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Confusion on electrical current vs charge flow

  1. Mar 14, 2006 #1

    es

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    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html#c3

    http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun2.html#c2

    (Using the diagram pointed to by the link)

    A hole wants to increase it's energy, right? So it should have no problem flowing right to left. Likewise, an electron wants to decrease it's energy so it should have no problem flowing left to right.

    So why is there no current flow in the direction of the electric field, right to left? (I almost can't even type that because I can just see that the depletion region is in the way, creating an open of sorts. But I would like to try to understand it from the band diagram point of view.)

    Is it that there are no electrons, other than the minority carriers, available to flow in the P-side? (i.e. they are all already in the lowest state.) If this is the case why doesn't the battery act as an electron source? (vice-versa for the holes) Shouldn't electrons at the negative terminal supplied by the battery create a current in the direction opposite to the "conduction direction" of the P-N junction? (i.e. travel from p-side to n-side, "conduction direction" in the diagram is defined as n-side to p-side)

    I guess the charge carriers don't flow like that from a battery. But if the load was a resistor, from the first quote the current is in the direction of the positive charge carriers, positive carriers flow in the direction of the electric field, and the electric field points right to left. So maybe it does?

    But I know this cannot be correct because this is a reverse biased diode, and I am confident only the leakage (minority current) exists inside a reverse biased diode.

    It seems like I almost have to be abusing one of these definitions. But which one?

    I found some links on batteries but it seems like it could be either way. Electrons flow from positive terminal (i.e. it's positive because the electrons are gone) or electrons flow into positive terminal (due to electrical field).

    I know I am making a mistake in my logic and I feel like this should be so obvious but I am just not getting it. I guess I have mental block or something. :)
     
    Last edited: Mar 14, 2006
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  3. Mar 14, 2006 #2

    Claude Bile

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    The lack of conduction due to the depletion zone only applies to a P-N junction without an external voltage applied to it (i.e. equilibrium). Once you forward bias the junction, current can move quite freely.

    Claude.
     
  4. Mar 16, 2006 #3

    es

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    Can't lack of conduction also apply to a P-N junction with an external field which points from n-side to p-side? This represents a reversed biased diode with a depletion region that has increased its width relative to thermal equilibrium, correct?. It conducts majority carries even less due to the fact that they need even more energy to move across the junction in the conduction direction (from n-side to p-side for electrons).

    So, if this thing appling this external feild is a battery, then does it have no electrons available on the p-side and no holes on the n-side?
     
  5. Mar 16, 2006 #4

    Claude Bile

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    Not if its reversed biased, no. If the junction is reversed biased, the battery wants to push all the electrons toward the N side and all the holes toward the P side.

    Claude.
     
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