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Improve a circuit's signal-to-noise ratio by reducing unwanted coupling and physical noise sources while retaining the signal bandwidth. Important methods include lowering appropriate resistances or temperature to reduce thermal electrical noise, controlling leakage and bias currents relevant to shot noise, minimizing pickup with shielding and small loop area, and selecting an operational amplifier for the source electrical impedance. Filtering limits integrated noise only if it preserves the information-bearing spectrum; a lock-in amplifier can move detection above a flicker noise region and select a narrow band. Uncorrelated noise variances add, whereas coherent pickup requires its phases and correlations to be considered. Repeated averaging reduces independent random error, but does not remove an arbitrary systematic error or slow drift.

For a dominant-pole voltage-feedback operational amplifier with $A(s)=A_0/(1+s/{\omega }_a)$ and $A_0\gg 1$, the unity-gain angular frequency is approximately ${\omega }_t=A_0{\omega }_a$. In the range where this pole dominates, a roughly constant closed-loop noise gain $G_N$ gives a bandwidth approximately $f_t/G_N$. This approximation depends on the amplifier model and feedback control circuit; it is not a universal statement about all amplifiers or all frequency-dependent feedback networks. ^[ref]

Electronic gain expresses how an electronic circuit scales a specified input signal quantity into an output quantity. For a linearized voltage amplifier around its operating point, the voltage gain is $A_v=\delta v_{\mathrm{out}}/\delta v_{\mathrm{in}}$. At a sinusoidal frequency it can be a complex phasor ratio, specifying both amplitude ratio and phase shift. Power gain instead compares the average power delivered to the load with the average signal power entering the amplifier:State the input/output quantities, frequency, load and operating conditions when specifying gain. A transimpedance amplifier relates output electric potential difference to input current and has a dimensional transfer ratio in ohms.

For an original circuit example, use an inverting operational amplifier with input resistor $R_1=10\mathrm{k\Omega }$, feedback resistor $R_2=100\mathrm{k\Omega }$ and load $R_L=1\mathrm{k\Omega }$. In its ideal negative-feedback linear regime,An input of $50\mathrm{mV}$ root mean square therefore gives $0.50\mathrm{V}$ root mean square at the output, with a phase inversion. The manufacturer's primary explanation derives this resistance-ratio gain using the input and feedback currents.^[ref] Suitable supplies, output-current capability and amplifier output range are assumptions of this worked model, not specifications of a particular device.

Here the input resistance is $R_1$ in the ideal model. For resistive ports and root-mean-square signal amplitudes,The example input power is $0.25\mu \mathrm{W}$ and load power $0.25\mathrm{mW}$, giving $G_P=1000$. The additional output energy comes from the amplifier's supply, controlled by the input signal. The impedance condition for converting voltage gain to power gain is essential: voltage magnitude alone does not specify load power.

Voltage gain is conventionally expressed as $20{\log }_{10}|A_v|$ decibels, while power gain uses $10{\log }_{10}{G}_P$. This example has voltage gain $20\mathrm{dB}$ and power gain $30\mathrm{dB}$; the two expressions agree when the resistive input and load impedances are equal. A negative voltage gain indicates phase inversion; its magnitude enters the logarithm.

A direct test drives a small sine wave into the input and measures the actual terminal signals with an oscilloscope. Sweeping frequency reveals the frequency response; increasing amplitude reveals amplifier clipping or other departure from the assumed linear gain. Keep the load and measurement impedances specified. Gain can amplify input noise along with signal, so the signal-to-noise ratio requires its own measurement.

An input bias current is a direct-current component drawn by an input terminal of a real operational amplifier. Through a source resistance it produces a voltage error, and through the feedback capacitor of an op amp integrator it can produce an output ramp. Its magnitude and sign depend on the amplifier technology and operating conditions.

The input offset voltage of an operational amplifier is the differential direct-current input voltage needed to make its output zero at a specified operating point. Its effect resembles a small unwanted input signal and can cause sustained output drift in an op amp integrator.

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.

A voltage-feedback operational amplifier with its noninverting input grounded and input and feedback resistances $R_i,R_f$ has ideal gain $-R_f/R_i$. Virtual short and negligible input current give the result by node-current balance.

An ideal operational amplifier with a grounded noninverting input, input resistor $R_i$, and parallel feedback resistor $R_f$ and capacitor $C$ implements a low-pass filter. The virtual short and Kirchhoff's current law give $V_{\mathrm{out}}/V_{\mathrm{in}}=-Z_f/R_i$, where $Z_f=R_f/(1+i\omega R_{f}C)$. The DC gain magnitude is $R_f/R_i$ and the cutoff frequency is $1/(2\pi R_{f}C)$. The capacitance reduces high-frequency feedback electrical impedance, causing the gain magnitude to fall.

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.

An ideal inverting operational amplifier with series $R_1,C_1$ input and parallel $R_2,C_2$ feedback has the displayed transfer function, where $K=R_2/R_1$ and ${\tau }_i=R_i{C}_i$. Its peak is at ${\omega }_0=1/\sqrt{{\tau }_1{\tau }_2}$ with voltage gain $K{\tau }_1/({\tau }_1+{\tau }_2)$. A wide input-high-pass/feedback-low-pass separation produces a broad plateau, not a narrow resonance. Validity requires stable negative feedback, sufficient loop gain and linear output operation.

Manufacturer's data sheet, revision D, October 2015.

Section 6.5, p. 5, gives typical input impedance $2\mathrm{M\Omega }$ at $25^{\circ }\mathrm{C}$ and typical large-signal voltage gain $200V/mV$. Section 7.4.2 describes closed-loop feedback. These are device-specific examples for interpreting a real operational amplifier.

A voltage-feedback operational amplifier driven at its noninverting input, with a feedback divider $R_f,R_g$ at its inverting input, has ideal gain $1+R_f/R_g$ under stable negative feedback.

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]

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.

University electronics course, Chapter 3: ideal high-gain negative feedback and practical operational amplifier limits.

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.

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.