= Pinch-off in a junction field-effect transistor
In an n-channel <junction field-effect transistor>, increasing drain voltage increases the local gate-channel reverse bias near the drain. The <depletion region> can meet there, ending the quasi-neutral channel before the drain. Carriers already in the channel are swept across the remaining high-field region, so drain-end pinch-off produces current saturation rather than zero current. A sufficiently negative gate voltage can instead deplete the whole channel and cut the device off. In the one-sided <depletion approximation>, a channel half-width $a$ and donor density $N_D$ give a depletion threshold voltage $U_P=eN_Da^2/(2\varepsilon_s)$; with built-in junction voltage $V_{\rm bi}$, drain-end pinch-off occurs approximately at $V_{DS}=U_P-V_{\rm bi}+V_{GS}$.
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