# 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 this happens physically speaking? How the current suddenly becomes negative?

• What do you mean "stocked"? – boyfarrell Apr 11 '15 at 13:14

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