Op amp integrator

An op amp integrator places a resistor $R$ between the input source and the inverting input of an operational amplifier, and a feedback capacitor $C$ between that input and the output, with the non-inverting input grounded. For an ideal operational amplifier, Kirchhoff's current law gives $v_{\mathrm{in}}/R=-Cd{v}_o/dt$, henceThe initial capacitor charge supplies the integration constant. A parallel feedback resistor $R_f$ replaces the transfer by $H(s)=-(R_f/R)/(1+s{R}_{f}C)$: the direct-current gain becomes finite while integration is recovered when $|s{R}_{f}C|\gg 1$. This limits offset-driven saturation. ^[ref]

Consider an op amp integrator with infinite input impedance, an internal controlled voltage $u=-A(s)v$ where $v$ is the inverting-input voltage, and a series output impedance equal to a resistance $r_o$. Assume no external output load. Writing $y$ for the output, Kirchhoff's current law givesEliminating $u$ and $v$ yieldsThe expression at $r_o=0$ is obtained by its limit. In particular, constant gain $A_0$ and $r_o=0$ give $H(s)=-A_0/[1+sRC(1+A_0)]$: a finite direct-current gain followed by an integrating regime. For the dominant-pole model $A(s)=A_0/(1+s/{\omega }_a)$ and $r_o=0$,If $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, there are separated turnover frequencies near $1/(A_{0}RC)$ and $A_0{\omega }_a$. The gain magnitude is respectively constant, proportional to ${\omega }^{-1}$ and proportional to ${\omega }^{-2}$. With the convention $e^{i\omega t}$ a continuous phase choice passes from $180^{\circ }$ through $90^{\circ }$ to $0^{\circ }$. If $r_o>0$ and $A(i\omega )\to 0$, capacitive feedthrough instead gives the ultimate limit $r_o/(R+r_o)$. Specifying finite gain alone therefore does not specify a unique high-frequency response.