Doped semiconductors

Semiconductors are often doped with atoms of some other element. The impurity atoms have a low concentration in the order of parts per million. There are two types of dopants, donors and acceptors. Donors add occupied states in the band gap just below the conduction band, and acceptors add empty states in the band gap just above the valence band. A dopant atom replaces one of the atoms in the crystal lattice. For instance, silicon forms the diamond crystal structure where every atom has four nearest neighbors. In the crystal, silicon has four valence electrons, which form four bonds to its nearest neighbors. Phosporus has five valence electrons and acts like a donor in silicon. When a phosporus atom sits on a silicon site, it forms four bonds with four of its valence electrons to its four silicon nearest neighbors. The fifth valence electron is easily ionized at room temperature and moves freely in the conduction band. Boron has three valence electrons and acts like an acceptor in silicon. The three boron valence electrons form three bonds with its four silicon nearest neighbors. At room temperature, an electron is easily excited from the valence band to the acceptor atom to form the fourth bond. This leaves behind a hole that can move freely in the valence band.

It is possible to dope a semiconductor with many of the elements of the periodic table. Some introduce states near the center of the band gap. Some elements introduce states at more than one energy in the band gap. Only those elements that introduce a single state within a few tens of meV below the conduction band are called donors, and this is called $n$-type. Only elements that introduce a single state a few tens of meV above the conduction band are called acceptors, and this is called $p$-type. It is not always possible to find suitable donors and acceptors for every semiconductor.

There is a temperature range near room temperature called the extrinsic range, where for $n$-type semiconductors, almost all of the donors are ionized, and the concentration of electrons in the conduction band is the same as the concentration of donors, $n\approx N_d$. The law of mass action, $pn=n_i^2$, also holds for extrinsic semiconductors, so the holes in an n-type semiconductor have a concentration $p \approx \frac{n_i^2}{N_d}$, which is typically orders of magnitude smaller than the concentration of electrons in an $n$-type semiconductor. For $n$-type, electrons are the majority carriers and holes are the minority carriers. Similarly, for a $p$-type semiconductor in the extrinsic range, the concentration of holes is almost equal to the concentration of acceptors, $p \approx N_a$, and the electron concentration is $n \approx\frac{n_i^2}{N_a}$. For $p$-type, holes are the majority carriers and electrons are the minority carriers. Semiconductor devices are typically operated in this extrinsic range of temperature, where the electron and hole concentrations are determined by the doping.

Charge neutrality
When dopants are added to a semiconductor, the semiconductor remains electrically neutral. In an $n$-type semiconductor, the donor atoms contribute mobile electrons to the conduction band. This leaves behind positively charged donor atoms that can't move because they are held by chemical bonds to their neighboring atoms in the crystal. Similarly, in a $p$-type semiconductor, the acceptors become negatively charged when they capture an electron from the valence band. This leaves positively charged mobile holes in the valence band. Charge neutrality requires that the number of positive charges equals the number of negative charges,

$$n+N_a^- = p+N_d^+,$$

where $n$ is the concentration of electrons in the conduction band and $p$ is the concentration of holes in the valence band,

$$n = \frac{\sqrt{\pi}D_c}{2}(k_BT)^{3/2}\exp{\left(\frac{\mu-E_c}{k_BT}\right)},\qquad p = \frac{\sqrt{\pi}D_v}{2}(k_BT)^{3/2}\exp\left(\frac{E_v-\mu}{k_BT}\right),$$

and $N_d^+$ is the concentration of ionized donors and $N_a^-$ is the concentration of ionized acceptors,

$$N_d^+=\frac{N_d}{1+2\exp\left(\frac{\mu-E_d}{k_BT}\right)},\qquad N_a^-=\frac{N_a}{1+4\exp\left(\frac{E_a-\mu}{k_BT}\right)}.$$

Here, $E_d$ is the donor energy level, and $E_a$ is the acceptor energy level. The formulas for the ionized donors and acceptors were derived using statistical physics. The asymmetry in the formulas appears because there is typically a single band at the minimum of the conduction band, while there are two degenerate bands (the light hole band and the heavy hole band) at the top of the valence band. If the total positive charge $p+N_d^+$, and the total negative charge $n+N_a^-$ are plotted as a function of the chemical potential, then the charge neutrality condition will be satisfied where the two lines cross. This condition can be used to determine the chemical potential of a doped semiconductor. Notice that if the donor concentration is only about 10% greater than the acceptor concentration, the chemical potential moves close to the conduction band. If the acceptor concentration is 10% greater than the donor concentration, the chemical potential moves close to the valence band.

log [1/cm³]
	$\mu$ [eV]