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AC skin effect in superconductors

Discussion in 'Electronic Design' started by Jamie M, Feb 24, 2013.

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  1. Jamie M

    Jamie M Guest


    Do any types of superconductors maintain zero electrical resistance
    as the AC frequency goes up? I was thinking about applications using
    superconducting plasmonic diodes like the ones on this page for solar

    Would it be possible for a superconductor to efficiently conduct THz and
    higher frequencies, I guess the current density will go up as the
    electrons get compressed on the surface, is that the limiting factor?

    Also is there a type of superconductor that only is superconductive on
    the surface, similar to how a topological insulator only conducts
    electricity on its surface?

  2. whit3rd

    whit3rd Guest

    Not really. Some linacs use superconductive resonators (resonant microwave
    cavities) that consist of thin lead plate on copper, that have very high
    Q at microwave frequencies, but it isn't infinite. Superconductors have some AC
    resistivity (or equivalent, there's energy losses).
  3. Jamie

    Jamie Guest

    Sure, our place of employment does that all week long and many times on
    the weekends.
    But we try to avoid the arcing part :)

  4. Guest

    Are superconductors unusually reflective at optical frequencies?

    Mark L. Fergerson
  5. No, there is a loss component at any frequency above dc. As
    a rule of thumb, the resistivity approaches the ballpark of the
    non-superconductive metal resistivity at the 'gap frequency',
    which for niobium is roughly 700 GHz. Below the gap frequency
    the conductivity increases IIRC ~ 1/f^2, which means that the
    losses become pretty damn small quite quickly as the frequency
    is lowered.

    There are materials with higher gap frequency, e.g. NbTiN
    is used in SIS mixers at >1 THz.

    More exotic materials such as YBCO show even higher
    gap frequency, but they are a bit messy. They don't obey the
    BCS theory, so they are not understood as well as the ordinary
    superconductors. Their gap is typically not isotropic so that
    the gap frequency is different depending on the direction of
    the field components w.r.t the lattice planes.
    All superconductors carry the current on their surfaces
    only, that is the distribution which minimizes the magnetic
    energy. But this is not the same thing as "being superconducive
    only at the surface". The effect has the same mechanism behind
    it as the ordinary skin effect, except that (i) it kicks in
    already at zero frequency, because the material conductivity
    is infinite; and (ii) the thickness of the current-carrying
    layer is not zero, but rather it equals the so-called London
    penetration depth (~90nm in Nb).

  6. No, they are extremely reflective only up to the gap frequency,
    which is below 1 THz for most ordinary superconductors. If someone
    in interested in details, the SC behaviour in the vicinity of the gap
    frequency is described by the Bardeen-Mattis theory.

  7. Tim Williams

    Tim Williams Guest

    Is that analogous to the plasma frequency in ordinary metals? Which I've
    heard is about where they start absorbing (most in the UV, except copper
    among others, which, as we know, looks pink because it's not such a great
    conductor up around blue). Eyeballing, is it coincidence that it's on the
    order of kT/q, I suppose taking T as the critical temperature?

    Interesting (in your other post) that you say it's roughly 1/f^2 (two pole
    lowpass :) ), whereas from gap theory (again, with ordinary materials) one
    might expect, say, an exponential response or something. Actually, the
    cutoff of ordinary materials isn't exponential with respect to frequency,
    in fact I've measured it for two semiconductors... I'll dig through my
    notes when I get back...

  8. The plasma frequency is due to the 'slowness' of the Thomas-
    Fermi screening, whereas the SC gap frequency is caused by photons
    starting to break the electron pairs. I don't see a connection
    offhand, but then again, maybe there is a way to visualize the
    pairing mechanism in such a way that an analogy can be recognized.
    It is not a coincidence. The supercurrent is carried by paired
    electrons, and you can break the pairs either by kicking them with
    strong-enough photons or by thermal agitation. The required energy
    is roughly the same in both cases.
    I think the 1/f^2 dependence comes from the two-fluid model, in
    case you want to look it up.

  9. Tim Williams

    Tim Williams Guest

    Hmm.. so to draw that analogy, one would have to know more about
    Thomas-Fermi screening. Which I don't. I'll have to read up!

    Now, by "break", does that actually mean a superconductor, say a
    superconducting film, becomes less conductive and can be "broken" by
    sufficiently intense light?

    That would make sense from a different standpoint, namely: a sufficiently
    intense B field is known to cause disruption (and, I would suppose, E as
    well, though a sufficiently strong surface E field might simply not be
    achievable in vacuum, i.e., it sparks first). If Bmax of "critical light
    flux" coincides with critical field strength, it would be very
    interesting. But then, such a classical explanation wouldn't depend on
    frequency, either.

    Ahh, physics...
    And, naturally, most stuff doesn't superconduct, because it's bathed least that much energy!
    Ah, the cutoff I was referring to:
    x axis is octaves in photon energy (i.e., log2(Ephoton), with E in eV), y
    axis is attenuation in dB. Passband attenuation is due to reflection and
    whatnot (0dB = no wafer in the apparatus). GaAs appears much sharper,
    which I'm sure is something to do with density of states around the
    bandgap or something like that. Evidently the silicon is ~100dB/octave,
    or around 16 poles, a pretty sharp cutoff (not related to 1/f^2). A
    pretty reasonable Butterworth filter, or something thereabouts.

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