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An integrating amplifier produces an output proportional to the time integral of its input over an intended operating range. A common circuit uses an operational amplifier, an input resistor $R$ and a feedback capacitor $C$. With the non-inverting input held at zero, ideal negative feedback maintains a virtual ground at the inverting input. Negligible amplifier input current then givesThe initial output records the capacitor's initial charge. Texas Instruments' circuit note derives this inverting relation and explains why real integration is limited by the circuit time constant and amplifier bandwidth.^[ref]

For an original example, choose $R=100\mathrm{k\Omega }$ and $C=10\mathrm{nF}$, so $RC=1\mathrm{ms}$. A constant0.20volt input lasting2milliseconds changes the output by $-0.40\mathrm{V}$. Repeating ten such pulses without intervening input or integrator reset changes it by $-4.0\mathrm{V}$ in the ideal model. The output retains information about accumulated pulse area, rather than only the present input value.

A small offset also accumulates. With the same components, a residual input-equivalent1millivolt gives a slope of $-1\mathrm{V}{\mathrm{s}}^{-1}$. Starting from zero, an idealized negative output limit at5volts would be reached in5seconds. The finite supply range causes amplifier clipping; an ideal integral extending indefinitely is not physically available.

Putting a resistor $R_f$ in parallel with the feedback capacitor supplies a DC path. Under ideal amplifier assumptions its zero-state transfer function isAt frequencies where $|s{R}_{f}C|\gg 1$, this approaches $-1/(sRC)$; at low frequency it instead approaches the finite gain $-R_f/R$. The resistor prevents unrestricted DC accumulation while changing the lowest usable integration frequency.^[ref]

An integrating amplifier can therefore turn a short detector pulse into a measurable accumulated output, but its initial condition, reset, baseline, frequency range and output limits belong to the measurement model. A charge amplifier is a related current-input arrangement in which collected charge is converted through feedback capacitance.

An integrator reset establishes the charge on an integrating amplifier's feedback capacitor before the next measurement. Without it, the next output includes the previous initial condition, and repeated input pulses or input offset voltage can eventually cause amplifier clipping. Resetting makes consecutive measurement intervals comparable; it does not reconstruct a pulse already lost to amplifier clipping.

Consider an inverting integrating amplifier with the input held at zero during reset. Place an analog switch across its feedback capacitor $C$. In the approximation of a stable ideal operational amplifier maintaining a virtual ground, a closed switch with constant on-state electrical resistance $R_{\mathrm{on}}$ gives the RC circuit relationThis exponential decay gives an integrator reset settling time for residual magnitude at most $\epsilon$:For stipulated values $C=10\mathrm{nF}$, $R_{\mathrm{on}}=10\mathrm{k\Omega }$ and $|V_{\mathrm{old}}|=0.50\mathrm{V}$, the reset time constant is $100\mu \mathrm{s}$. Reaching a residual of $1\mathrm{mV}$ requires about $621\mu \mathrm{s}$ in this model. Actual amplifier drive, stability and recovery from overload can change the response. A switch's control delay also differs from the full circuit's settling time.^[ref]

Opening the reset switch can itself disturb the initial condition. Analog switch charge injection transfers a small charge to the connected capacitor, producing a reset charge-injection pedestal of magnitudeFor a stipulated $10\mathrm{pC}$ deposited on $10\mathrm{nF}$, this step is $1\mathrm{mV}$. Its sign depends on the circuit and charge direction. A longer reset suppresses the preceding exponential residual but need not suppress the new step created when the switch opens. Analog Devices illustrates this capacitive coupling mechanism and the resulting output step separately from retained-charge settling.^[ref]

A practical experiment repeats zero-input measurements with several reset durations and reads the output at the same delay after switch opening. A decreasing residual indicates incomplete discharge; a persistent offset can instead include reset charge-injection pedestal, input offset voltage and subsequent drift. Repeated readings help separate the reproducible offset from noise. The ACF2101 provides a concrete historical implementation: separate hold and reset controls disconnect the input and discharge the integration capacitor, and its datasheet shows the sequence of hold, integration and reset states.^[ref]

The reset interval also creates integrator reset dead time: charge arriving while the circuit is being reset may not contribute to the next measured integral. Pulse timing relative to the reset therefore matters. A perfectly repeatable baseline alone does not establish that every input pulse was captured.

Connect a resistor $R_{\mathrm{in}}$ from the input source to the inverting input, a feedback resistor $R_f$ from output to that same input, and ground the non-inverting input. For ideal linear operation the inverting node is a virtual ground. Kirchhoff's current law gives $V_i/R_{\mathrm{in}}+V_o/R_f=0$, yielding the displayed voltage gain. The source sees input impedance $R_{\mathrm{in}}$, even though the amplifier input itself has infinite input impedance.

An ideal operational amplifier with a series input resistor and capacitor, resistive feedback and a grounded noninverting terminal realizes the displayed high-pass filter. Virtual ground and Kirchhoff's current law give $H=-R_f/[R_i+1/(i\omega C)]$. Its phase is $-\pi /2-{\tan }^{-1}(\omega R_{i}C)$ in the $e^{i\omega t}$ convention. Equal resistances do not make its phase frequency independent.

For an ideal operational amplifier with grounded non-inverting input, connect an input capacitor $C$ to the inverting input and a feedback resistor $R$ from output to that input. The virtual ground and Kirchhoff's current law give $C{\dot{V}}_{\mathrm{in}}=-V_{\mathrm{out}}/R$. The ideal response amplifies high-frequency noise, so practical differentiators restrict its operating band and maintain feedback stability.

The input source supplies $I_i=V_i/R_i$ into the virtual ground, so the complete circuit's input resistance is $R_i$. The ideal output is a zero-resistance controlled voltage source:These are properties of the closed-loop circuit, distinct from the infinite input resistance of the amplifier terminals themselves.

Ground the non-inverting input and take the ideal operational amplifier in stable, unsaturated negative feedback. Its inverting input is a virtual ground and draws no current. Kirchhoff's current law there givesThe minus sign is the phase inversion of the inverting amplifier.

Replace the feedback resistor by a capacitor $C_f$, retaining the input resistor $R_i$. The virtual ground makes the input charge flow onto the feedback capacitor rather than into the ideal amplifier. Integrating its current givesIf the calorimeter delivers $Q_{\mathrm{in}}=\kappa \mathcal{E}$ for deposited energy $\mathcal{E}$, then $\Delta V_o=-\kappa \mathcal{E}/C_f$. A reset establishes the initial output; a practical large resistor across $C_f$ limits long-time drift, but is omitted from the ideal integration interval.

An external source sees the input resistor between its terminal and the virtual ground. The input current is $I_i=V_i/R_i$, soThe infinite input resistance of the ideal operational amplifier itself is distinct from the finite input resistance of the complete inverting amplifier. Negative feedback makes the output approach an ideal voltage source; a real amplifier retains a small nonzero output resistance.

In the linear inverting amplifier, the non-inverting terminal is grounded and negative feedback connects the output to the inverting terminal. For large open-loop gain $A$, $V_o=A(V_{+}-V_{-})$; a finite unsaturated output therefore requires $V_{-}\simeq V_{+}=0$. ThusThis is a virtual ground: it has approximately ground potential, but it is not a conducting connection to ground. The ideal operational amplifier draws no input current, so input current flows through the feedback resistor. Saturation, loss of feedback, or insufficient bandwidth invalidates the approximation.

Use an ideal operational amplifier with its non-inverting terminal grounded, an input resistor $R$ and a feedback capacitor $C$, as in the upper circuit. Stable unsaturated negative feedback holds the inverting terminal at virtual ground, while the input current into the amplifier itself vanishes.

Applying Kirchhoff's current law at the inverting node givesThus the circuit is an inverting op amp integrator, with its integration constant set by the initial capacitor charge. A constant input makes the output ramp until a real amplifier reaches a supply rail. A practical low-frequency design usually adds a large feedback resistor to limit drift; the ideal integral relation applies where the feedback capacitor dominates.

Interchange the input and feedback components: put the capacitor $C$ between the input and the inverting terminal, and the resistor $R$ between the output and the inverting terminal. Keep the non-inverting terminal grounded, as in the lower circuit above. For an ideal operational amplifier under unsaturated negative feedback, the inverting node remains at virtual ground.

The input capacitor current is $I_C=Cd{V}_{\mathrm{in}}/dt$. It flows through the feedback resistor, givingThis is an inverting op amp differentiator. Its ideal gain magnitude increases with frequency, so it strongly amplifies high-frequency noise. Practical differentiators include components that restrict the useful band and stabilize feedback; that qualification does not change the displayed ideal low-amplitude relation within the operating band.

The original circuit has $R_1$ and $C$ in series between the input and the inverting node, with $R_2$ providing feedback and the noninverting node grounded. An ideal operational amplifier in stable linear negative feedback draws no input electric current and holds the inverting node at virtual ground. For phasors proportional to $e^{i\omega t}$, the input complex impedance is $Z_1=R_1+1/(i\omega C)$. Kirchhoff's current law gives ${\tilde{V}}_{\mathrm{in}}/Z_1=-{\tilde{V}}_{\mathrm{out}}/R_2$, henceThis is an inverting high-pass filter. Its phase can be chosen continuously asThus $V_{\mathrm{out}}(t)=V_0|H|\cos (\omega t+\arg H)$. The ratio above is a phasor ratio, not an instantaneous ratio at input zero crossings. Setting $R_1=R_2$ does not remove the capacitor's phase: the output lags by $90^{\circ }+{\tan }^{-1}(\omega R_{1}C)$, between $90^{\circ }$ and $180^{\circ }$. At $\omega R_{1}C=1$ the lag is $135^{\circ }$; only for $\omega R_{1}C\gg 1$ does it approach $180^{\circ }$. Being within the amplifier's own bandwidth does not specify the external RC corner frequency.

Use an ideal inverting summing amplifier, grounding the non-inverting input and connecting the two sources through resistances $R$ and $R/2$. A feedback resistance $R_f=2R$ gives the desired weights. Operational-amplifier negative feedback makes the inverting node a virtual ground, and negligible input current lets Kirchhoff's current law give

Replacing the feedback resistor by a capacitor $C$ makes a weighted op amp integrator:The initial capacitor charge determines the integration constant. A nonzero DC weighted input produces a ramp; a real operational amplifier eventually reaches its output-voltage limit. A parallel feedback resistance can limit drift while preserving integration over an appropriate frequency range.

For an ideal operational amplifier under negative feedback, the grounded non-inverting terminal makes the inverting terminal a virtual ground, and no electric current enters either input. The feedback resistor and capacitor have parallel electrical impedanceKirchhoff's current law at the inverting node gives ${\tilde{V}}_i/R_1+{\tilde{V}}_o/Z_f=0$, and henceWith the stated components, $R_{2}C=16.0\mu \mathrm{s}$ and $R_2/R_1=5$. Thus this inverting low-pass amplifier has exactly the same ideal cutoff frequency and normalized frequency response as the passive RC low-pass filter, with output multiplied by $-5$:Its amplitude ratios at the three stated frequencies are $4.99975$, $3.52616$, and $0.97560$. One principal choice of phase shift is $180^{\circ }-{\tan }^{-1}(\omega \tau )$: respectively $179.424^{\circ }$, $134.848^{\circ }$, and $101.252^{\circ }$. Adding or subtracting $360^{\circ }$ describes the same output; the extra inversion must be retained.

Use the following arrangement: an inverting operational amplifier with input resistor $R_a$ and feedback resistor $R_f$, followed by an op amp integrator with input resistor $R_i$ and feedback capacitor $C_i$. Ground both non-inverting inputs. Each ideal operational amplifier has infinite input impedance, so its inputs draw no electric current; negative feedback makes the inverting input a virtual ground.

Kirchhoff's current law at the first summing junction gives $V_1=-(R_f/R_a)V_{\mathrm{in}}$. At the second summing junction it gives $V_1/R_i+C_{i}d{V}_{\mathrm{out}}/dt=0$. Consequently the two inversions cancel:The initial output voltage records the initial capacitor charge. A reset switch across $C_i$ sets it to zero. The formula holds while both stages remain within their supply rails; a persistent input voltage otherwise produces amplifier saturation.

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]

Use an inverting operational amplifier. Ground the non-inverting input, connect the source to the inverting input through $R_{\mathrm{in}}$, and connect $R_f$ from the output back to that same input.

In linear operation the ideal operational amplifier draws no input electric current, and feedback holds the inverting input at the same electric potential as the grounded non-inverting input. This node is a virtual ground. Kirchhoff's current law there givesThe source supplies current $I_i=V_i/R_{\mathrm{in}}$, so its input impedance is $R_{\mathrm{in}}$. The required design is thereforeOnly these two resistors and the single ideal operational amplifier are needed in the signal circuit; its power supplies are implicit. The result assumes an input small enough to keep the output within its linear operating range.