By virtue of its transconductance, a MOSFET converts variations in its gate-source voltage to a small-signal drain current, which can pass through a resistor to generate an output voltage. Figure (a) shows the common-source CS stage which performs such an operation.

The input impedance of the circuit is very high at low frequencies.

If the input voltage increases from zero, M1 is off and Vout = Vdd as shown in figure (b). As Vin approaches Vth, M1 begins to turn on, drawing current from Rd and lowering Vout. If Vdd is not excessively low, M1 turns on in saturation and we have

$V_{out}=V_{DD}-R_D\frac{1}{2}\mu_nC_{0x}\frac{W}{L}(V_{in}-V_{TH})^2$

where channel length modulation is neglected. With furthur increase in Vin, Vout drops more and the transistor continues to operate in saturation until Vin exceeds Vout by Vth. At this point,

from which Vin1 - Vth and hence Vout can be calculated. For Vin > Vin1, M1 is in the triode region:

$V_{in1}-V_{TH}=V_{DD}-R_D\frac{1}{2}\mu_nC_{0x}\frac{W}{L}[2(V_{in}-V_{TH})V_{out}-V_{out}^2]$

If Vin is high enough to drive M1 into deep triode region, Vout << 2( Vin - Vth ) and from the equivalent circuit of figure (c).

$V_{out}=V_{DD}\frac{R_{on}}{R_{on}+R_D}=\frac{V_{DD}}{1+\mu_nC_{0x}\frac{W}{L}R_D(V_(in)-V_{TH})}$

Since the transconductance drops in the triode region, we usually ensure that Vout > Vin - Vth, operating to the left of point A in figure (b). Using the input - output characteristic and viewing its slope as the small signal gain, we have:

$A_v=\frac{\partial V_{out}}{\partial V_{in}}=-R_D\mu_nC_{0x}\frac{W}{L}(V_{in}-V_{TH})=-g_mR_D$