= 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 $C_{\rm ox}$, threshold voltage $V_T$, channel width $W$, length $L$, constant <charge carrier mobility> $\mu$ and $V_{
m ov}=V_{GS}-V_T>0$, the local sheet charge magnitude is $C_{\rm ox}(V_{
m ov}-V(x))$. Integrating the drift current along the channel gives
$$
\boxed{I_D=\mu C_{\rm ox}\frac WL\left(V_{
m ov}V_{DS}-\frac{V_{DS}^2}{2}\right),\quad0\leq V_{DS}\leq V_{
m ov}.}
$$
At $V_{DS}=V_{
m ov}$, the drain end reaches pinch-off. The ideal long-channel saturation current is $I_D=\mu C_{\rm ox}WV_{
m ov}^2/(2L)$; 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.
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