There is a feedback resistor Rf present in parallel with the feedback capacitor. The negative sign indicates that there is a phase shift of 180 degrees between input and output signals, as the input is provided at the inverting input of the op-amp.Ī practical opamp circuit is different from an ideal opamp.From the above equation, we can say that the output is times the integral of the input voltage.where C is the constant of integration, it is proportional to the output voltage at t = 0 seconds.The output voltage can be found by integrating the above equation.Substituting V2 in the above equation we get,.By using the concept of “virtual ground” we can write that V2 = 0.Substituting the value of I1 in the above equation,.The relation between current and voltage across the capacitor Cf is given by.As the input impedance, Ri is very high, IB is negligible hence,.Applying Kirchhoff’s current law at node V2, we get.

R1 is the input resistance and RL is load resistor. We assume that opamp used here is an ideal one. The above circuit is obtained by replacing the feedback resistor RF in the inverting amplifier configuration by a capacitor Cf. The output voltage is proportional to the amplitude and duration of the input signal.

The circuit in which output voltage waveform is an integration of the input signal is called as an integrator or op-amp integrator or integrating amplifier.

Icircuit op amp