Question 7.5: Calculate the junction capacitance of a pn junction. Conside...

The junction capacitance is found from Equation (7.42) as

\begin{array}{c}C^{\prime}=\left\{\frac{e \epsilon_{s} N_{a} N_{d}}{2\left(V_{b i}+V_{R}\right)\left(N_{a}+N_{d}\right)}\right\}^{1 / 2} \\ \end{array}     (7.42)

\begin{array}{c}C^{\prime}=\left\{\frac{\left(1.6 \times 10^{-19}\right)(11.7)\left(8.85 \times 10^{-14}\right)\left(10^{16}\right)\left(10^{15}\right)}{2(0.635+5)\left(10^{16}+10^{15}\right)}\right\}^{1 / 2}\end{array}

or

C^{\prime}=3.66 \times 10^{-9} \mathrm{~F} / \mathrm{cm}^{2}

If the cross-sectional area of the pn junction is, for example, A=10^{-4} \mathrm{~cm}^{2}, then the total junction capacitance is

C=C^{\prime} \cdot A=0.366 \times 10^{-12} \mathrm{~F}=0.366 \mathrm{pF}

Comment

The value of junction capacitance is usually in the \mathrm{pF}, or smaller, range.