The two rules:
- No Current Flows into the Input Terminals
- The Differential Input Voltage is Zero i.e. $V_1 = V_2 $
No current flows to the input of Op-amp, so $I_1 = I_f$.
Now, $I_C = C \frac{d V_C}{dt}$
$ \Rightarrow \frac{v_{in}-v_2}{R_1}=C \frac{d(v_2-v_o)}{dt}$
$ \Rightarrow \frac{v_{in}}{R_1} = - C \frac{dv_o}{dt}$ ( In an ideal op-amp, $v_1 = v_2 = 0$)
$ \Rightarrow \int_{0}^{t} \frac{v_{in}}{R_1} dt = - \int_{0}^{t} C \frac{dv_o}{dt} dt$
If the initial value of $v_o$ is assumed to be 0 V, this results in a DC error of:
$\qquad v_{o} = - \frac{1}{R_1 C} \int_{0}^{t} v_{in} dt$
AC Op-amp Integrator with DC Gain Control
AC is sinusoidal. AC integrator will produce another sinusoidal wave which will be $90^0$ out-of-phase with the input wave. This forms the basis of a Active Low Pass Filter with a corner frequency given as.
