Calculate the common-emitter current gain of a silicon npn bipolar transistor at T=300 \mathrm{~K} given a set of parameters.
Assume the following parameters:
\begin{aligned}D_{E} &=10 \mathrm{~cm}^{2} / \mathrm{s} & x_{B} &=0.70 \mu \mathrm{m} \\D_{B} &=25 \mathrm{~cm}^{2} / \mathrm{s} & x_{E} &=0.50 \mu \mathrm{m} \\ \tau_{E 0} &=1 \times 10^{-7} \mathrm{~s} & N_{E} &=1 \times 10^{18} \mathrm{~cm}^{-3} \\ \tau_{B 0} &=5 \times 10^{-7} \mathrm{~s} & N_{B} &=1 \times 10^{16} \mathrm{~cm}^{-3} \\ J_{r 0} &=5 \times 10^{-8} \mathrm{~A} / \mathrm{cm}^{2} & V_{B E} &=0.65 \mathrm{~V}\end{aligned}
The following parameters are calculated:
\begin{aligned}p_{E 0} &=\frac{\left(1.5 \times 10^{10}\right)^{2}}{1 \times 10^{18}}=2.25 \times 10^{2} \mathrm{~cm}^{-3} \\n_{B 0} &=\frac{\left(1.5 \times 10^{10}\right)^{2}}{1 \times 10^{16}}=2.25 \times 10^{4} \mathrm{~cm}^{-3} \\L_{E} &=\sqrt{D_{E} \tau_{E 0}}=10^{-3} \mathrm{~cm} \\L_{B} &=\sqrt{D_{B} \tau_{B 0}}=3.54 \times 10^{-3} \mathrm{~cm}\end{aligned}