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].

How reverse current suddenly increases at the break down voltage in case of a junction diode?

Passage II Diode current in a p-n junction diode is given by I-I_(0)[exp(eV)/(2kT)-1] where I_(0) is the reverse saturation current V is the voltage across the diode k is the boltzmann constant and T is the abosolute temperatuer For a given diode I_(0)=5 pA and T =300 k The forward current at a foward voltage V=0.6 V is

Passage II Diode current in a p-n junction diode is given by I-I_(0)[exp(eV)/(2kT)-1] where I_(0) is the reverse saturation current V is the voltage across the diode k is the boltzmann constant and T is the abosolute temperatuer For a given diode I_(0)=5 pA and T =300 k When the reverse bias voltage bias voltage changes from 1 v to 2 V the current will become equal to

When a p-n junction diode is reverse biased, the flow of current across the junction is mainly due to

When a p-n junction diode is forward biased, the flow of current across the junction is mainly.

When a p-n junction diode is reverse biased the flow of current across the junction is mainly due to