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.