 Solution

A field-effect transistor uses a gate voltage to control the conducting channel between source and drain. In a MOSFET, the gate is insulated from the semiconductor: an electric field changes the channel carrier population with little steady gate current. An n-channel device creates an semiconductor inversion layer when the gate sufficiently attracts electrons to a p-type surface. The drain supply provides the output power; the gate controls that flow.

The long-channel MOSFET current law follows from the gradual-channel approximation. With gate capacitance per area $C_{\mathrm{ox}}$, threshold voltage $V_T$, channel width $W$, length $L$, constant charge carrier mobility $\mu$ and $V_{mov}=V_{GS}-V_T>0$, the local sheet charge magnitude is $C_{\mathrm{ox}}(V_{mov}-V(x))$. Integrating the drift current along the channel givesAt $V_{DS}=V_{mov}$, the drain end reaches pinch-off. The ideal long-channel saturation current is $I_D=\mu C_{\mathrm{ox}}W{V}_{mov}^2/(2L)$; this assumes negligible channel-length modulation. A junction field-effect transistor instead changes the channel width through a reverse-biased depletion region.

Experimentally, measure $I_D$ against $V_{GS}$ at fixed drain bias to see gate control, and against $V_{DS}$ at several fixed gate biases to distinguish the approximately linear and saturation regimes. Simultaneously checking gate leakage tests the insulated-gate assumption.