 Solution

The positive charge per unit interface area on the n side and negative charge on the p side areUse the signed p-side charge $Q=Q_p$ because $V_{\mathrm{ext}}$ is the p-side voltage relative to n. Differentiating the width relation givesUsing part (e), the positive junction capacitance per unit area isIt decreases with increasing reverse-bias magnitude as $(V_b+|V_{\mathrm{ext}}|{)}^{-1/3}$. Equivalently,so a linear $C^{-3}$ versus voltage plot diagnoses this ideal linear grading and can determine $\alpha$.

For this linearly graded junction capacitance, the charge convention matters: if “stored charge” means the positive magnitude $Q_n$, then $d{Q}_n/d{V}_{\mathrm{ext}}=-{\epsilon }_s/w$; the positive measured capacitance is $-d{Q}_n/d{V}_{\mathrm{ext}}=d{Q}_n/d{V}_R$ for $V_R=-V_{\mathrm{ext}}$. The total net depletion charge is zero and is not the stored terminal charge. At substantial forward bias, minority carrier storage and the failure of the sharp depletion approximation prevent extrapolation of the apparent divergence at $V_{\mathrm{ext}}=V_b$.