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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}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

 (a)  What will be the forward current at a forward voltage of 0.6\; V\; ?

Answers (1)


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 0.6V:

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


Hence forward current is 0.0625A

Posted by

Pankaj Sanodiya

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