1. Tardigrade
  2. Question
  3. Physics
  4. 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 ]

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

Semiconductor Electronics: Materials Devices and Simple Circuits

Solution:

Current in junction diode, $I=I_0\left(e^{e V / k T}-1\right)$
In forward biasing, $V$ is positive ; In reverse bias $V$ is negative. Then $I _{ r }= I _0$
$\frac{ I _{ F }}{ I _{ r }}=\frac{ I _0\left( e ^{ eV / kT }-1\right)}{ I _0}=\left( e ^{ eV / kT }-1\right)$