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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 diodeI_{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?

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 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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