image
Figure 1.10 Schematic sketch of the charge in the depletion region.

In our case assuming:

dEdx={qNdεs0<x<xnqNaεsxp<x<0 (1.234)

image (1.234)

then

E={qNdεs(xnx)0<x<xnqNaεs(xp+x)xp<x<0 (1.235)

image (1.235)

The electric field is continuous and it also equals zero outside the junction region, thus:

E(0+)n=E(0)pE(xn)=E(xp)=0 (1.236)

image (1.236)

which yields: ...

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