A silicon p–n junction has acceptor concentration $N_A=5\times 10^{17}{\text{cm}}^{-3}$ on the p‑side and donor concentration $N_D=1\times 10^{15}{\text{cm}}^{-3}$ 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 $V_R=15\text{V}$ 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 $\mu {\text{F/cm}}^2$. (Take $q=1.6\times 10^{-19}\text{C}$)

Step-by-step solution

1. Calculate the semiconductor permittivity: ${\epsilon }_s={\epsilon }_r{\epsilon }_0=11.7\times 8.85\times 10^{-14}=1.035\times 10^{-12}\text{F/cm}$.
2. Add built‑in and reverse bias voltages: $V_T=V_{bi}+V_R=0.85+15=15.85\text{V}$.
3. Use the depletion‑width formula for an abrupt junction: $W=\sqrt{\frac{2{\epsilon }_s{V}_T}{q}\frac{N_A+N_D}{N_A{N}_D}}$.
4. Compute the factor $\frac{2{\epsilon }_s{V}_T}{q}=\frac{2\times 1.035\times 10^{-12}\times 15.85}{1.6\times 10^{-19}}\approx 2.05\times 10^8{\text{cm}}^2$.
5. Compute the concentration term $\frac{N_A+N_D}{N_A{N}_D}=\frac{5\times 10^{17}+1\times 10^{15}}{5\times 10^{17}\times 1\times 10^{15}}\approx 1.00\times 10^{-15}{\text{cm}}^3$.
6. Multiply the two factors: $2.05\times 10^8\times 1.00\times 10^{-15}=2.05\times 10^{-7}{\text{cm}}^2$.
7. Take the square root: $W=\sqrt{2.05\times 10^{-7}}\approx 4.53\times 10^{-4}\text{cm}=4.53\mu \text{m}$ (rounded to $4.5\mu \text{m}$).
8. Now find capacitance per unit area: $C^{\prime }={\epsilon }_s/W=1.035\times 10^{-12}\text{F/cm}/4.53\times 10^{-4}\text{cm}\approx 2.29\times 10^{-9}{\text{F/cm}}^2$.
9. Convert to $\mu {\text{F/cm}}^2$: $2.29\times 10^{-9}\text{F}=2.29\times 10^{-3}\mu \text{F}$.
10. State the final answers.

Concept tested: Depletion width and junction capacitance

Published 21 September 2026 · How this solution was prepared and checked