Design a pn junction to meet maximum electric field and voltage specifications.
Consider a silicon pn junction at T=300 \mathrm{~K} with a p-type doping concentration of N_{a}=2 \times 10^{17} \mathrm{~cm}^{-3}. Determine the \mathrm{n}-type doping concentration such that the maximum electric field is \left|\mathrm{E}_{\max }\right|=2.5 \times 10^{5} \mathrm{~V} / \mathrm{cm} at a reverse-biased voltage of V_{R}=25 \mathrm{~V}.
The maximum electric field is given by Equation (7.36). Neglecting V_{b i} compared to V_{R}, we can write
\begin{array}{c}\mathrm{E}_{\max }=-\left\{\frac{2 e\left(V_{b i}+V_{R}\right)}{\epsilon_{s}}\left(\frac{N_{a} N_{d}}{N_{a}+N_{d}}\right)\right\}^{1 / 2} \\ \end{array} (7.36)
\begin{array}{c}\left|\mathrm{E}_{\max }\right| \cong\left\{\frac{2 e \mathrm{~V}_{R}}{\epsilon_{s}}\left(\frac{N_{a} N_{d}}{N_{a}+N_{d}}\right)\right\}^{1 / 2}\end{array}
or
2.5 \times 10^{5}=\left\{\frac{2\left(1.6 \times 10^{-19}\right)(25)}{(11.7)\left(8.85 \times 10^{-14}\right)}\left[\frac{\left(2 \times 10^{17}\right) N_{d}}{2 \times 10^{17}+N_{d}}\right]\right\}^{1 / 2}
which yields
N_{d}=8.43 \times 10^{15} \mathrm{~cm}^{-3}
Conclusion
A smaller value of N_{d} results in a smaller value of \left|\mathrm{E}_{\max }\right| for a given reverse-biased voltage. The value of N_{d} determined in this example, then, is the maximum value that will meet the specifications.