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The saturation current of a P-N junction germanium at \mathrm{27^{\circ}C} is \mathrm{10^{-5}amp}. The potential required to be applied in order to obtain a current of 250mA in forward bias will be:

Option: 1

0.57 V


Option: 2

0.48 V


Option: 3

0.26 V


Option: 4

0.63 V


Answers (1)

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\mathrm{I_f=I_s\left[e^{q V / K T}-1\right] or e^{\frac{q V}{k t}}=\frac{I_t}{I_s}+1=\frac{250 \times 10^{-3}}{10^{-5}}+1=25001}

\mathrm{\frac{\mathrm{qV}}{\mathrm{KT}}=10.126 \therefore V=\frac{10.126 \times 1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}}=0.26 \mathrm{~V}

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Gaurav

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