13 Jul 2026

Op-Amp Differentiator and Integrator

practical pg-ii op-amp differentiator integrator

Aim

To study the waveforms produced by practical op-amp differentiator and integrator circuits.

Apparatus

741 op-amp, dual DC supply, resistors, capacitors, function generator, CRO, breadboard, and connecting wires.

Experimental arrangement

The differentiator uses an input capacitor and feedback resistor; interchanging those two elements gives the integrator.

Theory

An operational amplifier produces $V_o=A(V_+-V_-)$, where the open-loop gain $A$ is very large. With negative feedback and an unsaturated output, only a minute input difference is needed, so $V_-\approx V_+$. Since the non-inverting terminal is grounded, the inverting node is at virtual ground. The op-amp input current is negligible, so Kirchhoff’s current law can be applied directly at this node.

For the differentiator, the capacitor current is $i=C_i\,dV_i/dt$. The same current passes through $R_f$, giving

\[V_o=-R_fC_i\frac{dV_i}{dt}.\]

A constant input gives zero output, while the constant slopes of a triangular wave give two constant output levels. A square wave changes abruptly, so its edges produce positive and negative pulses. A practical differentiator includes frequency-limiting components to prevent excessive amplification of high-frequency noise.

For the integrator, the input current is $i=V_i/R_i$ and this charges the feedback capacitor. Since $i=-C_f\,dV_o/dt$,

\[V_o=-\frac{1}{R_iC_f}\int V_i\,dt+V_o(0).\]

A square input therefore gives a triangular output. A resistor is normally placed in parallel with $C_f$ to limit the DC gain and prevent drift into saturation.

Observations

Input waveform	Differentiator output	Integrator output
Square	Positive and negative spikes	Triangular waveform
Sine	Cosine-like waveform	Negative cosine-like waveform
Triangular	Square waveform	Parabolic segments

Calculation

For a differentiator with $R_f=10$ k$\Omega$, $C_i=0.01\,\mu$F, and input slope $dV_i/dt=0.40$ V ms$^{-1}$,

\[V_o=-(10^4)(0.01\times10^{-6})(400)=-0.040\,\text{V}.\]

For an integrator with $R_i=10$ k$\Omega$, $C_f=0.10\,\mu$F, and a constant $1$ V input applied for 1 ms,

\[\Delta V_o=-\frac{(1)(1\times10^{-3})}{(10^4)(0.10\times10^{-6})}=-1.0\,\text{V}.\]

Result

The differentiator produces an output proportional to the rate of change of the input, while the integrator produces an output proportional to the time integral of the input.

Viva Questions

1. Why is the op-amp called inverting here? The signal is applied to the inverting terminal and the non-inverting terminal is grounded.
2. What limits a practical differentiator? High-frequency noise and the finite bandwidth of the op-amp.
3. Why is a capacitor used at the input of an integrator? Its impedance causes the input current to be proportional to the time integral of voltage.

Maxima Code

Download the PG-II electronics calculation file.