In a typical semiconducting material, we start with silicon (group IV) and introduce atoms of group V or group III depending on whether we are constructing an n- or p-type semiconductor respectively.

Why is it optimal to choose doping atoms one group away? Why not introduce atoms from groups VI or II, for instance?

  • $\begingroup$ This is a great question, but it would benefit from a chemist. My intuition is that the elements one group away, once one electron is shed, have an orbital structure which is identical to the Si, so they sit in the lattice well, creating minimal defect, allowing their electron to join the conduction band without as much stress in the lattice. If you make a +2/-2 ion the nuclear charge will probably bind an electron in the neighborhood, and the lattice will be distorted more, and perhaps the electrons won't even enter the conduction band, or you'll make a disordered insulator. Maybe, maybe not. $\endgroup$
    – Ron Maimon
    May 26, 2012 at 2:30

3 Answers 3


Look at the II-VI options from a chemical perspective.

For silicon, Group III boron p-doping would be replaced by Group II beryllium doping, and Group V phosphorus n-doping would be replaced by Group VI sulfur doping.

The beryllium would tend to act like a Lewis base that donates a pair of electrons to some nearby silicon atom, forming a stronger and more localized bond than is possible with the odd electron count created by boron doping. Similarly, the sulfur would tend to act like a Lewis acid and form a localized sulfide bond with some nearby silicon atom, versus the odd electron count created by phosphorus.

So, the main reason for picking elements immediately "next door" (off by one electron) is to avoid introducing paired electrons, which on average will bond more strongly and more locally via a Lewis base-acid process.

That said, there's enough complexity going on there that it's still possible that you could get II-VI or maybe even I-VII doping to work in the right situations.

  • $\begingroup$ Ok, this is a helpful start. I understand now that, for instance, V/III has ionic properties, and hence probably less desirable than V/IV. But there are many semiconductor materials that are ionic; e.g., IV/VI are in use. Hence there is still a question: why not V/III, given that ionic compounds are in use? $\endgroup$
    – Simon S
    Jun 2, 2012 at 18:10
  • $\begingroup$ @SimonS, sorry, sort of lost track of this one when it moved. I... don't really understand your question, specifically what you mean by "ionic" in this context. If you mean the Lewis acid-base pair idea, that's not necessarily a strongly ionic bond; it's more like a covalent bond for which both electrons came from the same atom. So I was just saying that if you create essentially a free radical, an incomplete bond with no easy way to get a local partner. But it's clearly not a necessity. Beyond that you'd have to get a lot deeper into the modeling, and I'll bow out for that. $\endgroup$ Jun 5, 2012 at 23:39

This is a great question, and one that I also have and would like to see a better explanation for the answer, but I can offer what I know (I would comment, but I do not have that usability at the moment).

When you add a dopant to say, Si, you are essentially shortening the distance between the valence and the conduction band by adding either an accessible conduction (empty) band just above the original valence band (in the case of p-type like B) or are making the original conduction band more accessible by adding a higher energy valence band (in the case of n-type like N).

My guessed answer to your question is one of two things:

  • either the group III/V elements are optimal in shortening this band gap, and that II produces excess conduction too high and VI excess valence too low, or...
  • the change in valence of the dopant too greatly affects the lattice stabilization of the Si crystal. Essentially by causing too many localized vacancies/defects that the lattice strain negatively affects the overall conductivity of the Si.

Semiconductor elements have half-filled outer-most shells so if we doped group VI or II elements which have 6 and 2 electrons in the outer-most shell and they form covalent bonds with each other just like Si doped to Si and that would result in a no-free-electron insulator because there is no vacancy there .


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