One-sided depletion voltage
= One-sided depletion voltage
{title2=$V_{\mathrm{dep}}=eN_Dd^2/(2\epsilon)$}
For a uniformly doped <p-n junction> whose depleted thickness $d$ lies almost entirely on the donor side, <Poisson's equation> gives $dE/dx=eN_D/\epsilon$. The minimum field obeys $E(d)=0$. Integrating the triangular <electric field> profile gives $V_{\mathrm{dep}}=eN_Dd^2/(2\epsilon)$. The external reverse bias is this total voltage minus the built-in <Built-in potential>[junction potential].