A silicon p‑n junction at $300\text{K}$ has an acceptor concentration $N_A=5\times 10^{16}{\text{cm}}^{-3}$ on the p‑side and a donor concentration $N_D=2\times 10^{15}{\text{cm}}^{-3}$ 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 $n_i^2=np$ for thermal equilibrium.
2. For the p‑side, minority carriers are electrons: $n_p=n_i^2/N_A$.
3. Compute $n_i^2=(1.5\times 10^{10}{)}^2=2.25\times 10^{20}{\text{cm}}^{-6}$.
4. Then $n_p=2.25\times 10^{20}/(5\times 10^{16})=4.5\times 10^3{\text{cm}}^{-3}$.
5. For the n‑side, minority carriers are holes: $p_n=n_i^2/N_D$.
6. Thus $p_n=2.25\times 10^{20}/(2\times 10^{15})=1.125\times 10^5{\text{cm}}^{-3}\approx 1.13\times 10^5{\text{cm}}^{-3}$.

Concept tested: Mass‑action law (minority carrier concentration)

Published 16 September 2026 · How this solution was prepared and checked