Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about semiconductor physics

  1. Sep 21, 2004 #1
    Hello,

    my question is the following:
    we know from books on semiconductor physics that the energy levels of the electrons in a pure semiconductor cristal form in fact bands of energy.

    I was wondering why these energy levels are in the form of bands an not simple distinct energy levels as the ones of a single atom for example.
    In fact, what is the physical quantum system these energies are solution to?
    Put in another way, what is the potential energy a single electron in the cristal is object to?

    I have been told that Pauli exclusion principle played a certain role in the formation of the energy bands.Is this correct?

    thank you
     
  2. jcsd
  3. Sep 21, 2004 #2

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    An atom in isolation (or in a gas) does not behave the same way as when they are close to other atoms in a solid. The wavefunction of each atom then overlaps with its neighbors, forming a completely different type of phenomena than when it is isolated. The periodicity of the potential of the solid in real space adds a very significant factor to the symmetry of the situation (example: see Bloch wavefunction).

    Zz.
     
  4. Sep 21, 2004 #3

    Claude Bile

    User Avatar
    Science Advisor

    Okay, lets start with a conductor, as they are simpler than semiconductors.

    The characteristic property of a conductor is the presence of non-localised electrons, that is, electrons that are free to move within the lattice. While the picture that is often given is one of an electron zipping around inside an atomic lattice, this picture is not entirely accurate.

    When one solves the Schrodinger equation for these electrons, one gets a plane wave (Why? Beacuse a periodic potential must yield a periodic solution due to the symmetry, this is Bloch's theorem). This implies that these non-localised electrons actually exist everywhere within the lattice.

    This is where the Pauli Exclusion Principle applies. Since there are many non-localised electrons, all existing throughout the entire lattice and the fact that electrons are fermions, each electron has a slightly different energy so as not to violate the Exclusion Principle.

    The difference between semiconductors and conductors is that semiconductors have a conduction band that is populated due to thermal excitation. Also, the presence of dopants within the semiconductor can create extra energy levels within the forbidden energy gap.

    Claude.
     
  5. Sep 22, 2004 #4
    Claude's explanation is fine, but a different way of seeing the origin of the bands in crystals is too imagine larger and larger molecules and clusters.

    When two atoms approach eachother, symmetric and asymmetric solutions of the wave equation start to have different energies. The symmetric solution with high electron-density between the nuclei has a lower energy than the antisymmetric wave function with a node of the electron density between the atoms. These are the bonding and the anti-bonding orbitals.

    If you add more atoms, each one adds a new level, but because the distance increases, the energy of the new level falls between the extremes of bonding and antibonding of the closest pairs. With about 10^23 atoms in a crystal (Avogadros number), this results in a continuous band of levels.
     
  6. Sep 22, 2004 #5
    How should I picture visually the orbital of an electron in this cristal?

    I also find it rather difficult to admit than the electron "feels" the potential energy of the far nucleus.
    What happens visually in terms of the electron orbital when we apply an electric field to the cristal ?
     
  7. Sep 22, 2004 #6

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    There are no strict "orbitals" for the valence electron in a metal, for example. This is where you need to know a bit of solid state physics, and a LOT of quantum mechanics.

    When the valence band of each atom significantly overlap with another (or more), then via the indistinguishibility principle, you no longer are able to detect when two of them interchange position. This is when the Fermi-Dirac statistics kicks in. This indistinguishibility also means they can easily hop from one location to another. QM model this as a superposition of plane waves. If you want to know more, you will have to open a solid state physics text.

    More on this at this link: http://www.mtmi.vu.lt/pfk/funkc_dariniai/quant_mech/bands.htm

    Zz.
     
  8. Sep 23, 2004 #7
    So is the conduction in such material due to electrons (travelling between atoms?) thus changing between energetically close states in the conduction band, or should the electrons be totally unbound to the atoms in order to have an electric current?

    I am asking this because I want to know if when we apply an electric field, the electrons are simply given the ability to change between allowed states in the conduction band. Or should we stick with the classical view we have been told in school, where the current is due to "free" electrons circulating in one direction imposed by the electric field?

    Said in another way, what causes an electric current in a semiconductor?

    Thanks again guys
     
    Last edited: Sep 23, 2004
  9. Sep 23, 2004 #8

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Try "weakly bound". The degree of bounding depends on what kind of approximation you want to explain whatever it is you want to explain. For models of electrons in the conduction band that are completely unbound, you would get a free electron gas. In a conductor, this is equivalent to the Drude model of the conduction band. This is what gives you Ohm's law, for example. However, very quickly you will realize that this model will break down when you examine other experimental measurements. Then you start inserting an "almost free" electron model, where you have a series of Bloch potential, but still, very weak potential. So you start adding in more and more "corrections" till you get a description that is accurate enough for whatever it is that you want.

    Zz.
     
  10. Sep 23, 2004 #9
    But can we say that the "weekly bound" electrons of the conduction band that are accelerated by the electric field are the origin of the current in the band theory?

    What is the difference in energy between the conduction band energy and the first ionisation energy?
     
  11. Sep 23, 2004 #10
    ZapperZ, I am sorry.I would like to know exactly how is current generated in a N doped material in the band theory perspective.
     
  12. Sep 23, 2004 #11

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Come again?

    A "current" simply means you have applied a potential difference across it, and the presence of conduction electrons allows for the existence of charge flow.

    This is getting confusing because there is an ABRUPT jump between general semiconducting band (which is what this thread is about till now) and now suddenly we're talking about electrical transport, and not only that, electrical transport in a specific, n-type semiconductor! Let me clearly mention that you cannot get a lesson in band theory on here, even from me. If it were that easy, there would be no reason to have books on it and classes to attend.

    Zz.
     
  13. Sep 23, 2004 #12

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    The connection between band structure and electrical transport, whether the material is doped or not has been the subject of many a dissertation, mine included, hence the name Dr Transport. To date, electrical transport has been figured out in exactly two semiconductors, p-type Si and Ge to any degree of accuracy without resorting to Monte Carlo integration of the coefficients (sorry Zapper but I have got concerns with Monte Carlo) or relaxation time approximation. I took the first steps in calculating transport coefficients in anisotropic materials, i.e. pseudo-III-V materials and it took me 5 years of calculation and and theorizing about symmetries in crystals to get there and I only included hole-acoustic phonon scattering mechanism. The approach is exact, but cumbersome and takes time to get right. (Five more years and I might have it all completed and correct if someone would pay me to do it full time, and I'd be able to get all of the III-V's in the process. Hint hint, if someone out there is a funding agent get ahold of me and we can talk.)


    General band structure calculations come in many different flavors, [tex] \vec{k} * \vec{p} [/tex] is a perturbative expansion about a known point int the Brillouin zone, usually the [tex] \Gamma [/tex] point or the center of the Brillouin zone. To do full band structure calc, Augmented Plane Wave (APW) , Orthogonal Plane Wave (OPW) and a bunch of other ab-initio methods have been used to sokme degree of accuracy given good inputs. Everyday a new band structure calculation comes out, better than the previous. In addition, if the matrial is exotic, ie.e like a chalcopyrite, years of work needs to be done to even get in the ball park, so we couldn't even scratch the surface here. I guggest that you talk to your solid state professor and see what they have to say, it woul dtake too much time and effort to get even the basics explained in a coherent way in th is forum.

    dt
     
  14. Sep 24, 2004 #13
    Conduction can be understood at different levels. Simplest is the Drude model, which is OK for doped semiconductors at room temperature. Conduction is possible because for the conduction electrons there are unoccupied states accessible with infinitesimal increases of energy.

    For simple metals, the quantum free electron gas is more accurate. In that approximation, the electrons are in a continuum of states up to the Fermi energy. In an electric field all valence electrons acquire some extra momentum, and the "Fermi sphere" is displaced a bit from the origin in momentum space. Also here unoccupied states are accessible with infinitesimal increases of energy.

    So an electric field accelerates electrons into states with higher momentum, but this acceleration does not contnue indefinitely. It is limited by inelastic scattering. After some relaxation time [tex]\tau[/tex] the momentum distribution is reset to be centered around zero.
     
    Last edited: Sep 24, 2004
  15. Sep 24, 2004 #14

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Hey, I'm assuming your inference to me was due to my admission in another thread that I did some monte carlo simulations. :) No need to be sorry about since I don't use Monte Carlo often either. Besides, I was given the model, and then a boatload of money to write a simulation for it. When you're a graduate student still looking for an assistantship, you'll sell your soul to the devil if he can support you through school! :)

    Zz.
     
  16. Sep 24, 2004 #15
    If I have well understood, in semiconductors under an electric field, the conduction results from conduction band electrons gaining a slight energy from the electric field and going to slightly different energy levels in the conduction band?
     
    Last edited: Sep 24, 2004
  17. Sep 24, 2004 #16

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    Now, again there is a a reference to the relaxation time, a convenient explanation, with some physical basis because it is easy to solve. If you do the problem correctly, you don't need the relaxation time. Calculations have been done to higher order obtaining better results. For n-type materials, the explanations are correct as given, technically they could be better. In the conduction band, if your band structure is calculated correctly, there should be some interaction between the conduction and valence bands and the wave function for the electron would indicate this. Scattering cross sections would not only depend on the angle between the incident and scattered [tex] \vec{k} [/tex] vectors but their orientation relative to the Brillouin Zone. In the relaxation-time approximation, the scattering is assumed to be isotropic and the scattering rates depend only on the angle between the two vectors. Sadly, for many n-type materials, this is ignored, the Fermi surface, or surface of constant energy, is spherical or nearly spherical so the essential physics is ignored. In silicon and germanium, the conduction band is not parbolic, but slightly non-parabolic, people ignore this and do the first order calculation, it may be good enough, but it isn't completely correct.

    If the problem is electronic transport o in the valence band or in hole transport, the situation becomes much much more complicated. Not only is the conduction band taken into account, but the heavy, light and spin orbit bands. Each are non-parabolic and highly anisotropic in nature. For any point away from the very center of the Brillouin zone, the wave functions for the valence bands are a combination of the valence band wave functions at the center. Again, scattering is not isotropic, but very anisotropic in nature. There are more bands to worry about and there is scattering between the valcence bands further complicationg things. The best way to calculate the transport coefficients is by expanding the scattering rates in terms of products of spherical harmonics, then the Boltzmann Equation can be converted from a intego-differential equation into a matrix equation. The transport coefficients are then found quite easily, the lowest order angular momenta pair of harmonics then gives the relaxation time approximation scattering rate.

    As you can see, the relaxation time approximation can give you a 90-95% solution and in some cases, that is good enough. If not, this other formalism must be employed and then complete characterization of the material can be accomplished. This has been done for silicon as I have said before, this hasn't been done for any of the other semiconductors.


    Zapper, you refered to a Monte Carlo calculation in your past. Nothing personal, all in jest, I know you have your reservations now as do many others in the community.
     
  18. Sep 26, 2004 #17
  19. Sep 26, 2004 #18

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    seen it, downloaded the pictures for my screen saver and posted them on the wall.........
     
  20. Sep 26, 2004 #19

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, yes. The pictures really help explain the concepts quite nicely.
     
  21. Sep 26, 2004 #20
    yes, i share the same impression...

    She is definitely a nice illustration of a p-junction...wouldn't you say ?

    marlon :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question about semiconductor physics
Loading...