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An ideal operational amplifier has infinite input impedance and open-loop gain, zero output impedance and no offsets or response limits. Its input currents vanish. Stable operational-amplifier negative feedback, a finite output and unsaturated linear operation then imply a virtual short, $v_{+}=v_{-}$. Real amplifiers have finite frequency-dependent gain, input bias currents, offset voltage, output loading, finite bandwidth, slew rate and supply-rail limits. Equal input voltages cannot be assumed in an open-loop or saturated circuit.

Here the noninverting input is grounded. In the ideal-feedback limit the inverting node is a virtual ground, so the resistor current charges the feedback capacitor:The initial capacitor charge fixes the integration constant. This is an op amp integrator, with $H(i\omega )=-1/(i\omega RC)$.

A tiny direct-current input, input bias current or offset can drive the output towards saturation. A parallel feedback resistor $R_f$ provides a direct-current return path, giving $H(s)=-(R_f/R)/(1+s{R}_{f}C)$; choose $\omega R_{f}C\gg 1$ over the intended integration band. A reset switch sets initial charge; low-offset components and suitable signal amplitude reduce drift and saturation. The useful band must also remain below the amplifier's bandwidth and slew-rate limits.

For the stipulated finite but constant real gain $A=100$, infinite input impedance and negligible output impedance, write the inverting-node voltage as $v=-V_{\mathrm{out}}/A$. With phasors proportional to $e^{i\omega t}$,ThereforeThe low-frequency gain is $-100$. The corner is ${\omega }_c=1/(101RC)$; above it the amplitude falls as $100/(101\omega RC)$ and the phase approaches $+\pi /2$. phase $+\pi$ and $-\pi$ describe the same low-frequency inversion. This finite-gain circuit is a low-pass response with an integrating high-frequency regime, not an ideal integrator down to zero frequency. A genuinely frequency-dependent $A(i\omega )$ would introduce additional high-frequency limitations beyond this specified model.

An ideal operational amplifier draws no current at either input. With stable operational-amplifier negative feedback in its unsaturated linear range, infinite open-loop gain also gives a virtual short, $V_{+}=V_{-}$. The amplifier adjusts its output to enforce that equality; it is not a conducting connection between the inputs. Here $V_{+}=0$, so the ideal summing node is a virtual ground. Kirchhoff's current law gives $V_{\mathrm{in}}/R_1=-V_{\mathrm{out}}/R_2$, hence the ideal gain is $-R_2/R_1$.

For the finite-input-resistance inverting amplifier let the inverting-node voltage be $v$, and continue to neglect output impedance. The controlled output and node-current balance areEliminating $v$ givesDividing the denominator by $R_2$ gives the requested resistance-sum form. The ideal result follows whenever $|A|\gg 1+R_2/R_1+R_2/R_{\mathrm{in}}$; taking $R_{\mathrm{in}}\to \infty$ alone is insufficient if $A$ is small. At finite frequency $A$ and the gain are phasor ratios, subject to feedback stability.

With equal resistors and sufficiently large loop gain, $V_{\mathrm{out}}\simeq -V_{\mathrm{in}}$. This gives a controlled sign reversal or 180-degree phase inversion, useful for subtraction and opposite-phase signal paths, with amplitude set by a resistor ratio rather than uncertain open-loop gain. The source still sees approximately $R_1$, so this is not a high-input-impedance voltage follower. For the ideal-input, single-pole model, the noise gain of an operational amplifier is two and the closed-loop bandwidth is approximately half the gain-bandwidth product. Output swing, slew rate and load constraints must also permit linear operation.

In an operational-amplifier negative feedback loop, an operational amplifier has large open-loop gain $A$, with $V_o=A(V_{+}-V_{-})$. Feeding a fraction $\beta V_o$ back to the inverting input opposes an output increase. For a non-inverting signal input,Its sensitivity to open-loop gain changes is reduced by $d\ln A_{\mathrm{cl}}/d\ln A=1/(1+A\beta )$. Thus resistor ratios can set a reproducible gain despite device variation; feedback also suppresses internal distortion and reduces output impedance. Gain is traded for usable bandwidth, subject to the amplifier's frequency response.

For the non-inverting connection, place $R_f$ between output and inverting input and $R_g$ between that input and ground. Then $\beta =R_g/(R_f+R_g)$, giving gain $1+R_f/R_g$. For an inverting connection, ground the non-inverting input, inject the signal through $R_{\mathrm{in}}$ and return $R_f$ from output to the inverting node. Small differential input and negligible input current then give gain $-R_f/R_{\mathrm{in}}$. The near-zero node is a virtual ground, not a physical ground connection.

The feedback must remain stable at the frequencies where loop gain matters. Extra phase lag can turn the intended opposition into oscillation; adequate phase margin, supply headroom, input common-mode range, output-current capability and slew rate are required. Feedback neither prevents clipping nor removes all input noise.^[ref]