A silicon p–n junction at $300\text{K}$ has an acceptor concentration $N_A=2\times 10^{16}{\text{cm}}^{-3}$ 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 the minority‑carrier concentration of electrons on the p‑side and of holes on the n‑side.

Step-by-step solution

1. The relation for minority carriers in thermal equilibrium is $n_p=\frac{n_i^2}{N_A}$ and $p_n=\frac{n_i^2}{N_D}$.
2. Calculate $n_i^2={\left(1.5\times 10^{10}\right)}^2=2.25\times 10^{20}{\text{cm}}^{-6}$.
3. Electron concentration on the p‑side: $n_p=\frac{2.25\times 10^{20}}{2\times 10^{16}}=1.125\times 10^4{\text{cm}}^{-3}$.
4. Hole concentration on the n‑side: $p_n=\frac{2.25\times 10^{20}}{1\times 10^{15}}=2.25\times 10^5{\text{cm}}^{-3}$.
5. Round to appropriate significant figures.

Concept tested: Minority carrier concentration in a p‑n junction

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