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InterviewSolution
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In a p-n junction diode, the current I can be expressed I=I_(0) exp[(eV)/(k_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_ is the Boltzmann contant (8.6xx10^(-5)eV//K) and T is the absolute temperature. If for a given diode, I_(0)=5xx10^(-12)A and T=300K thenWhat will be the forward current at a forward voltage of 0.6V? What will be the increase in the current if the voltage across the diode is increased to 0.7 V? What is dynamic resistance? What will be the current if reverse bias voltage changes from 1V to 2V? |
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Answer» Solution :Here, `I_(0)=5xx10^(-12)A, T=300K, k_(B) =8.6xx10^(-5)eVK^(-1)=8.6xx10^(-5)xx1.6xx10^(-19)JK_(-1)` If `V=0.6V`, then`(eV)/(k_T)=(1.6xx10^(-19)xx0.6)/(8.6xx10^(-5)xx1.6xx10^(-19)xx300)=23.26` `I=I_(0)[EXP((eV)/(k_T)-1)]=5xx10^(-12)[e^(23.26)-1]=5xx10^(-12)[1.259xx10^(10)-1]` `=5xx10^(-12)xx1.259xx10^(10)=0.063A` If `V=0.7V, then (eV)/(k_T)=(1.6xx10^(-19)xx0.7)/(8.6xx10^(-5)xx1.6xx10^(-19)xx300)=27.14` `I=I_(0)[exp((eV)/(k_T)-1)]=5xx10^(-12)[e^(27.14)-1]=5xx10^(-12)[6.07xx10^(11)-1]` `=5xx10^(-12)xx6.07xx10^(11)=3.035A` :. Increase in current, `DeltaI=(3.035-0.063)=2.972A` Since, `DeltaI=2.972A, DeltaV=0.7-0.6=0.1V` Dynamic resistance`=(DeltaV)/(DeltaI)=(0.1V)/(2.972A)=0.0336Omega` As the voltage CHANGES from 1 to 2 V, the current I will be almost EQUAL to `I_(0)(=5xx10^(-12)A)`, SHOWING that the diode possesses practially infinite dynamic resistance in the reverse bias. |
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