A silicon p–n junction at $300,\text{K}$ has an acceptor concentration $N_A=4\times 10^{16},{\text{cm}}^{-3}$ on the p‑side and a donor concentration $N_D=1\times 10^{15},{\text{cm}}^{-3}$ on the n‑side. The intrinsic carrier concentration of silicon at this temperature is $n_i=1.5\times 10^{10},{\text{cm}}^{-3}$. Calculate (a) the minority‑electron concentration on the p‑side and (b) the minority‑hole concentration on the n‑side.

Step-by-step solution

1. Use the mass‑action law: $n_i^2=N_A{p}_n$ on the n‑side and $n_i^2=N_D{n}_p$ on the p‑side.
2. Compute $n_i^2=(1.5\times 10^{10}{)}^2=2.25\times 10^{20},{\text{cm}}^{-6}$.
3. Minority‑electron concentration on the p‑side: $n_p=n_i^2/N_A=2.25\times 10^{20}/(4\times 10^{16})=5.625\times 10^3,{\text{cm}}^{-3}$, round to $5.6\times 10^3$.
4. Minority‑hole concentration on the n‑side: $p_n=n_i^2/N_D=2.25\times 10^{20}/(1\times 10^{15})=2.25\times 10^5,{\text{cm}}^{-3}$, round to $2.3\times 10^5$.

Concept tested: Minority carrier concentration

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