Question 4.14: The diode current is 0.6 mA when the applied voltage is 400 ...

The diode current,                  I=I_{o}\left(e^{\frac{qV }{\eta KT} }-1 \right)

Therefore,                     0.6\times 10^{-3} =I_{o}\left(e^{\frac{qV }{\eta KT} }-1 \right)=I_{o}e^{\frac{qV }{\eta KT} }

=I_{o}e^{\frac{400 }{25\eta} }=I_{o}e^{\frac{16 }{\eta} }         (1)

Also,                               20\times 10^{-3}=I_{o}e^{\frac{500 }{25\eta} }=I_{o}e^{\frac{20 }{\eta} }                               (2)

Dividing Eq. (2) by Eq. (1), we get

\frac{20\times 10^{-3} }{0.6\times 10^{-3} } =\frac{I_{o}e^{\frac{20 }{\eta} }}{I_{o}e^{\frac{16 }{\eta} }}

Therefore,                             \frac{100}{3} =e^{\frac{4}{\eta } }

Taking natural logarithms on both sides, we get

\log _{e}\frac{100}{3} =\frac{4}{\eta }

3.507=\frac{4}{\eta }

Therefore,                                      \eta=\frac{4}{3.507} =1.14