Abrupt-junction depletion width and capacitance ($W=\sqrt{2{\epsilon }_s(V_{\mathrm{bi}}-V)(N_A+N_D)/(e{N}_A{N}_D)}$)

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 $N_A{w}_p=N_D{w}_n$. Integrating Poisson equation gives $V_{\mathrm{bi}}-V=e(N_A{w}_p^2+N_D{w}_n^2)/(2{\epsilon }_s)$, with positive $V$ denoting forward bias. Therefore $W=w_n+w_p$ has the displayed value, $w_n=N_{A}W/(N_A+N_D)$ and $w_p=N_{D}W/(N_A+N_D)$. The positive depletion capacitance per area is $C_A={\epsilon }_s/W$. It equals $d{Q}_p/dV=-d{Q}_n/dV$ when $Q_p=-e{N}_A{w}_p$ and $Q_n=e{N}_D{w}_n$ 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.