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Amplifier clipping occurs when an electronic amplifier cannot follow the output amplitude predicted by its small-signal voltage gain. Its finite amplifier output range truncates the waveform. For a deliberately simple, instantaneous model with positive gain $G$ and fixed lower and upper output limits,Here $v_i$ and $v_o$ are input and output electric potential differences. The flat portions express a nonlinearity: doubling an already excessive input need not change the output. Real output limits depend on the supply, load and output stage. They need not equal the supply rails. Texas Instruments' example integrating amplifier uses separate supply and output-range design goals and explains how unrestricted DC integration eventually reaches saturation.^[ref]

Amplifier clipping can corrupt an area measurement even when the waveform still looks easy to integrate. Suppose $G=10$, $V_{\mathrm{lo}}=-2\mathrm{V}$, $V_{\mathrm{hi}}=2\mathrm{V}$, and a rectangular input is $0.4\mathrm{V}$ for $1\mathrm{ms}$ and zero otherwise. Its true area is $0.4\mathrm{V}\mathrm{ms}$, but the ideal unclipped output of $4\mathrm{V}$ is reduced to $2\mathrm{V}$. Dividing the clipped waveform area by the gain gives50% below the true input area. Any rectangular input at or above $0.2\mathrm{V}$ with the same duration produces this same clipped plateau in the stipulated model. Subsequent integration cannot recover which of those input amplitudes was present.

A useful bench experiment feeds a repeatable pulse into an electronic amplifier and records both input and output on an oscilloscope. Increase the input amplitude while holding the gain, pulse duration and load fixed. The output first scales with the input, then ceases to do so near its limits. Reducing the gain tests whether an apparent input plateau is actually a receiver limit. Comparing measured areas before and after that change reveals the clipped waveform area bias directly.

The instantaneous clipping formula does not describe the internal dynamics of a saturated integrating amplifier. Once its feedback assumptions fail, recovery can take additional time. The ACF2101's historical datasheet explicitly distinguishes output overload from ordinary reset and describes extra recovery delay; the relevant behavior must be characterized for the actual circuit.^[ref] An integrator reset establishes a new starting state, while a suitable input range prevents information being lost before that reset.

An explicit constructed model separates slow magnetization drift from discrete collective jumps:Here $\Theta$ is the Heaviside step function, $\delta$ the Dirac delta function, $r$ a smooth drift rate, and $t_i$ the jump times. The ideal discontinuities stand for events shorter than the resolved timescale; actual magnetic domain walls move over finite times. This equation stipulates jump sizes rather than deriving them from atomic spin. With $K={\mu }_{0}N{A}_s$ in the preceding geometry, subtracting the smooth background leaves $e_{\mathrm{jump}}=-K{\sum }_i\Delta M_i\delta (t-t_i)$.

Let a stable linear filter have causal impulse response $h(t)$, normalized by $\int h(t)dt=1$, and let $G$ be a constant dimensionless voltage gain. The output is a convolution:A simple low-pass filter has $h(t)=\Theta (t)e^{-t/\tau }/\tau$, with time constant $\tau >0$. Its output pulse has peak proportional to $\Delta M_i/\tau$ and signed area $-GK\Delta M_i$. Integrating the convolution proves area preservation because $\int h=1$; a wider filter response lowers peaks and lengthens tails. This ideal filter models resolution, rather than claiming the cited instrument had exactly this response.

The nearby positive jumps in the figure have additive negative pulse areas even when their tails overlap. For $\tau =0.50\mathrm{ms}$ the first of the pair still has $e^{-0.2/0.5}=e^{-0.4}\approx 0.67032$ of its initial height when the second arrives. Thresholding can merge those responses into one detected event; limited signal bandwidth also obscures their separation. The total signed area still records the sum if the complete response is integrated, but cannot identify how that sum was partitioned into events. Equal simultaneous jumps are exactly indistinguishable from a single jump of their summed size in this linear measurement. Oppositely signed simultaneous changes can cancel.

Area preservation requires the stated electronics. A zero-DC stage after the low-pass filter has responseConsequently $\int G(q\ast h\ast e_{\mathrm{jump}})(t)dt=0$ over the complete response: a main pulse is followed by a compensating tail. Integrating only a threshold-selected lobe, clipping the record, or saturating the amplifier breaks the simple pulse-area inference. Recovering a magnetization change from such a readout needs the calibrated frequency response and an appropriate reconstruction. Neither the audible crackle nor the displayed voltage is a direct staircase of magnetization.

Let $K(f)$ be the complete current-to-voltage frequency response from detector input to readout, in $V/A$. For a real stable linear time-invariant system, the shot contribution to the output voltage variance isIf $K(f)=R_{F}GH(f)$ with feedback electrical resistance $R_F$, a reference voltage gain $G$, and dimensionless normalized frequency response $H$, thenThis is the effective noise bandwidth for unit reference magnitude. The transimpedance amplifier factor $R_F$ converts electric current to voltage; $G$ multiplies voltage. Neither is an additional noise source merely because it amplifies an existing fluctuation. In contrast, a real amplifier adds input voltage noise and input current noise, and its feedback resistor contributes thermal electrical noise. Those terms need their own transfers to the output. In particular, input voltage noise generally has a different transfer from an injected detector electric current. Simply multiplying every input specification by $R_{F}G$ is incorrect.

For a unit-gain exponential low-pass filter, with time constant $\tau >0$,The cutoff frequency $f_c$ is not $B_{\mathrm{n}}$. An independent time-domain derivation sums independent event contributions to giveThe same reasoning for a rectangular averaging gate has $h=1/T$ on $[0,T]$ and zero elsewhere, hence effective noise bandwidth $1/(2T)$ and variance $qI/T$, agreeing with the shot noise counting model. Equal numerical durations $T=\tau$ do not make these two different linear filters equivalent.

With the noninverting input grounded, a differential input impedance $R_{\mathrm{in}}$ draws current from the summing node. For zero output impedance, $V_o=-Av$ and Kirchhoff's current law gives $(v-V_i)/R_1+(v-V_o)/R_2+v/R_{\mathrm{in}}=0$. Eliminating $v$ yields the displayed voltage gain. The resistor-ratio result requires $|A|\gg 1+R_2/R_1+R_2/R_{\mathrm{in}}$ as well as stable, unsaturated operation. The same expression applies to small-signal phasors with complex $A(f)$.

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.

A voltage gain determines a power gain only when the electrical conditions at both ports are included. The familiar equality between $20{\log }_{10}|A_v|$ and $10{\log }_{10}{G}_P$ requires a particular impedance relation; it is not a conversion between arbitrary voltage and power ratios.

Use phasors whose moduli are root mean square amplitudes, at one sinusoidal frequency. At a load with electrical impedance $Z$ and electrical admittance $Y=1/Z$, the current is $I=YV$. Its complex electrical power is $V{I}^{\ast }$, with the complex conjugate arising from the relative phase of the two sinusoids. Averaging $v(t)i(t)$ over a cycle gives the active powerTo see the averaging step, write $v(t)=\sqrt{2}|V|\cos (\omega t+{\varphi }_V)$ and similarly for $i(t)$. The product's cycle average is $|V||I|\cos ({\varphi }_V-{\varphi }_I)$, the real part of $V{I}^{\ast }$. Peak-amplitude phasors instead require an extra factor $1/2$.

Let $Y_{\mathrm{in}}$ describe the actual input impedance and $Y_L$ the actual load impedance. Take $\mathrm{Re}{Y}_{\mathrm{in}}>0$ and $\mathrm{Re}{Y}_L>0$ with nonzero terminal signals, and define currents into the amplifier at the input and into the load at the output. With the measured loaded voltage gain $A_v=V_L/V_{\mathrm{in}}$,Thus $G_P=|A_v{|}^2$ requires equality of the relevant electrical conductances, not merely equality of impedance moduli. For example, input $Z_{\mathrm{in}}=50\Omega$ and load $Z_L=(30+40i)\Omega$ both have modulus $50\Omega$, but their electrical admittances have real parts $0.020$ and $0.012\mathrm{S}$. The multiplier is $0.6$, so $G_P=0.6|A_v{|}^2$.

For real resistive ports this becomesTake an original example with measured $|A_v|=0.5$, $R_{\mathrm{in}}=10\mathrm{k\Omega }$ and $R_L=50\Omega$. The voltage gain is $-6.02\mathrm{dB}$, while the power gain is $50$, or $16.99\mathrm{dB}$. An active buffer can reduce the signal's electric potential difference while delivering more signal power than it accepts at its input; the supply provides the additional energy.

The amplifier's output impedance is distinct from the connected load impedance. To show why that distinction matters, model its output by a Thévenin equivalent with open-circuit voltage $A_0{V}_{\mathrm{in}}$ and series resistance $R_o$. The voltage divider givesWith $A_0=0.6$ and $R_o=10\Omega$, the $50\Omega$ load produces the preceding $A_v=0.5$ example. Differentiating $R_L/(R_o+R_L{)}^2$ gives $(R_o-R_L)/(R_o+R_L{)}^3$, so delivered power is largest at $R_L=R_o$, as in the maximum power transfer theorem. Here its power gain is $90$. Increasing $R_L$ much further makes $A_v$ approach $A_0$ but reduces delivered power. That high-load-impedance regime is impedance bridging, useful for sensing the source's voltage with little loading.

A bench demonstration uses a function generator, a known termination and an oscilloscope. The Keysight guide explains that an expected-load setting changes how amplitude is displayed for generators with a fixed series output impedance; a high-impedance load can receive twice the voltage expected for a matched $50\Omega$ termination.^[ref] As an independent low-frequency resistive calculation, an open-circuit source of $1\mathrm{V}$ root mean square with $50\Omega$ series resistance delivers $0.5\mathrm{V}$ and $5\mathrm{mW}$ to $50\Omega$. Into $1\mathrm{M\Omega }$ it delivers about $0.99995\mathrm{V}$ but only $1.00\mu \mathrm{W}$. The larger terminal reading accompanies much less delivered power. Specify the actual load, frequency and amplitude convention when comparing gain measurements.

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 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.

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]

The closed-loop gain from an equivalent differential input-voltage disturbance to the output is the noise gain. For an ideal-input inverting amplifier, $\beta =R_1/(R_1+R_2)$ and $G_{\mathrm{n}}=1+R_2/R_1$, although the voltage gain is $-R_2/R_1$. A dominant-pole amplifier has approximate closed-loop bandwidth $f_T/G_{\mathrm{n}}$ when loop gain is large at low frequency and the circuit is stable.

A non-inverting operational amplifier applies the signal to its non-inverting input and returns part of the output to its inverting input through negative feedback. Connect resistor $R_f$ from the output to the inverting input, and resistor $R_g$ from that input to a reference circuit node at electric potential $V_{\mathrm{ref}}$.

In stable, unsaturated ideal operation, the inputs draw no electric current and their electric potentials are equal. Writing the signal as $V_{\mathrm{in}}$, Kirchhoff's current law at the inverting input givessoFor a grounded reference circuit node, the voltage gain is $1+R_f/R_g$. For example, $R_f=20\mathrm{k\Omega }$ and $R_g=10\mathrm{k\Omega }$ give gain three. The equal-input-potential approximation requires sufficiently large loop gain; amplifier saturation or unstable feedback invalidates it.^[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.

The non-inverting input is grounded, so the finite voltage gain gives $V_{\mathrm{OUT}}=-A{V}^{-}$, or $V^{-}=-V_{\mathrm{OUT}}/A$. Infinite input impedance makes Kirchhoff's current law at the inverting inputSubstitute $V^{-}=-V_{\mathrm{OUT}}/A$ and multiply by $R_1{R}_2$:Hence the inverting operational amplifier hasAs $A\to \infty$, the voltage gain tends to $-R_2/R_1$. Setting $V^{-}=0$ before taking that limit would discard precisely the correction being calculated.

Use an RC low-pass filter: put the resistor in series with the input and the capacitor between the output node and the return conductor. Measure the output voltage across the capacitor, with negligible output loading. The voltage divider then has transfer functionAt the high frequency, the attenuation requirement isAt the low frequency, approximately unchanged amplitude means $2\pi f_{l}RC\ll 1$, soThe upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-frequency amplitude would sharpen the upper bound to approximately $68.7\mu \mathrm{F}$.

A practical illustrative choice is $C=22\mu \mathrm{F}$. Its voltage gain is approximately $0.99896$ at $1\mathrm{Hz}$ and $0.001096$ at $20\mathrm{kHz}$: the latter amplitude is reduced by a factor of about $912$. Its cutoff frequency is $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.

The ratio of output to input signal electric potential difference for specified source and load conditions. A signed voltage gain records whether the output reverses polarity; its value can depend on frequency.