Integrator circuit (or) Write short notes on
integrator?
[APRIL/MAY-2011]
VO
Vi
LINEAR INTEGRATED CIRCUITS
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 A circuit in which the output voltage wave is the integral of the input voltage waveform
is the Integrator or Integration Amplifier.
Fig(i) Ideal Integrator
 The nodal equation at node A is,
Vin  V A
d (V0  V A )
 CF
0
R1
dt
 Since node A is at virtual ground, VA= 0

Vin
dV0
 CF
0
R1
dt

dV0
1

Vin
dt
R1C F
 Integrating on both sides we get,
1
V0  
R1C F
t
V
in
dt  C
---------------- (A)
0
where, C is the integration constant. Thus the output voltage is directly proportional to
negative integral of input voltage and inversely proportional to time constant R1CF.
 In Op-amp integrator the effective input capacitance by Millers theorem is
Cf (1-Av).
Where, AV 
gain of the Op = Amp
 Gain AV is infinite for ideal Op-Amp, so effective time of Op-amp becomes large which
results in perfect integration.
 The equation (A) written in phasor notation,
1
V0 ( s)  
.VIN ( s)
sR1C F
In steady state, s = j
 A
V0 ( s)
1
1


Vin ( s)
SR1C F
 jR1C F
 A
1
R1C F
LINEAR INTEGRATED CIRCUITS
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Page 35
 A 
fb
f
where,
where fb is the frequency where gain is 0dB.
ERRORS IN IDEAL INTEGRATOR:
 At low frequencies (0), the gain becomes infinite (saturates).
 Input offset voltage gets amplified and appears as error voltage which causes
saturation.
 It is difficult to pull integrator out of saturation, hence perfect integration is not
possible.
 Limited bandwidth.
Fig (ii) Practical Integrator.
The demerits of ideal circuit are overcome by practical integrator circuit, in which
feedback resistor RF is connected across CF. It reduces the low frequency gain of the Op-Amp.
It is also called LOSSY INTEGRATOR.
Nodal equation at node A gives,
Vin (s)
V ( s)
 sC F V0 ( S )  0
0
R1
RF




1

.V ( s )
 V0 ( s )  

R1  in
 sR1C F 

RF 


V0 ( s)
 RF / R1

Vin ( s)
1  sRF C F
Put s = j ,
 A
V0 ( j )
 RF / R1
 RF / R1


Vin ( j ) 1  jRF C F
1  j 2fRF C F
 A 
 R F / R1
1 j
f
fa
Where,
LINEAR INTEGRATED CIRCUITS
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1
, is the break frequency at which gain is (0.707) RF /R1
2RF C F
 The value of fa is selected such that fa<fb
 fb, the input frequency should be 10 times. fa. i.e fb = 10 fa.
fa 
 For perfect integration, time period T of input signal should be, T  RFCF.
APPLICATIONS:
 Analog computers, ADC’s
 Wave shaping circuits.
 Ramp generators.
.
15.Write short notes on sine wave Oscillator. (or)
[April 2005]
[ANS: You can write any one of the Oscillator, either Wien bridge or
RC-phase shift Oscillator.]
a. Write short notes on RC-phase shift Oscillator. [April 2008]
 RC- phase shift oscillator consists of Op-amp as amplifier stage and three cascaded RC
networks as feedback network.
 Op-Amp provides 180o phase shift as it is in inverting node. Additional phase shift of 180o is
provided by RC feedback network.
 The transfer function of RC feed-back n/w is given by

Vf
V0

1
1  6 / SRC  5 / S R 2 C 2  1 / S 3 R 3C 3
2
--------- (1)
 In steady state put s = j in equation (1)
 

1  5

1
  f   f 3 
f1 
  j 6 1    1  
f 
  f   f  
2
--------- (2)
Where,
1
2RC
 For satisfying Barkhausen’s Criterion, i.e., AV   1  should be real. So imaginary term in
equation (2) should be equated to zero.
f1 
LINEAR INTEGRATED CIRCUITS
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f  f 
 6 1    1 
 f   f 
3
2
f 
 6   1 
 f 
f
 1  6
f
Hence, the frequency of oscillation f0 is given by,
f0  f 
1
----------------- (A)
6 2RC 
Also loop gain AV. =1
Av
1

2
 f1 
1  5 
 f0 
2
 
 f 
 Av  1  5 1   1  5 6
 f0 
2
 29
 Av  29
 Av

Rf
R1
 29
 RF =29 R1
 Thus the circuit will produce a sinusoidal waveform of frequency f0. If the gain is atleast 29
and the total phase shift is 360o.
 If the AV   1, due to variations in circuit parameters, oscillations will die out.
16.
Explain the operation of Monostable Multivibrator with neat circuit
diagram?
[NOV/DEC-2008]
 It is also know as One-shot Multivibrator or Monoshot since it produces a single pulse of
specified duration in response to each external trigger signal.
 There is only one stable state. When an external trigger is applied, the output changes its
state to a new state, called quasistable state.
LINEAR INTEGRATED CIRCUITS
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 The circuit remains in this state for a particular fixed time interval. After sometime, it returns
back to its original stable state.
 In fact an internal trigger is generated to drive the circuit to its original state. Changing and
discharging of capacitor provides the internal trigger signal.
 The diode D1 clamps capacitor voltage to 0.7 V, when the output is at +Vsat
 A negative going pulse of magnitude V1 is applied to differentiator R4C4 and diode D2 gives
negative going triggering impulse to (+) input terminal.
 Assume in stable state that V0 is at + Vsat
 Diode D1 conducts and capacitor voltage, VC gets clamped to 0.7 V. Voltage at (+) input
terminal through R1R2 is +.Vsat .
 Negative trigger of magnitude V1 is applied to (+) terminal so effective signal at this terminal
is less than 0.7 V  Vsat  (V1 )  0.7v, the output of Op-Amp will switch from +Vsat
to –Vsat.
 The diode will now get reverse biased and the capacitor starts charging exponentially to –Vsat
through R.
 When capacitor voltage become more negative then -Vsat, the output switches to +Vsat
 Now capacitor, C, changes towards +Vsat until VC is 0.7 V as it gets clamped to the voltage.
 The general solution for a single time constant low pass RC circuit with Vi and Vf as initial
and final values is
v0  vf  vi  v f et / RC
 For this circuit, V f  Vsat , Vi  V0 .Hence output VC is given by
Vc  Vsat  VD  Vsat e t / RC
 At t = T, Vc  Vsat
  Vsat  Vsat  V0  Vsat e T / RC
 Vsat 1     v D  v sat e T / RC
 e T / RC 

vsat 1   
v D vsat
T  RC . ln
1  VD / Vsat e T / RC
LINEAR INTEGRATED CIRCUITS
1 
------------- (A)
SCE/DEPT. OF ECE/II YEAR
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Where,
R2
R1  R2
Equation (A) is the pulse width of Monostable multivibrator. If Vsat>>V0 and R1 = R2,  = 0.5,
then

T = 0.69RC
------------- (B)
Trigger pulse width, Tp should be less than T capacitor voltage, VC reaches VD at T   T
17. Draw and explain the operation of Triangular wave generator?
[April/MAY 2008]
 A triangular wave can be simply obtained by integrating a square wave. The frequency of
square wave and triangular wave is the same.
 The amplitude of square wave is constant at  Vsat. But the amplitude of triangular wave
decreases as the frequency increases. This is because the reactance of capacitor C2 decreases
at high frequencies.
 A resistance R4 is connected across C2 to avoid saturation problems at low frequencies.
 Another circuit with less number of components containing a two level Comparator & an
Integrator is shown below.
 Output of comparator A1 is a square wave of amplitude  Vsat and is applied to (-) terminal of
integrator A2 producing a triangular wave output.
 This triangular wave is feedback as input to the comparator A1 through a voltage divider R2
and R3.
LINEAR INTEGRATED CIRCUITS
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 At time t = t1, when negative going ramp reaches -Vramp, effective voltage at P is slightly less
then 0 V. Thus the output switches from + Vsat to -Vsat. When V0 is -Vsat, the V0 increases in
positive direction.
 At time t = t2, V0 reaches Vramp, thus effective voltage at P becomes just above 0V, thereby
switching the output of A1 from -Vsat to +Vsat. This cycle repeats and thus triangular wave is
generated.
 The amplitude of triangular wave depends on RC value of integrator A2 and output level of
comparator A1.
 The effective voltage at point P, when output of A1 is at +Vsat is
 Vramp 


R2
 Vsat   Vramp 
R2  R3
----------- (1)
 At t = t1, the voltage at P becomes zero,
 Vramp 
 R2
 Vsat 
R3
----------- (2)
 Similarly at time t = t2, when o/p of A1 switches from t = -Vsat to +Vsat,
 Vramp 
 R2
 Vsat   R2 .Vsat
R3
R3
----------- (3)
 Peak to peak amplitude of triangular wave is
V0 ( p  p )  Vramp   Vramp 
V0( p  p )  2.
R2
.Vsat
R3
(3) - (2).
----------- (4)
 Output switches from -Vramp to +Vramp in time period T/2. Putting the values in basic
integrator equation, we get.
1
v0 
Vi dt
RC 
LINEAR INTEGRATED CIRCUITS
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v0 ( p  p ) 

1
R1C1
T /2
  V dt
sat
0
Vsat T / 2
V T 
dt  sat  

R1C1 0
R1C1  2 
 T  2 R1C1
V0( p  p )
Vsat
 2.R2 / R3 Vsat 
 2 R1C1 

Vsat



T
from equation (4)
4R1C1 R2
R3
-------------- (A)
 Hence the frequency of oscillation is
f0 
-------------- (B)
R3
1

T 4C1 R1 R2
PROBLEMS:
1. Design a second order Butterworth LPF having upper cut-off frequency 1
kHz. Determine the frequency response. [Nov 2003]
Solution:
Given:
fH = 1 kHz.
We know that, f H 
1
2RC
Assume C = 0.1F.
 R  1.6k
For second order fitter  = 1.414.
Then Pass band gain A0 = 3- 
= 3 – 1.414
A0 = 1.586
The Transfer function of II Order butterworth LPF is,
LINEAR INTEGRATED CIRCUITS
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1.586
s  1.414 s  1
2
Now
A0  1 

Rf
R1
Rf
R1
 1.586
 0.586
Assume Rf = 5.86 k
 Ri  10k
The circuit is as follows
The gain equation is
 20 log
V0
 20 log
Vi
A0
 f 
1   
 fh 
4
For calculating frequency response the frequency is taken from 0.1 fh to 10 fh i.e, 100 Hz
to 10 kHz as fh = 1 kHz.
Frequency, f in Hz.
100
Gain magnitude in dB.
4.00
200
4.00
500
3.74
1000
1.00
5000
-23.95
10000
-35.99
LINEAR INTEGRATED CIRCUITS
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2. R1 = 10k, Rf = 100k, Vi = 1V. A load of 25 k, is connected to the
output terminal. Calculate i) i1 ii)V0 iii). iL and total current i0 into the output
pin.
Solution:
a).
i1 
vi
1V

 0.1mA.
R1 10k
b).
vo 
Rf
R1
vi 
100k
(1V )  10V .
10k
V0
10V
vi 
 0.4mA
RL
25k
The direction of i2 is shown in figure.
c).
iL 
Total current i0 = i1 + iL
i0 = 0.1 mA + 0.4 mA
 i0  0.5mA.
3.
An input of 3V is fed to non-inverting terminal of an Op-Amp. The
amplifier has a Ri of 10k and Rf of 10k. Find output voltage
[April 2004]
Solution:
Non-inverting amplifier gain is
R
ACL  1  F
R1
10k
 1
10k
 ACL  2
Also,
V
ACL  0
Vi
 V0  ACL .Vi
= 2 (3V)
 V0  6.Volts
LINEAR INTEGRATED CIRCUITS
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4.
Design a phase shift oscillator to oscillate at 100Hz.[Nov
2003]
Solution:
Frequency of oscillation, f0 is
1
f0 
6 2RC 
Assume C1 = 0.1 F
R
Use

1

6 2  (100)  0.1 10 6
R = 6.49 k.
R = 6.5 k.

To prevent loading of amplifier by RC network, R1  10 R

Let R1 = 10 R = 65 k
 R f  29R1
 R f  29  65k
 R f  1885k
The circuit is as follows. Where,
LINEAR INTEGRATED CIRCUITS
R = 6.5 k, C = 0.1 F……..
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