Hey friends, welcome to Kohiki.com – ALL ABOUT ELECTRONICS. So, in this article, we will see, how to use an op-amp as a differentiator With an **op-amp differentiator solved problems**.

- Op-Amp Differentiator

## Op-Amp Differentiator

**Op-Amp as a Differentiator**

Now, here do not get confused between the differential amplifier and the differentiator. So, in the case of this differential amplifier, this differential amplifier (Op-Amp Differentiator) the difference between the two input signals, while in the case of this differentiator, the output will be the differentiation of the input signal.

So, using this differentiator circuit we can get an idea of the rate at which the input signal is changing.

So, now suppose if you apply the DC signal to the differentiator, then the output of this differentiator will be equal to zero. Because here, the signal is not at all changing with the time. (Op-Amp Differentiator) And on the other end, suppose if you apply the square wave which is having a very fast rise time, then at the output you will get spikes where this very fast transition is happening.

- So, in this way, this
**differentiator circuit can be used to find the high-frequency components**of the input signal. Or we can use this differentiator circuit for edge detection purposes. (Op-Amp Differentiator) So, now let’s understand how we can design this differentiator circuit using the op-amp. - So, now in the last Article of the op-amp, we have seen that how to use this op-amp as an integrator. So, now in that circuit, suppose if we interchange the position of this capacitor and the resistor, then the same circuit can be used as a simple differentiator.

**Derivation of Op-Amp Differentiator Circuit**

**So, now let’s understand how this circuit will act as a differentiator**. So, let’s say to this circuit we have applied some input signal and the current Ic is flowing through this capacitor. (Op-Amp Differentiator) (**op-amp differentiator solved problems**) And let’s say, current If is the current which is flowing through this resistor Rf. And here, we are assuming that our op-amp is the ideal op-amp.

**So, no current is flowing inside this op-amp.**Now, because of this negative feedback, this node will act as a virtual ground. So, the potential at this node will be equal to zero Volt.**So, now let’s apply KCL at this node**, and let’s find the output voltage in terms of the input voltage. So, applying KCL at this node we can write, current Ic that is equal to current If. Now, in the last article, we have already seen that this capacitor current Ic can be given by the expression, C times dVc/dt.

Where Vc is the voltage across this capacitor. (Op-Amp Differentiator) So, we can write this expression as C dVc/dt, which is equal to this current If. And this current If, will be equal to 0 minus V out divided by this feedback resistor Rf. Now, here this capacitor voltage Vc is nothing but the input voltage Vin. Because this node will act as a virtual ground.

**so, we can write this expression as C times(vin)/dt,**which is equal to minus Vout divided by Rf. And if we simplify it then we will get output voltage (Op-Amp Differentiator) Vout as minus Rf times this capacitor C, multiplied by this d(vin)/dt.- So, in this way, if you see this circuit, in this
**circuit the output voltage**will be the differentiation of the input signal. (Op-Amp Differentiator) And it will get multiplied by this factor. So, using this differentiator circuit, we can differentiate the different types of input signals.

**Output of differentiator for the different input signals**

So, as we have seen so far, if we apply the DC signal, then the output will be equal to zero.

And suppose, if we apply the square wave to the (Op-Amp Differentiator) differentiator then in the output we will get the spikes at the transition points. Similarly, let’s say, if we apply the sinusoidal signal that is sin(wt), then at the output we will get -cos(wt).

And likewise, (Op-Amp Differentiator) if we apply the triangular wave at the input, then at the output of this differentiator, we will get a square wave pulse. And the amplitude of these different output signals depends upon the value of this feedback resistor Rf and the value of capacitor C.So, we will see more about it when we will take different examples based on this differentiator.

**Limitations of the simple differentiator circuit**

So, now let’s see some limitations of this simple differentiator. (Op-Amp Differentiator) That means, suppose if you are designing the practical differentiator, then what are the things which you need to take care of.

So, now before we understand, let’s write this output voltage Vout in terms of the impedance of this resistor and the capacitor. so, we can write this output voltage Voutas minus Rf divided by Xc times this input voltage Vin.

Where this Xc is the reactance of this capacitor. (Op-Amp Differentiator) And it can be given by the expression 1 divided by 2*Pi times fc. So, if we put this value, then we can write out divided by vin, that is the gain of this differentiator will be equal to minus Rf times 2*Pi*f*c.

**So, as you can see, the gain of this differentiator is directly proportional to the frequency f**. So, as the frequency increases, the gain of this differentiator will also increase.

**Frequency Responce Of Simple Differtiator**

So, now if you see the frequency response of this simple differentiator, then it will look like this. So, at zero frequency, the gain of this differentiator will be equal to zero. (Op-Amp Differentiator) And as the frequency increase, the gain of this differentiator will also increase. Now, at one particular frequency, the gain of this differentiator will be equal to zero dB. And that frequency is known as zero dB frequency or simply f0.

So, this f0 will be given by the expression,1 divided by 2*Pi times, this Rf multiplied by this capacitor C.So, at this frequency, the gain of this differentiator will be equal to 0 dB. And as we go beyond this 0 dB frequency, then the gain of this differentiator will also increase. Now, here, the gain of this differentiator can not increase indefinitely.

And the gain of this differentiator will be restricted by the (Op-Amp Differentiator) open-loop gain of the op-amp. So, the maximum gain which you can attain with this differentiator will be the intersection point of this differentiator and the open-loop gain response of the op-amp.

- So,
**if you see the frequency response of the simple differentiator**, then the simple differentiator response will look like this. Now, here in the case of this differentiator circuit, at 0Hz or at DC level, the gain of this differentiator is zero.

**So, here, we will not face the problem of input offset voltage**that we have faced during the integrator circuit. But in the case of this differentiator circuit, as the frequency will increase, the gain of this differentiator will also increase. So, we can say that this differentiator is very sensitive to high-frequency noise.**So, even if your input signal frequency is a low-frequency signal,**but still the high-frequency noise will see more gain than this input signal frequency. Apart from that, if you see the input impedance of this circuit, then the input impedance will be equal to the reactance of this capacitor. And as the frequency increases, the reactance of this capacitor will reduce. And because of that, the input impedance that is seen through the source will also reduce.**So, we can say that there are two limitations**of this simple differentiator. First, as the frequency will increases, the input impedance of this circuit will reduce. And second, this simple differentiator is very sensitive to high-frequency noise. So, now let’s see, how we can avoid these problems.

**Practical Op-Amp differentiator**

So, now suppose, if we introduce one more resistor in series with this capacitor, then we can avoid the problem of very low impedance and high-frequency sensitivity. (Op-Amp Differentiator) So, now, if you see this circuit, the input impedance of this circuit not only depends upon the value of the capacitor, but it also depends upon the value of a resistor.

Or precisely we can say that the input impedance of this circuit will be equal to under root R square plus Xc square. And as the frequency will increases, the value of this capacitive reactance will reduce. But because of this resistor R, your input impedance of the circuit will be always greater than the value of resistor R.

**So, in this way, we can ensure that even though we are operating at high frequencies**, the input impedance that is seen through the source will be equal to R. (Op-Amp Differentiator) Apart from that if you see, at high frequencies the gain of this signal will be equal to -Rfdivided by R. Because at high frequencies the value of this capacitive reactance Xc will be very low.**So, in this way, by introducing this series resistor**, we can restrict the high-frequency gain of this differentiator.And moreover, that, suppose if we introduce this feedback capacitor Cf then we can further increase the stability of the output at high frequencies. So, this is how the practical differentiator circuit will look like.

- So, if you closely look at this circuit, this circuit contains two low pass filters. So, this resistor-capacitor pair will act as one low pass filter and this feedback resistor and capacitor pair will act as another lowpass filter. And because of that, we will have two cut-off frequencies.

**So, here the cut-off frequency**because of this input resistor R and C is equal to 1 divided by 2*Pi times R *C. And similarly, the cut-off frequency because of this feedback resistor Rf and Cf will be equal to 1 divided by 2*Pi times Rf*Cf. Now, here, let’s assume that f2 is greater than f1.- So, now let’s see, how the response of this practical differentiator will look like.
**So, here, as the input signal increases**, the gain of this differentiator will also increase at the rate of 20 dB/decade. And once you reach the first cut-off frequencyf1, then because of this low-pass filter action, the gain will also reduce at the rate of -20dB/decade.**So, this +20 dB/decade and -20 dB/**decade will cancel out each other.**So, because of that, you will see a flat response**. And once again, when you reach the cut-off frequency f2, then you will see the reduction in the gain at the rate of -20 dB/decade. (Op-Amp Differentiator) So, if you see the frequency response of practical differentiator, then it will look like this.

**Frequency Responce Of Practical Op-Amp differentiator**

So, now when f1=f2, in that case, if you see the response of this practical differentiator then it will look like this. So, to the cut-off frequency f1, the gain will increase at the rate of 20 dB/decade and if you go beyond this cut-off frequency then the gain will decrease at the rate of -20 dB /decade.

**So, if you want to use this circuit as a differentiator**, then your input signal should be between this 0 dB frequency f0 and this cut-off frequencyf1.**So, whenever this f1 is not equal to f2**, in that case, your input signal frequency should be between this f0 and whatever is minimum between this f1 and f2. (Op-Amp Differentiator) Now, for accurate differentiation, your input signal frequency fs should be at least 10 times less than the upper cut-off frequency of this differentiator. And this point will get clear to you when we will take one example.

So. let’s take one example based on this practical differentiator.

**op-amp differentiator solved problems**

So Here We Are Discussing With Some Example Based On **op-amp differentiator solved problems**

**Example 1**

op-amp differentiator solved problems

So, here we have given the circuit for the practical differentiator and we have been asked to find the useful range for the differentiation. So, if you see here, the value of R is 100ohm, the value of C is 10 nF, and the value of Cf is 100 pF.

While if you see the value of Rf, that is equal to 5-kilo-ohm. (op-amp differentiator solved problems) So, here we will have two cut-off frequencies,f1 and f2. So, let’s, first of all, find the f1 and f2and based on that we will decide the useful range of the differentiation.

So, the cut-off frequency f1 because of this input resistor R and C will be equal to 1 divided by (2*Pi)times R*C. And if we put the value of R as 100 ohms and the value of C as 10 nF, (op-amp differentiator solved problems) then the value of this first cut-off frequency f1 will be equal to 159.2 kHz.

Similarly, the second cut-off frequency because of this feedback resistor and the capacitor will be equal to 1 divided by 2* Pi times*Cf (op-amp differentiator solved problems).

- And if we put the value of this Rf and Cf, then the value of f2 will come out as 318.3 kHz.
- So, in this way, we found the value of f1as 159.2 KHz and the value of f2 as 318.3 kHz.
- So, here f1 is less than f2. So, for proper differentiation, your input signal fs should be less than this cut-off frequency f1.
- And for accurate differentiation, your signal differentiation fs should be at least 10 times less than this upper cut-off frequency f1 (op-amp differentiator solved problems).

So, here, for proper differentiation, your signal frequency should be less than 15.9 kHz. So, now based on this let’s see some more examples of this practical differentiator.

op-amp differentiator solved problems

**Example 2**

op-amp differentiator solved problems

Now, in this example, we have been given the same circuit but now our input signal is a sinusoidal signal. And we have been asked to find the output signal. (op-amp differentiator solved problems) So, for this circuit, we have already found the upper cut-off frequency. And the value of this upper cut-off frequency is 159.2 kHz.

- if you see here, the signal frequency, the signal frequency fs is 3 kHz.
- So, it is way below this upper cut-off frequency.
- So, we can ensure that your input signal will properly get differentiated.
- Now, let’s find the output signal. So, the output signal Vout can be given by the expression, – Rf times C, multiplied by the dVin/dt.
- That is the differentiation of the input signal. Now, here the value of Rf is 5 kilo-ohm, while the value of C is 10 nF.

And our input signal is 2 times sin(2*pi*3000t)So, now if we put all these values then we can write output voltage Vout as minus 10to the power minus 4, d/dt sin(2*Pi*3000t) And if we take the (op-amp differentiator solved problems) differentiation then we can write, minus 10 to the power minus 4, multiplied by (2*Pi*3000*cos(2*Pi*3000t)And if we further simplify it then our output voltage Vout will be equal to -1.885*cos(2*pi*3000t)So, this will be the output voltage.

**So, as you can see here, if we apply the sinusoidal signal**, then the output of this circuit will be equal to minus of cosine signal. So, here the output signal amplitude depends upon the value of this feedback resistor Rf, capacitor C, and the input signal frequency fs.**So, if we change any of these three parameters**, then based on that your output signal amplitude will also change. (op-amp differentiator solved problems) So, now suppose if we operate the same circuit at 6 kHz, then at 6KHz, your output will be getting doubled.

Now, one more thing which I need to tell you is that you can apply this expression to this practical differentiator only when your input signal frequency is way below this upper cut-off frequency. So, now based on this let’s see one more example based on this practical differentiator.

op-amp differentiator solved problems

**Example 3**

op-amp differentiator solved problems

So, nowhere, the circuit is the same but instead of the sinusoidal signal, we are applying the triangular signal. And the frequency of this triangular signal is 4 kHz. So, for the given circuit, we have already found that the upper cut-off frequency is 159.2 kHz. While here our signal frequency fs is 4KHz. (op-amp differentiator solved problems) So, as our signal frequency is far below this upper cut-off frequency, so it is ensured that our input signal will properly get differentiated. So, now here let’s find the waveform and the amplitude of the output signal. So, here the frequency of the input signal is 4 kHz.

- So, in the time period, if you see, it will be equal to 250 microseconds. That means this time period will be equal to 250 microseconds. (op-amp differentiator solved problems) So, we can say that the halftime will be equal to 125 microseconds.
- Now, in the first 125 microseconds, the input signal is varying from -3 V to +3V. So, we can say that the slope of the input signal for the first 125 microseconds is 6V divided by 125 microseconds.
- That means, 48000 V/s. (op-amp differentiator solved problems) Now, we know that for the given circuit, the output voltage Vout will be equal to – Rf times C, multiplied by dVin/dt, that is the slope of the input signal.
- Now, here we know that the value of Rf is5 Kilo-ohm and the value of C is 10 nF.
- And we have already found the slope of this input signal for the first 125 microseconds.
- And that is equal to 48000 V/sec.

So, if we simplify it, then we will get the output voltage Vout as -2.4V. So, for the first 125 microseconds, if you see, our output voltage will be equal to -2.4V. So, now if you see for the next 125 microseconds, our slope is exactly the opposite.

So, if you calculate the output voltage for these 125 microseconds, then you will find the output voltage Vout as +2.4 V. So, in this way, (op-amp differentiator solved problems) if you see the output voltage, our output voltage will vary between these -2.4V and +2.4V. So, in this way using this differentiator circuit, we can perform the differentiation on any given input signal.

op-amp differentiator solved problems

**FAQ (People also ask)**

### What is an op amp differentiator?

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### How does a differentiator work?

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### Why capacitor is used in op amp?

### Why capacitor is used in differentiator?

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### Why is the differentiator called a high pass filter?

**You Tube Video**

Here Are YouTube Video Based On Op-Amp Differentiator Which Was Uploaded By EC Academy

**Last World / Summary**

So, I hope in this Article you understood how to design the differentiator using the op-amp (Op-Amp Differentiator). And whenever you are designing the practical differentiator, then what are the things which you need to take care of.

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 – for more such Articles.