For a junction diode, the ratio of forward current `(I_(f))` and reverse current is.
[`I_(e)` = electronic charge,
`V` = voltage applied across junction,
`k` = Boltzmann constant
`T` = temperature in kelvin].

Options

A. `e^(-v//kT)`
B. `e^(V//kT)`
C. `(e^(eV//kT))`
D. `(e^(V//kT)-1)`

Answer

C

Explanation

Current in junction diode, `l_(f)-l_(0)(e^(eV//kT)-1)` In forward biasing, `V` is positive and in reverse bias `V` is negative Then,
`I_(r)=I_(0)`
`(I_(f))/(I_(r))=(I_(0)(e^(eV//KT)-1))/(I_(0))`
`=(e^(eV//kT)-1)`.