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 $V_{bi}=0.90,\text{V}$ and a reverse bias $V_R=10,\text{V}$ is applied. Find (a) the total depletion width $W$ (in $\mu \text{m}$) and (b) the junction capacitance per unit area $C^{\prime }={\epsilon }_s/W$ (in $\mu {\text{F/cm}}^2$).

Step-by-step solution

1. Calculate ${\epsilon }_s={\epsilon }_{r}varepsilo{n}_0=11.7\times 8.85\times 10^{-14}=1.03545\times 10^{-12},\text{F/cm}$.
2. Compute the concentration factor: $(N_A+N_D)/(N_A{N}_D)=\frac{1.01\times 10^{18}}{1\times 10^{34}}=1.01\times 10^{-16},{\text{cm}}^3$.
3. Use the depletion width formula $W=\sqrt{\frac{2{\epsilon }_s}{q}\frac{N_A+N_D}{N_A{N}_D}(V_{bi}+V_R)}$ with $q=1.602\times 10^{-19},\text{C}$.
4. Calculate the term inside the square root: $\frac{2{\epsilon }_s}{q}=1.292\times 10^7,{\text{V}}^{-1}{\text{cm}}^{-1}$; multiply by $1.01\times 10^{-16},{\text{cm}}^3$ and $10.9,\text{V}$ to get $1.423\times 10^{-8},{\text{cm}}^{-1}$.
5. Take the square root to obtain $W=1.193\times 10^{-4},\text{cm}=1.19,\mu \text{m}$.
6. For capacitance per unit area, $C^{\prime }={\epsilon }_s/W=1.03545\times 10^{-12}/1.193\times 10^{-4}=8.676\times 10^{-9},{\text{F/cm}}^2=8.68\times 10^{-3},\mu {\text{F/cm}}^2$.

Concept tested: Depletion width and junction capacitance

Published 18 September 2026 · How this solution was prepared and checked