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Reverse Recovery Current of a diode

  1. Apr 10, 2015 #1

    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?

    Thank you
  2. jcsd
  3. Apr 11, 2015 #2
    Hello again,

    I am talking about this graph, which force is driving the excess of minority charges when the negative voltage is applied?

    Attached Files:

  4. Apr 14, 2015 #3
    Electric field force, F = qE. Minority carriers get stored in the depletion zone under forward bias. When the bias suddenly reverses, the E field of the external source points in a direction that forces the minority carriers back to their native material. Holes stored on the n side are forced to transit back to the p side and vice-versa. Similarly, electrons in the p side are forced back to the n side. Once the stored charge has been removed from the depletion zone and returned to its origin, the "tail current" ceases. Make sense?

  5. Apr 24, 2015 #4


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    Gold Member

    When the potential is reversed the minority charges must be cleared out. This occurs both by dffusion and by recombination (holes in the n region recombine with electrons in the n region, and likewise with electrons in the p region). The negative current denotes that the sense of flow is opposite to that during forward bias.
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