Numerical 1: Mass Action Law — Finding Minority Carrier Concentration

Problem: In a pure silicon crystal at 300 K, the intrinsic carrier concentration n_i = $1.5 \times 10^{16}$ $m^{-3}$. The crystal is doped with phosphorus to give an electron concentration n_e = $4.5 \times 10^{22}$ $m^{-3}$. Find the hole concentration n_h and identify majority/minority carriers.

Step 1 — Identify the law: $n_e \times n_h = n_i^2$

Units check: $[\text{m}^{-3}] \times [\text{m}^{-3}] = [\text{m}^{-6}]$ on both sides ✓

Step 2 — Rearrange for n_h: $n_h = \frac{n_i^2}{n_e}$

Step 3 — Substitute values: $n_h = \frac{(1.5 \times 10^{16} \text{ m}^{-3})^2}{4.5 \times 10^{22} \text{ m}^{-3}}$

$n_h = \frac{2.25 \times 10^{32} \text{ m}^{-6}}{4.5 \times 10^{22} \text{ m}^{-3}}$

Step 4 — Calculate: $n_h = \frac{2.25}{4.5} \times 10^{32-22} \text{ m}^{-3} = 0.5 \times 10^{10} \text{ m}^{-3} = 5.0 \times 10^9 \text{ m}^{-3}$

Step 5 — Identify carrier types:

• n_e = $4.5 \times 10^{22}$ $m^{-3}$ >> n_h = $5.0 \times 10^{9}$ $m^{-3}$
• Majority carriers: electrons (n-type semiconductor, phosphorus = pentavalent)
• Minority carriers: holes
• Ratio n_e/n_h = $4.5 \times 10^{22}$/$5.0 \times 10^{9}$ ≈ 10^{13} — enormous asymmetry confirms heavy doping