Question 4.8: Determine the temperature at which 90 percent of acceptor at...

Find the ratio of holes in the acceptor state to the total number of holes in the valence band plus acceptor state. Taking into account the Boltzmann approximation and assuming the degeneracy factor is g=4, we write

\frac{p_{a}}{p_{0}+p_{a}}=\frac{1}{1+\frac{N_{v}}{4 N_{a}} \cdot \exp \left[\frac{-\left(E_{a}-E_{v}\right)}{k T}\right]}

For 90 percent ionization,

\frac{p_{a}}{p_{0}+p_{a}}=0.10=\frac{1}{1+\frac{\left(1.04 \times 10^{19}\right)\left(\frac{T}{300}\right)^{3 / 2}}{4\left(10^{16}\right)} \cdot \exp \left[\frac{-0.045}{0.0259\left(\frac{T}{300}\right)}\right]}

Using trial and error, we find that T=193 \mathrm{~K}.

Comment

This example shows that at approximately 100^{\circ} \mathrm{C} below room temperature, we still have 90 percent of the acceptor atoms ionized; in other words, 90 percent of the acceptor atoms have “donated” a hole to the valence band.