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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?

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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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