Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Lesson: Applications of Operational Amplifier: Nonlinear Circuits,
Lesson-I: Integrator
Lesson Developer: Dr. Arun Vir Singh
College/Department: Shivaji College, University of Delhi
Institute of Lifelong Learning, University of Delhi
page no.1
Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Table of Contents
Applications of Operational Amplifier: Nonlinear Circuits:Lesson-I: Integrator
1.1 Nonlinear Circuits
1.2. Integrator
1.2.1 Basic Integrator
1.2.2 Input and Output Wave Forms
1.2.3 Limitations
1.2.4 Frequency Response of Basic Integrator
1.3 Practical Integrator
1.3.1 Frequency Response of Practical Integrator
1.4 Applications of Integrator
Summary
Exercise
Glossary
References
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
1.1 Nonlinear Circuits
In nonlinear circuits output of op-amp does not follow the input. Integration and
differentiation are two arithmetic operations performed by using nonlinear op-amp circuits.
In these operations op-amp is used with negative feedback. Other nonlinear applications
are comparators, voltage level detectors, zero-crossing detectors. In nonlinear applications
op-amp is generally used in open-loop configuration. In this lesson, we shall discuss
integrator.
1.2 Integrator
Integrator is a circuit that performs a mathematical operation called integration. Output
voltage of such a circuit is equal to the integral of the input. The integrator is very useful in
many applications which require the generation or processing of analog signals. Such as
triangular wave, square, and saw tooth wave generators. The most popular application is to
generate a linearly increasing or decreasing voltage, which is known as a ramp.
1.2.1 Basic Integrator
Such a circuit is obtained by using operational amplifier in the inverting configuration with
the feedback resistor Rf replaced by a capacitor, Cf. as shown in figure1.
Fig: 1. Basic integrator
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Developed by :ILLL
This is a basic integrator circuit. Input to this integrator can be time varying (sin, cosine,
triangular, and square wave) or a step/ a dc signal and it is applied at input resistor R1.
Output is taken across the capacitor. Notice that the capacitor is feedback element and
forms an RC circuit with the input resistor.
The expression for output voltage is obtained by using Kirchhoff’s current law at node N.
[1.1]
Since IB is very very small. So the current flowing through the input resistor R1 is equal to
the current flowing the feedback element /capacitor.
[1.2]
The current through the capacitor is related to voltage by the relation
[1.3]
is the voltage across the capacitor, given by
,
Current through the input resistor
[1.4]
[1.5]
But
[1.6]
Substituting Eq.[1.6] in Eq.[1.2]
[1.7]
Substituting Eqs.[1.5] and [1.3] in Eq.[1.7]
[1.8]
Substituting for
from Eq.[1.4] in Eq.[1.8]
[1.9]
By using the concept of virtual ground
in Eq.[1.9]
[1.10]
Output voltage can be obtained by integrating Eq. [1.10 ]
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
=
[1.11]
Where C is the constant of integration and is proportional to the output voltage
at time
t=0. Hence output is (i) directly proportional to negative integral of the input signal voltage.
So the circuit is an integrating circuit.
(ii) Inversely proportional to the time constant
.
1.2.2 Input and Output Wave Forms
Step Input
A step input is shown in figure 2a. Let its magnitude is A unit. Mathematically it can be
expressed as
Fig:2(a).Step input signal of magnitude (+A)and 2(b) Negative going ramp with slope (-A).
Developed by :ILLL
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
If
=1, and C=0 , then from Eq.[1.11]
[1.12]
The output is a straight line with a slope of –A. The output is shown in Fig.[2b]
Example
Consider a positive step input
=1 Volt,
to the integrator circuit
shown in Fig.[1] . Calculate the output voltage and draw the output waveforms.
Solution
Let the power supply voltage =±15 V. Calculated output for the circuit of Fig. 1 is
This is a negative ramp voltage dropping at a rate (slope) of 10,000 V/s. This ramp voltage
will drop from +15 V to -15 V in 30 V/10,000 sec or 3×
sec(3 ms). After 3 ms output
changes from +15 V to -15 V and gets saturated at -15 V. Figure 3 shows the input step
waveform (3a) and the resulting output ramp waveform (4b).
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Fig. 3 Input step waveform (3a) and output ramp waveform (3b).
Developed by :ILLL
So we can conclude if input signal is a step function, output voltage is a ramp and is
opposite in polarity to the input voltage and is multiplied by the factor
. Output is
saturated at voltage level slightly < ± supply Voltage. This is because of the voltage drop
across the op-amp.
Value Addition :FAQ
Why is the output of step input linear?
Body Text:
The capacitor charging in a simple RC circuit is not linear but is exponential. But here it is
linear. The key thing about using an op-amp with an RC circuit to form an integrator is that
the capacitor’s charging current is made constant, thus producing a straight-line (linear)
voltage rather than an exponential voltage.
Suggested Reading:
Electronics Devices and Circuit Theory by Robert. L Boylestad and L. Nashelsky,8th Edition.
Op-Amps and Linear Integrated Circuits : Ramakant A. Gayakwad, 3rd Edition
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Square Wave Input
The square wave input signal is shown in figure 4a. We can say that it is made of two step
inputs (i) a positive step between time period 0 to T/2 and (ii) a negative step for time
period T/2 to T. Mathematically it can be expressed as
Fig: (4a)Square wave input (4b) output wave triangular wave
Developed by :ILLL
As discussed earlier the output for the positive step input is a straight line with a slope of
negative A. So for the period 0 to T/2 output will be straight line with negative slope (-A).
For the period T/2 to T, output will be straight line with positive slope (+A). Mathematically
output can be expressed as
[1.13]
Output wave form is shown in figure (3b). This is called triangular wave. So integral of
square wave is triangular wave, and also shown in animation 1.
Animation1
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Animation 1: Integrated output of square wave from.
Developed by: ILLL.
Sine Wave Input
Sine wave input signal is represented by expression
[1.14]
Output is given by substituting Eq.[1.13] in Eq.[1.11] for
and C=0
Output is cosine wave and is shown along with the input signal in animation 2
Animation2
Animation 2: Integration of sine wave from.
Developed by: ILLL.
Triangular Wave Input
For triangular wave input, integrated output wave form is a sine wave and is shown in
animation 3.
Animation3
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Animation 3: Animated output waveform for triangular input wave form.
Developed by: ILLL
1.2.3 Limitations
When input voltage to the integrator is zero, (
) the integrator works as an open loop
amplifier. This is because the capacitor
acts as an open circuit to the input offset voltage
. This input offset voltage and the part of the input current charging capacitor Cf produce
the error voltage at the output of the integrator.
1.2.4 Frequency Response of Basic Integrator
As already mentioned for dc signals, the capacitor
behaves as an open circuit element
and there is no negative feedback. The op-amp thus operates in an open loop, resulting in
an infinite gain. In practice, output never becomes infinite. The plot in figure 5 shows the
frequency response of the basic integrator circuit. The gain decreases with a constant rate
over the entire range of the plot. The gain goes down 20 db for every decade (power of 10)
with
Fig.5. Frequency response of basic integrator.
Developed by :ILLL
increase in frequency .As frequency continues to increase the gain will continue to diminish
at the same rate. The frequency at which gain in decibel is zero, is called higher cut off
frequency
and is given by the equation
[1.15]
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
As frequency decrease, the gain will continue to increase.
frequencies produces a practical limit for using this circuit.
This increasing gain to lower
Value addition :FAQ
Why is decrease in gain of basic integrator with increase in frequency?
Body Text:
Gain of integrator is
We can write the components of basic integrator shown in Fig.1.in S domain as under
(S),
.where S=jω=2πf
, and
,where
where
.
and
Therefore A=
.
Magnitude of gain is
ω
ω
, so for a fixed value of
we can say that gain decreases with increase in
frequency f.
The frequency (f= at which the magnitude of the gain of basic integrator is 1 (or 0dB)
can be obtained by equating the above equation to 1.
Hence
(Cutoff frequency)
Suggested reading:
Op-Amps and Linear Integrated Circuits : Ramakant A. Gayakwad, 3rd Edition.
Linear Integrated Circuits :D.Roy Choudhury and Shail B. Jain,2nd Edition.
1.3 Practical Integrator
The limitations of ideal integrator can be minimized in practical integrator to limits the low
frequency gain and hence minimizes the variations in the output voltage circuit. In this
circuit a resistor Rf is connected across the feedback capacitor Cf to reduce the error voltage
at the output. The circuit for the practical integrator is shown in figure 6.The addition of the
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
resistor also corrects the stability and low frequency roll-off problems. Gain of this circuit if
given by ratio of
. A practical op-amp integrator approximates the characteristics of
the ideal circuit, but has a fixed gain equal to ratio of Rf and R1.
Fig:6. Practical integrator
Developed by :ILLL
Output is given by
[1.16]
1.3.1 Frequency response of Practical Integrator
Frequency response is shown in figure 7. In this plot the dc gain of the integrator is 10
(20 dB). Notice that the circuit’s gain is relatively constant until 30Hz and then starts to
decrease.
The point where the gain begins to decrease is called the cutoff frequency. This is defined
as the point where the gain is down 3 db from the gain in the pass band. A 3dB reduction in
gain corresponds to a 0.707 reduction in the output voltage from the level in the pass band.
This point is defined by the gain limiting frequency fa and is given by
[1.17]
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
After the cutoff point is reached, the gain of the circuit falls at a rate of -20 dB/decade; just
as in the ideal integrator circuit. It eventually takes the shape of the ideal integrator below
the cutoff frequency.
To use this circuit as an integrator, the lowest frequency expected to be encountered must
fall into this part of the circuit response. The lowest frequency is the frequency at which
Fig:7. Frequency response of practical integrator
Source: Self/ Developed by ILLL.
gain is 0dB and is given by
[1.18]
The circuit acts as an integrator in the frequency range fa to fb. the value of fa and in turn
and Rf Cf values should be selected such that fa<fb.
Input frequencies, f<fa passes to output without integration. So integrator acts as a low pass
filter. In this range gain (
constant.
For proper integration, the time period T of the input signal has to larger than or equal to
That is,
where
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Value Addition: Did you know
How an infinite gain of ideal integrator, gets limited to a finite value in practical integrator.
Body Text:
Gain of practical integrator shown in fig.6 in terms of frequency using S domain is
.where
Magnitude of gain is
At f=0,
i.e 20 log
, S=2πf
, where
.
Thus an infinite dc gain of ideal integrator, gets limited to
integrator.
(a finite value) in practical
When f =fa
i.e 20 log
20 log
=20 log
=20 log(0.707)+20 log
=- 3dB + d.c gain.
Thus the dc gain remains constant for all frequencies less than fa, and the gain drops by 3dB
at the frequency f = fa, which is the break frequency .And for frequency f>fa gain reduces
at the rate of 20dB/decade.For integration ,frequency response must be straight line of
slope -20 dB/decade, which is possible for the frequency f>faand f<fb..
Suggested Reading
Op-Amps and Linear Integrated Circuits : Ramakant A. Gayakwad, 3rd Edition.
Linear Integrated Circuits :D.Roy Choudhury and Shail B. Jain,2nd Edition.
Example
Let
,
and
circuit shown in Fig.[6]. Calculated values of
(square wave) for the practical integrator
,
,gain A and
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
(i) Gain limiting frequency
(ii) Cutoff frequency
, using Eq.[1.17]
, using Eq.[1.18]
(iii) Gain
Frequency response of the circuit and input and output wave forms are obtained using
probe. Frequency response is already shown in Fig.[7]. It can be seen from the response
curve, that in frequency range (f <fa) gain of integrator (20 dB) is constant up to 30 Hz.
Calculated and obtained values of , and gain are approximately equal. After fa gain
decreases at the rate of ≈20dB/decade. At fb gain is 0.
Response of the practical integrator for square wave at three different values of frequencies
is illustrated in figures 8 to 10.
(i) f<fa . f=5 Hz
Input and output wave forms are illustrated in figure 8.
Fig:8. Input and output wave form for f=5Hz < fa.
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Developed by :ILLL
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Output wave is inverted having magnitude (Vo) equal to 10V. So gain is equal to 10 or
20dB. Comparing with the input signal we can say that output wave form is same as that of
input. It is not integrated. Hence input signal is passed by the integrator.
(ii) f<fa . f=50 Hz
Input and output wave forms are shown in figure 9. At this frequency, gain decreased
slightly as observed from the response curve. It can be seen that output wave is not
perfectly square. We can say that circuit starts integrating input wave.
Fig:9. Input and output wave form for f=50Hz < fa.
Developed by :ILLL
However, magnitude of output is 10 V, which is equal to the gain of integrator. Comparing
with the input signal we can say that input signal is not integrated properly.
(iii) f>fa but < fb , f=500 Hz .
Input and output wave forms are shown in figure 10. Frequency of the input signal is
chosen, such that it lies between fa and fb ie.( fa,<f,<fb). Output wave is triangular as
expected. So we can say that input signal is integrated by the integrator. Magnitude of
output wave decreased. Output is inverted.
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
So for integration of any type of input signal, frequency should be between gain limiting
frequency and cutoff frequency.
Fig: 10. Input and output wave forms for f=500Hz, T=2ms
Developed by :ILLL
1.4 Applications Of Practical Integrator
Integrator circuits are used in following applications
(i) In analog computer
(ii) Analog-to- digital conversion (ADC)and wave shaping circuits.
(iii) In ramp generators.
(iv) In solving differential equations.
Summary: After studying this chapter, you should be able to
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator





Explain and analyze the operation of integrators.
Discuss the ideal integrator and compare it to practical integrator.
Show the various output wave forms.
Explain the limitations of ideal integrator.
Explain the frequency response curve of ideal and practical integrator.
EXERCISES
Question Number
Type of question
1
Multiple choice questions
(1) In an ideal integrator, the feedback element is a
Question Number
Type of question
(a) resistor(b) capacitor (c) zener diode (d) voltage divider
2
Fill in the blanks
(2)The rate of change of an integrator’s output voltage in response to a step input is set
by
(a) the RC time constant (b) the amplitude of the step input(c) the current through the
capacitor (d) all of these
(3) For a step input, the output of an integrator is
(a) a pulse (b) a triangular waveform (c) a spike (d) a ramp
(4) For a triangular wave input, the output of an integrator is
(a) a pulse (b) a sine wave(c) a spike (d) a ramp
(5) For proper integration, the time period T of the input signal has to
(a) larger than
(b) equal to
(c) smaller than
(d) both a and b
Correct answers
(1)
(2)
(3)
(4)
(5)
b
a
d
b
d
Type of question
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
2
Fill in the blacks
(1) Integration of a constant is -------------.
(2) Integrator acts as a ------------filter.
(3) Gain of the practical integrator decreases in the frequency range ---- to----at the rate of ----------.
(4) Output of integrator is ----------- phase with respect to input.
(5) The scale multiplier (factor) of a basic integrator is --------.
a
Correct answers
(1)
(2)
(3)
(4)
(5)
Straight line
low pass
fa to fb, 20dB/decade
out of phase
-
Question Number
Type of question
3
Subjective questions
1. Discuss the operation of basic op-amp integrator.
2. Mention the limitations of basic integrator.
3. What do you mean by gain limiting frequency?
4. What are the differences between basic and practical op-amp integrator?
5. Discuss the frequency response curve of practical op-amp integrator.
6.What are the applications of integrating circuit?
7. Which type of filter we get from integrator.
Type of question
4
Unsolved questions
1. Calculate the lower frequency limit for the integrator shown in figure 6 .Given
.
2 For an integrator circuit shown in figure 1 how long does it take for the output to reach
saturation. Given Vin= +10 mV,
and power supply voltage ±15V.
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
3.Determine the rate of change of output voltage in response to a single pulse input of
amplitude 1V, to the integrator shown in figure 1. Given
.
Solution
(1)
,
(2)
, Vin=+10 mV
,
Saturated voltage =- 15V ,
(3)
,
GLOSSARY
Integrator: A circuit that produces an output which approximates the area under the curve
of input curve.
Differentiator: A circuit that produces an output which approximates the instantaneous
rate of change of input function.
Node: A node is any junction wherein two or more two-terminal components meet.
KCL: Kirchhoff’s Current Law simply states that the currents entering a node are equal in
magnitude to the currents leaving that same node, so
Time constant: It is equal to the product of the circuit resistance (in ohms) and the circuit
capacitance (in farads).
Ramp: A linearly increasing or decreasing voltage.
Input offset voltage: The differential dc voltage required between the inputs for the
output to be zero.
Cut of frequency: This is defined as the point where the gain is down 3 db from the gain
in the pass band(where gain is constant).
Decible (db): Unit of gain. A logarithmic measure of the ration of one power to another or
one voltage to another,
Decade: When frequency changes from f1 to f2 such that f1=10f2, it is referred to as
decade.
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Applications of Operational Amplifier: Nonlinear Circuits Lesson-I: Integrator
Roll off. The decrease in voltage gain of op-amp with frequency is called as frequency roll
off.
Bode Plot: When the gain is represented in db and the frequency is plotted on a
logarithmic scale, a Bode plot is produced. Bode plots are used to determine the stability of
control systems and the frequency response of filter circuits
Virtual Ground: Means the node N shown in Fig.1.1 is at 0V but it is not mechanically
ground. So no current flows from point N to ground.
Low pass filter:A circuit which passed the low frequency.
References/ Bibliography/ Further Reading Source:
Op-Amps and linear Integrated Circuits :Ramakant A. Gayakwad,3rd Edition.
Electronic Principles :A.P. Malvino ,6th Edition.
Electronics Devices and Circuit Theory by Robert.L Boylestad and L. Nashelsky.,2nd Edition.
Electronic Devices by Thomas L. Floyd,6th Edition.
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