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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.

Loaded voltage gain is the ratio of an amplifier's output to input signal electric potential differences with a specified load connected. In a small-signal model, let the open-circuit output be $v_{\mathrm{oc}}=A_0{v}_{\mathrm{in}}$ and let the complete amplifier have output impedance $Z_o$. A load $Z_L$ then givesThis is the voltage divider result from Thévenin's theorem at the output terminals. The university source explains why a linear circuit can be replaced at specified terminals by a source and series impedance.^[ref] The output impedance used here must describe the complete amplifier at the stated operating point and frequency, including its feedback.

The input terminal matters too. For a generator $v_s$ with source impedance $Z_s$ driving amplifier input impedance $Z_i$, the corresponding source-to-load ratio in a unilateral model isThis includes both input and output loading; it differs from the terminal-to-terminal voltage gain defined above.

For an original resistive example, choose $A_0=12$, $Z_o=100\Omega$, $Z_L=900\Omega$, $Z_i=1\mathrm{k\Omega }$ and $Z_s=50\Omega$. The loaded terminal gain is $10.8$, while the source-to-load gain is $72/7\approx 10.286$. Reducing the load to $100\Omega$ makes the terminal gain six, even though the stipulated open-circuit gain remains twelve. These are model parameters, not device specifications.

A direct test holds the actual amplifier input signal fixed, changes a known load and measures the output with an oscilloscope whose probe impedance is included in the effective load. Use small enough signals to retain linear operation and respect the amplifier output range. If the output-to-input ratio changes with loading, an unloaded gain alone is insufficient to predict the circuit's behavior.

At nonzero frequency the impedances and gain can be complex. A reactive-load phase shift in a finite-output-impedance amplifier can change both magnitude and phase, while input-output coupling or nonlinear current limiting can invalidate this simple unilateral divider model. The ideal feedback resistance-ratio examples in the manufacturer's introduction apply within their corresponding operating assumptions.^[ref]

A reactive-load phase shift in a finite-output-impedance amplifier occurs because the load and output impedance form a frequency-dependent divider. In the model of loaded voltage gain, assume a positive real output resistance $R_o$, an ideal capacitive load $C_L$ and open-circuit gain $A_0$ approximately constant over the band of interest. With $\omega =2\pi f$,The divider follows from the terminal equivalent described by the university circuit notes.^[ref]

Putting $\tau =R_o{C}_L$ givesThus even an ideal capacitor can change both amplitude and phase when the source driving it has finite resistance. Its reactive current and energy exchange are described by complex electrical power; zero average dissipation in the ideal capacitor does not mean zero output-current demand.

For an original example, choose $R_o=100\Omega$, $C_L=10\mathrm{nF}$ and $A_0=10$. Then $\tau =1\mu \mathrm{s}$ and the load pole is $f_p=1/(2\pi \tau )\approx 159.2\mathrm{kHz}$. At that frequency the gain magnitude is $10/\sqrt{2}\approx 7.071$ and the extra phase lag is $45^{\circ }$. A measured input of $10\mathrm{mV}$ root mean square gives about $70.7\mathrm{mV}$ root mean square at the capacitor, requiring approximately $0.707\mathrm{mA}$ root-mean-square capacitive current.

A bench test holds the actual amplifier input fixed while sweeping frequency and compares loaded and unloaded output amplitudes and phases. Include probe capacitance in $C_L$. The group delay introduced by a capacitive amplifier load also matters when comparing pulse waveforms.

This single-pole calculation assumes the stated constant open-circuit gain and resistive output impedance. The complete amplifier may have additional frequency dependence and feedback effects. The manufacturer's ideal feedback-gain examples are defined within their operating assumptions.^[ref] Check the actual output-current and linear operating range before interpreting a departure from this model as a capacitor-only effect.