OP amp Applications

1. LINEAR APPLICATIONS OF OP-
AMPS

2. SYLLABUS
Applications of Op-Amps: Addition,
Subtraction, Multiplication, Current to
Voltage and Voltage to current conversion
operations , Integration, Differentiation,
Active Filters, Comparators, Schmitt
trigger, Instrumentation Amplifier, f to V
converter.

3. LINEAR APPLICATIONS
• Adder
• Subtractor
• Voltage follower
• Current to voltage converter
• Voltage to current converter
• Integrator
• Differentiator
• Active filters

4. Non-linear applications
• Comparators
• Logarithmic amplifiers
• Exponential amplifiers
• Peak detectors
• Precision rectifiers
• Waveform generators
• Clippers & clampers

5. If Rf = R, then Vout = v1+v2+v3

6. Non-inverting
Summing
Amplifier
Therefore, using the
superposition theorem, the
voltage V2 = V1
Vb & Vc = 0. Net resistance
= R+R/2
If RF=2R1, 1+RF/R1=3
Vo= Va+Vb+Vc

7. Determine the output voltage
=12

8. Averaging Amplifier
• A summing amplifier can be made to produce the
mathematical average of the input voltages. The
amplifier has a gain of Rf/R, where R is the value of
each input resistor. The general expression for the
output of an averaging amplifier is
• VOUT=-(Rf /R)(VIN1+VIN2+…+VINn)
• Averaging is done by setting the ratio Rf/R equal to
the reciprocal of the number of inputs (n); that is ,
Rf/R=1/n.

9. Problem
Show that the amplifier in produces an output whose magnitude is the
mathematical average of the input voltages.

11. ADDER-SUBTRACTOR
It is possible to perform addition and subtraction simultaneously with a single
op-amp using the circuit shown above.
The output voltage Vo can be obtained by using superposition theorem. To find
output voltage Vo1 due to V1 alone, make all other input voltages V2, V3 and V4
equal to zero. The simplified circuit is shown in Fig.b. This is the circuit of an
inverting amplifier and its output voltage is,

12. 4.3 (b) Op-amp adder- subtractor, (c)
Simplifier circuit for V2 = V3 = V4 = 0,
(d) Simplified circuit for V1 = V2 = V4 = 0
Similarly V02 = -V2
Now, the output voltage
Vo3 due to the input
voltage signal V3 alone
applied at the (+) input
terminal can be found by
setting V1=V2=V4=0. The
circuit now becomes a
non-inverting amplifier as
shown in Fig. 4.3 (d). The
voltage Va at the non-
inverting terminal is
Applying Thevinons theorem at
inverting terminal, we get
=R/2
=1/3 R
=V3

13. So, the output voltage Vo3 due to V3 alone is
Similarly, it can be shown that the output
voltage V04: due to V4 alone is
Vo4 = V4 (4.14)
Thus, the output voltage Vo due to all four
input voltages is given by

14. Scaling Adder
• A different weight can be assigned to each
input of a summing amplifier by simply
adjusting the values of the input resistors. The
output voltage can be expressed as
• VOUT=((Rf/R1)VIN1+(Rf/R2)VIN2+…+(Rf/Rn)VINn)

15. Problem
Determine the weight of each input voltage
for the scaling adder in Figure 7-16 and find
the output voltage.

16. Voltage to Current Converter (Transconductance Amplifier)
In many applications, one may have to convert a voltage signal to a
proportional output current. For this, there are two types of circuits
possible.
•V-I Converter with floating load
•V-I Converter with grounded load
Figure 4.8 (a) shows a voltage to current converter in which load ZL is
floating. Since voltage at node 'a' is vi therefore,
Fig. 4.8 Voltage to current converter
Vi
iL proportional to Vi
is controlled by R1
(b) Grounded load
(a) Floating load
iL = vi /R
Vi = iLR1

17. That is the input voltage vi is converted into an output current of vi /R1, it may be seen that
the same current flows through the signal source and load and,
therefore, signal source should be capable of providing this load current.
Since the op-amp is used in non-inverting
mode, the gain of the circuit is 1 + R/R= 2. The
output voltage is,
vo = 2v1 = vi + vo – iLR
Vi = iLR
iL = vi /R
A voltage-to-current converter with grounded load is shown in Fig. 4.8 (b). Let vi
be the voltage at node 'a'. Writing KVL, we get

18. As the input impedance of a non-inverting amplifier is very high, this circuit has the
advantage of drawing very little current from the source.
A voltage to current converter is used for low voltage dc and ac voltmeter, LED and
zenar diode tester.

19. Current to Voltage Converter (Trans-resistance Amplifier)
Photocell, photodiode and photovoltaic cell give an output
current that is proportional to an incident radiant energy or light.
The current through these devices can be converted to voltage by
using a current-to-voltage converter and thereby the amount of
light or radiant energy incident on the photo-device can be
measured.
Figure 4.9. shows an op-amp used as I to V converter. Since the
(-) input terminal is at virtual ground, current ii flows through the
feedback resistor Rf. Thus the output voltage v0 = - iiRf. It may be
pointed out that the lowest current that this circuit can measure
will depend upon the bias current of the op-amp. This means that
for 741 (bias current is 3 nA) can be used to detect lower
currents. The resistor Rf is sometimes shunted with a capacitor Cf
to reduce high frequency noise and the possibility of oscillations.

20. Inverting Current to Voltage Converter

21. Non-inverting Current to Voltage Converter.

22. INTEGRATORS AND
DIFFERENTIATORSIntegrators produce output voltages that are
proportional to the running time integral of the input
voltages. In a running time integral, the upper limit of
integration is t. Passive Integrator Gain is Less than
Unity. Attenuation is more. For perfect integration,
large values of RC are required.
Gain decreases with increase of frequency.

23. where C is the integration constant and is proportional to the value
of the output voltage vo at time t = 0 seconds.
For example, if the input is a sine wave, the output will be a cosine
wave.
If the input is a square wave, the output will be a triangular wave,
as shown in Figure 7-23(c) and (b), respectively.
Equation-7-23

24. When vin = 0, the integrator of Figure 7-23(a)
works as an open-loop amplifier. This is because
the capacitor CF acts as an open circuit (XCF = ∞)
to the input offset voltage Vio. In other words, the
input offset voltage Vio and the part of the input
current charging capacitor CF produce the error
voltage at the output of the integrator. Therefore, in
the practical integrator shown in Figure 7-25, to
reduce the error voltage at the output,
a resistor RF is connected across the feedback
capacitor CF. Thus, RF limits the low-frequency
gain and hence minimizes the variations in the
output voltage.

25. A practical Integrator

26. The frequency response of the basic integrator is shown in
Figure 7-24. In this figure, fb, is the frequency at which the
gain is 0 dB and is given by
Both the stability and the low-frequency roll-off problems can be
corrected by the addition of a resistor RF as shown in the
practical integrator of Figure 7-25. The frequency response of
the practical integrator is shown in Figure 7-24 by a dashed line.
In this figure, f is some relative operating frequency, and for
frequencies up to fa the gain RF/Ri is constant. However, after
fa, the gain decreases at a rate of 20 dB/decade. In other
words, between fa, and fb, the circuit of Figure 7-25 acts as an
integrator. The gain-limiting frequency fa is given by
Frequency Response

27. If the op-amp was ideal, an integrator as shown in above Figure would require just one
resistor, R, and one capacitor, C, and the relation between the output and input voltages
would be given by
fa < fb. If fa = fb/10, then RF = 10R1. The input signal will be integrated properly, if
time period T of the input signal
=fa =fb
The Integrating range is in between fa & fb

28. Frequency Response of an Ideal & Lossy Integrator
=fa =fb
Integrating Range
20db/decade

29. Square waves are integrated to triangles, triangles to parabolas etc.

30. EXAMPLE 7-15
In the circuit of Figure 7-23, R1CF = 1 second, and the input is a step (dc)
voltage, as shown in Figure 7-26(a). Determine the output voltage and
sketch it. Assume that the op-amp is initially nulled.

31. SOLUTION The input function is constant beginning at t = 0 seconds. That
is, Vm = 2 V for 0 < / =s 4. Therefore, using Equation (7-23),
The output voltage waveform is drawn in Figure 7-26(b); the waveform is
called a ramp function. The slope of the ramp is -2 V/s. Thus, with a
constant voltage applied at the input, the integrator gives a ramp at the
output.

32. Determine the rate of change of the output voltage in response to
the input square wave, as shown for the ideas integrator in Figure 7-
22(a). The output voltage is initially zero. The pulse width is 100μs.
Describe the output and draw the waveform.

33. Differentiator Circuit
Figure 7-27 Basic
differentiator, (a) Circuit,
(b) Frequency response.
fc = Unity Gain BW
Gain increases with frequency
Practical
Ideal
fc = Unity Gain BW
Figure 7-28
Figure 7-27b

34. Since IB = 0,
ic = iF
Since v1= v2 = 0 V,

35. fa is the frequency at which the gain is 0db and is given by
At fa gain is 0db
fC is the unity gain-bandwidth of the op-amp, and f is some relative operating
frequency.
Both the stability and the high-frequency noise problems can be corrected by
the addition of two components: R1 and CF, as shown in Figure 7-28(a), This
circuit is a practical differentiator, the frequency response of which is shown in
Figure 7-27(b) by a dashed line. From frequency f to fb, the gain increases at 20
dB/decade. However, after fb the gain decreases at 20 dB/decade. This 40-dB/
decade change in gain is caused by the R1C1 and RFCF combinations. The gain-
limiting frequency fb is given by (7-29)
where R1C1= RFCF , help to reduce significantly the effect of high-frequency
input, amplifier noise, and offsets. Above all, it makes the circuit more stable by
preventing the increase in gain with frequency. Generally, the value of fb and in
turn R1C1 and RFCF values should be selected such that
fa< fb< fc (7-30)

36. But v1= v2 = 0 V, because A is very large. Therefore,
Thus the output v0 is equal to the RFC times the negative
instantaneous rate of change of the input voltage vin with time.
Since the differentiator performs the reverse of the integrator's
function, a cosine wave input will produce a sine wave output, or a
triangular input will produce a square wave output. However, the
differentiator of Figure 7-27(a) will not do this because it has some
practical problems. The gain of the circuit
(RF /XC1) increases with increase in frequency at a rate of 20
dB/decade. This makes the circuit unstable. Also, the input
impedance XC decreases with increase in frequency, which makes
the circuit very susceptible to high-frequency noise. When
amplified, this noise can completely override the differentiated
output signal. The frequency response of the basic differentiator is
shown in Figure 7-27(b). In this figure, fa is the frequency at which
the gain is 0 dB and is given by

37. Figure 7-28 Practical differentiator, (a) Circuit, (b) Sine wave input and resulting cosine
wave output, (c) Square wave input and resulting spike output.

38. Steps For the Design of Practical Differentiator:
1. Select fa equal to the highest frequency of the input
signal to be differentiated. Then, assuming a value of C1
< 1 μF, calculate the value of RF .
2. Choose fb = 20 fa and calculate the values of R1, and CF
so that R1C1 = RFCF .
The differentiator is most commonly used in wave
shaping circuits to detect high-frequency components
in an input signal and also as a rate-of-change detector
in FM modulators.

39. EXAMPLE 7-16
Design a differentiator to differentiate an input signal that varies in frequency from
10 Hz to about 1 kHz. If a sine wave of 1 V peak at 1000 Hz is applied to the
differentiator of part (a), draw its output waveform.
SOLUTION (a) To design a differentiator, we simply follow the steps outlined
previously:

41. Problem
Determine the output voltage of the ideal op-amp
differentiator in Figure 7-26 for the triangular-wave input
shown.

42. Comparison Between Integrator &
Differentiator.
The process of integration involves the
accumulation of signal over time and hence sudden
changes in the signal are suppressed. Therefore an
effective smoothing of signal is achieved and hence,
integration can be viewed as low-pass filtering.
The process of differentiation involves
identification of sudden changes in the input signal.
Constant and slowly changing signals are supressed
by a differentiator. It can be viewed as high-pass
filtering.

43. INSTRUMENTATION AMPLIFIER
In a number of industrial and consumer applications, one is required to measure
and control physical quantities. Some typical examples are measurement and
control of temperature, humidity, light intensity, water flow etc. These physical
quantities are usually measured with the help of transducers. The output of
transducer has to be amplified so that it can drive the indicator or display system.
This function is performed by an instrumentation amplifier. The important features
of an instrumentation amplifier are:
(i) high gain accuracy
(ii) high CMRR
(iii) high gain stability with low temperature coefficient (iv) low dc offset
(v) low output impedance
There are specially designed op-amps such as (µA725 to meet the above stated
requirements of a good instrumentation amplifier. Monolithic (single chip)
instrumentation amplifier are also available commercially such as AD521, AD524,
AD620, AD624 by Analog Devices, LM-363.XX (XX -»10,100,500) by National
Semiconductor and INA1O1,1O4, 3626, 3629 by Burr-Brown.

50. Non Linear Applications:
Precision rectifiers.
•The major limitation of ordinary diode is that it cannot rectify
voltages below vγ (~ 0.6 V), the cut-in voltage of the diode.
• A circuit that acts like an ideal diode can be designed by placing
a diode in the feedback loop of an op-amp as in Fig. 4.10 (a).
Here the cut-in voltage is divided by the open loop gain A0L (~
104) of the op- amp so that vγ is virtually eliminated. When the
input Vi > Vγ /AOL then voA, the output of the op-amp exceeds Vγ
and the diode D conducts. Thus the circuit acts like a voltage
follower for input vi > Vγ /A0L (i.e., 0.6/104 = 60μv) and the
output v0 follows the input voltage vi during the positive half
cycle as shown in Fig. 4.10 (b).

51. •When vi is negative or less than Vγ /A0L, the diode D is off and
no current is delivered to the load RL except for small bias
current of the op-amp and the reverse saturation current of the
diode. This circuit is called the precision diode and is capable of
rectifying input signals of the order of mill volt. Some typical
applications of a precision diode discussed are:
•Half-wave Rectifier
•Full-Wave Rectifier
•Peak-Value Detector.
•Clipper.
•Clamper.

52. Fig. 4.10 (a) Precision diode, (b) Input and output
waveforms

53. When Vi is +ve, Voa is –ve. D2 is reverse biased and output is zero. When Vi is -ve,
Rf=R1, Voa is +ve, and D2 conducts even when the input is < 0.7V. The op-amp
should be a high speed version as it alternates between open loop & closed loop
operations. High Slew rate is required as the input passes through zero, Voa must
change from 0.6V to -0.6V. If the diodes are reversed, -ve output occurs.

54. An inverting amplifier can be converted into an ideal half-wave rectifier by adding
two diodes as shown in Fig. 4.11 (a). When vi is positive, diode D1 conducts
causing v0A to go to negative by one diode drop (~ 0.6 V). Hence diode D2 is
reverse biased. The output voltage v0 is zero, because, for all practical purposes,
no current flows through Rf and the input current flows through D1.
For negative input, i.e., vi < 0, diode D2 conducts and D1 is off. The negative input
vi forces the op-amp output v0A positive and causes D2 to conduct. The circuit
then acts like an inverter for Rf =R1 and output v0 becomes positive.
The input, output waveforms are shown in Fig. 4.11 (b). The op-amp in the circuit
of Fig. 4.11 (a) must be a high speed op-amp since it alternates between open
loop and closed loop operations. The principal limitation of this circuit is the slew
rate of the op-amp. As the input passes through zero, the op-amp output voA must
change from 0.6 V to - 0.6 V or vice-versa as quickly as possible in order to
switch over the conduction from one diode to the other. The circuit of Fig. 4.11(a)
provides a positive output. However, if both the diodes are reversed, then only
positive input signal is transmitted and gets inverted. The circuit, then provides a
negative output.

55. Full-wave Rectifier
Fig. 4.12 (a) Precision full wave rectifier, (b) Equivalent circuit for vi > 0; D1 is on
and D2 is OFF; op-amp A1 and A2 operate as inverting amplifier
A full wave rectifier or absolute value circuit is shown in Fig. 4.12 (a). For positive
input, i.e. vi > 0, diode D1 is on and D2 is off. Both the op-amps A1 and A2 act as
inverter as shown in equivalent circuit in Fig. 4.12 (b). It can be seen that v0 = vi

56. For negative input, i.e. vi< 0, diode D1 is off and D2 is on. The equivalent circuit
is shown in Fig. 4.12 (c). Let the output voltage of op-amp A1 be v. Since the
differential input to A2 is zero, the inverting input terminal is also at voltage v.
KCL at node 'a' gives
The equivalent circuit of Fig. 4.12 (c) is a non-inverting amplifier as shown in
Fig. 4.12 (d). The output v0 is,
(4.31)
(4.32)
Hence for vi < 0, the output is positive. The input and output waveforms are
shown in Fig. 4.12 (e). The circuit is also called an absolute value circuit as
output is positive even when input is negative. For example, the absolute
value of | +2 | and | -2 | is +2 only. It is possible to obtain negative outputs for
either polarity of input simply by reversing the diodes.

57. Fig. 4.12 (c) Equivalent circuit for v, < 0, (d) Equivalent circuit of (c)
Fig. 4.12 (e) Input and output waveforms

58. LOG -AMPLIFIER

59. Log Amplifiers
The basic log amplifier produces an output voltage as a
function of the logarithm of the input voltage; i.e.,
Vout = -K ln(Vin), where K is a constant.
Recall that the a diode has an exponential characteristic up to
a forward voltage of approximately 0.7 V. Hence, the
semiconductor pn junction in the form of a diode or the base
emitter junction of a BJT can be used to provide a logarithm
characteristic.

61. There are several applications of log and antilog amplifiers.
•Antilog computation may require functions such as In x,
log x or sinh x. These can be performed continuously with
log-amps.
•One would like to have direct dB display on digital
voltmeter and spectrum analyzer. Log-amp can easily
perform this function.
•Log-amp can also be used to compress the dynamic
range of a signal.
Log Amplifier
The fundamental log-amp circuit is shown in Fig. 4.34 (a)
where a grounded base transistor is placed in the feedback
path. Since the collector is held at virtual ground and the base
is also grounded, the transistor's voltage-current relationship
becomes that of a diode and is given by,

63. Since, IC = IE for a grounded base transistor
IS = emitter saturation current = 10-13 A
k = Boltzmann's Constant
T = absolute temperature (in 0K)
Since IE = IC
Taking natural Log on both sides,
we get
From Fig.4.18(a)

64. The circuit, however, has one problem.
•The emitter saturation current IS varies from transistor to
transistor and with temperature. Thus a stable reference
voltage Vref cannot be obtained.
•This is eliminated by the circuit given in Fig. 4.18 (b). The
input is applied to one log-amp, while a reference voltage is
applied to another log-amp. The two transistors are
integrated close together in the same silicon wafer. This
provides a close match of saturation currents and ensures
good thermal tracking.

65. V1
V2 Thermistor

66. Assume, Is1 = Is2 = Is (4.39)
and then, V1
=
4.41
4.42
4.43
4.44
Thus reference level is now set with a single external voltage source. Its dependence on
device and temperature has been removed. The voltage Vo is still dependent upon
temperature and is directly proportional to T. This is compensated by the last op-amp
stage A4 which provides a non-inverting gain of (1 + R2/RTC ). NOW, the output voltage is,
Where RTC IS A PTC THERMISTOR.

67. Vin is converted in to current Ic = IEBOeVin/K

69. Antilog Amplifier
The circuit is shown in Fig. 4.19. The input V; for the antilog-amp is fed into the
temperature compensating voltage divider R2 and RTC and then to the base of
Q2. The output Vo of the antilog-amp is fed back to the inverting input of A1
through the resistor R1. The base to emitter voltage of transistors Ql and Q2 can
be written as

72. OP-AMP COMPARATORS

73. Non-Inverting Comparator.
(b) Input and Output wave-
forms when Vref is +ve
(c) Input and Output wave-
forms when Vref is -ve

74. Inverting Comparator.
b) Input and Output Wave
Forms when Vref is +ve and
c) Input and Output Wave
Forms when Vref is –ve

75. WINDOW COMPERATOR
Used in A.C Voltage Stabilizers.

77. Input(Volts) LED3 LED2 LED1
Less than 2V ON OFF OFF
Less than 4V &
More than 2V
OFF ON OFF
More than 4V OFF OFF ON
If Vcc = 6V

78. Fig. 5.5 (a) Zero crossing detector and (b) Input and output waveforms
(a)
(b)

79. (a) Inverting Schmitt Trigger circuit (b)} (c) and (d) Transfer Characteristics of
Schmitt Trigger
INVERTING SCHMITT TRIGGER
The input voltage vi triggers the output vo every
time it exceeds certain voltage levels, VLT & VUT
If Vref = 0, then the voltage at the junction of R1
& R2 will form will determine VUT & VLT .
If Vi < VLT, Vo = +Vsat
Vi > VLT, Vo = -Vsat
ZERO
if Vref
is
Zero.

81. (a) Input and Output waveforms of Schmitt Trigger and (b) Output v0 versus Vi
plot of the hysteresis voltage.
If a sine wave frequency f=1/T is applied, a symmetrical square wave is obtained
at the output. The vertical edge is shifted in phase by  from zero crossover
Where sin  = VUT/Vm and Vm is the peak sinusoidal voltage.

82. NON-INVERTING SCHMITT TRIGGER
The input is applied to the non-inverting input terminal of the op-amp. To understand the
working of the circuit, let us assume that the output is positively saturated i.e. at +Vsat.
This is fedback to the non-inverting input through R1. This is a positive feedback.
Now though Vin is decreased, the output Continues its positive saturation level unless and
until the input becomes more negative than VLT. At lower threshold, the output changes
its state from positive saturation + Vsat to negative saturation - Vsat. It remains in negative
saturation till Vin increases beyond its upper threshold level VUT.
Now VA = voltage at point A =IinR2 = VUT
As op-amp input current is zero, I in entirely passes through R1.

84. Chattering can be defined as production of multiple output transitions the input
signal swings through the threshold region of a comparator. This is because of
the noise.
Eliminates Comparator Chatter.

85. S.No. Schmitt Trigger. Comparator.
1. The feedback is used. No feedback is used.
2. Op-amp is used in closed loop mode. Used in open loop mode.
3. No false triggering. False Triggering.
4. Two different threshold voltages exists as
VUT & VLT
Single reference voltage Vref or –Vref.
5. Hysteresis exists. No Hysteresis exists.
Comparison.

86. Square & Triangular
waveform generation

87. Square Wave Generator

89. Vref = β Vsat
Let V0 initially be + Vsat. The capacitor charges
through R to + β Vsat. Then V0 goes to – Vsat . The
cycle repeats and output will be a Square Wave.
Where β = R2/(R1+R2)

90. VSat
+ _
Triangular/rectangular wave generator.

91. Operation of the Circuit
Let the output of the Schmitt trigger is + Vsat. This forces current + Vsat/R1
through C1, charging C1 with polarity positive to left and negative to right.
This produces negative going ramp at its output, for the time interval t1 to t2.
At t2 when ramp voltage attains a value equal to LTP of Schmitt trigger, the
output of Schmitt trigger changes its stage from
+ Vsat to -Vsat,
Now direction of current through C reverses. It discharges and recharges in
opposite direction with polarity positive to right and negative to left. This
produces positive going ramp at its output, for the time interval t2 to t3. At t3
when ramp voltage attains a value equal to UTP of Schmitt trigger, the
output of Schmitt trigger changes its state from - Vsat to + Vsat and cycle
continues.
The circuit acts as free running waveform generator producing
triangular and rectangular output waveforms.

92. Vo’ =+Vsat = Vin
-Vramp
+ -
0V

93. Vo(PP) = 2Vramp
Vin = VSat
Substitute V(pp) from Eqn. (4)