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Question about majority and minority concentrations in diodes

  1. Feb 19, 2014 #1
    In my notes package it shows that for a REV biased diode, in the p type side, the concentration of majority carriers (pp0 holes) increases as we get closer to the junction. However, I don't get how this makes sense, since the metal contact between the negative terminal of the source and the p-type region is negatively charge, shouldn't we expect the holes to be attracted to the metal contact? Thus, in the graph shown below, the concentration of pp0 should be larger at the beginning and decrease as it gets closer to the junction?

    PS: feel free to ignore some of the things I wrote :P

  2. jcsd
  3. Feb 21, 2014 #2
    The np product at that point is given by [itex]n^{2}_{i}e^{F_n-F_p/kT}[/itex].

    If none of these quantities could have changed at a given point, then it must be constant. As you see in the diagram the [itex]n_{p0}[/itex] concentration drops to zero near the depletion region as those electrons have been are swept across by the large electric field.

    So if the np product must be constant, and n decreases, p has to increase to account for it.

    Conceptually, you could also think of holes being quickly swept from the other side and "piling up" at the end of the depletion region.
  4. Feb 21, 2014 #3
    Hmm, that makes a lot of sense when you put it that way. But what about the other way to think of it? The fact that there is a negative charge on the left contact and thus the positive holes should be attracted to it. This should result in an increase in concentration of holes near the metal contact. In other words, there are less holes close to the pn junction and thus the potential barrier is increased?

    Also, can anyone explain this then:

    Why would the minority concentrations increase near the junction?
    Last edited: Feb 21, 2014
  5. Feb 21, 2014 #4
    Look first of all, holes can't really do anything even if they are attracted by terminal. Look it the other way, its the electrons that move and generate holes. Now having said that first look at the frwd bias condition.

    Now we know that the depletion region is formed of two polarities of atoms negative and positive. The positive atom or the donor exists on the N side and negative atom exists on the P side.
    Now as we forwards bias the PN junction the free electrons on the N side will penetrate the depletion region and impart energy to the immobile positive atoms to release an electron, this way a hole will be created together with an electron. As holes don't really move the electron will move towards the P side whereas the hole will be left behind thus reducing the depletion region.

    Same will happen on the P side, some of the incoming electrons will collide with the negative immobile atom and this will eject an electron from the atom thus creating a hole and electron simultaneously, thus reducing the junction width.

    As we move towards the metal contact in P side the hole concentration will become constant whereas the excess electron concentration will also become constant after recombination as electron hole pairs are created in pairs.

    Same applies for the reverse bias as a free electron on P side strikes the immobile negative atom a electron and hole will be generated simultaneously and this accounts for the decreased concentration of electrons and increased concentration of holes as electrons generated will move towards the N side leaving the hole.
    Last edited: Feb 21, 2014
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