Q. In $PN$-junction diode the reverse saturation current is $10^{- 5}amp$ at $27{^\circ}C$. The forward current for a voltage of $0.2\ volt$ is [exp$(7.62) = 2038.6,\ k = 1.4 \times 10^{- 23}J/K$]
A) $2037.6 \times 10^{- 3}amp$
B) $203.76 \times 10^{- 3}amp$
C) $20.376 \times 10^{- 3}amp$
D) $2.0376 \times 10^{3}amp$
Explanation
(c) The forward current
$i = i_{s}(e^{eV/kT} - 1) = 10^{- 5}\left\lbrack e^{\frac{1.6 \times 10^{- 19} \times 0.2}{1.4 \times 10^{- 23} \times 300}} - 1 \right\rbrack$
$= 10^{- 5}\lbrack 2038.6 - 1\rbrack = 20.376 \times 10^{- 3}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.