The figure (a) shows a common gate (CG) stage which senses the input at the source and produces the output at the drain. The gate is connected to a dc voltage to establish proper operating conditions. The bias current of M1 flows through the input signal source. Alternatively, as shown in figure (b), M1 can be biased by a constant current source, with the signal capacitively coupled to the circuit.

Let us assume that Vin decreases from a large positive value. For $V_{in}\geq V_b - V_{TH}$ , M1 is off and Vout = Vdd. For lower values of Vin, we can write

$I_D = \frac{1}{2} \mu_nC_{ox}\frac{W}{L}(V_b - V_{in}-V_{HT})^2$

if M1 is in saturation, as Vin decreases, so does Vout, eventually driving M1 into the triode region if

$V_{DD} - \frac{1}{2} \mu_nC_{ox}\frac{W}{L}(V_b - V_{in}-V_{HT})^2 R_D = V_b - V_{TH}$

The input-output characteristics is shown. If M1 is saturated, we can express the output voltage as

$V_{out} = V_{DD} - \frac{1}{2} \mu_nC_{ox}\frac{W}{L}(V_b - V_{in}-V_{HT})^2 R_D$

Obtaining a small signal gain of

$\frac{\delta V_{out}}{\delta V_{in}}= -\mu_nC_{ox} \frac{W}{L}(V_b - V_{in}-V_{HT})(-1-\frac{\delta V_{TH}}{\delta V_{in}})R_D$

Since, $\frac{\delta V_{out}}{\delta V_{in}}= -\mu_nC_{ox} \frac{W}{L}(V_b - V_{in}-V_{HT})(1+\eta)$

we have

$\frac{\delta V_{out}}{\delta V_{in}}= -\mu_nC_{ox} \frac{W}{L}R_D(V_b - V_{in}-V_{HT})(1+\eta)$

$=g_m(1+\eta)R_D$