In a p-n junction diode, the current I can be expressed as
$I=I_{0} \exp \left(\frac{e V}{2 k_{B} T}-1\right)$
where $I_{0}$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, $k_{B}$ is the Boltzmann constant $\left(8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}\right)$ and $\mathrm{T}$ is the absolute temperature. If for a given diode $\mathrm{l}_{0}=5$ $\times 10^{-12} \mathrm{~A}$ and $\mathrm{T}=300 \mathrm{~K}$, then
(a) What will be the forward current at a forward voltage of $0.6$ V?
(b) What will be the increase in the current if the voltage across the diode is increased to $0.7 \mathrm{~V}$ ?

The expression for current in a p-n junction diode, is given as

$I=I_{0} \exp \left(\frac{e V}{2 k_{B} T}-1\right)$

Here, $l_{0}=5 \times 10^{-12} \mathrm{~A}$

$\mathrm{T}=300 \mathrm{~K}$

$\mathrm{k}_{\mathrm{B}}$ is the Boltzmann constant with a value of $8.6 \times 10^{-5} \mathrm{eV} / \mathrm{k}=8.6 \times 10^{-5} \times 1.6 \times 10^{-19}=1.376 \times 10^{-23} \mathrm{~J} / \mathrm{K}$

(a) Forward voltage is given as $V=0.6 \mathrm{~V}$

$I=5 \times 10^{-12} \exp \left(\frac{1.6 \times 10^{-19} \times 0.6}{2 \times 1.376 \times 10^{-23} \times 300}-1\right)$

$I=5 \times 10^{-12} \exp (22.36)$

$=0.0256 \mathrm{~A}$

(b) Voltage across the diode is given to get increased to $0.7 \mathrm{~V}$

$I^{\prime}=5 \times 10^{-12} \exp \left(\frac{1.6 \times 10^{-19} \times 0.7}{2 \times 1.376 \times 10^{-25} \times 300}-1\right)$

$I^{\prime}=5 \times 10^{-12} \exp (26.25)$

$l^{\prime}=1.257 \mathrm{~A}$

Change in current will be $\Delta \mathrm{l}=\mathrm{l}^{\prime}-\mathrm{I}$

$=1.257-0.0256$

$=1.23 \mathrm{~A}$

Change is voltage will be $0.7-0.6=0.1 \mathrm{~V}$