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

Question

For a junction diode the ratio of forward current ( I f . ) and reverse current ( I r ) is [e = electronic charge, V = voltage applied across junction, k = Boltzmann constant, T = temperature in kelvin ] (A) e - V / k T (B) e V / kT  (C) ( e - eV / kT +1) (D) (ee V / k T-1)

Tardigrade Admin · Accepted Answer

Current in junction diode, I=I0(ee V / k T-1) In forward biasing, V is positive ; 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)

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

$e^{- V / k T}$

$e^{V / k T}$

$(e^{- e V / k T} + 1)$

$(e^{e V / k T} - 1)$

Current in junction diode, $I = I_{0} (e^{e V / k T} - 1)$
In forward biasing, $V$ is positive ; In reverse bias $V$ is negative. Then $I_{r} = I_{0}$
$\frac{I _{F}}{I _{r}} = \frac{I _{0} ( e ^{e V / k T} - 1 )}{I _{0}} = (e^{e V / k T} - 1)$

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

e−V/kT

eV/kT

(e−eV/kT+1)

(eeV/kT−1)