 Solution

Treat the $p^{+}$ strips as much more strongly doped than the $n$ bulk, so the depletion layer lies predominantly in the latter. Fully ionized donors leave uniform space charge $\rho =e{N}_D$. In one dimension,At the depletion boundary $x=d$, the minimum-bias solution has $E(d)=0$. Thus $E(x)=e{N}_D(x-d)/{ϵ}_{\mathrm{Si}}$, and the magnitude of the potential drop isFor $N_D=10^{18}{\mathrm{m}}^{-3}$, $d=300\mu \mathrm{m}$, and ${ϵ}_r=11.6$, this isThe applied reverse bias is approximately this value; more precisely the built-in junction potential contributes to the total depletion voltage and is subtracted from the required external bias. It is small on the scale of seventy volts and is not specified. Here the stated free-carrier density is used as the effective donor density of the neutral bulk before depletion.