Determine the impurity doping concentrations in a \mathrm{p}^{+} \mathrm{n} junction given the parameters from Figure 7.11.
Assume that the intercept and the slope of the curve in Figure 7.11 are V_{b i}=0.725 \mathrm{~V} and 6.15 \times 10^{15}\left(\mathrm{~F} / \mathrm{cm}^{2}\right)^{-2}(\mathrm{~V})^{-1}, respectively, for a silicon \mathrm{p}^{+} \mathrm{n} junction at T=300 \mathrm{~K}.
The slope of the curve in Figure 7.11 is given by 2 / e \epsilon_{s} N_{d}, so we may write
N_{d}=\frac{2}{e \epsilon_{s}} \cdot \frac{1}{\text { slope }}=\frac{2}{\left(1.6 \times 10^{-19}\right)(11.7)\left(8.85 \times 10^{-14}\right)\left(6.15 \times 10^{15}\right)}
or
N_{d}=1.96 \times 10^{15} \mathrm{~cm}^{-3}
From the expression for V_{b i}, which is
V_{b i}=V_{t} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}}\right)
we can solve for N_{a} as
N_{a}=\frac{n_{i}^{2}}{N_{d}} \exp \left(\frac{V_{b i}}{V_{t}}\right)=\frac{\left(1.5 \times 10^{10}\right)^{2}}{1.963 \times 10^{15}} \exp \left(\frac{0.725}{0.0259}\right)
which yields
N_{a}=1.64 \times 10^{17} \mathrm{~cm}^{-3}
Comment
The results of this example show that N_{a} \gg N_{d}; therefore the assumption of a one-sided junction was valid.