Minority carriers suppressed in n-type

In an n-type semiconductor, donors (P, As, Sb) supply extra electrons, so electrons are the majority carriers. Minority carrier suppression means that the hole concentration becomes much smaller than the intrinsic value $n_i$ when donors are added.

Proof

1. Mass action law (valid in thermal equilibrium for a non-degenerate semiconductor):

$$np = n_i^2$$

because $n = N_ce^{-(E_c-E_F)/kT}$ and $p = N_ve^{-(E_F-E_v)/kT}$, so $np = N_cN_ve^{-E_g/kT} = n_i^2$, independent of $E_F$ (i.e. of doping).

2. Charge neutrality:

$$n = p + N_d^+ \approx p + N_d$$

since all donors are ionised at room temperature.

3. Solve for n: substitute $p = n_i^2/n$:

$$n^2 - N_dn - n_i^2 = 0 \;\Rightarrow\; n = \frac{N_d}{2}+\sqrt{\frac{N_d^2}{4}+n_i^2}$$

For $N_d \gg n_i$:

$$n \approx N_d,\qquad p = \frac{n_i^2}{N_d}$$

4. Compare with intrinsic:

$$\frac{p}{n_i} = \frac{n_i}{N_d} \ll 1$$

So $n$ is raised above $n_i$ by the factor $N_d/n_i$, and $p$ is pushed below $n_i$ by the same factor. The minority holes are suppressed.

Physical reason

The large number of donor electrons greatly increases the chance that a thermally generated hole meets an electron and recombines. The recombination rate is proportional to $np$, while the generation rate depends only on temperature; equilibrium then requires $np$ to stay at $n_i^2$, so a large $n$ forces a small $p$.

Numerical example (Si, 300 K)

Since $p \ll n$, the conductivity is $\sigma \approx eN_d\mu_e$: it is controlled by the donor concentration only.