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A p.n. junction diode is operated in forward biasing mode at 300K temperature. The voltage (in mV) at which the forward bias current equals the reverse saturation current would be:(Take, Boltzmann's constant (k)=8.62\times 10^{-5} \frac{eV}{K}log_e2=0.693)

 

Option: 1

17.92


Option: 2

1.792


Option: 3

179.2 


Option: 4

0.179


Answers (1)

best_answer

We know forward biasing current in a p.n diode is given by

 I=I_0(e^{\frac{eV}{KT}}-1)

When I=I_0I_0=I_0(e^{\frac{eV}{KT}}-1)

\Rightarrow2=e^{\frac{eV}{KT}}

Taking natural logs on both sides 

\Rightarrow log_e2=\frac{eV}{KT}

\Rightarrow V=\frac{KT}{e}log_e2

            =\frac{8.62\times 10^{-5}(\frac{eV}{K})\times 300 K}{e}\times 0.693

            =17.92 \times 10^{-3}V

            =17.92 mV

Posted by

Rishabh

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