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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.

An input bias current is a direct-current component drawn by an input terminal of a real operational amplifier. Through a source resistance it produces a voltage error, and through the feedback capacitor of an op amp integrator it can produce an output ramp. Its magnitude and sign depend on the amplifier technology and operating conditions.

The input offset voltage of an operational amplifier is the differential direct-current input voltage needed to make its output zero at a specified operating point. Its effect resembles a small unwanted input signal and can cause sustained output drift in an 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]

Let $v$ be the inverting-input potential. The non-inverting input is grounded, so the ideal operational amplifier rules give $v=0$ and no input current. Kirchhoff's current law at that node therefore givesIntegrating from an initial time $t_0$,This op amp integrator performs inverted time integration; the initial voltage across the capacitor supplies the integration constant. For zero initial conditions its transfer function is $H(s)=-1/(sRC)$. In sinusoidal steady state, with the convention $e^{i\omega t}$,For example, a cosine input produces a negative sine output after its constant component is set. The additional inversion makes the phase $+90^{\circ }$ modulo $360^{\circ }$, rather than the $-90^{\circ }$ of non-inverting integration.