Assuming x_{B 0} \ll L_{B}, the excess minority carrier electron concentration in the base can be approximated by Equation (12.15b), which is
\begin{aligned}\delta n_{B}(x) & \approx \frac{n_{B 0}}{x_{B}}\left\{\left[\exp \left(\frac{e V_{B E}}{k T}\right)-1\right]\left(x_{B}-x\right)-x\right\} \end{aligned} (12.15b)
The collector current is
\left|J_{C}\right|=e D_{B} \frac{d\left[\delta n_{B}(x)\right]}{d x} \cong \frac{e D_{B} n_{B 0}}{x_{B}} \exp \left(\frac{V_{B E}}{V_{t}}\right)
The value of n_{B 0} is found as
n_{B 0}=\frac{n_{i}^{2}}{N_{B}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{5 \times 10^{16}}=4.5 \times 10^{3} \mathrm{~cm}^{-3}
For V_{C B}=2 \mathrm{~V}, we find (see the following Exercise Problem Ex 12.8)
x_{B}=x_{B 0}-x_{d B}=0.70-0.0518=0.6482 \mu \mathrm{m}
and
\left|J_{C}\right|=\frac{\left(1.6 \times 10^{-19}\right)(25)\left(4.5 \times 10^{3}\right)}{0.6482 \times 10^{-4}} \exp \left(\frac{0.60}{0.0259}\right)=3.195 \mathrm{~A} / \mathrm{cm}^{2}
For V_{C B}=10 \mathrm{~V}, we find (see the following Exercise Problem Ex 12.8)
x_{B}=0.70-0.103=0.597 \mu \mathrm{m}
and
\left|J_{C}\right|=\frac{\left(1.6 \times 10^{-19}\right)(25)\left(4.5 \times 10^{3}\right)}{0.597 \times 10^{-4}} \exp \left(\frac{0.60}{0.0259}\right)=3.469 \mathrm{~A} / \mathrm{cm}^{2}
We now can find, from Equation (12.45a)
\begin{aligned}\frac{d I_{C}}{d V_{C E}} \equiv g_{o}=& \frac{I_{C}}{V_{C E}+V_{A}}=\frac{1}{r_{o}} \\ \end{aligned} (12.45a)
\begin{aligned} \frac{d J_{C}}{d V_{C E}}=\frac{\Delta J_{C}}{\Delta V_{C B}}=\frac{J_{C}}{V_{C E}+V_{A}}=\frac{J_{C}}{V_{B E}+V_{C B}+V_{A}}\end{aligned}
or
\frac{3.469-3.195}{8}=\frac{3.195}{0.60+2+V_{A}}
The Early voltage is then determined to be
V_{A}=90.7 \mathrm{~V}
Comment
This example indicates how much the collector current can change as the neutral base width changes with a change in the B-C space charge width, and it also indicates the magnitude of the Early voltage.