**Definition: Reverse saturation current**, often denoted by πΌ_{π}β or πΌ_{0}β, is a small current that flows through a diode even when it is reverse-biased. This current is generally due to the drift of charge carriers (electrons and holes) across the junction without an external forward voltage.

### Reasons for Reverse Saturation Current:

**Thermal Generation of Electron-Hole Pairs:**In semiconductors, thermal energy can cause the breaking of covalent bonds, generating electron-hole pairs. Due to the ambient temperature, these carriers are present without an external voltage.**Minority Carrier Injection:**In the reverse bias condition, the minority carriers in the p-type and n-type regions gain enough energy from the thermal environment to cross the depletion region, contributing to the reverse current.

**Formula for Reverse Saturation Current:**

The reverse saturation current in a pn-junction diode can be expressed by the formula:

Where:

- π΄ is the diode area
- π΄
^{β}(often called Richardson’s constant) depends on the material properties - π is the absolute temperature (in Kelvin)
- πΈ
_{π}β is the bandgap energy of the semiconductor - π is the Boltzmann constant
- π is the base of the natural logarithm

This formula shows that πΌ_{0}β increases with temperature and depends inversely on the bandgap energy of the semiconductor.

**Solved Example: **

Calculate the **reverse saturation current **for a silicon diode at 300K given that the diode’s area is 1Γ10^{β6} m^{2}, Richardson’s constant for silicon is 120A/cm^{2}β
K^{2}, and the bandgap energy is 1.1 eV. Also Calculate the reverse saturation current at 350K.

Solution:

1 eV=1.6Γ10^{β19} Joules

Eg= 1.1Γ1.6Γ10^{β19} Joules

Diode Area(A)= 1Γ10^{β6} m^{2}

A= 1Γ10^{β6} Γ100 Γ100 cm^{2}

A= 10^{-2} cm^{2}

Reverse current at 350K,

This calculation illustrates how significantly the reverse saturation current can increase with just a 50 K increase in temperature, highlighting the exponential sensitivity of semiconductor devices to temperature changes

### Diode current under Reversed Bias

The diode current (πΌ*I*) under any bias condition can be expressed by the Shockley diode equation:

Where:

- πΌ is the diode current.
- πΌ
_{0}β is the reverse saturation current. - π is the voltage across the diode.
*q*is the charge of an electron (1.6Γ10^{β19}Coulombs).*k*is Boltzmann’s constant (1.38Γ10^{β23}J/K).*T*is the absolute temperature in Kelvin.*n*is the ideality factor, typically ranging from 1 to 2, which accounts for the deviation of the diode from ideal behavior.

In the reverse bias condition (π<0), therefore V has a negative value,

Putting the value of equation (2) in equation(1),

Thus, *I*ββ*I*_{0}β means the diode current approximately equals the reverse saturation current but with the opposite sign. This shows πΌ_{0β} directly defines the magnitude of the leakage current in reverse bias.