 Solution

Assume three-dimensional parabolic band edges, with density-of-states masses $m_c,m_h$ and one valley per edge, each with spin degeneracy two. Their effective densities of states areFor the semiconductor mass-action law, integrating the square-root density of states against the Boltzmann tail gives $n=N_c{e}^{(\mu -E_c)/(k_{B}T)}$ and $p=N_v{e}^{(E_v-\mu )/(k_{B}T)}$. Multiplication cancels the chemical potential, givingIf there are additional valley degeneracies, multiply by their product; anisotropic bands use the appropriate density-of-states masses. In the later equal-mass model, $A=4m^3/(2\pi {\hslash }^2{)}^3$. No numerical mass or valley multiplicity has been invented.

When $\mu =E_c=0$, the conduction-edge occupation is exactly $1/2$, so the nondegenerate approximation for electrons fails at every positive temperature. Define the Normalized complete Fermi-Dirac integralFor the carrier densities with chemical potential pinned to a band edge, with the two specified equal masses, put $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Direct Fermi-Dirac integration then givesThus $n\propto T^{3/2}$ exactly within the parabolic model. At low temperature, $p\simeq N(T)e^{-E_g/(k_{B}T)}$ is thermally activated; at $k_{B}T\gg E_g$, $p/n\to 1$ from below. Both vanish at zero temperature and increase at fixed chemical potential. The other states that pin $\mu$ can supply the compensating charge; one must not additionally impose $n=p$.

For the Pauli reduction of the pinned-edge mass-action product, the exact product relative to the classical law isIt approaches $f_0$ at low temperature and $f_0^2\simeq 0.58545$ in the formal high-temperature limit. The low-temperature activation factor is unchanged but its prefactor is reduced; the classical mass-action formula overestimates the product. The high-temperature limit here refers to the given unbounded parabolic bands, not a claim that real semiconductor band structures remain unchanged at arbitrarily high temperature.^[ref]

Finally, for a fixed positive $\mu$, the zero-temperature Fermi sphere has radius $k_{\mu }=\sqrt{2m\mu }/\hslash$, soThe low-temperature expansion at fixed $\mu$ gives the fixed-chemical-potential Sommerfeld density increase:The monotonic density at fixed chemical potential in a square-root density of states initially rises quadratically from $n(0)$. For $k_{B}T\gg \mu$, its leading behaviour crosses to $f_{0}N(T)\propto T^{3/2}$. The fixed chemical potential is what allows the particle number to change.