# Reverse recovery current of a diode

When the diode is in forward bias phase and suddenly we apply a reverse bias voltage, the diode takes time to be fully blocked because of the excess of minority charges that are stored in P and N region. Thus a reverse current appears in the diode for a short time before it is blocked. My question is, is this current a diffusion of the excess of minority carriers that are stocked? If yes, why does this happen physically speaking? How does the current suddenly become negative?

• What do you mean "stocked"? Commented Apr 11, 2015 at 13:14

## 1 Answer

In transient conditions you also need to consider the displace current which flows in response to a time varying electric field. This will happen as you reverse the bias applied across the diode,

$$J = j_e + j_h + \epsilon_0 \frac{\partial\bf{E}}{\partial t}$$

where $j_e$ and $j_h$ are the currents due to flow of particles (the drift and diffusion of electrons (e) and holes (h)) and the third term is the displacement current.

• Are you saying the temporary current is due to finite capacitance of the PN junction? Commented Jan 22, 2019 at 20:12
• This additional transient term comes from application of Maxwell equations. I have not looked into it further, but i guess it causes a delayed current to flow and would seem to be a capacitance like term. I don’t know how this is related to junction capacitance as we normally think about it. Commented Jan 22, 2019 at 20:29
• It's not very clear. In ideal capacitor, you can have displacement current inside while zero real current there, so I do not think that presence of the first necessitates presence of the second. One way in which your answer could make sense is if the transient is due to charging up the PN capacitor to final reverse voltage. Commented Jan 22, 2019 at 20:33
• A few years ago I was trying to do transient simulations of PN junctions. So I could not use conventional drift diffusion approaches. This equation came out of that research. I never got to the bottom of it. But from what I remember, to correctly solve in the time domain that term is needed. Commented Jan 22, 2019 at 20:37