Question 4.7: Determin e the fraction of total electrons still in the dono...

Using Equation (4.55), we find

\begin{aligned} \frac{n_{d}}{n_{d}+n_{0}}= \frac{1}{1+\frac{N_{c}}{2 N_{d}} \exp \left[\frac{-\left(E_{c}-E_{d}\right)}{k T}\right]} \\ \end{aligned}     (4.55)

\begin{aligned} \frac{n_{d}}{n_{0}+n_{d}}=\frac{1}{1+\frac{2.8 \times 10^{19}}{2\left(10^{16}\right)} \exp \left(\frac{-0.045}{0.0259}\right)}=0.0041=0.41 \% \end{aligned}

Comment

This example shows that there are very few electrons in the donor state compared with the conduction band. Essentially all of the electrons from the donor states are in the conduction band and, since only about 0.4 percent of the donor states contain electrons, the donor states are said to be completely ionized.