Determine the forward-biased B-E voltage required to achieve a recombination factor equal to \delta=0.9967.
Consider an npn bipolar transistor at T=300 \mathrm{~K}. Assume that J_{r 0}=10^{-8} \mathrm{~A} / \mathrm{cm}^{2} and that J_{s 0}=10^{-11} \mathrm{~A} / \mathrm{cm}^{2} .
The recombination factor, from Equation (12.44), is
\delta=\frac{1}{1+\frac{J_{r 0}}{J_{s 0}} \exp \left(\frac{-e V_{B E}}{2 k T}\right)} (12.44)
We then have
0.9967=\frac{1}{1+\frac{10^{-8}}{10^{-11}} \exp \left(\frac{-e V_{B E}}{2 k T}\right)}
We can rearrange this equation and write
\exp \left(\frac{+e V_{B E}}{2 k T}\right)=\frac{0.9967 \times 10^{3}}{1-0.9967}=3.02 \times 10^{5}
Then
V_{B E}=2(0.0259) \ln \left(3.02 \times 10^{5}\right)=0.654 \mathrm{~V}
Comment
This example demonstrates that the recombination factor may be an important limiting factor in the bipolar current gain. In this example, if V_{B E} is smaller than 0.654 \mathrm{~V}, then the recombination factor \delta will fall below the desired 0.9967 value.