Questions tagged [depletion-region]

When a p–n junction is reverse-biased (that is, “p” is under negative voltage and “n” is under positive voltage), or if the p–n voltage is very low, then the current through the junction is weak. It is said that the junction is “turned off”. This effect is due to depletion region that forms on both sides of the junction. It is depleted of charge carriers, either majority or minority. Under the reverse bias the depletion region has a significant space charge: positive one at the “n” side and negative one on the “p” side. This forms a doubly charged layer, whereas the semiconductor away of the depletion zone is quasi-neutral. The resulting electric field deters majority carriers from entering and pushes them out of the zone. Minority carriers are thrown into the opposite side, where they become majority carriers and are pushed out. That’s why carriers that form in the region (by thermal excitations or other effects) don’t accumulate.

A simple model of the depletion region is the full depletion. It assumes that the boundary between the depleted and the quasi-neutral regions is abrupt. It also neglects the width of the transition between “p” and “n” conduction types. The result is two uniform regions of a space charge of opposite signs. Magnitudes of these charges are determined by concentration by respective doping donor/acceptor atoms. This approximation is usable under significant reverse voltage, when the junction behaves similarly to a capacitor.

In a real semiconductor there is no sharp depletion–quasi-neutral boundary, and the transition between “p” and “n” types has some width as well. These factors are not significant under the reverse voltage much exceeding the potential differences of the electronic band structure, but they should be accounted for under a low voltage.