A silicon diode at $300\text{K}$ has a reverse saturation current $I_s=2\times 10^{-14}\text{A}$ and an ideality factor $n=1$. When a forward bias of $0.7\text{V}$ is applied, the measured forward current is $I_F=10\text{mA}$. Assuming the minority‑carrier lifetime $\tau =0.5\mu \text{s}$ and a diode cross‑sectional area $A=1\times 10^{-4}{\text{cm}}^2$, calculate the diffusion capacitance $C_D$ of the diode. (Use $V_T=25.85\text{mV}$.)

Step-by-step solution

1. The diffusion capacitance of a forward‑biased diode is given by $C_D=\frac{I_F\tau }{V_T}$.
2. Convert all quantities to SI units: $I_F=10\text{mA}=0.01\text{A}$, $\tau =0.5\mu \text{s}=0.5\times 10^{-6}\text{s}$, $V_T=25.85\text{mV}=25.85\times 10^{-3}\text{V}$.
3. Calculate the numerator: $I_F\tau =0.01\times 0.5\times 10^{-6}=5\times 10^{-9}\text{C}$.
4. Now divide by $V_T$: $C_D=\frac{5\times 10^{-9}}{25.85\times 10^{-3}}\approx 1.934\times 10^{-7}\text{F}$.
5. Express the result in a convenient unit: $1.934\times 10^{-7}\text{F}=1.934\times 10^{-7}\text{F}=193\text{nF}$ (rounded to $190\text{nF}$).

Concept tested: Diffusion capacitance of a forward‑biased diode

Published 11 September 2026 · How this solution was prepared and checked