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Metal-semiconductor junction with p-type and n-type and varying work functions

  1. Nov 1, 2012 #1
    Hi All,

    I am trying to get my head around the energy band diagrams associated with forming a metal-semiconductor junction for both p and n type s/c. I am also consider for each type, the case of [itex]\phi[/itex]m > [itex]\phi[/itex]n (work function of metal greater than work function on s/c) and [itex]\phi[/itex]m < [itex]\phi[/itex]n (work function of metal less than work function of s/c).

    I have come up with energy band diagrams (attached) for both cases with n-type s/c. The left shows both materials separated in a vacuum, and the right shows the bending of the bands when combined and under thermal equilibrium.

    I have drawn partially the case of p-type with [itex]\phi[/itex]m < [itex]\phi[/itex]p with what I believe to be the correct bending of the bands.

    However I am unsure where the Fermi level should equalise to for the p-type cases. I *think* that the charge carriers here are holes and as such there is a depletion region for [itex]\phi[/itex]m < [itex]\phi[/itex]p.

    Any guidance in terms of drawing the band diagrams for both p-type cases would be much appreciated!

    Attached Files:

  2. jcsd
  3. Nov 4, 2012 #2
  4. Nov 5, 2012 #3
    You seem to have the correct picture for the first two diagrams in terms of the flow of carriers and the direction band bending. In the third diagram, however, it doesn't make sense to talk about flow of holes. Since in the metal there is no electronic excitation gap it does not make sense to talk about holes. The part where you drew an arrow from the metal to the p-type semiconductor you should replace the "h" with an "e." In this diagram, the band bending would look like the one in the second figure. Note that the side which has a positive charge has a lower vacuum level. Now, in the fourth diagram (which you did not draw) for the ##\phi_m > \phi_p## case, the band bending would look like the first diagram and your valence band would cross the Fermi level (similar to how the conduction band crosses the Fermi level in the second diagram).

    Attached Files:

    Last edited: Nov 5, 2012
  5. Nov 7, 2012 #4
    thanks so much for clearing that up, really appreciate the time you took to draw the diagram as well :)
  6. Nov 20, 2012 #5
    My two cents worth of comments...

    Schottky contacts are always dirty. Band diagrams work badly for them. You don't get (by far) the contact potentials estimated from work functions, doping etc.

    Putting the Fermi level near the conduction (in N type) or valence band (in P type) would make the diagrams more convincing, as this would impose to bend the bands.

    I don't know if you can get the metal's level outside the gap. Metals and semiconductors exist that should do it, but I haven't heard of such a practical case.

    I suggested to use such a situation (metal below valence band, 3rd diagram of 1st drawing) to inject a huge density of holes into silicon using - was it platinum, tungsten...? This would be a new test for silicon superconduction. The first was obtained (at Orsay, hi there) by brutal P overdoping of silicon and cooling to few mK. A metal contact would bring the carriers without the dopant, hence keep true silicon instead of a 10% boron alloy... Better: said metal does not become a superconductor (at least when it keeps its holes).

    An other superconduction test in silicon would involve a heterojunction with P-doped GaP which, from its wider gap, would also inject dense holes in silicon without dopant there.

    I suggested it few years ago on a forum, no idea if this was made meanwhile.

    Marc Schaefer, aka Enthalpy
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