Saved Bookmarks
| 1. |
In a p-n junction diode , the current I can be expressed as , I=I_(0)[e^(eV//kT)-1] where I_(0) is called 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 constant (8.6xx10^(-5)eVK^(-1)) and T is the absolute temperature. If for a 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 current, if the voltage across the diode is increase to 0.7V? What is the dynamic resistance? What will be the current, if the reverse bias voltage changes from 1V to 2V? |
|
Answer» Solution :Here, `k=8.6xx10^(-5) eVK^(-1)` `=8.6xx10^(-5)xx1.6xx10^(-19)JK^(-1)`, `I_(0)=5xx10^(-12)A, T=300K`. If `V=0.6V`, then forward current is `I=I_(0)[e^(eV//kT)-1]` `=5xx10^(-12)[e^(((1.6xx10^(-19))xx0.6)/((8.6xx10^(-5)xx1.6xx10^(-19))xx300))-1]` `=5xx10^(-12)[e^(23.255)-1]` `=5xx10^(-12)[(2.718)^(23.255)-1]=0.063A` If `V=0.7V`, then `I_(1)=5xx10^(-12)[e^(((1.6xx10^(-19xx0.7)))/((8.6xx10^(-5)xx1.6xx10^(-19))xx300))-1]` `=5xx10^(-12)[e^(27.1318)-1]=3.035A` `:.` Increase in current `DeltaI=I_(1)-I=3.035-0.063=2.972A` Now `DeltaV=0.7-0.6=0.1V, DeltaI=2.972A` Dynamic resistance, `=(DeltaV)/(DeltaI)=0.1/2.972=0.0336Omega` For a change in VOLTAGE from 1 to 2 V, the current will remain to saturation stae, `I_(0)=5xx10^(-12)A`. It shown that the diode possesses practically infinite resistance in REVERSE biased. |
|