# 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, $\inline k_{B}$  is the Boltzmann constant $\inline (8.6\times 10^{-5}eV/K)$  and $\inline T$ is the absolute temperature. If for a given diode $\inline I_{0}=5\times 10^{-12}A$  and $\inline T=300\; K,$ then (a)  What will be the forward current at a forward voltage of $\inline 0.6\; V\; ?$

P Pankaj Sanodiya

As we have

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

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

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 forword current is 0.0625A

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