A silicon p‑n junction has an acceptor concentration $N_A=1\times 10^{18}{\text{cm}}^{-3}$ on the p‑side and a donor concentration $N_D=1\times 10^{16}{\text{cm}}^{-3}$ on the n‑side. The relative permittivity of silicon is ${\epsilon }_r=11.7$, ${\epsilon }_0=8.85\times 10^{-14}\text{F/cm}$, the built‑in potential $V_{bi}=0.85\text{V}$ and a reverse bias $V_R=20\text{V}$ is applied. (a) Find the total depletion width $W$ (in $\mu \text{m}$). (b) Determine the junction capacitance per unit area $C^{\prime }={\epsilon }_s/W$ (in $\mu {\text{F/cm}}^2$).

Step-by-step solution

1. Calculate the semiconductor permittivity: ${\epsilon }_s={\epsilon }_r{\epsilon }_0=11.7\times 8.85\times 10^{-14}=1.03545\times 10^{-12}\text{F/cm}$.
2. Use the depletion‑width formula for an abrupt junction under bias: $W=\sqrt{\frac{2{\epsilon }_s}{q}\frac{N_A+N_D}{N_A{N}_D}(V_{bi}+V_R)}$.
3. Compute the factor $\frac{N_A+N_D}{N_A{N}_D}=\frac{1\times 10^{18}+1\times 10^{16}}{(1\times 10^{18})(1\times 10^{16})}=1.01\times 10^{-16}{\text{cm}}^3$.
4. Calculate $\frac{2{\epsilon }_s}{q}=\frac{2\times 1.03545\times 10^{-12}}{1.6\times 10^{-19}}=1.2943\times 10^7\text{Vcm}$.
5. Multiply by the previous factor: $1.2943\times 10^7\times 1.01\times 10^{-16}=1.3072\times 10^{-9}$.
6. Add the voltages: $V_{bi}+V_R=0.85+20=20.85\text{V}$.
7. Now $W=\sqrt{(1.3072\times 10^{-9})(20.85)}=\sqrt{2.725\times 10^{-8}}=5.22\times 10^{-4}\text{cm}=5.22\mu \text{m}$.
8. Capacitance per unit area is $C^{\prime }={\epsilon }_s/W=1.03545\times 10^{-12}/5.22\times 10^{-4}=1.983\times 10^{-9}{\text{F/cm}}^2$.
9. Convert to microfarads: $1.983\times 10^{-9}\text{F}=1.983\times 10^{-3}\mu \text{F}$.
10. Thus $C^{\prime }=1.98\times 10^{-3}\mu {\text{F/cm}}^2$.

Concept tested: Depletion width and junction capacitance

Published 16 September 2026 · How this solution was prepared and checked