= Abrupt-junction depletion width and capacitance
{title2=$W=\sqrt{2\varepsilon_s(V_{\rm bi}-V)(N_A+N_D)/(eN_AN_D)}$}
For a planar abrupt <p-n junction> with ionized densities $N_A,N_D$, p-side width $w_p$ and n-side width $w_n$, zero external field requires $N_Aw_p=N_Dw_n$. Integrating <Poisson equation> gives $V_{\rm bi}-V=e(N_Aw_p^2+N_Dw_n^2)/(2\varepsilon_s)$, with positive $V$ denoting forward bias. Therefore $W=w_n+w_p$ has the displayed value, $w_n=N_AW/(N_A+N_D)$ and $w_p=N_DW/(N_A+N_D)$. The positive depletion capacitance per area is $C_A=\varepsilon_s/W$. It equals $dQ_p/dV=-dQ_n/dV$ when $Q_p=-eN_Aw_p$ and $Q_n=eN_Dw_n$ are signed space charges per area. Differentiating a positive charge magnitude with respect to forward voltage instead gives a negative derivative. The square-root law assumes partially depleted regions, negligible mobile charge and no substantial injected-carrier storage.
Back to article page
