A silicon diode at $300\text{K}$ has cross‑sectional area $A=2\times 10^{-4}{\text{cm}}^2$, minority carrier lifetime $\tau =1.0\mu \text{s}$, and diffusion coefficient for holes $D_p=12{\text{cm}}^2{\text{s}}^{-1}$. Under a forward bias of $0.6\text{V}$, the forward current measured is $I_F=5\text{mA}$. Assuming ideality factor $n=1$, calculate the diffusion capacitance $C_D$ of the diode using $C_D=\frac{I_F\tau }{V_T}$, where $V_T=25.85\text{mV}$.

Step-by-step solution

1. Write the diffusion‑capacitance expression: $C_D=\frac{I_F\tau }{V_T}$.
2. Convert all quantities to SI units: $I_F=5\text{mA}=5\times 10^{-3}\text{A}$, $\tau =1.0\mu \text{s}=1.0\times 10^{-6}\text{s}$, $V_T=25.85\text{mV}=25.85\times 10^{-3}\text{V}$.
3. Substitute: $C_D=\frac{(5\times 10^{-3})(1.0\times 10^{-6})}{25.85\times 10^{-3}}$.
4. Multiply the numerator: $5\times 10^{-3}\times 1.0\times 10^{-6}=5\times 10^{-9}$.
5. Divide by $25.85\times 10^{-3}$: $\frac{5\times 10^{-9}}{25.85\times 10^{-3}}=\frac{5\times 10^{-9}}{2.585\times 10^{-2}}=1.934\times 10^{-4}\text{F}$.
6. Round to three significant figures: $C_D=1.93\times 10^{-4}\text{F}$.

Concept tested: Diffusion capacitance of a forward‑biased diode

Published 14 September 2026 · How this solution was prepared and checked