A diode is a non-linear component of an electrical circuit. That allows current in forward biasing and block current in reverse biasing. The behavior of a diode can be identified using VI characteristics. The diode current depends upon the voltage across the diode. The diode current can be expressed in the form of a diode current equation.

## Derivation of Diode Current Equation

The diode current equation 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 $K=C+273\deg$

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 $e^{-\frac{V}{\eta V_{T}}}\ll 1$. So, $I\approx I_{o}$, is valid up till external voltage is below the breakdown 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 decreases as the material quality increases.

## Diode Current Equation 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:

How to calculate diode current with the given data in the example

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

$V=0.15v$

So, the diode current 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.

**Diode Equation Calculator:**

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