Diode is non-linear component of an electrical circuit, which allow current in forward biasing and block current in reverse biasing. The behavior of a diode can be identified using VI characteristic. Current of the diode depends upon the voltage across the diode. The diode current can be expressed in the form of diode current equation.

## DIODE CURRENT EQUATION

The current equation of the diode relates the current with the voltage across the diode. Suppose the voltage $V$ across and current $I$ flowing through the diode. The current $I$ can be expressed as

$I=I_{o}\lbrack e^{\frac{V}{\eta V_{T}}}-1\rbrack \ldots (1)$

Where

$I$ – Diode Current

$I_{o}$- Diode reverse saturation current at room temperature

$V $- External Voltage applied to the diode

$\eta $ – A constant, two for Silicon and one for Germanium

$V_{T}$ – $\frac{kT}{q}=\frac{T}{11600}$ Volts-equivalent of temperature, thermal voltage

By putting the value of $V_{T}$, we may get the following equation

$I=I_{o}\lbrack e^{\frac{qV}{\eta kT}}-1\rbrack $

Where

$k$ – Boltzmann’s constant, $1.38066\times 10^{-23} J/K $

$q $- charge of an electron, $1.60219\times 10^{-19}C$

$T $- temperature of the diode junction $K=C+273deg$

At the room temperature K=300, the thermal voltage $V_{T}=26mV$. Put these values in the current equation (1), we get

$I=I_{o}\lbrack e^{40\frac{V}{\eta }}-1\rbrack $

As the constant $\eta $, is one for germanium and two for silicon, so the above equations will become

For Germanium

$I=I_{o}\lbrack e^{40V}-1\rbrack $

For Silicon

$I=I_{o}\lbrack e^{20V}-1\rbrack $

For reverse bias current the sign of the voltage applied $V$ is changed and the equation for reverse bias will be

$I=I_{o}\lbrack e^{\frac{-V}{\eta V_{T}}}-1\rbrack $

As $V\gg V_{T}$, so the term $e^{-\frac{V}{\eta V_{T}}}\ll 1$. Therefore $I\approx I_{o}$, which is called reverse saturation current and is valid up till external voltage is below the break down voltage. The diode reverse saturation current is also called dark saturation current. It depends upon the rate of recombination and quality of the material. It is also notable that the dark current increases as the temperature increases. And it decrease as the material quality increases.

## Example:

A reverse bias is applied to the germanium PN junction diode and noted the reverse saturation current of $0.3\mu A$ at room temperature. What will be the current at room temperature when the forward bias is $0.15v

$.

### Solution:

Given data for finding diode current in the example

$I_{0}=0.3\times 10^{-6}A$

$V=0.15v$

Therefore, the current flowing through the diode will be

$I=I_{o}\lbrack e^{40V}-1\rbrack $

$I=0.3\times 10^{-6}(e^{40\times 0.15}-1)$

$I=120.73\mu A$

The above will be the forward bias current of the diode for the given

situation.