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Forward biased diode operation

  1. Feb 21, 2011 #1
    Hello Forum,

    trivial questions: physics phenomena are usually explained via simple models. In DC electronics the water analogy is often adopted but breaks down when we move to AC.

    When a simple diode is forward biased DC current is able to flow in the circuit powered by a constant voltage source. Holes meet the electrons and continue to recombine.

    Naively, I would wonder if at one point in time all the excess electrons available in the N doped part of the diode would meet up with the excess holes in the P-doped part and the process of electrical conduction would stop since there is no more movement of charge carriers.

    I know that physically that does not happen and the diode continues to work as a closed switch. What is wrong with the low level explanation of holes-electron recombination? Why does conduction not stop? Doping is the addition of a finite number of charge carriers.

  2. jcsd
  3. Feb 21, 2011 #2


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    Well when you say forward biased, it means there is a potential there to keep the action going.


    http://hyperphysics.phy-astr.gsu.edu/hbase/solids/diod.html" [Broken]
    Last edited by a moderator: May 5, 2017
  4. Feb 21, 2011 #3
    In a DC circuit the simple picture has electrons moving from one terminal of a battery to the other. In reality electrons themselves move rather slowly and randomly (read about "drift speed" here: http://en.wikipedia.org/wiki/Electric_current ), and what we mean by "current" is the average number of electrons that cross a particular plane in 1 second, not the number of electrons that move from one terminal all the way to the other. Think of a long, narrow trough filled with water. At the very far end there is an overflow tube connected to a one gallon can such that one drop of water added to the trough would force one drop of water to exit the overflow tube. If you took an eye-dropper and dropped 1 gallon of water stained with red dye into the trough one drop at a time, it is unlikely any of the red dye would end up at the far end in the overflow bucket, and yet the overflow bucket would be full.

    In an AC circuit the simple picture has electrons moving back and forth as the current reverses direction, but in reality we are again only measuring the average number of electrons that cross a plane during a time interval. By placing a diode in an AC circuit all we are doing is limiting the general direction of electron flow. When the source voltage reverse biases the diode, it becomes a very high impedance path (measured as "leakage"), when the voltage source forward biases the diode it becomes a very low impedance path with a fairly fixed voltage drop. During the transition from forward bias to reverse bias the diode actually allows some current to flow in the reverse direction before it becomes a high impedance path (referred to as reverse recovery time).

    So, to answer your OP, the electrons that cross the diode junction are always replaced by other electrons, and the "holes" always pass their electrons forward, so, No, the diode will not stop conduction at some future time unless there is some mechanical, electrical or thermal stress that induces failure.

    Hope that helped,

  5. Feb 22, 2011 #4
    Thank you Fish.

    somehow I envisioned holes and electrons deriving from the doping of the semiconductors running into each other and annihilating themselves. Being doping finite I was lead to think that the process had finite duration.

  6. Feb 22, 2011 #5


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    Fish4funs wiki link is okay but you might be interested in these as well.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html" [Broken]

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html" [Broken]
    Last edited by a moderator: May 5, 2017
  7. Mar 20, 2011 #6
    That is the end-point within the depletion layer - it's why the depletion layer has finite thickness at a given junction voltage. And the time it takes to do that is part of the diode's switching time.
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