Mass-action law

The mass-action law: in a semiconductor at thermal equilibrium, the product of the free-electron concentration $n$ and the hole concentration $p$ is a constant fixed only by temperature, no matter how the material is doped.

$n\cdot p=n_i^2$

Here $n$ is the electron concentration (per cm³), $p$ the hole concentration, and $n_i$ the Intrinsic carrier concentration, the value $n$ and $p$ each take in pure (intrinsic) material, where $n=p=n_i$ so $np=n_i^2$ trivially. Even after doping changes $n$ and $p$ individually by orders of magnitude, their product still equals $n_i^2$.

Why it holds

It comes straight from the balance of generation and recombination. Thermal generation breaks bonds at a rate set by temperature alone; it doesn’t care whether the silicon is doped. Recombination destroys an electron and a hole together, at a rate proportional to how many of each are available, i.e. proportional to $np$. At equilibrium the two rates are equal:

$\text{(generation rate, fixed by }T)=(\text{constant})\times np$

So $np$ is forced to a fixed value at a given temperature. Evaluating it for the intrinsic case (where $n=p=n_i$) pins that value at $n_i^2$. Doping shifts $n$ and $p$ in opposite directions but can’t change the product, because the generation rate it must balance hasn’t changed.

Consequence

Since $np$ is pinned, raising one carrier population forces the other down by the same factor. If Doping increases the electron concentration by a factor of a million, then to keep $np=n_i^2$ the hole concentration drops by a factor of a million:

$n\uparrow \times 10^6⟹p=\frac{n_i^2}{n}\downarrow \times 10^6$

This is how doping works, and why doped silicon has one carrier type that overwhelmingly dominates (Majority and minority carriers). In n-type silicon with donor concentration $N_D$, essentially every donor contributes a free electron so $n\approx N_D$, and the law gives the minority-hole concentration:

$p=\frac{n_i^2}{N_D}$

For example, silicon doped with $N_D=10^{16}{\text{cm}}^{-3}$ has $n\approx 10^{16}{\text{cm}}^{-3}$ but only $p\approx (1.5\times 10^{10}{)}^2/10^{16}\approx 2.25\times 10^4{\text{cm}}^{-3}$ holes, a billion-to-one imbalance. Symmetrically for p-type material: $p\approx N_A$, $n\approx n_i^2/N_A$. This one relation lets you write down the minority-carrier concentration in any doped sample, which sets the Reverse saturation current of a junction.