Question 7.2: Calculate the space charge width and electric field in a pn ...

In Example 7.1, we determined the built-in potential barrier as V_{b i}=0.635 \mathrm{~V}. From Equation (7.31), the space charge width is

\begin{aligned}W &=\left\{\frac{2 \epsilon_{s} V_{b i}}{e}\left[\frac{N_{a}+N_{d}}{N_{a} N_{d}}\right]\right\}^{1 / 2}     (7.31) \\&=\left\{\frac{2(11.7)\left(8.85 \times 10^{-14}\right)(0.635)}{1.6 \times 10^{-19}}\left[\frac{10^{16}+10^{15}}{\left(10^{16}\right)\left(10^{15}\right)}\right]\right\}^{1 / 2} \\&=0.951 \times 10^{-4} \mathrm{~cm}=0.951 \mu \mathrm{m}\end{aligned}

Using Equations (7.28) and (7.29), we can find x_{n}=0.8644 \mu \mathrm{m}, and x_{p}=0.0864 \mu \mathrm{m}.

\begin{array}{l}x_{n}=\left\{\frac{2 \epsilon_{s} V_{b i}}{e}\left[\frac{N_{a}}{N_{d}}\right]\left[\frac{1}{N_{a}+N_{d}}\right]\right\}^{1 / 2} \\ \end{array}     (7.28)

\begin{array}{l}x_{p}=\left\{\frac{2 \epsilon_{s} V_{b i}}{e}\left[\frac{N_{d}}{N_{a}}\right]\left[\frac{1}{N_{a}+N_{d}}\right]\right\}^{1 / 2} \end{array}     (7.29)

The peak electric field at the metallurgical junction, using Equation (7.16) for example, is

\begin{array}{c}\mathrm{E}=\frac{-e N_{d}}{\epsilon_{s}}\left(x_{n}-x\right) \quad 0 \leq x \leq x_{n} \\ \end{array}     (7.16)

\begin{array}{c}\mathrm{E}_{\max }=-\frac{e N_{d} x_{n}}{\epsilon_{s}}=-\frac{\left(1.6 \times 10^{-19}\right)\left(10^{15}\right)\left(0.8644 \times 10^{-4}\right)}{(11.7)\left(8.85 \times 10^{-14}\right)}=-1.34 \times 10^{4} \mathrm{~V} / \mathrm{cm}\end{array}

Comment

The peak electric field in the space charge region of a pn junction is quite large. We must keep in mind, however, that there is no mobile charge in this region; hence there will be no drift current. We may also note, from this example, that the width of each space charge region is a reciprocal function of the doping concentration: The depletion region will extend further into the lower-doped region.