Transistor Amplifier

Transistor current under saturation is dependent upon $v_{GS}$, thus making it a transconductance amplifier. A load resistor can be added to turn the transistor into a voltage amplifier.

$v_o=v_{DS}=V_{DD}-i_D{R}_D$
$v_{DS}=V_{DD}-\frac{1}{2}{k}_n{R}_D(v_{GS}-V_t{)}^2$

At saturation triode boundary
$V_{GS}{|}_B=V_t+\frac{\sqrt{2k_n{R}_D{V}_{DD}+1}-1}{k_n{R}_D}$

Saturation
$V_{DS}=V_{DD}-\frac{1}{2}{k}_n{R}_D(V_{GS}-Vt{)}^2$

$A_v=\frac{\partial v_{DS}}{\partial v_{GS}}{|}_{v_{GS}=V_{GS}}$
$A_v=-k_n(V_{OV})R_D$

$I_D=\frac{k_n}{2}{V}_{OV}^2$
$A_v=-\frac{I_D{R}_D}{V_{OV}/2}$
$|A_{v,max}|=\frac{V_{DD}-V_{DS}{|}_B}{V_{OV}{|}_B/2}=\frac{V_{DD}-V_{OV}{|}_B}{V_{OV}{|}_B/2}$

Small signal

$i_D=\frac{k_n}{2}(V_{GS}+v_{gs}-Vt{)}^2$
$=\frac{k_n}{2}(V_{GS}-V_t{)}^2+k_n(V_{GS}-V_t)v_{gs}+\frac{k_n}{2}{v}_{gs}^2$
Neglect nonlinear term with $v_{gs}\ll 2V_{OV}$
$g_m=k_n{V}_{OV}=\sqrt{2k_n{I}_D}=\frac{2I_D}{V_{OV}}$
$A_v=-g_m{R}_D$
$r_o=\frac{1}{\lambda I_D}$

Configurations

Ground one of three terminals, creating two port network with common ground. Common-source most popular.