Q. For a junction diode the ratio of forward current $(I_{F})$ and reverse current $(I_{r})$ is [$I_{e} =$electronic charge, $V =$voltage applied across junction, $k =$ Boltzmann constant, $T =$temperature in kelvin]
A) $e^{- V/kT}$
B) $e^{V/kT}$
C) $(e^{- eV/kT} + 1)$
D) $(e^{V/kT} - 1)$
Explanation
(d) Current in junction, $V$ is positive; In reverse bias $V$ is negative. Then $I_{r} = I_{0}$
$\frac{I_{F}}{I_{r}} = \frac{I_{0}(e^{eV/kT} - 1)}{I_{0}} = (e^{eV/kT} - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.