Question

In a p-n junction diode, the current $I$ can be expressed as $I=I_0(\exp \left({\displaystyle \frac{eV}{K_{B}T}}\right)-1)$, 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 $I_0=5\times {10}^{-12}A$ 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 the reverse bias voltage changes from $1V$ to $2V$?

Answer 1

Hint: This problem can be solved by using the expression for the current in the p-n junction diode as given in the question and using it to solve the part questions one by one by plugging in the proper information in the equation.

Formula used:
$R_{dynamic}={\displaystyle \frac{\mathrm{\Delta }V}{\mathrm{\Delta }I}}$

Complete answer:
It is given that the current $I$ can be expressed as
$I=I_0(\exp \left({\displaystyle \frac{eV}{K_{B}T}}\right)-1)$ --(1)
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.
Also,
$I_0=5\times {10}^{-12}A$
$T=300K$
Therefore, let us solve the question parts one by one.
(a) Given forward voltage $V=0.6V$ and we have to find $I$.
Using (1), we get
$$I=(5\times {10}^{-12})(\exp \left({\displaystyle \frac{1\times 0.6}{8.6\times {10}^{-5}\times 300}}\right)-1)=(5\times {10}^{-12})(\exp \left(23.256\right)-1)=(5\times {10}^{-12})(1.26\times {10}^{10}-1)$$
$\therefore I\approx 0.063A$
(b) Now the voltage is increased to $V=0.7V$, we have to find out the increase in current.
$$I=(5\times {10}^{-12})(\exp \left({\displaystyle \frac{1\times 0.7}{8.6\times {10}^{-5}\times 300}}\right)-1)=(5\times {10}^{-12})(\exp \left(27.132\right)-1)=(5\times {10}^{-12})(6.07\times {10}^{11}-1)$$
$\therefore I\approx 3.035A$
Therefore, the increase in current is
$\mathrm{\Delta }I=3.035-0.063=2.972A$
(c) The dynamic resistance $R_{dynamic}$ is given by
$R_{dynamic}={\displaystyle \frac{\mathrm{\Delta }V}{\mathrm{\Delta }I}}$ --(2)
Where $\mathrm{\Delta }V,\mathrm{\Delta }I$ are the change in voltage and current respectively.
The change in voltage when going from $V=0.6V$ to $V=0.7V$ is $\mathrm{\Delta }V=0.7-0.6=0.1V$.
The change in current for both the cases as found out in (b) is $\mathrm{\Delta }I=3.035-0.063=2.972A$.
Hence, using (2), we get
 $R_{dynamic}={\displaystyle \frac{0.1}{2.972}}=0.0336\mathrm{\Omega }$
Therefore, the dynamic resistance is $0.0336\mathrm{\Omega }$.
(d) When the voltage is changed from $1V$ to $2V$ in reverse bias, the current will stay the same, that is, the reverse saturation current $I_0=5\times {10}^{-12}$.
This is because in the reversed bias state, the p-n junction diode is said to have infinite resistance and hence, from (2), we can see that if the resistance is infinite, the change in current will be zero.

Note: A p-n junction diode is a non-ohmic device which means that it does not follow Ohm’s law and therefore, the resistance of the device does not remain constant. Hence, we cannot find out the resistance by simply dividing the voltage by the current. We can only get the dynamic resistance of the diode for a certain change in voltage that brings about the respective change in current.