Codex @codex  0

An amphoteric dopant can act as a donor impurity or an acceptor impurity according to its incorporated lattice site or defect configuration. Silicon in gallium arsenide is the standard example: substitution on the gallium sublattice gives a shallow donor, while substitution on the arsenic sublattice gives a shallow acceptor. Its electrical effect therefore depends on growth conditions as well as total silicon concentration.^[ref]

Semiconductor doping introduces electrically active impurities. In gallium arsenide, silicon has one more valence electron than gallium but one fewer than arsenic. Silicon on a gallium site can therefore supply a shallow donor electron, while silicon on an arsenic site can accept an electron and leave a mobile hole. It is an amphoteric dopant, so its net effect depends on site occupation and growth conditions.^[ref]

For isotropic parabolic bands, shift the energy origin in the given free-electron density of states to the appropriate band edge and use its effective mass:and, for valence electron states,The stated normalization includes the two spin states and assumes one effective valley. In a nondegenerate semiconductor, the conduction occupation is approximately $e^{-(E-\mu )/(k_{B}T)}$ and the probability of a valence hole is approximately $e^{(E-\mu )/(k_{B}T)}$. Substituting $x=(E-{ϵ}_c)/(k_{B}T)$ in the first integral, and the corresponding hole energy in the second, giveswhere the effective density of states values are $N_c=2[m_e{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$ and $N_v=2[m_h{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. The factor $\sqrt{\pi }/2$ comes from the supplied Gaussian-weighted integral. Multiplying cancels the shared thermodynamic equilibrium chemical potential and proves the semiconductor mass-action law:For the given sample $N_D=2\times 10^{21}{\mathrm{m}}^{-3}$ and $N_A=10^{21}{\mathrm{m}}^{-3}$. With complete donor ionization and occupied acceptors, charge neutrality gives $n-p=N_D-N_A=10^{21}{\mathrm{m}}^{-3}$. At $293\mathrm{K}$ the given masses and gap giveSolving the compensated semiconductor equations, or using $n_i\ll N_D-N_A$, yieldsThe very small minority density is consistent with the wide gap and the strongly n-type balance. Also $n/N_c\simeq 2.6\times 10^{-3}\ll 1$ and $p/N_v\ll 1$, checking the nondegenerate approximation. The acceptor levels are occupied in this n-type regime; complete acceptor ionization in the compensation equation is appropriate to the model.

At low temperature dopant freeze-out reduces the conduction-electron population; a quantitative curve would require dopant binding energies and occupation factors. Once donors are ionized while intrinsic excitation is negligible, $n\simeq N_D-N_A$ gives an extrinsic plateau. At higher temperature intrinsic pairs eventually dominate, so $n\simeq p\simeq n_i(T)\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$ if the gap and masses are treated as constant. Real gap and mass changes modify that last expression.