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Band Gap in Insulators

  1. Sep 10, 2009 #1
    Hello all, I'm just getting to grips with conduction in solids and am trying to find some values for the band gap between the valence and the conduction bands. All I can find anywhere are typical values for semiconductors (or about an eV). I also found that carbon in the diamond configuration has a band gap of 5.5eV, but I've heard elsewhere that diamond is in fact a semiconductor (I know there's a thin line between the two, but I'm unsure that this counts as an insulator). I found this value on http://hyperphysics.phy-astr.gsu.edu/hbase/solids/band2.html

    I only ask because if 5.5eV is standard for insulators, I can't work out why they would insulate at anything much more than 5V (as this would give the electrons enough energy to move to the conduction band and therefore conduct). If possible, could you reply with where you found the information out, cause without that info, I can't be sure the answers are right.

    Thanks in advance!

    Edit: I just read (on wikipedia: http://en.wikipedia.org/wiki/Insulator_(electrical)#Breakdown so don't give it too much credit) that the breakdown voltage is indeed equal to the band gap energy. Although this makes me feel that my understanding is getting close, can I just confirm that usual band gaps are much higher (the voltage in my mains cables is at 230V, so that would imply a band gap of *at least* 230eV)
    Last edited: Sep 10, 2009
  2. jcsd
  3. Sep 11, 2009 #2
    The nomenclature surrounding the terms "insulator" and "semiconductor" can be a bit confusing, because it's often not used consistently. If one wanted to be really strict, one might reserve the term "semiconductor" for an insulator with a bandgap of less than, say, 10kT, or a doped semiconductor. But often times the terms are used interchangeably with materials with band gaps less than 2 eV.

    5.5 eV is fairly large for an insulator. Most insulators have considerably smaller gaps, but diamond is on the larger side. A band gap of 230 eV would be outrageous. Usually breakdown for insulators is quoted in terms of the electric field, because the size of the system the voltage is applied to is relevant, and typical units are kV/mm.
  4. Sep 11, 2009 #3
    Thanks for giving some meaning to the word semiconductor from a practical point of view. I'm still not sure I understand exactly why the applied voltage has to be so high. Actually, thinking about it some more, it's beginning to make sense.

    If I imagine a wire (which is surrounded by an insulator) at a potential of 230V, then an electron outside the insulator would like to get to the lower potential inside the wire. However, it doesn't have enough energy, so the only way it could get through is via quantum tunnelling, with a fairly tiny probability of doing so. However, that doesn't explain the breakdown of an insulator..

    Is it possible to calculate this breakdown voltage per meter (or kV/mm) in any way, from basic principles? And what is the mechanism behind this breakdown?
  5. Sep 11, 2009 #4
    Well, I don't know how to calculate it. The mechanism is tunneling. A simple picture is if you graph V(x) for the conduction and valence states. If you have an electric field, then going in one direction, the energies of those states will gradually decrease. See my crude drawing below. Eventually, if you travel far enough, the conduction states will be lowered to the point where they have the same potential as valence states further back. Then electrons can tunnel. But if that distance is very far, then the tunneling probability will be very low, and breakdown won't be observed. The larger the electric field, the steeper the slope of V(x) will be, so that will increase the probability of breakdown tunneling.

    Code (Text):

              -----                       conduction states
    -----    ----------->    -----
              -----                       valence states
    x ->
  6. Sep 11, 2009 #5
    Haha, as crude as your drawing may be, it's incredibly elucidating! Thanks a bunch, I think I've cracked it. As for the calculation of the tunnelling probability, I'm guessing it has an analogy of tunnelling through a saw tooth potential, whose width corresponds to the point where the conduction band energy is cancelled out by the field. Thus, if the insulator is 1mm thick, the distance is 1mm, an enormous value from a quantum point of view, so merely equalling the band gap energy will not do. However, make the saw tooth steep enough and the distance isn't quite so large. Now the electron has a higher probability of conduction. I read that the breakdown for air is 784 V/mm. Thus the distance that a band gap of 5eV will be spanned in will equal about 60,000 angstroms. This still sounds like a large distance to tunnel through though, but there are a *lot* of electrons in the material, so it may cancel out. Does anyone have any ideas if my logic is logical?
  7. Sep 11, 2009 #6
    For diamond, the breakdown field is 1000 kV/mm (http://www.kobelco.co.jp/p047/products/np0802e/np08022e.htm" [Broken]), so it would take a voltage of 5.5 eV applied across 5.5 nm (about 15 unit cells) to cause breakdown. For breakdown to happen through air, it's a bit more difficult to analyze because one it will depend on the work functions of the materials between which conduction takes place.
    Last edited by a moderator: May 4, 2017
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