A MOSFET can operate as a small signal resistor if its gate and drain are shorted called as 'diode connected' device in analogy with its bipolar counterpart as in figure (a).

This configuration exhibits a small signal behavior similar to a two terminal resistor. The transistor is always in saturation because the drain and the gate have the same potential. Using the small signal equivalent as shown in figure (b) to obtain the impedance of the device, we write V1 = Vx and Ix = Vx/ro = 1/gm. If body effect exists, we can use the circuit as shown below to write V1 = -Vx, Vbs = -Vx and

$(g_m + g_{mb})V_x +\frac{V_x}{R_0} = I_x$

It follows that

$\frac{V_x}{I_x} = \frac{1}{g_m+g_{mb}+r_0^-1}$

$=\frac{1}{g_m+g_{mb}}||r_0$

$\approx \frac {1}{g_m+g_{mb}}$

For negligible channel length modulation, the above equation can be substituted in the voltage gain of resistive load CS amplifier for load impedance, yielding

$A_v = -g_{m1}\frac{1}{g_{m2}+g_{mb2}}$

$=-\frac{g_{m1}}{(g_{m2})(1+\eta)}$

where $\eta = \frac{g_{mb2}}{g_{m2}}$

Expressing gm1 and gm2 in terms of device dimensions and bias currents, we have

$A_v = -\frac{\sqrt{2\mu_nC_{ox}(\frac{W}{L})_1I_{D1}}}{\sqrt{(2\mu_nC_{ox}(\frac{W}{L})_2I_{D2}}} \frac{1}{(1+\eta)}$

Since, $I_{d1} = I_{d2}$,

$A_p = -\sqrt {\frac{(\frac{W}{L})_1}{(\frac{W}{L})_2}} \frac{1}{(1+\eta)} $

Hence, if the variation of with the output voltage is neglected, the gain is independent of the bias currents and voltages as long as M1 stays in saturation.