Hey friends, welcome to the Kohiki ALL ABOUT ELECTRONICS. So, in this article, we will see, how to use an op-amp as the integrator.
Op-Amp Integrator Now, in the earlier Articles of an op-amp, we have seen how to use an op-amp as a summing amplifier and differential amplifier. And in this configuration, so far we have used resistors with this op-amp.
So, now suppose, if we use the reactive elements like inductor or capacitor along with this op-amp then we can use this op-amp as integrator or differentiator.
Op-Amp as an integrator (Derivation)
Op-Amp Integrator So, in this article, we will see, how to use this op-amp as an integrator by using the resistors and capacitors along with this op-amp. So, first of all, just look at this inverting op-amp configuration.
Op-Amp Integrator And in this configuration, the feedback resistor is connected between the output terminal and the inverting input terminal.
Op-Amp Integrator So, now suppose, if we replace this feedback resistor with a capacitor then the same circuit can be used as an integrator.
So, let us understand, how this circuit will act as an integrator.
So, in this circuit let’s assume that the current that is entering into this resistor R is let’s say In.
Op-Amp Integrator And the current that is going out of this node is let’s say current Ic, that is the current which is flowing through this capacitor.
Op-Amp Integrator And as this capacitor is connected in the feedback, so let’s say this capacitor is Cf.
So, now, here, because of the feedback, this node will act as a virtual ground.
So, at this node, the potential will be equal to zero.
So, let’s apply KCL at this node.
So, if we apply KCL then we can write it as I in that is equal to Ic, which is the current that is flowing through this capacitor Cf.
Now, this current Iin will be equal to Vinminus zero divide by this resistor R.
Op-Amp Integrator And that will be equal to this current Ic. Now, if you know about the capacitor current then the current IC can be given by the expression, C times dVc/dt. Where Vc is the voltage across this capacitor.
Now, we have already seen this expression in the earlier Article on transient analysis right! But if you don’t know about it then let us quickly go through it. Op-Amp Integrator So, the charge across the capacitor can be given by the expression C times the voltage across the capacitor.
So, if you take the derivative of this expression then we will get, the rate of change of charge that is equal to current, that is equal to times dVc/dt.
Here we are assuming that the capacitance is not changing with time.
So, in this way, we got this expression.
So, now let’s use this expression and put this value in this expression.
So, if you put this value then we will get, Vin divide by R, which is equal to Cf times dVc/dt.
Where Vc is the voltage across this capacitor.
So, we can write this expression as Cf times/dt (0-Vout) Which will come out as Vin divide by R that is equal to (-Cf )times d(vout)/dt.
So, if we rearrange this expression, then we can write it as d(Vout)/dt, which is equal to [-1/(R*Cf)]*VinAnd if we take integration on both sides then we will get the output voltage as Vout that is equal to -1 divide by R*Cf times integration of the input voltage.
Here we are assuming that we have no output voltage at this terminal before we have started the integration. Suppose we have initial voltage at this terminal before the integration then we have one more term that is equal to Vout (0+).
Op-Amp Integrator That is the initial output voltage before we have started the integration. So, this is the expression of the output voltage in terms of the input voltage. Op-Amp Integrator So, as you can see here, the output voltage is the integration of the input voltage. And here this term R*Cf represents the integration time of this integrator. So, using this circuit we can integrate any input signal.
How is an op-amp used as an integrator: The operational amplifier integrator is an electronic integrated circuit. Based on an operational amplifier (operational amplifier), it can perform mathematical operations that integrate time; that is, its output voltage is proportional to the input voltage integrated over time.
How does an integrator circuit work: The integrator circuit outputs the integral of the input signal in a frequency range according to the circuit time constant and the bandwidth of the amplifier. The input signal is applied to the inverting input, so the output is inverted with respect to the polarity of the input signal.
What is integrator and differentiator: The differentiator circuit produces a constant output voltage for a constantly changing input voltage. The integrator circuit generates a stable output voltage for a constant input voltage.
The output of Integrator for the different input signals
So, suppose if you apply 1V of DC signal at the input then at the output you will get this signal. So, as you can see initially, the output voltage will increase in a negative direction and at a certain point, the output voltage will get saturated.
Because the output voltage cannot go beyond the saturation voltage right!! Now, here the rate at which this output voltage reaches the saturation point depends upon the value of this R and feedback capacitor Cf. Op-Amp Integrator So, by changing the value of this R and Cf, we can change the slope of this output signal.
What is the output of an integrator: The output of the integrator is 180° out of phase with respect to the input because the input is applied to the inverting input of the op-amp. An integrator circuit is usually used to generate a ramp wave from a square wave input.
What is the output of the integrator when input is a square wave: If a square wave is provided as the input of an integrating amplifier, the resulting output will be a triangular wave or a sawtooth wave. In this case, the circuit is called a ramp generator. In a square wave, the voltage level changes from low to high or from high to low, which causes the capacitor to be charged or discharged.
What does an Integrator do to a signal: The integrator circuit outputs the integral of the input signal in a frequency range according to the circuit time constant and the bandwidth of the amplifier. The input signal is applied to the inverting input, so the output is inverted with respect to the polarity of the input signal.
What is the output differentiator and integrator: The differentiator circuit produces a constant output voltage for a constantly changing input voltage. The integrator circuit generates a stable output voltage for a constant input voltage.
Similarly, if we apply the sinusoidal signal at the input, that is sin(wt), then at the output, we will get a cosine signal. That is Cos(wt).
Op-Amp Integrator likewise, if you apply the square wave at the input, then at the output you will get a triangular wave. Now, here the amplitude of the output signal depends upon the value of R and Cf.
Op-Amp Integrator And that will be get cleared to you when we will take a few examples based on this integrator circuit. So, now so far we have seen this simple integrator circuit and we understood that using Op-Amp Integrator this circuit we can integrate any input signal.
But in real life, if you use this integrator, then you may not get the desired output which you should get using the theoretical calculations Op-Amp Integrator.
Limitations of the simple integrator circuit
So, let’s understand the limitations of this simple integrator circuit, and let’s also see how we can overcome these limitations. So, to understand that let’s write the output voltage in terms of the impedance of this capacitor and the resistor. So, now we know that the reactance of this capacitor can be given by the expression 1 divide by wC right !! Which is equal to 1 divided by (2 Pi times f*C).
So, we can write the output voltage Vout as-(Xc)divide by the R times the input voltage Vin.
Or we can say that the gain of this op-amp will be equal to Xc divide by R.
that is equal to -1 divided by (2 pi times*C*F) So, from this expression, you can see that zero Hertz or at very low frequencies, the gain of the op-amp will be very high and ideally, it will be equal to infinite.
But practically the gain of the op-amp will be limited by the open-loop gain of the op-amp.
So, at zero Hertz, the gain of the op-amp will be equal to the open-loop gain of the op-amp.
Because at zero Hertz, or at DC level, this capacitor will act as an open circuit.
because of that, the input voltage will be get amplified by the open-loop gain of the op-amp.
So, the output will always get saturated. And as the frequency increases, the reactance of this capacitor will reduce.
because of that, the output voltage will also reduce.
So if you see the frequency response of a simple integrator, then it will look like this. So, at zero Hertz or at DC level, the gain of this integrator will be equal to the open-loop gain of this op-amp. And this open-loop gain used to be in the range of 10 to the power 5 to the 10 to the power 6 Op-Amp Integrator.
So, in decibel, we can say that it is equal to 100dB. And as the frequency increases, the gain will reduce. Now, suppose if you are using this op-Amp at let’s say this frequency, then you should not face the problem of the very high gain of this integrator.
But even if you use the op-amp at this frequency, then also it is quite possible that your output will get either distorted or saturated. Now, the reason for this saturation or distortion is the input offset voltage of the op-amp.
Op-Amp Integrator Because in real-life scenarios, the op-amps which we are using are not ideal op-amps. And every practical op-amp has some sort of input offset voltage. So, even though this input off-set voltage is of a few million volts. But still, it can saturate the output of this op-amp.
So, let’s say a few million volts of input offset voltage is present at this terminal. So, this input offset voltage will be get integrated by this capacitor. And eventually, the output of this op-amp will go towards saturation.
What are the limitations of the basic integrator: Due to this error voltage, the output waveform may be distorted. Another limitation of the ideal integrator is its small bandwidth. Therefore, an ideal integrator can only be used for a small input frequency range. Due to all these limitations, an ideal integrator is not actually used.
How can you overcome the limitations of a basic integrator: A practical integrator overcomes the limitations of an ideal integrator using resistors Rf and Cf in parallel. The basic circuit is shown in the figure. A compensation resistor has been added to compensate for the effect of the bias current. The resistance Rf will reduce the low-frequency gain of the operational amplifier.
What are the problems in an ordinary op-amp integrator: The problem of the ordinary operational amplifier differentiator is instability and high-frequency noise. In the actual differentiator, a resistor is added in series with the capacitor at the input, and a capacitor is connected in parallel with the resistor in the feedback circuit to eliminate the above-mentioned problems.
Practical Op-Amp integrator
So, in this way, even though you are operating at a higher frequency, but because of this input offset voltage, it is quite possible that your output will be get saturated. So, we can avoid this problem by introducing one more resistor in the feedback in parallel with this capacitor. So, in this way, by introducing one more resistor, in parallel with this capacitor, we can avoid the problem of saturation of the output voltage.
Because now if you see at low frequencies this capacitor will act as an open circuit.
So, at DCvoltage or at low frequencies, the gain which is being seen by this op-amp will be equal to -Rf/R1.
So, even though few, millivolts of input offset voltage is present at the input terminal, it will only get multiplied by this factor.
So, in this way, we can avoid the problem of saturation.
So, if you see any practical integrator, you will find this feedback resistor in parallel with this capacitor.
So, moreover that if you see any practical integrator, you will also find the resistor which is being connected at the non-inverting terminal.
Op-Amp Integrator So, basically, this resistor is connected to avoid the effect of input bias currents. So, we will see more about this input offset voltage and the input bias currents in a separate Article.
So, now because of this feedback resistor Rf, if you see the frequency response of any practical integrator, then it will look like this. So, as you can see here, at low frequencies, Op-Amp Integrator the gain which is being seen by the op-amp will be equal to constant and this is equal to Rf divide by R1 (V/F Ratio Is Kept Constant in VFD).
As we go beyond a certain frequency, then you will see a reduction in the output.
Op-Amp Integrator if you see the response of this integrator, it) is very similar to the response of the low pass filter.
Op-Amp Integrator, in fact, will act as a low pass filter. Op-Amp Integrator So, let us say here the 3 dB frequency of the cut-off frequency of this integrator is fL.
So, now if you apply any input signal, which is having a frequency lesser than this cut-off frequency fL, then op-amp will act as inverting op-amp and it will provide the constant gain of -Rf divided by R1.
So, to use this op-amp as an integrator input signal should have a frequency larger than this cut-off frequency fL.
Op-Amp Integrator And this cut-off frequency fL can be given by the expression 1 divide by 2 Pi times Rf *Cf.
So, as you go beyond this cut-off frequency fL, you will see a reduction in the output voltage.
Op-Amp Integrator And at one particular frequency, the gain of this integrator will be equal to zero dB. So, let’s say this frequency is zero dB frequency of let’s say f0.
So, this f0 can be represented as 1 divided by 2 Pi times R*Cf.
So, with proper integration of the input signal, the input frequency should be larger than this cut-off frequency fL.
Op-Amp Integrator And for proper integration, this signal frequency should be at least 10 times this cut-off frequency fL.
So, to integrate any in-out signal properly, the input signal frequency should be in between this cut-off frequency fL and zero dB frequency.
Well practically you can also go beyond this zero dB frequency, but if you go beyond this zero dB frequency then you will see the attenuation in the output signal.
So, for the proper integration of any input signal, it should lie between this cut-off frequency fL and f0.
So, now let’s take some examples based on this integrator so that the concept will be cleared to you.
So, here we have a practical integrator circuit and in this circuit, we have been asked to low frequency limit for this integration. So, here the value of this feedback capacitor is 10 nF and the value of this feedback resistor is 100K, while the value of this resistor R is 1Kilo ohm.
so, we know that the expression of this cut-off frequency fL is equal to 1divide by 2 Pi times Rf*Cf. So, if we put the value of this Rf and Cf, then the value of cut-off frequency fL will come out as 159 Hz.
Op-Amp Integrator So, for proper integration, the input signal frequency should be larger than this cut-off frequency fL. And for proper integration, it should be atleast10 times this cut-off frequency fL.
So, in this way, based on your input signal range, you can design your integrator. So, now let’s see the second example. So, in this example, the circuit is the same circuit that we have seen in the last example, but nowhere, the input signal is the sinusoidal signal. And if you see it here, the frequency of the input signal is 5KHz.
Op-Amp Integrator Now, we have already seen that the cut-off frequency of this integrator is 159 Hz. And for proper integration, the signal frequency should be at least 10 times this cut-off frequency.
Which comes out as, 1.59 kHz. Now, here if you see, the input frequency, it is 5KHz. so, this signal will properly get integrated. So, let’s find the expression for the output voltage. Op-Amp Integrator So, we know that the output voltage Vout can be given by the expression -1 divide by R*Cf, multiplied by the integration in it.
Now, here the value of R is 1Kilo ohm and the value of Cf is 10 nF. So, let’s put all the values into this expression. So, now if you put all these values, and integrate this signal, then you will get 10 t the power 5, divide by 2 Pi times 5000, multiplied by cos (2*Pi*5000t) So, if you simplify it then you will get3.18 *cos ( 2*Pi*5000t) So, this will be the output voltage Vout after the integration.
So, as you can see here, the amplitude of the output voltage depends upon the value of R, Cf, and input signal frequency. So, now let’s see one more example-based on this integrator circuit.
So, in this example, the circuit is the same, but now the input signal has been changed from the sinusoidal signal to the squire wave. And we need to find the output voltage at this end.
Op-Amp Integrator Now, we know that whenever we apply any squarewave to the integrator, the output voltage will be equal to the triangular wave. So, let’s find the amplitude of this triangular wave, and let’s also see whether this signal fulfills the criteria of our integration.
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Op-Amp Integrator So, if you see the output waveform, the output waveform will look like this. But just calculate the value of this output voltage and do let me know that value in the comment section below.
So, I hope in this Article, you understood, how to design the integrator using the op-amp or Op-Amp Integrator.
So, if you have any questions or suggestions, do let me know in the comment section below. If you like this article, hit the like button and subscribe to the kohiki for more such Articles.