How it works
The MOSFET small-signal model consists of a gate-to-source voltage vgs, a dependent current source gm·vgs between drain and source, and output resistance ro = 1/λID in parallel with the drain. Transconductance gm = 2ID/(VGS − VTN) = √(2·kn·ID). At ID = 1 mA with kn = 2 mA/V², gm = 2 mA/V. For a common-source amplifier with RD = 10 kΩ, voltage gain Av = −gm·(RD ∥ ro). Gate input impedance is theoretically infinite at DC, making MOSFETs ideal for op-amp input stages. The body effect introduces an additional current source gmb·vbs when source is not tied to body, adding to complexity.
Key points to remember
Unlike the BJT, the MOSFET draws zero DC gate current, so IB = 0 and there is no rπ in the model — the entire input impedance looking into the gate is determined by external bias resistors. gm for a MOSFET is typically lower than a BJT at the same bias current, which is why MOSFET amplifiers generally have lower voltage gain than BJT stages at identical quiescent currents. The channel-length modulation parameter λ (typically 0.01 to 0.1 V⁻¹) accounts for the finite output resistance ro. Body effect threshold shift ΔVT = γ(√(2φF + VSB) − √(2φF)) must be considered in multistage CMOS circuits where source is not grounded.
Exam tip
Every Anna University analog electronics exam asks you to derive the voltage gain of a common-source amplifier using the small-signal model — draw the model first, mark vgs clearly, then write the KCL at the drain node before simplifying.