= Impedance condition for converting voltage gain to power gain
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 $VI^*$, with the <complex conjugate> arising from the relative phase of the two sinusoids. Averaging $v(t)i(t)$ over a cycle gives the <active power>
$$
P=\operatorname{Re}(VI^*)=|V|^2\operatorname{Re}Y.
$$
To see the averaging step, write $v(t)=\sqrt2|V|\cos(\omega t+\phi_V)$ and similarly for $i(t)$. The product's cycle average is $|V||I|\cos(\phi_V-\phi_I)$, the <real part> of $VI^*$. Peak-amplitude phasors instead require an extra factor $1/2$.
Let $Y_{\rm in}$ describe the actual <input impedance> and $Y_L$ the actual <load impedance>. Take $\operatorname{Re}Y_{\rm in}>0$ and $\operatorname{Re}Y_L>0$ 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> $A_v=V_L/V_{\rm in}$,
$$
\boxed{G_P=\frac{P_L}{P_{\rm in}}
=|A_v|^2\frac{\operatorname{Re}Y_L}{\operatorname{Re}Y_{\rm in}}.}
$$
Thus $G_P=|A_v|^2$ requires equality of the relevant <electrical conductances>, not merely equality of impedance <moduli>. For example, input $Z_{\rm in}=50\,\Omega$ 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 becomes
$$
G_P=|A_v|^2\frac{R_{\rm in}}{R_L},\qquad
10\log_{10}G_P=20\log_{10}|A_v|+10\log_{10}\frac{R_{\rm in}}{R_L}.
$$
Take an original example with measured $|A_v|=0.5$, $R_{\rm in}=10\,\mathrm{k\Omega}$ 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 $A_0V_{\rm in}$ and series resistance $R_o$. The <voltage divider> gives
$$
A_v=A_0\frac{R_L}{R_o+R_L},\qquad
G_P=|A_0|^2R_{\rm in}\frac{R_L}{(R_o+R_L)^2}.
$$
With $A_0=0.6$ 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.
\Image[/voltage-and-power-gain-loading.png]
{title=Voltage transfer and delivered power respond differently to the same load resistance}
{description=Original Thévenin amplifier model with open-circuit voltage gain 0.6, output resistance 10 ohms and input resistance 10 kilohms. The measured voltage gain increases toward 0.6 as the load grows; power gain peaks at a 10-ohm load. The marked 50-ohm point has voltage gain 0.5 and power gain 50.}
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.<Output Load Termination in the Keysight BenchVue Function Generator documentation>[Output Load Termination, fixed output impedance and the 50-ohm/open-circuit examples]{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 $1.00\,\mathrm{\mu W}$. The larger terminal reading accompanies much less delivered power. Specify the actual load, <frequency> and amplitude convention when comparing gain measurements.
Back to article page
