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Schottky Barrier

  1. Jun 26, 2013 #1
    Hello Everyone,

    We know how a schottky barrier forms with a depletion region of width 'W' and contact potential
    'qV' as shown in below figure,
    The depletion width 'W' depends mainly on doping concentration
    of the semi-conductor and contact potential, for example for a given doping concentration the width is 1μm.

    Now question is what happens to the contact potential if i took a semiconductor of thickness or width only 0.1μm, without changing the doping concentration.

    Any answers would be greatly appreciated!

  2. jcsd
  3. Jun 26, 2013 #2
    But what's on the other side? If you just have a floating piece of semiconductor, then it can just adjust its chemical potential as needed. More likely, you would have a complete circuit, which means there's another metal contact on the other side of the 0.1um semiconductor.

    The Schottky barrier (as shown) is a case of electrons from the semiconductor transferring to the metal, leaving behind positive ions. The width of the depletion region is the amount of material it takes before the number of positive ions matches the number of transferred electrons (i.e. charge balance for overall neutrality). Higher doping means a higher density of ions, so a smaller amount of material (lower W) is necessary.

    The band bending at the interface is because of the electric field from the electrons that transferred to the metal. Once you've matched that with the number of positive ions, all the field lines are terminated and you're back to flat bands. Now if your piece of semiconductor was less than W, then the band bending is not finished at the other end -- this means there's an electric field still present.

    Exactly how the metal there would react to this electric field is hard to say because realistically it would form its own Schottky barrier. There's some symmetric self-consistent solution to this that maintains charge neutrality. (If this second contact was ideal, i.e. no Schotkky barrier, then the semiconductor E-field would repel surface electrons in the metal and leave behind positive ions, making up for the fact that semiconductor didn't have enough positive ions of its own).
  4. Jun 27, 2013 #3
    Hello sir, Thank you for your reply..
    But really i can't understand what happens when the second contact is ideal, i.e. no schottky barrier,
    You say's that " the semiconductor E-field would repel surface electrons in the metal and leave behind positive ions ", How this is happening.? Can you please explain me in brief or in bond bending (graphical) form..

    What i am thinking is that the semiconductor fermi level is still higher than metal so electrons flow from semiconductor to metal until both fermi levels align.but thin semiconductor consist no such amount of electrons .
    so what is the contact potential or bond bending..??
  5. Jun 27, 2013 #4
    You need to draw it with the vacuum level to see it properly. Basically, the semiconductor needs to have a depletion width of positive ions to match the charge of the electrons in the Schottky metal. If there isn't enough "space" to do this because the semiconductor is terminated before reaching the depletion width length, then something else needs to make up for those missing positive ions. The second ideal metal does this.

    Graphically: look at the bottom right picture and cut off the semiconductor at some length shorter than W. Then add the second ideal metal. With just this you can't see what's going on, but you can if you add the vacuum level. For this you were to solve the Poisson equation to get the potential, you would see a discontinuous change in the vacuum level right at the ideal semiconductor-metal interface. That discontinuity is caused by a dipole - i.e. an atomically sharp plane of charge.

    Going from left to right you have this:
    1) Many, many electrons inside the Schottky metal right at the metal's surface with the semiconductor.
    2) A small density of positive ions in the semiconductor (donors who lost their electrons to the Schottky metal and so they aren't neutral).
    3) As you go further right you cumulatively pass more and more positive charge, but the total is less than the number of electrons at the metal surface.
    4) You reach the second ideal metal. Its electrons have been repelled because of the field from the electrons of the Schottky metal (there weren't enough semiconductor ions to screen the field because the length is shorter than W). Because these electrons have been repelled, the ideal metal has positive ions at its surface. The number will be equal the difference between the Schottky metal's negative electrons and the semiconductor's positive ions.

    So in total, the number electrons at the Schottky surface equals the number of semiconductor ions plus the number of ideal metal ions. This gives overall neutrality (amount of negative charge = amount of positive charge) for the whole metal-semiconductor-metal structure. Local non-neutrality arises from the different work functions.
  6. Jun 28, 2013 #5
    Hello sir,
    Thank You very much for your valuable reply,
    I can understand a lot..Still i have a number of sub-questions regarding the same.

  7. Jun 28, 2013 #6
    I'm doubtful if that's the best model. It seems this application of Poisson's equation is somewhat arbitrarily set up. How do you have excess electrons? The electrons are supplied by the donor atoms, after all. If Nd is limited to a certain total number by the semiconductor volume, then Poisson's equation cannot be simplified by the approximation Phi=0 at x far into the n-type material because there IS no far into the n-type material.

    Another way to view this situation is that the electric field built should be accumulated by the equilibrium condition whereby diffusing electrons are balanced with the field-accelerated electrons. Diffusion would be smaller because it's proportional to the number electrons that are there to diffuse. Therefore the electric field would also be smaller, and thus the equilibrium condition would be established with a smaller depletion width. It's outside the standard model.

    One thing that seems to be missing from my explanation: how does the Fermi level align at the Schottky contact? The potential difference at the barrier should be the same regardless of the amount of material. However, there is commonly an interface perturbation involved. Perhaps this is related to that - the Fermi level of the semiconductor would be effectively reduced because of geometry?
  8. Jun 28, 2013 #7
    If Fermi level of semiconductor reduced by geometry, then the Fermi levels align at the Schottky contact with smaller potential difference.
    because of smaller volume a small amount of electron diffusion(and hence the small potential difference)
    can bring the Fermi level of semiconductor to align with metal..??
  9. Jun 28, 2013 #8
    If I'm reading you correctly, I think that's what I'm trying to say. Essentially, you're building a Schottky diode with an ideality factor far from 1.
  10. Jun 28, 2013 #9
    Seems logic and strange.

    With out changing the Fermi levels of materials,they brought into equilibrium even at low contact potentials
    just with their smaller geometry/volume alone...
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