In a p-n junction diode, the current I can be expressed as $I=I_{0}exp\left(\frac{eV}{2K_{B}T}-1\right)$, 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 $(8.6\times 10^{-5}eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_o=5\times 10^{-12}$$A$ and $T=300K$, then
(a) What will be the forward current at a forward voltage of $0.6V$?(b) What will be the increase in the current if the voltage across the diode is increased to $0.7V$?(c) What is the dynamic resistance?(d) What will be the current if reverse bias voltage changes from $1V$ to $2V$?

(a)

In a p-n junction diode, the expression for current is $I=I_{0}exp[\frac{eV}{2k_{B}T}-1]$

where $I_0$ is the reverse saturation current = $5\times 10^{-12}A$

$T=$Absolute temperature = $300K$

$V$ is the voltage across diode.

For forward voltage $V=0.6V$ 

$I=5\times 10^{-12}\times exp[\frac{1.6\times 10^{-19}\times 0.6}{1.376\times 10^{-23}\times 300}-1]=0.0256A$

(b)

In a p-n junction diode, the expression for current is $I=I_{0}exp[\frac{eV}{2k_{B}T}-1]$

where $I_0$ is the reverse saturation current = $5\times 10^{-12}A$

$T=$Absolute temperature=$300K$

$V$ is the voltage across diode.

For forward voltage, $V=0.7V$ 

$I=5\times 10^{-12}\times exp[\frac{1.6\times 10^{-19}\times 0.7}{1.376\times 10^{-23}\times 300}-1]=1.257A$

Hence, increase in current = $1.257A-0.0256A=1.23A$

(c)

In a p-n junction diode, the expression for current is $I=I_{0}exp[\frac{eV}{2k_{B}T}-1]$

where $I_0$ is the reverse saturation current = $5\times 10^{-12}A$

$T=$Absolute temperature=$300K$

$V$ is the voltage across diode.

For forward voltage $V=0.6V$ 

$I=5\times 10^{-12}\times exp[\frac{1.6\times 10^{-19}\times 0.6}{1.376\times 10^{-23}\times 300}-1]=0.0256A$

For forward voltage $V=0.7V$ 

$I=5\times 10^{-12}\times exp[\frac{1.6\times 10^{-19}\times 0.7}{1.376\times 10^{-23}\times 300}-1]=1.257A$

Hence increase in current=$1.257A-0.0256A=1.23A$

Dynamic resistance = Change in voltage /Change in current

$=\frac{0.7-0.6}{1.23}=0.081\Omega$

(d)

If reverse bias voltage changes from $1V$ to $2V$, the current will be almost constant, because dynamic resistance in the reverse bias is infinite.