 Full-depletion capacitance of a symmetric junction slab

Let each side of a symmetric abrupt p-n junction have thickness $d$, area $\mathcal{A}$ and doping density $N$. Partial depletion gives $w^2={\epsilon }_s(V_{\mathrm{bi}}-V)/(eN)$. Both sides become fully depleted at $V_{\mathrm{fd}}=V_{\mathrm{bi}}-eN{d}^2/{\epsilon }_s$. Beyond that point, the ionized dopant charges are fixed, but an incremental terminal voltage changes the electric field uniformly across the full thickness $2d$. Gauss's law then gives $d{Q}_{\mathrm{terminal}}={\epsilon }_s\mathcal{A}dV/(2d)$, up to terminal orientation. The measured positive capacitance approaches $C_{\mathrm{fd}}$ rather than vanishing. This geometrical result assumes ideal contacts, negligible mobile carriers, no breakdown and no other series capacitances.