Common-gate amplifier

The common-gate amplifier drives the input at the source, takes the output at the drain, and holds the gate at AC ground. Non-inverting, low input resistance, and the natural current buffer of the MOSFET family.

Why the gain is positive

Hold the gate at AC ground and drive the source with $v_s$. The gate-source voltage is gate minus source, and with the gate fixed at zero AC, $v_{gs}=-v_s$. The drain current is then

$i_d=g_m{v}_{gs}=-g_m{v}_s$

That current flows in the drain branch. Taking the usual convention that drain current flows into the drain, the AC drain voltage is

$v_d=-i_d{R}_D=-(-g_m{v}_s)R_D=+g_m{R}_D{v}_s$

so

$A_v=\frac{v_o}{v_s}=+g_m{R}_D\text{(non-inverting)}$

Same gain magnitude as the Common-source amplifier, but a $+$ sign instead of $-$. The double negative cancels out: the gate-source voltage inverts the input, and the drain inverts again.

Input at source, output at drain, gate AC-grounded; non-inverting, low $R_{in}=1/g_m$.

Low input resistance, high output resistance

Looking into the source you see the transistor’s source terminal, which presents

$R_{in}=\frac{1}{g_m}$

typically only a few hundred ohms. That low value is the defining trait of the common gate. The MOSFET T-model makes it fall straight out: the $1/g_m$ element sits exactly between gate and source, so it is what you look into from the source. This $1/g_m$ result neglects the transistor’s output resistance $r_o$. Including it, the exact input resistance is $R_{in}=(R_D+r_o)/(1+g_m{r}_o)$, which actually depends on the drain load $R_D$. For the usual case $g_m{r}_o\gg 1$ with a modest $R_D$ it collapses back to $\approx 1/g_m$, which is why the simple form is used at this level. The output resistance at the drain is high by contrast, much higher than the common-source case.

What it is good for

A low input resistance is bad for voltage amplification from a high-impedance source, since you lose most of the signal in input loading. But it makes the common gate a good current buffer: it takes a current at its low-impedance input and delivers it to a high-impedance output essentially unchanged. Its real use is as the top half of the cascode (a Common-source amplifier feeding a common gate), where the CS stage’s transconductance and the CG stage’s high output resistance combine to build the very high-gain stages inside op-amps.