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The closed-loop gain is the input-to-output transfer after closing the feedback connection. In the elementary noninverting voltage model it tends to $1/\beta$ for large stable loop gain, rather than retaining the uncertain open-loop gain $A$. A frequency-dependent transfer must be checked for stability before that approximation is used.

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.

The open-loop gain of an operational amplifier is the ratio of its output voltage to its differential input voltage before an external feedback control connection is applied. A common small-signal model is $A(s)=A_0/(1+s/{\omega }_a)$, with a large direct-current gain $A_0$ and a dominant pole at $-{\omega }_a$. Above ${\omega }_a$ its magnitude falls approximately inversely with angular frequency.

An operational amplifier is a differential electronic amplifier whose unloaded voltage is modeled in its linear regime by $v_o=A(s)(v_{+}-v_{-})$. Its large open-loop gain, large input impedance and small output impedance make negative feedback useful for setting a reproducible transfer function with external components. Supply voltages bound the available output swing; the linear model does not apply after saturation.

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]

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.

The open-loop gain is large and roughly constant at low frequency, then falls as internal poles become important. A dominant-pole approximation isAbove $f_p$, but before higher poles dominate, $|A|\propto f^{-1}$, a slope of $-20\mathrm{dB}$ per decade, with increasing phase lag. The approximate unity-gain frequency is $f_T=A_0{f}_p$. A 741-class amplifier has typical low-frequency gain about $2\times 10^5$; this is the open-loop gain, distinct from the feedback-set signal gain. The finite-gain formula at the parent level explains why the closed-loop response eventually rolls off.^[ref]

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.

The ideal operational-amplifier rules are zero input currents and, when stable negative feedback maintains unsaturated linear operation, equal input voltages. Infinite open-loop gain motivates the latter; it must not be assumed to guarantee asymptotic stability for every external network.

For the numerical sketch and usefulness discussion, use the transfer function derived in part (c). The chosen values give $K=5$ and $\tau =RC=10^{-4}\mathrm{s}$. With $x=\omega \tau$,The formal amplitude falls from 5 to zero, with high-frequency slope $-40$ dB per decade. Its unwrapped phase rises from zero through $+90^{\circ }$ at $\omega =10^4{\mathrm{s}}^{-1}$ to $+180^{\circ }$; the corresponding characteristic frequency is 1.59 kHz. At that frequency the amplitude is 2.5.

These are formal sinusoidal particular-solution curves, not a stable settled filter response. For the formal sinusoidal response of an unstable linear system, replacing $i\omega$ by the Laplace variable $s$ gives $G(s)=5/(1-s\tau {)}^2$, with a repeated pole at $s=+1/\tau$. A homogeneous output contains $(A+Bt)e^{t/\tau }$ and grows, so real perturbations lead to supply-limited nonlinear behaviour. With equal positive resistors and capacitors in the frequency-setting network, the general denominator is $1+(3-K)s\tau +s^2{\tau }^2$: the Sallen-Key equal-component stability bound is $K<3$. For example $K=3-\sqrt{2}$ gives a Butterworth filter response. Thus the Sallen-Key low-pass filter architecture is useful for active filtering with suitable gain and damping, but the stated $K=5$ component set is unsuitable for linear filtering. The component values are retained in this diagnosis.^[ref]

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]

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.

A non-ideal amplifier is not specified by infinite input impedance alone: its open-loop gain and output impedance must also be modeled. Take its internal controlled output to be $u=-A(s)v$, where $v$ is the inverting-input potential, and place a resistance $r_o$ between $u$ and the actual output $y=V_{\mathrm{out}}$. The non-inverting input remains grounded; assume no external load. This gives an explicit finite-gain operational-amplifier integrator model.

Writing Kirchhoff's current law at the input and output nodes givesThe second equation gives $y=(sC{r}_o-A)v/(1+sC{r}_o)$. Substitution in the first gives the general transfer functionFor sinusoidal phasors, set $s=i\omega$. The numerator and denominator are complex, so the magnitude gives the amplitude ratio and their arguments give the relative phase.

The simplest model takes $r_o=0$ and a constant finite gain $A_0>0$. It givesThus below ${\omega }_l=[(1+A_0)RC{]}^{-1}$ the amplifier has gain magnitude $A_0$ and inversion. Above that frequency it approaches an integrator with magnitude $A_0/[(1+A_0)\omega RC]\simeq 1/(\omega RC)$ and phase $+90^{\circ }$. Constant finite gain by itself predicts no further high-frequency turnover.

For a more realistic sketch, use the standard dominant-pole open-loop gain $A(s)=A_0/(1+s/{\omega }_a)$, still with $r_o=0$. ThenThe exact frequency response isAssuming $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, the turnover frequencies are well separated: ${\omega }_l\simeq 1/(A_{0}RC)$ and ${\omega }_h\simeq A_0{\omega }_a$, the approximate unity-gain angular frequency of the amplifier. The regimes arewith magnitude slopes $0,-20,-40\mathrm{dB}$ per decade and smooth phase transitions. This is the solid curve in the figure; $180^{\circ }$ and $-180^{\circ }$ denote the same phase.

If finite $r_o$ is retained and $A(i\omega )\to 0$, the general expression instead approaches $r_o/(R+r_o)$ at very high frequency: the capacitor couples the input to the output through a passive voltage divider. The dashed curve illustrates that eventual feedthrough. Other amplifier poles or an external load would alter the response again. No particular values of $A_0,{\omega }_a,r_o$ are given in the paper, so the numerical sketch is illustrative and the general expression states the model dependence explicitly.

Under stable negative feedback and within its linear operating range, an ideal voltage-feedback operational amplifier has equal input voltages. This relation is a consequence of large open-loop gain, and does not imply current flow between the inputs.