In a `p-n` junction diode, the currect `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 formward 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

`I_(o) = 5 xx 10^(-12) A, k = 8.6 xx 10^(-5) eVk^(-1)`
`= 8.6 xx 10^(-5) xx 1.6 xx 10^(-19) Jk^(-1)`
(a) `I =I_(0) (e^(.^(ev)//_(2kT))-1)`,
For `V = 0.6V`,
`I = 5 xx 10^(-12) ((1.6 xx 10^(-19) xx 0.6)/(e^(2 xx 8.6 xx10^(-5)xx1.6 xx10^(-19) xx300))-1)`
=`5 xx 10^(-12) (e^(23.52) -1)`
=`5 xx 10^(-12) (1.256 xx 10^(10) -1) = 0.0628.4`
(b) For `v=0.7 v`, we have
`I = 5 xx 10^(-12) ((1.6 xx 10^(-19) xx 0.7)/(e^(2 xx 8.6 xx10^(-5)xx1.6 xx10^(-19) xx300))-1)`
`I = 5 xx 10^(-12) (e^(27.32) -1)`
=` 5 xx 10^(-12) (6.054 xx 10^(11) -1) = 3.0271 A`
`:. Delta I = 3.271 -0.0628 = 2.9643 A`
( c) `Delta I =2.9643, Delta v = 0.7 -0.6 = 0.1 V`
dynamic resistance `= (Delta v)/(Delta I) = (0.1)/(2.9643) = 0.0337 Omega`
(d) For change in voltage from `1` to `2 v`, the current will remain equal to `I_(0) = 5 xx 10^(-12) A`. It shows that the diode possesses practically infinite resistance in reverse biasing.