what is reason for saturation current in reversed biased diodes
The saturation current in reverse-biased diodes is mainly due to the minority charge carriers present in the semiconductor material. Let's see how this happens.
In a reverse-biased diode, the positive terminal of the battery is connected to the N-side (which contains excess electrons) and the negative terminal is connected to the P-side (which has an excess of holes or positive charge carriers).
When a reverse bias voltage is applied to the diode, it creates an electric field that repels the minority charge carriers (electrons in the P-side and holes in the N-side) away from the junction. This electric field effectively blocks the flow of majority charge carriers (electrons in the N-side and holes in the P-side).
However, there are always some minority charge carriers present in the semiconductor material even without any bias. These carriers can diffuse across the junction due to thermal energy and a phenomenon called "minority carrier injection." As a result, a small amount of current flows through the reverse-biased diode. This current is called the saturation current.
To calculate the saturation current, you can use the diode equation:
I = Is * (e^(Vd / (n * Vt)) - 1)
Where:
- I is the diode current
- Is is the saturation current
- Vd is the applied voltage across the diode
- n is the ideality factor (typically around 1 for most diodes)
- Vt is the thermal voltage (approximately 26 mV at room temperature)
The saturation current is typically a very small value, on the order of nanoamperes or picoamperes. It is important to note that the saturation current increases with temperature due to an increase in the number of thermally generated minority carriers.
In summary, the reason for saturation current in reverse-biased diodes is the presence of minority charge carriers that can diffuse across the junction, resulting in a small current flow.