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Q.14.10     In a p-n junction diode, the current I can be expressed as

   $I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

where $I_{0}$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}V/K)$ and   $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and   $T=300\; K,$ then

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

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best_answer

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When reverse voltage is 1V, V= -1

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-1)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

When the reverse voltage is -2V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-2)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

In both case current is very small and approximately equal to the reverse saturation current, hence their difference is negligible which causes dynamic resistance of infinity.

Posted by

Pankaj Sanodiya

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