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Basically in almost every semiconductor texts, there will be all these concepts concerning electrons, holes, dopants, fermi-levels.

However, I have been always confused about the picture of hole transport in semiconductor device, say, a simple PN junction.

With a specific acceptor level and dopant concentration, we have some "holes" in the valence bands, which are in fact the absence of some electrons having gone to the acceptor levels, then all theses books seem to assume in the valence bands of the p-type region, there will be only holes that conduct current?

My questions:

  • aren't there still many electrons in the valence band? though they have some negative effective masses, but still do contribute to the transport?

  • under a certain external force(say,E), the electron and holes in the valence band are moving in the same directions since electrons have negative effective mass and holes have positive one, so the corresponding currents cancel with each other in a P-type semiconductor?

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  • $\begingroup$ It is possible to think of electronic transport solely as the motion of electrons (no holes). No big surprise there, since truly there are only electrons. However the idea of a negative effective mass at the top of the valence band is a bit awkward, hence the concept of the hole. But be careful, once you introduce the concept of the hole you are also saying that the valence band is 101% full of particles: 100% full of electrons that collectively are inert, with an extra 1% of holes that carry charge. $\endgroup$
    – Nanite
    Commented Nov 11, 2013 at 22:18
  • $\begingroup$ “… electrons have negative effective mass…” Sorry, no further reading. $\endgroup$
    – Incnis Mrsi
    Commented Oct 23, 2014 at 6:50
  • $\begingroup$ @IncnisMrsi I don't think you understand the point here, either... $\endgroup$
    – Lorniper
    Commented Oct 27, 2014 at 8:12

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There are two alternative ways to calculate the current from the valence band:

(A) Add up the current of all the electrons in the valence band. The unoccupied states are unoccupied; there's nothing there, they don't carry current, you should ignore them, obviously.

(B) Ignore the current from all the electrons in the valence band. At the same time, pretend that the unoccupied states are actually occupied by something called "holes", with positive effective mass and positive charge. Calculate the current due to the holes alone.

Of these (A) is obviously a correct way to do the calculation, but it's mathematically and intuitively a very difficult calculation to do. The surprising and important fact (explained here) is that the (B) calculation gives the exact same result as the (A) calculation. Since (A) gives the correct answer, so does (B). Since (B) is a simpler calculation than (A), and easier for visualization and intuition, everybody always calculates and describes valence-band current using the (B) approach.

Your proposal for calculating the valence-band current is to track the current of the holes in the unoccupied states, and simultaneously track the current of all the electrons in the occupied states. That is neither (A) nor (B), it's doing both at once and adding them. So it will give you the wrong answer, off by a factor of 2.

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  • $\begingroup$ let me try to summarize, in a P-type region. the electrons at CBM is far less than the holes at VBM, which is again far less than the electrons at VBM. e@CBM move in the same direction as h@VBM while h@VBM move at opposite direction to e@vbm. We can sum up the current: e@CBM(very small)+ h@VBM(dominant).. Am I correct? $\endgroup$
    – Lorniper
    Commented Nov 12, 2013 at 16:54
  • $\begingroup$ No. In (B) you "Ignore the current from all the electrons in the valence band." Yes, all the electrons in the valence band, I really mean it. Ignore the electrons at the valence band maximum, ignore the electrons at the valence band minimum, ignore the electrons at the valence band middle ... $\endgroup$
    – Steve Byrnes
    Commented Nov 12, 2013 at 17:33
  • $\begingroup$ thanks, the hole density at VBM is much smaller than the "realistic" electron density at VBM, is it ok ? $\endgroup$
    – Lorniper
    Commented Nov 12, 2013 at 18:08
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As far as I understand, the electrons in the valence band do contribute to the transport, but only because they can move to holes, and this is equivalent to hole movement. There is no other place for these electrons to move to but to holes.

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  • $\begingroup$ if these valence electrons move together with holes, since they have negative effective masses, they should response to the external force(E) to the opposite direction, i.e. move towards cathod, this is wierd, because they cancel the hole current... $\endgroup$
    – Lorniper
    Commented Nov 11, 2013 at 7:27
  • $\begingroup$ @Lorniper: I did not say the electrons move "together with holes", I said they "move to holes", therefore, in the direction opposite to that of hole motion: if an electron in point A moved to a hole in point B, this is equivalent to movement of the hole from point B to Point A, i.e., in the opposite direction. $\endgroup$
    – akhmeteli
    Commented Nov 11, 2013 at 7:36
  • $\begingroup$ well, there is something illusive here.. Do you agree, under a certain external force(say,E), the electron and holes in the valence band are moving in the same directions since electrons have negative effective mass and holes have positive one?(clearly this is true at valence band maximum) $\endgroup$
    – Lorniper
    Commented Nov 11, 2013 at 10:55
  • $\begingroup$ @Lorniper: I don't quite see why they move in the same direction. Could you give a reference? $\endgroup$
    – akhmeteli
    Commented Nov 11, 2013 at 12:48
  • $\begingroup$ basically I agree with this answer physics.stackexchange.com/questions/10800/…, though did not ask my question $\endgroup$
    – Lorniper
    Commented Nov 11, 2013 at 21:35
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The lack of electrons in the structure of the substrate offer a resistance to electrons traveling its direction.And a doped substrate between two pn junctions can act as a control for electron flow.

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  • $\begingroup$ Please edit rather than making a new post. Also, please take care that you "answers" actually offer an answer to the question. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 11, 2013 at 19:46

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