D.30 Number of conduction band electrons

This note finds the number of electrons in the conduction band of a semi­conductor, and the number of holes in the valence band.

By definition, the density of states ${\cal D}$ is the number of single-particle states per unit energy range and unit volume. The fraction of electrons in those states is given by $\iota_{\rm {e}}$. Therefore the number of electrons in the conduction band per unit volume is given by

$$i_{\rm e} = \int_{{\vphantom' E}^{\rm p}_{\rm c}}^{{\vph... ... top}} {\cal D}\iota_{\rm e} {\,\rm d}{\vphantom' E}^{\rm p}\$$

where ${\vphantom' E}^{\rm p}_{\rm {c}}$ is the energy at the bottom of the conduction band and ${\vphantom' E}^{\rm p}_{\rm {top}}$ that at the top of the band.

To compute this integral, for $\iota_{\rm {e}}$ the Maxwell-Boltzmann expression (6.33) can be used, since the number of electrons per state is invariably small. And for the density of states the expression (6.6) for the free-electron gas can be used if you substitute in a suitable effective mass of the electrons and replace $\sqrt{{\vphantom' E}^{\rm p}}$ by $\sqrt{{\vphantom' E}^{\rm p}-{\vphantom' E}^{\rm p}_{\rm {c}}}$.

Also, because $\iota_{\rm {e}}$ decreases extremely rapidly with energy, only a very thin layer at the bottom of the conduction band makes a contribution to the number of electrons. The integrand of the integral for $i_{\rm {e}}$ is essentially zero above this layer. Therefore you can replace the upper limit of integr­ation with infinity without changing the value of $i_{\rm {e}}$. Now use a change of integr­ation variable to $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{({\vphantom' E}^{\rm p}-{\vphantom' E}^{\rm p}_{\rm {c}})/{k_{\rm B}}T}$ and an integr­ation by parts to reduce the integral to the one found under “!” in the notations section. The result is as stated in the text.

For holes, the derivation goes the same way if you use $\iota_{\rm {h}}$ from (6.34) and integrate over the valence band energies.