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Understanding the band profile diagram in a junction

  1. Sep 10, 2005 #1
    I am confused here and hope to get help from you physicians..the assistant in the uni could not help me..
    when we have a junction between two materials with different fermi levels, for example a metal and a semiconductor, we say the fermi level of the semiconductor gets shifted till it's equal to the work function of the metal and as a result the band diagram of the semiconductor gets bent. and now the conduction band for example through the semiconductor far from the contact is shifted by the value of the barrier.
    now is this true?
    I suppose the energy levels of the semiconductor far from the contact shall not be aware of this contact and it makes no sense that the energy levels through the material, how ever long it is, will be different from a neutral semiconductor without a contact..
    can somebody tell me what is going on?

    Attached Files:

  2. jcsd
  3. Sep 10, 2005 #2
    No this is not right.

    Both Fermilevels will align themselves to reach thermodynamical equilibrium (local charge neutrality at the interface region). This is done by charge transfer from one material to the other. In the left figure you have a n-type semiconductor and a metal. Electrons flow from the semiconductor (SC) to the metal. Positive donor ions remain at the SC-surface region and this charge yields the band bending (Poisson's equation). The Schottky barrier (the energy difference between the metal fermilevel and the n-type SC conductionband edge) arises due to the transferred charge. That is what happens.

    Indeed, in this case, the metal workfunction is equal to the conductionband edge of the SC but this is not always the case. For example, when there is a voltage over the interface region, you get another situation.

    If you wanna know more, i suggest you look for R. Tung's website :

    Look for the Schottky barrier tutorial

    He does a great job at explaining this...
    I also recommend to look for the socalled fermi Level Pinning fenomenon (i work on this for my phd). It expresses the fact that the metal workfunction changes in magnitude when you replace SiO2 by HfSiO4 in a metal/SiO2/SC-stack, irrespective of the metal itself (material,etc). This problem disturbs the basic MOSFET-operation because of several reasons...


  4. Sep 10, 2005 #3
    Tung will teach you that in case of an interfacial region between two materials in a stack, there does not have to be charge transfer from one material to the other. This is the case when there is an insulator between the two materials like the metal/SiO2/SC stack. In that case, the Fermilevels will still align for the same reason. The potential difference over the interface equals the surface charge at the SC-surface and its image charge at the metal surface. The big question is what system generates this surface charge ? You have several models like the Metal induced Gap states where the metal wavefunctions penetrate through the interface (SiO2 layer) and they generate a certain amount of energylevels in the SC-bandgap. These energylevels (and hence the electrons occupying them) can be seen as a metal of which the Fermi level is called the charge neutrality level. Depending on the position of the SC-Fermilevel with respect to the charge neutrality level (the SC-surface-Fermi Level) the SC surface acquires the required net-charge. "required" means corresponding to lowest potential energy of the material-stack. This is just one of the several models that are out there (look at Tung's website for this). The biggest problem with this model is the fact that it does not depend on the interfacial chemistry and Tung corrected this in his socalled Bond-Polarization-model...But this is probably i bridge too far for now...Feel free to ask more questions on this, if you want

  5. Sep 10, 2005 #4
    hi marlon, thanks for your reply..
    that was good what you said, in the first reply, it's indeed the change of charge density near the contact what causes the band bending, so in the case of the n-semiconductor the left positive donors raise the energy of the energy levels at that region. What's confusing to me is that in the diagrams it's drawn as if the energy levels at the contact do not rise, but instead they have the same value as before making the contact with the metal, and as you move deeper in the SC they fall until they reach the same distance from the fermi level they had before..I hope you see what I mean?
    it is seen in the attachment, metal and SC before and after contact, how the energy levels at the contact are constant and the shift happens inside the SC instead..

    thanks for the website, I was at there before indeed, but I found it complicated, or not with enough definitions to start with as I'm new to this subject.
    anyway, at that time I could not understand what's the charge neutrality level (cnl), your explanation gave me an insight and I went back to that site, but I see tung explains it as a characteristic for the material surface independent whether it's metal or vacuum on the other side..what's the connection with the metal wavefunctions?
    though I think too it's still far for now, but any comments are welcome!
    I'm most concerned about the first question,
    thanks a lot..

    Attached Files:

  6. Sep 11, 2005 #5
    Don't mind too much about it. If you compare the Fermi Level of the metal before and after contact is made, they look indeed to be aligned and do not change. However, the metal-Fermi level will shift a bit with respect to the FL of the metal without contact with the semiconductor. It is not depicted in this figure but it must happen.This is not too important though. The important part is to realize that the FL-will be aligned throughout the entire contact. In order to achieve that, there has to be charge transfer and thermodynamic equilibrium corresponds to the situation where the diffuse-current (majoritycarriers) is equal to the driftcurrent (minority-carriers).

  7. Sep 11, 2005 #6
    what about the conduction and valence band levels deep in the semiconductor. they seem to be shifted in the figure, so their value before is different from their value after making the contact, though no charge density change happens their. is it really the case?
  8. Sep 11, 2005 #7
    They are shifted because the SC fermi level is shifted towards to metal fermi level. But again, they are only different at the surface. In the bulk medium the energy bands do not change before or after you make contact with the SC because only the surface will "feel" the contact and change it's electrostatics.

    So basically, the difference in fermi level between two materials prior to contact is equal to the socalled interface dipole. This is the dipole that arises because of the SC-spacecharge and the metal-image charge. This potential difference (only at the surfaces) is equal to the difference in FL before the contact is made between two materials. So the bulk properties will not change (same energybands as before contact) and the difference between the two materials will be compensated at the interface (ie the interface dipole)

    Ofcourse, it is a fact that the entire energy band diagram is shifted because there is one aligned vacuumlevel after contact is made. With respect to the bulk properties, you can say that the workfunction is indeed shifted and thus changed but that only expresses that the materials are in a "new situation". This does not imply that the bulk properties have changed because the energy differences between every band stays the same, only the reference level (ie the vacuumlevel) shifts. The only difference after contact is that the energy bands are bent with respect to the bulk energy bands

    Last edited: Sep 11, 2005
  9. Sep 11, 2005 #8
    so the diagram is wrong because in the diagram the conduction and valence band energies remain constant at the interface, and decay as you go deeper in the bulk!
  10. Sep 11, 2005 #9
    Yes i get the point now. Indeed the difference of the energybands with respect to the vacuum level DOES NOT change at the interface. In order to align the FL's, the bands of the SC undergo bending as outlined before. This is indeed the very base of the famous Mott-Schottky model for metal/semiconductor contacts. This model is indeed wrong for several reasons like the Fermi level Pinning fenomenon, the fact that no interface chemistry is taken into account (for example, the metal workfunction depends on the crystallographic direction of the surface), no interfacial region, etc etc...

    Historically this is just the first model that attempts to explain the electrostatics at M/SC-interfaces.

  11. Sep 11, 2005 #10
    here I found this diagram, see attachment, so that I can put my question in a clear way.
    this is a pn-junction, on both sides is the same material but different doping, p and n, so before we bring them in contact only the fermi levels are different but the energy levels are equal. after the contact, the fermi levels get aligned and we get a band bending.
    now the question is, we see in the diagram that the energy levels far from the contact are not equal, and this is what's making my confusion.

    Attached Files:

  12. Sep 11, 2005 #11
    They are not equal to what ? You mean to each other ? This cannot be the case since the energy bands of a p-SC are not at the same position with respect to the Fermi level as those of a n-type SC.

  13. Mar 4, 2007 #12
    Hi yous guys, I tuned into this because I was searching google and found this conversasion. Marlon, if you do not mind, would you take the picture that IMAS attached of a pn junctiion and the band levels (2 28 pnq), and drew in the vacuum levels , and workfunction, on both the p type and n type.
    I appreciate it, I am not near a University to ask any one and this stuff was not taught in my pre-Schottky graduate days.

    Thanks, Nelson
  14. Mar 5, 2007 #13
    Hi Marlon, better still than answering the above question for me, tell me where I can find the periferal explalination of the thumbnail that is in the #4 entry above by imod on 09-10-2005 ast 8:52 pm. It shows the work potential. I would like the have the write-up with an explaination?

  15. Mar 6, 2007 #14

    There you go.

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