The basic **Op-amp Differentiator** circuit is the exact opposite to that of the Integrator Amplifier circuit that we looked at in the previous tutorial.

Here, the position of the capacitor and resistor have been reversed and now the reactance, Xc is connected to the input terminal of the inverting amplifier while the resistor, Rƒ forms the negative feedback element across the operational amplifier as normal.

This operational amplifier circuit performs the mathematical operation of **Differentiation**, that is it “*produces a voltage output which is directly proportional to the input voltage’s rate-of-change with respect to time*“. In other words the faster or larger the change to the input voltage signal, the greater the input current, the greater will be the output voltage change in response, becoming more of a “spike” in shape.

Related Products: Analog Divider and Multiplier | CATV Amplifier

As with the integrator circuit, we have a resistor and capacitor forming an RC Network across the operational amplifier and the reactance ( Xc ) of the capacitor plays a major role in the performance of a **Op-amp Differentiator**.

The input signal to the differentiator is applied to the capacitor. The capacitor blocks any DC content so there is no current flow to the amplifier summing point, X resulting in zero output voltage. The capacitor only allows AC type input voltage changes to pass through and whose frequency is dependant on the rate of change of the input signal.

At low frequencies the reactance of the capacitor is “High” resulting in a low gain ( Rƒ/Xc ) and low output voltage from the op-amp. At higher frequencies the reactance of the capacitor is much lower resulting in a higher gain and higher output voltage from the differentiator amplifier.

However, at high frequencies an op-amp differentiator circuit becomes unstable and will start to oscillate. This is due mainly to the first-order effect, which determines the frequency response of the op-amp circuit causing a second-order response which, at high frequencies gives an output voltage far higher than what would be expected. To avoid this the high frequency gain of the circuit needs to be reduced by adding an additional small value capacitor across the feedback resistor Rƒ.

Ok, some math’s to explain what’s going on!. Since the node voltage of the operational amplifier at its inverting input terminal is zero, the current, i flowing through the capacitor will be given as:

The charge on the capacitor equals Capacitance x Voltage across the capacitor

The rate of change of this charge is:

but dQ/dt is the capacitor current,i

from which we have an ideal voltage output for the op-amp differentiator is given as:

Therefore, the output voltage Vout is a constant -Rƒ.C times the derivative of the input voltage Vin with respect to time. The minus sign indicates a 180^{o} phase shift because the input signal is connected to the inverting input terminal of the operational amplifier.

One final point to mention, the **Op-amp Differentiator** circuit in its basic form has two main disadvantages compared to the previous operational amplifier integrator circuit. One is that it suffers from instability at high frequencies as mentioned above, and the other is that the capacitive input makes it very susceptible to random noise signals and any noise or harmonics present in the source circuit will be amplified more than the input signal itself. This is because the output is proportional to the slope of the input voltage so some means of limiting the bandwidth in order to achieve closed-loop stability is required.

Related Products: Comparator | OP Amp

If we apply a constantly changing signal such as a Square-wave, Triangular or Sine-wave type signal to the input of a differentiator amplifier circuit the resultant output signal will be changed and whose final shape is dependant upon the RC time constant of the Resistor/Capacitor combination.

The basic single resistor and single capacitor op-amp differentiator circuit is not widely used to reform the mathematical function of **Differentiation** because of the two inherent faults mentioned above, “Instability” and “Noise”. So in order to reduce the overall closed-loop gain of the circuit at high frequencies, an extra resistor, Rin is added to the input as shown below.

Adding the input resistor Rin limits the differentiators increase in gain at a ratio of Rƒ/Rin. The circuit now acts like a differentiator amplifier at low frequencies and an amplifier with resistive feedback at high frequencies giving much better noise rejection. Additional attenuation of higher frequencies is accomplished by connecting a capacitor Cƒ in parallel with the differentiator feedback resistor, Rƒ. This then forms the basis of a Active High Pass Filter as we have seen before in the filters section.

Error! Please fill all fields.

your input output graph seems to be wrong. Exactly opposite of what it should be.

Here’s the right one http://www.electronicshub.org/wp-content/uploads/2015/01/Input-and-Output-Waveforms-for-Square-Wave.jpg

You’re right. Output voltage should have opposite sign to the input as this is an inverting op-amp

Why do you assume there’s is the right one?

how to design a differentiator to obtain a1d(v1)/dt+a2d(v2)/dt-a3d(v3)/dt

hi i have 5v constant and i using one 555 timer but only need one pulse to activate the 555, i saw the OpAmp can have only one pulse, can this works?

There are lots of different ways to create a single pulse or a train of pulses, but I would not mix an analogue op-amp with a digital circuit unless you had to.

how differential amplifier can be used for rise time measurement

Differentiator generates square wave when input is triangular but the vice versa is not true, why???

The integrator and differentiator perform mathematical functions. If you differentiate a function that has a slope (up hill or down hill) in maths, you get a horizontal line (square waves). And if you integrate a horizontal line function in mathematics, you get a slope (which forms your triangular waves)

Integrator generates square wave when input is triangular but the vice versa is not true, why???

Prove that, v(out)= -RfC(dv in/dt)

thank you for your all subjects about amplifier.It is easy to understand in my opinion.

improve your information more and more student can’t undrestand by this

I’m good feel…….