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An ideal operational amplifier has infinite open-loop gain and input impedance, zero output impedance, unlimited response speed and zero input offset voltage and input bias current. Its input currents vanish. In a stable linear circuit with negative feedback and a finite output, its differential input voltage tends to zero as the open-loop gain tends to infinity. Thus $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$ are the familiar conditional circuit rules. The equality of the input voltages cannot be imposed on an open-loop, unstable or saturated circuit.

Input impedance is the ratio of a small applied input-voltage phasor to the resulting input-current phasor, with the output termination and operating point specified. A finite input impedance loads a voltage source. For an ideal operational amplifier the input impedance is infinite, so neither input terminal draws current.

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]

An ideal operational amplifier has infinite differential open-loop gain, infinite input impedance, zero output impedance, unlimited bandwidth and slew rate, and zero input offset voltage, input bias current and intrinsic noise. The idealized model has no finite output-swing or output-current limit.

The usual circuit rules areThe first follows from infinite input impedance. The second follows because a finite output requires $v_{+}-v_{-}=v_o/A\to 0$ as the open-loop gain tends to infinity. It is a virtual equality of potentials, not a conducting connection between the inputs. It cannot be imposed for open-loop operation or after an amplifier reaches its supply rails.

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.