The op amp integrator is used in numerous applications, especially in analog computers.
An integrator is an op amp circuit whose output is proportional to the integral of the input signal.
If the feedback resistor \(R_f\) in the inverting amplifier is replaced by a capacitor, we obtain an ideal integrator.
At node \(a\), \(i_R = i_C\) for an ideal op amp. Hence,
\(\displaystyle i_R = i_C = \frac{v_i}{R} = -C\,\frac{dv_o}{dt}\)
\(\displaystyle dv_o = -\frac{1}{RC}\,v_i\,dt\)
Integrating both sides
\(\displaystyle \int_{0}^{t}dv_o = -\frac{1}{RC}\int_{0}^{t}v_i\left(\tau\right)\,d\tau\)
\(\displaystyle v_o\left(t\right) - v_o\left(0\right) = -\frac{1}{RC}\int_{0}^{t}v_i\left(\tau\right)\,d\tau\)
Assuming \(v_o\left(0\right) = 0\)
\(\boxed{v_o\left(t\right) = -\frac{1}{RC}\int_{0}^{t}v_i\left(\tau\right)\,d\tau}\)
Solving for the output voltage \(v_o\) using the impedance \(\mathbf{Z}_C\) of the capacitor \(C\).
\(\displaystyle \mathbf{Z}_C = \frac{1}{j\omega C}\)
In s-domain, \(\displaystyle s = j\omega \to \mathbf{Z}_C = 1 / sC\)
\(\displaystyle i_R\left(s\right) = i_C\left(s\right) = \frac{v_i\left(s\right)}{R} = -\frac{v_o\left(s\right)}{\mathbf{Z}_C} = -\frac{v_o\left(s\right)}{1 / sC} = -sC\,v_o\left(s\right)\)
\(\displaystyle \frac{v_i\left(s\right)}{R} = -sC\,v_o\left(s\right)\)
\(\displaystyle v_o\left(s\right) = -\frac{1}{RC}\cdot\frac{v_i\left(s\right)}{s}\)
Find the inverse Laplace transform to obtain the time domain function.
\(\boxed{v_o\left(t\right) = -\frac{1}{RC}\int_{0^-}^{t}v_i\left(\tau\right)\,d\tau}\)
To ensure that \(V_o\left(0\right) = 0\), it is always necessary to discharge the integrator’s capacitor prior to the application of signal.
In practice, the op amp integrator requires a feedback resistor to reduce DC gain and prevent saturation. Care must be taken that the op amp operates within the linear range so that it does not saturate.