Worked example: phosphorus-doped silicon

A silicon sample is doped with $N_D = 10^{16}\;\text{cm}^{-3}$ phosphorus atoms. Find the majority-carrier density, the minority-carrier density (using the mass-action law with $n_i = 10^{10}\;\text{cm}^{-3}$), and label the type.

1. Given. Donor concentration $N_D = 10^{16}\;\text{cm}^{-3}$, intrinsic carrier density $n_i = 10^{10}\;\text{cm}^{-3}$ at 300 K. Phosphorus is a group V atom, so it acts as a donor — every ionised P contributes one free electron.
2. Equation. Mass-action law and the full-ionisation approximation:
   $$n \cdot p = n_i^2$$
   $$n \approx N_D \quad (\text{since } N_D \gg n_i)$$
3. Substitution. Take $n \approx 10^{16}$ cm⁻³ directly from the donor count, then rearrange the mass-action law for $p$:
   $$p = \frac{n_i^2}{n} = \frac{(10^{10})^2}{10^{16}} = \frac{10^{20}}{10^{16}}$$
4. Result. $n = 10^{16}\;\text{cm}^{-3}$ (electrons, majority) and $p = 10^{4}\;\text{cm}^{-3}$ (holes, minority). This sample is strongly n-type.
5. Sanity-check. Electrons outnumber holes by a factor of $10^{12}$, so essentially all the current is carried by electrons — exactly what we expect from an n-type material. And the minority concentration $10^4$ cm⁻³ is six orders of magnitude below the intrinsic value, just as the mass-action law demands when $n$ is six orders above it.