Acceptor dopants and σ = peμh

Acceptors and holes in the valence band

A trivalent impurity (B, Al, Ga) in Si has only three valence electrons, so one of its four covalent bonds is incomplete. This empty bond creates a localized energy level $E_a$ just above $E_v$ (about 0.045 eV for B in Si). At room temperature, an electron from a neighbouring Si–Si bond (the valence band) easily gains this small energy and moves into the empty bond. The boron becomes a fixed negative ion $B^-$, and the missing electron in the valence band is a free hole that can move through the crystal.

 Ec -------------------------
 
 Ea  - - - - (-) (-) (-) - -   ionized acceptors
 Ev -----o-----o-----o-------  holes in VB

Each ionized acceptor gives one hole and no free electron, so when $N_a \gg n_i$: $p \approx N_a$ and $n = n_i^2/N_a \ll p$.

Proof of $\sigma = pe\mu_h$

Let an electric field $E$ act on a p-type sample of cross-section $A$. Holes drift along $E$ with drift velocity $v_{dh} = \mu_h E$. In time $dt$, the holes within a length $v_{dh}dt$ cross the area, carrying charge $dQ = e\,p\,A\,v_{dh}\,dt$. So the hole current density is

$$J_h = \frac{1}{A}\frac{dQ}{dt} = e\,p\,v_{dh} = e\,p\,\mu_h E$$

The few electrons contribute $J_e = e\,n\,\mu_e E$ in the same way, so

$$\sigma = \frac{J_h + J_e}{E} = e\,p\,\mu_h + e\,n\,\mu_e$$

Since $n \ll p$ in a p-type semiconductor, the electron term is negligible:

$$\sigma \approx p e \mu_h \approx N_a e \mu_h$$