A semiconductor with permittivity occupies the space z 0. One dopes such a semiconductor by

A semiconductor with permittivity ε occupies the space z ≥ 0. One “dopes” such a semiconductor by implanting neutral, foreign atoms with uniform density N_D in the near-surface region 0 ≤ z ≤ d. Assume that one electron from each dopant atom ionizes and migrates to the free surface of the semiconductor. The final result (illustrated by the diagram) is a region with uniform positive charge density _eN_D and a layer of negative charge with density σ localized at z = 0.

(a) Find and sketch the electric field E₊(z) at every point in space produced by the volume charge.

(b) Find σ and the electric field E₋(z) produced by σ. Sketch E₋ on the same graph used to sketch E₊ in part (a).

(c) Sketch the total electric field and check that your graph is consistent with integrating Gauss’ law from z = −∞to z=∞.

Transcribed Image Text:

eND + O + + z = 0 + + + + + + + + z =d - Z