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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 I0 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 $(8.6\times 10^{-5}\; eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{12}A$ and $T=300\; K,$ then

(b) What will be the increase in the current if the voltage across the diode is increased to $0.7 \; V?$

Answers (1)

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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 the forward voltage is 0.7V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.7}{1.376*10^{-23}*300}-1 ]=3.029A$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence the increase in the forward current is

$I(whenv=0.7) - I(whenv=.6)$ $= 3.029- 0.0625 = 2.967A$

Posted by

Pankaj Sanodiya

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