reverse recovery of diode

Thread Starter


Joined Sep 11, 2020
When excess charge carriers are settled in power diode while in reverse bias, why the voltage across depletion layer is still 0.7 V. Why reverse recovery of diode happens can some one explain.


Joined Aug 21, 2008
The diode can be revers biased but all junctions are different. Some can only be reverse biased by 1 or 2 volts before they begin to break down, some hundreds of volts.

When a diode is the state of conducting forward current the voltage across the junction is about 0.7 volts (it really varies a lot). To switch it to blocking current flow you need to reverse the bias. The reverse bias voltage will draw current as the minority carriers are swept out of the junction. The more current that can be provided the faster the junction will be depleted, that is until the junction breaks down, in which case that is as fast as it can get.

When the junction becomes reverse biased and the charge is depleted it is in the blocking state. The amount of time to get from some particular current through the junction in the forward direction to the time when reverse leakage becomes the dominant source of current flowing through the junction is the reverse recovery time.