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I've been studying the effects of surface states in semiconductors (in the absence of interfaces, so just semiconductor-vacuum), and am struggling with the concept of depletion/accumulation layers. I've read quite a few texts on the subject, and each of them always deals with n-type or p-type semiconductors. But I don't know what would happen in the case of the undoped semiconductor.

Take for example the case of an n-type semiconductor with acceptor surface states as depicted in the figure below, not so graciously taken from Solid Surfaces, Interfaces and Thin Films by Lüth. One has acceptor states around the middle of bandgap that are partially filled, resulting in a negative surface charge. This negative charge is then compensated by the depletion layer, consisting of ionized donors. enter image description here

Conversely, one could take partially filled donor like states near the bandgap middle and a p-type semiconductor; this results in a positively charged surface and a negative depletion layer due to occupied bulk acceptors.

But now my question comes: how do these scenario's work for undoped semiconductors? Surface charges will still form, and they still need to be compensated by space charge layers, but in the absence of donor/acceptor dopants to be depleted/occupied, how does this happen?

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  • $\begingroup$ If the surface states are really mid-gap, and the Fermi level in the intrinsic material is mid-gap, what is the problem? The Fermi levels line up just fine without requiring any formation of depletion layers to get the levels to line up. $\endgroup$ – Jon Custer Mar 24 '17 at 15:21
  • $\begingroup$ @JonCuster Okay, that's true, and a flaw in my question. What if they are not truly midgap? Take InSB, where the surface states are close to the valence band (just inside the gap). Or InAs, where they are above the conduction band. $\endgroup$ – user129412 Mar 24 '17 at 15:25
  • $\begingroup$ So that will pin the Fermi level at the surface. Now what? Well, just because there isn't doping doesn't mean there aren't carriers around to move. $\endgroup$ – Jon Custer Mar 24 '17 at 15:30
  • $\begingroup$ Okay, but that becomes a bit tricky right? I mean, I can see how a normal lattice atom (instead of a dopant) in the depletion region might lose one of its electrons, but the opposite already sounds stranger; it would have to host an extra one. Or well, perhaps that is not the right way to look at it. I could see an accumulation layer form in the case of InAs, where more electrons than usual are present around the surface, causing a net negative charge layer, and perhaps in the case of a depletion layer it just means that less electrons are there. $\endgroup$ – user129412 Mar 24 '17 at 15:37
  • $\begingroup$ But still this becomes quite tricky, as all the expressions in textbooks (schottkey barriers and such) always use $N_D$, the donor doping concentration, for the thickness of such layers. What determines the thickness in undoped systems then? $\endgroup$ – user129412 Mar 24 '17 at 15:38

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