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Schottky barrier formation

  1. Feb 22, 2016 #1
    The usual way to explain the formation of a Schottky barrier between a metal and semiconductor is the following:

    Suppose a metal and a semiconductor with different work functions are put into contact. To compensate for the work function difference ΔW of the two materials a charge layer will build up at the interface creating an electrostatic potential φ(z) s.th.

    -eφ(z=0) = ΔW

    Where z=0 is at the interface. Now what I don't understand is this. Since the metal is a perfect conductor we must have that φ(0) = 0 at the interface between the metal or at least that φ tends to a constant. The usual way to solve this problem is to employ the method of images, i.e. we mirror the charge distribution in the semiconductor at the interface, since this effectively is the same as solving the problem of having a charge distribution above an infinitely grounded plane at z=0, in which case (V(z=0)=0, V(∞)=0) are the appropriate boundary conditions.

    But are these two requirements compatible? For me it seems that the latter requires that there can be no correction to the conduction band energy at the metal-semiconductor interface, which is exactly what we want when ΔW≠0.
    Last edited: Feb 22, 2016
  2. jcsd
  3. Feb 22, 2016 #2


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    Your equation is correct. You are also correct saying that the potential within the metal is constant. It is not because it is a perfect conductor, rather, the electron density in a metal is so large than any practical external field is screened within less than an atomic layer thickness.
    The potential drop is all within a thin depletion layer of the semiconductor.
    You don't need to use anything as complicated as the mirror image of the charge distribution because the system is essentially one-dimensional. All you have to do is consider charge, field and voltage as a function of a distance from the metal-semiconductor interface.To a very good approximation, all the carries within the depletion layer are gone. That leave a net charge density equal to the dopant concentration. So you have a uniform charge density ## \rho ## over thickness ## z##. That gives you a linearly decreasing field into the semiconductor and a quadratic potential barrier.
  4. Feb 23, 2016 #3
    Okay I still need to understand why the image method is wrong to use. To recap: The method of images is employed to solve the problem of finding the potential from a charge distribution above and infinite conducting plane. Isn't this exactly the problem I have? The infinite conducting plane being the metal-semiconductor interface.
    I don't understand the argument "It is not because it is a perfect conductor, rather, the electron density in a metal is so large than any practical external field is screened within less than an atomic layer thickness." - Can you elaborate on that?
    Also it seems that all this assumes that my heterostructure is doped. What if it isn't? Will a barrier not form in any case?
    Last edited: Feb 23, 2016
  5. Feb 23, 2016 #4


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    Hi aaaa202

    I didn't say that image method is wrong to use, it is just not useful. Typical problem of electrostatic is to find potential field given boundary conditions like potentials at certain surface and some volume charge distribution. The solution is to solve a 3-dimensional second order partial differential equation (Poisson's equation) which is, in general, not easy. I did a number of this calculations using Finite Element Analysis tools.
    The image method is a mathematical trick that allows you to calculate potential field created by a 3-dimensional charge distribution over an infinite conductive plane without solving the Poisson's equation: you just simply add potential of the given charge and that of the mirror image of the given charge (of the opposite polarity, of course). The boundary condition at the surface (V = 0) are satisfied automatically. Of course, the actual charge of the conductive plane is located at the surface of the plane.
    In the case of a Schottky barrier, the charge distribution is one-dimensional and (almost) constant and the solution can be obtained by simple integration. Using the image method doesn't save you anything. It just doubles your integration volume.

    As for your second question, conductivity of a material is a product of carrier density * their mobility. The screening of the static electric field does not involve movements of electrons and their mobility is irrelevant. All that matters is electron (or hole) density. There is a bit info in Wikipedia at https://en.wikipedia.org/wiki/Electric-field_screening

    Your last question. Yes, all practical Schottky diodes are made with doped semiconductor, usually n-type. However, the barrier is always created even if the semiconductor is intrinsic. When two conductive materials are in contact, there is always a potential barrier equal to the difference in the work function. It is just that in some cases you won't get a diode only an ohmic contact.
  6. Feb 24, 2016 #5
    Okay so as far as I can tell you are saying that it is not wrong to use the image method, but it is much easier to simply assume a constant charge density and demand the boundary condition of the potential described above.
    What I'm doing currently is simulating what happens when a heterostructure and metal are put together by solving the Schrödinger and Poisson equation self-consistently. Doing so I would like to see the Schottky barrier come out naturally from my approach. This is the reason why I have employed the image method. But how can the Schottky barrier come out naturally when the potential I perturb with (coming from the image method and solving the schrödinger equation in the semiconductor) is zero and the metal-semiconductor interface.
    I hope you understand my question. I have an attached an example of how a simulation result could look like. To the far right there is a thin layer of aluminium, which is why I have cut of the electron density before zero because it explodes close to the metal interface.

    Attached Files:

  7. Feb 24, 2016 #6
    I think the problem is that I start with a common chemical potential for the heterostructure and metal. Instead I should start with different ones and let them align by running the self-consistent scheme.
  8. Feb 24, 2016 #7


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    I apologize, I gave you only a partial answer. I guess this is an effect of a bias - I've been living in a macroscopic world since left academia.
    What I told you is good to understand how a commercial Schottky diode works and it is valid only if you have a doped semiconductor with the depletion layer at the interface.
    In general, to get the potential distribution you need to solve Poisson-Boltzmann equation https://en.wikipedia.org/wiki/Poisson–Boltzmann_equation
    that solves the Poisson equation while the charge distribution is given by Boltzmann statistics.
    Either approach will give you a 'mean field' answer, that is the value of a potential averaged over thousands of unit cells ant this is the value of the potential an electron will 'see' when away from the metal.
    Closer to the metal, there will be an additional potential due to attraction between an electron and its mirror image. One note though, the image will not be perfect.
    The Coulomb potential in reciprocal space is ## V(k) \sim \frac 1 {k^2} ## and electrons can't screen potentials for ## k \gt 2k_F ##
    See for example an article https://en.wikipedia.org/wiki/Thomas–Fermi_screening.
    As for the need to find the actual field. Typical field of in a depletion layer of a Schottky diode is ## 1 V/\mu m##. For an intrinsic semiconductor, the field will be much lower. Maybe you can just assume that on the scale of your heterojunction the potential is constant and equal to the difference in work functions?

    Hope this helped
  9. Feb 25, 2016 #8
    Yes I think that will be my tactic for now. Start with a chemical potential for the heterostructure and for the metal. Then allows electrons to flow from the metal until the difference between the fermi energy of the metal and the chemical potential far into the bulk of the heterostructure equals the difference in work functions. Thank you for your replies. They have been very helpful.
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