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In amplifier saturation, the requested output lies outside the amplifier's linear operating range, such as when supply rails limit output voltage. The negative feedback virtual short approximation then need not hold. Output swing, current drive and recovery behaviour depend on the device and load; finite slew rate is a separate dynamic limit even when the output has not reached a rail.

Return a fraction of the output to the inverting input so that it opposes the input difference. Large stable loop gain makes the closed-loop gain depend mainly on the external feedback ratio rather than the uncertain open-loop gain. Frequency-dependent phase lag can destabilize the loop; saturation and finite slew rate also limit the approximation.^[ref]

The passive RC low-pass filter is simple, inexpensive, requires no power supply, and has no operational amplifier saturation or finite bandwidth limitation. It offers no voltage gain; its output impedance is finite, and both source electrical resistance and loading can change the observed frequency response. Its resistor also produces thermal noise.

The active inverting low-pass amplifier provides a low-frequency voltage gain of magnitude five and, within the usable bandwidth, a small output impedance, so it can drive a load with much less change in frequency response. The input electrical resistance is approximately $R_1=1\mathrm{k}\Omega$ and may itself load the source. It needs a power supply and adds electrical noise and offset; finite operational amplifier gain, bandwidth, slew rate, available output swing, and output electric current limit the ideal formula. Its inversion must also suit the application. The best choice depends on required voltage gain, source and load electrical impedances, and the signal's frequency and amplitude.

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.

An ideal operational amplifier has infinite differential open-loop gain, infinite input impedance, zero output impedance and unlimited bandwidth and output capability, and zero input offset voltage and input bias current. In a stable negative-feedback circuit operating linearly, its input currents vanish and $V_{-}=V_{+}$. This equality is a feedback consequence, not a rule for a saturated or open-loop amplifier. Here the grounded non-inverting input makes the inverting node a virtual earth.

The inverting two-pole band-pass filter has a series resistor and capacitor input, while the feedback arm is a parallel resistor and capacitor. With Laplace variable $s$,Kirchhoff's current law at virtual earth gives the minus sign. Substituting $s=i\omega$ yieldsIt vanishes at zero and infinite frequency, with low-frequency $|H{|}^2\simeq (\omega R_2{C}_1{)}^2$ and high-frequency $|H{|}^2\simeq 1/(\omega R_1{C}_2{)}^2$. Differentiation gives the single maximumFor ${\tau }_1\gg {\tau }_2$, the broad interval $1/{\tau }_1\ll \omega \ll 1/{\tau }_2$ has $|H|\simeq R_2/R_1$. For ${\tau }_2\gg {\tau }_1$, the interval $1/{\tau }_2\ll \omega \ll 1/{\tau }_1$ instead has $|H|\simeq C_1/C_2$. The same rising and falling asymptotes apply outside either plateau.

Writing $a={\tau }_1+{\tau }_2$ and $b={\tau }_1{\tau }_2$, the half-power frequencies and quality factor areEqual time constants give the narrowest fractional pass-band achievable here, with ${\omega }_{\pm }=(\sqrt{2}\pm 1){\omega }_0$ andThis topology cannot make a genuinely sharp, high-$Q$ band-pass. Its poles are real, and the selectivity bound for a two-real-pole band-pass prevents arbitrarily narrow selection. A more selective resonant filter needs additional circuit structure. The gain above answers the component-ratio question for the sharpest available setting without assuming an impossible limit.

Real amplifiers have finite frequency-dependent gain and phase lag, so virtual earth becomes inaccurate, the response changes and stability must be checked. Gain-bandwidth limits, slew rate for large signals, output loading and amplifier saturation also matter. Component tolerances move the corners; capacitor parasitics, stray capacitance, resistor noise and input noise become significant. The ideal high-frequency tail should not be extrapolated beyond the amplifier and lumped-component operating ranges.

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]

The ideal voltage-feedback Operational amplifier has zero input currents and, while stable negative feedback holds it in its linear operating range, equal input voltages: $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$. The first rule follows from infinite input impedance; the second is the large-open-loop-gain limit of $v_o=A(v_{+}-v_{-})$. Equal input electric potential is a virtual short, not an electrically conducting connection.

For an inverting amplifier with the noninverting input grounded, $v_{-}=0$ and node-current balance gives $v_i/R_i=-v_o/R_f$, hence $v_o=-R_f{v}_i/R_i$. For a noninverting amplifier, the feedback divider gives $v_{-}=v_o{R}_g/(R_f+R_g)=v_i$, hence $v_o=(1+R_f/R_g)v_i$.

One should check output swing, input common-mode range, slew rate, Gain-bandwidth product and stability before using these limits in a real circuit. Saturation, positive feedback or operation beyond bandwidth invalidates the equal-voltage rule; finite input bias currents and offset voltage can also matter.^[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.

A real operational amplifier has large but finite, frequency-dependent open-loop gain; the gain falls and its phase changes at higher frequencies. Finite bandwidth and slew rate constrain small-signal and large-signal responses respectively. There are also finite input impedance and output impedance, nonzero input bias currents and input offset voltage, and internally generated noise. Offset and bias can drift with temperature.

The output is limited by the supply voltages and output-current capability, and the inputs have a permitted common-mode range. Rejection of common-mode and supply changes is finite. Consequently $v_{+}-v_{-}$ need not vanish exactly, and neither the virtual-ground approximation nor linear gain formulas apply after saturation. For the calculation below the stipulated infinite input impedance removes the input loading currents, while finite gain and optionally finite output resistance are retained.

The slew rate bounds the large-signal output-voltage change per unit time. It is distinct from a small-signal gain-bandwidth product. A sinusoidal output of amplitude $V_p$ and angular frequency $\omega$ needs $\omega V_p$ below the available slew rate to avoid this source of distortion.