"John Larkin" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> I theory, LED voltage is the log of current, so at some very low
> current there won't be enough voltage across the junction to make a
> photon of anywhere near the expected wavelength. It could be that
> materials defects will kill things before that point.
Well, everything is radiating, although at 26meV and no bias, there's damned
little all the way out at 2eV. You're looking at the tail end of two
decades of exponent there.
There's no reason why, for instance, you can't get 2.000eV photons from a
semiconductor with 1.984V across it. You can electrolyze water the same
way -- it does proceed below the reaction voltage, it's just endothermic and
slow as hell. You don't get something for nothing, so there's still current
draw, with an electron plopping through the bandgap for every photon
radiated, it's just falling through a slightly lower voltage.
Question for Phil: does the process of photon production cause a blip in the
diode voltage? This should be detectable as shot noise on the diode's
terminal voltage -and- on a photodiode directly in front of the LED, and
there should be perfect correlation between the two effects (minus quantum
efficiency, so maybe you'll only detect 1 in 5 events at the photodiode).
Is this measurable? I think it should be.
Tim
--
Deep Friar: a very philosophical monk.
Website:
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