= Transistor small-signal amplifier
{title2=$A_v=-g_m(R_D\parallel r_o\parallel R_L)$}
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 $g_mv_i$ 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> give
$$
v_i=\frac{r_i}{R_s+r_i}v_s,\qquad
\frac{v_o}{R_D}+\frac{v_o}{r_o}+\frac{v_o}{R_L}+g_mv_i=0.
$$
Consequently the port <voltage gain>, overall source-to-output <voltage gain>, and load <current gain> magnitudes are
$$
A_v=\frac{v_o}{v_i}=-g_mR_P,\qquad
\frac{v_o}{v_s}=-g_mR_P\frac{r_i}{R_s+r_i},\qquad
|A_i|=\frac{|v_o|/R_L}{|v_i|/r_i}=g_mr_i\frac{R_O}{R_O+R_L}.
$$
This <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> $P_L=v_o^2/R_L$ is
$$
G_P=\frac{P_L}{P_i}
=g_m^2r_i\frac{R_O^2R_L}{(R_O+R_L)^2}.
$$
Unlike $|A_i|$, this tends to zero as either $R_L\to0$ or $R_L\to\infty$. Differentiation gives
$$
\frac{dG_P}{dR_L}=g_m^2r_iR_O^2\frac{R_O-R_L}{(R_O+R_L)^3},\qquad
G_{P,\max}=\frac{g_m^2r_iR_O}{4}\quad\text{at }R_L=R_O.
$$
Thus 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 $P_i=0.100\,\mathrm{\mu W}$ and $P_L=0.181406\,\mathrm{\mu W}$: $G_P=1.81406$. 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>.
\Image[/transistor-controlled-source-load-and-gain.png]
{title=Chosen <transistor> model: <transconductance>, load and supply}
{description=Original calculated illustration. Top: incremental <electrical circuit>. Lower left: the no-external-load <direct current> load line intersects three chosen local device curves. Lower right: port <voltage gain>, load <current gain> and <power gain> as $R_L$ changes. All parameters are stipulated teaching values.}
{source=transistor-controlled-source-load-and-gain.py}
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 $V_Q=V_{DD}-R_DI_Q$. Locally extend the model as $I=I_Q+g_mv_i+(V-V_Q)/r_o$. With the external <electrical load> isolated at <direct current>, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_mR_Ov_i$. 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>.
Back to article page
