Calculate the width of the space charge region in a pn junction when a reversebiased voltage is applied.
Again consider a silicon pn junction at T=300 \mathrm{~K} with doping concentrations of N_{a}= 10^{16} \mathrm{~cm}^{-3} and N_{d}=10^{15} \mathrm{~cm}^{-3}. Assume that n_{i}=1.5 \times 10^{10} \mathrm{~cm}^{-3} and V_{R}=5 \mathrm{~V}.
The built-in potential barrier was calculated in Example 7.1 for this case and is V_{b i}=0.635 \mathrm{~V}. The space charge width is determined from Equation (7.34). We have
\begin{array}{c}W=\left\{\frac{2 \boldsymbol{\epsilon}_{s}\left(V_{b i}+V_{R}\right)}{e}\left[\frac{N_{a}+N_{d}}{N_{a} N_{d}}\right]\right\}^{1 / 2} \\ \end{array} (7.34)
\begin{array}{c}W=\left\{\frac{2(11.7)\left(8.85 \times 10^{-14}\right)(0.635+5)}{1.6 \times 10^{-19}}\left[\frac{10^{16}+10^{15}}{\left(10^{16}\right)\left(10^{15}\right)}\right]\right\}^{1 / 2}\end{array}
so that
W=2.83 \times 10^{-4} \mathrm{~cm}=2.83 \mu \mathrm{m}
Comment
The space charge width has increased from 0.951 \mu \mathrm{m} at zero bias to 2.83 \mu \mathrm{m} at a reverse bias of 5 \mathrm{~V}.