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For a junction diode, the ratio of forwa...

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].

A

`e^(-v//kT)`

B

`e^(v//kT)`

C

`(e^(eV//kT)-1)`

D

`(e^(V//kT)-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Current in junction diode, `l_(f)=l_(0)(e^(eV//kT)-1)` In forward baising, 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)`
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