In a `p-n` junction diode, the current `I` can expressed as `I=I_(0) exp((eV)/(2k_(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 xx 10^(-5) eV//K)` and `T` is the absolute temperature. If for a given diode `I_(o) = 5 xx 10^(-12) A` and `T = 300 K`, then
(a) What will be the forward current at a forward voltage of `0.6V` ?
(b) What will be the increase in the current if the voltage across the diode is increased to `0.7 V` ?
( c) What is the dynamic resistance ?
(d) What will be current if reverse bias voltage changes from `1 V` to `2 V` ?

Explanation

In a p-n junction diode, the expression for current is given as:
`I=I_(o)exp((eV)/(2k_(B)T)-1)`
Where,
`I_(0)` = Reverse saturation current `= 5 xx 10^(−12)`A
T = Absolute temperature = 300 K
`k_(B)` = Boltzmann constant `= 8.6 xx 10^(−5) eV//K = 1.376 xx 10^(−23) J K^(−1)`
V = Voltage across the diode
(a) Forward voltage, V = 0.6 V
`=5xx10^(-12)[exp((1.6xx10^(-19)xx0.6)/(1.376xx10^(-23)xx300))-1]`
`therefore` Current, I
`=5xx10^(-12)xxexp[22.36]=0.0256 A`
Therefore, the forward current is about 0.0256 A.
(b) For forward voltage, V’ = 0.7 V, we can write:
`I=5xx10^(-12)[exp((1.6xx10^(-19)xx0.7)/(1.376xx10^(-23)xx300)-1)]`
`=5xx10^(-12)xxexp[26.25]=1.257` A
Hence, the increase in current, `DeltaI = I' − I`
= 1.257 − 0.0256 = 1.23 A
`=("CHange in voltage")/("CHange in current")`
(c) Dynamic resistance
`(0.7-0.6)/(1.23)=(0.1)/(1.23)=0.081Omega`
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.