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Forward Bias

For a junctio...

Question

For a junction diode the ratio of forward current ( $I_{f}$ ) and reserve current ( $I_{r}$ ) is
[ $e =$ electronic charges,
$V =$ voltage applied across junction,
$K =$ Boltzmann constant,
$T =$ temperature in Kelvin]

A

e^{- V / K T}

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B

e^{V / K T}

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C

(e^{- e V / K T} + 1)

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D

e^{e V / K T} - 1

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Solution

The correct option is D $e^{e V / K T} - 1$
Current in junction diode, $I = I_{0} (e^{e V / K T} - 1)$
In forward biasing, $V$ is positive ; In reverse bias $V$ is negative. Then $I_{r} = I_{0}$
$\frac{I_{F}}{I_{r}} = \frac{I_{0} (e^{e V / K T} - 1)}{I_{0}} = (e^{e V / K T} - 1)$

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