= Solution
An n-channel <MOSFET> has two heavily n-doped source and drain regions in a p-type body, with an insulated gate above the intervening surface. A positive gate voltage first depletes surface <electron holes> and then induces an <electron> <semiconductor inversion layer>. Above threshold $V_T$ this layer connects source to drain. Gate voltage controls the channel charge through the oxide capacitance, while the insulating gate permits a very small steady gate current.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2004-mosfet-cross-section.png]
{title=An induced electron channel connects the source and drain of an n-channel MOSFET}
{description=Original schematic cross-section in the on state. The oxide separates the gate electrically from the body; the thin surface channel is an inversion layer rather than a permanently n-doped bridge.}
For the <long-channel MOSFET current law>, let $C_{\rm ox}$ be oxide capacitance per area, $W,L$ channel width and length, and use constant mobility $\mu_n$. In the gradual-channel approximation,
$$
I_D=\mu_nC_{\rm ox}\frac WL\left[(V_{GS}-V_T)V_{DS}-\frac{V_{DS}^2}{2}\right],\qquad0<V_{DS}<V_{GS}-V_T.
$$
At larger drain voltage the ideal channel pinches off near the drain and the current saturates at
$$
\boxed{I_{D,\rm sat}=\frac12\mu_nC_{\rm ox}\frac WL(V_{GS}-V_T)^2.}
$$
Carriers still flow through the pinch-off region; pinch-off does not mean zero drain current. These formulas neglect channel-length modulation, short-channel effects and velocity saturation. A p-channel device reverses the carrier type and operating polarities. Gate control makes MOSFETs useful as electronic switches and amplifiers.
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