Answers: 3
![100\ A = (2\ x\ 10^{-15}\ A)(e^{[\frac{1.6\ x\ 10^{-19} C\ V_{D}}{(1)(1.38\ x\ 10^{-23})(300\ k)}]}-1)\\\\\frac{100\ A}{2\ x\ 10^{-15}\ A} = e^{[(38.64 V^{-1})(V_{D})]}-1](/tpl/images/0969/5696/caa19.png)
![ln(5\ x\ 10^{16}) = ln[e^{(38.6\ V^{-1}\ V_{D})} - 1]\\38.45 = (38.6\ V^{-1})(V_{D}) - 1\\\\V_{D} = \frac{38.45}{38.6\ V^{-1}} + 1\\\\](/tpl/images/0969/5696/0540d.png)
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