 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 $U_P=e{N}_D{a}^2/(2{\epsilon }_s)$; with built-in junction voltage $V_{\mathrm{bi}}$, drain-end pinch-off occurs approximately at $V_{DS}=U_P-V_{\mathrm{bi}}+V_{GS}$.