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Formation of Schottky Diode

  1. Oct 23, 2006 #1
    Here is my question in the formation of Schottky Diode:

    "Why does conduction band is lower than the original value in the semiconductor side?"

    And I think it is lower because the postitve charged donor ions provide a attractive force ,which makes the electrons in semiconductor more easy to escape from bonding and go to conduction band.

    Am I right?
  2. jcsd
  3. Oct 26, 2006 #2


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    The conduction band bends down (lower energy) for a metal against p-type semiconductor (SC) contact. It bends up, naturally, for n-type.

    The band bending does not arise from donor atoms--why would it be different at the interface than in the bulk material? Instead it is the nature of the metal/SC interface. Imagine that a planar metal sheet is brought up to the surface of the SC. The Fermi energies must become equal, so Ef of the metal decreases to match Ef of the SC. In the process electrons leave the metal and penetrate the SC, depressing its conduction band. The charge density decays exponentially in a distance characterized by the Debye depth, so the band bends back up to the bulk value.

    Most SC books cover this. For an exhaustive treatment, starting from a simple description in ch. 1, see
    Henisch, "Semiconductor Contacts", Oxford U Press
  4. Nov 6, 2006 #3
    I've read about schottky diode that , such a metal-semiconductor combination is taken in which Ef of metal lies between the bottom of conduction band and top of valence band of the semiconductor. Why is it so?
  5. Nov 6, 2006 #4


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    marcus, your post is a little confusing to me! It seems like you are describing band bending on the metal side of the interface! There isn't a Debye length for a semiconductor (you need lots of free charges for debye screening, so the Debye depth applies to the metal side of the interface), and the charge density on the semiconductor side doesn't have an exponential decay profile - in fact, the profile is pretty close to the depletion profile you see in a regular PN junction.

    jasum, you're partially correct in that if you have an n-type Schottky interface, the field on the SC side is primarily due to the ionized donors within the "depletion region". However, this additional potential doesn't make it "more easy to escape from bonding". What you describe would be seen as a reduction in the bandgap, not a bending of the bands. The bending is caused, as marcus explained, by the matching of chemical potentials across the interface. The bending profile, however, is a function of the total charge density, but typically, it goes approximately quadratically within the depletion region (assuming a nearly uniform density of ionized donors here).

    savi, you want any "diode" to have the ability to rectify (conduct in one direction only). With a Schottky diode, you do that by putting the conduction band (of the SC) just above the Fermi energy (ofnthe metal). This way, a small positive bias voltage (exceeding the Schottky barrier height) results in a large conductivity, but any negative bias voltage (short of breakdown) leaves you with an essentially insulating barrier. That's what a diode needs to be able to do.
    Last edited: Nov 6, 2006
  6. Nov 7, 2006 #5


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    Oh you're right, my wording was confusing. :blushing: Only the SC (semiconductor) bands bend, downwards for p-type and upwards for n-type.

    I'm not a semiconductor guy, Gokul, so please correct any errors in the following. As I understand it, the potential V(x) falls parabolically as one moves away from the metal contact and into the SC of a classic Schottky barrier. The barrier width L_0 through which the potential is nonzero is related to the built-in potential across the contact and to the Debye length L_D, where (a) The built-in potential depends on properties of the metal and of the doped SC and (b) The Debye length exists (by definition) for any material with free charge carriers, including a doped semiconductor. Typical numbers listed in the books for a silicon device are L_D = 100 nm and L_0 = 4 * L_D.

    The electron density n(x) for a parabolic potential in n-type material rises from a low value (the barrier is depleted, as you say) to the bulk value according to the Boltzmann distribution exp(-eV(x) / kT). Thus log(n(x)) rises parabolically as one moves away from the contact, but the density n(x) itself changes exponentially.

    ZapperZ notes the difference between Fermi energy in a metal and chemical potential in a SC:
    http://https://www.physicsforums.com/showthread.php?t=133914" [Broken]
    Accordingly I should have said that Ef of the metal matches the chemical potential of the SC in my earlier post.
    Last edited by a moderator: May 2, 2017
  7. Nov 9, 2006 #6
    but I thought ,Marcus, that SC band bends downwards for N type and upwards for Ptype ?!!
  8. Nov 9, 2006 #7


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    Nope. Here's an illustration from wikipedia
    http://en.wikipedia.org/wiki/Image:Ohmic2.png" [Broken]
    and you should see the same thing in any book on semiconductor physics.

    Edit: It's actually possible to choose a metal and a doped semiconductor combination that make the bands bend the other way, but this is the usual situation.
    Last edited by a moderator: May 2, 2017
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