Integrated Nanophotonic Devices, 2nd Edition

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

In our case assuming:

$\frac{\text{d}E}{\text{d}x}=\{\begin{array}{c}\frac{q{N}_d}{{\epsilon }_s}0<x<x_n\\ \frac{-q{N}_a}{{\epsilon }_s}-x_p<x<0\end{array}$ (1.234)

(1.234)

then

$E=\{\begin{array}{c}\frac{-q{N}_d}{{\epsilon }_s}(x_n-x)0<x<x_n\\ \frac{-q{N}_a}{{\epsilon }_s}(x_p+x)-x_p<x<0\end{array}$ (1.235)

(1.235)

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

$\begin{array}{l}E{(0^{+})}_n=E{(0^{-})}_p\\ E(x_n)=E(-x_p)=0\end{array}$ (1.236)

(1.236)

which yields: ...