If I enclose an LED in "tin foil" so that the photons it gives off are reflected back to it, would the power consumed by the LED go down?

  • $\begingroup$ Why do you think it should (as you tend to think in that way) go down?-This helps others to understand you and help you early. $\endgroup$
    – Sensebe
    Dec 13 '14 at 11:09
  • $\begingroup$ It's an easy enough experiment. Why don't you try and see what happens? $\endgroup$
    – CuriousOne
    Dec 13 '14 at 11:20
  • $\begingroup$ @CuriousOne it's not really an easy experiment. Applying a high-reflectivity layer to the outside of an LED is difficult enough (tin foil will be full of folds which act as absorbers); getting those photons to hit the PN junction instead of the substrate or the potting material is next to impossible. The question here, I think, is whether reabsorption of photons aka photoelectric effect will reduce the net current draw from the power source. $\endgroup$ Dec 13 '14 at 12:44
  • $\begingroup$ Are you thinking that, if photons go back into the LED, there is no need for recombinations of electron-hole to create some new, so you could decrease the current and therefore the power? If that's the case, remember that in order to absorb photons you'd need a photodiode, which is basically the same p-n junction used in an LED but in reverse bias. $\endgroup$
    – SuperCiocia
    Dec 13 '14 at 12:52
  • $\begingroup$ @CarlWitthoft: Place the LED in the focal point of a convex lens and place a mirror behind the lens. Most of the light from the LED will be reflected back onto the chip. Alternatively you can use an enlarging beauty mirror. LED chips are fairly large and it's not hard, at all, to reflect the light back on the chip. Good luck! $\endgroup$
    – CuriousOne
    Dec 13 '14 at 12:54

I guess the question is whether the LED will also act as a photodiode, i.e. whether incident light will excite electrons and therefore generate a current. If so, then reflect the diodes light back onto it will indeed reduce the current flowing in the diode by generating an opposing EMF.

This turns out to be surprisingly hard to Google, but Wikipedia suggests that LEDs will indeed act as photodiodes, and it cites a couple of papers to back up the claim.

However I'd guess that the photodiode efficiency will be low because LEDs are normally designed to have a direct band gap, which means excited electrons will recombine very fast and either just emit light again or dissipate their energy as heat. So while I'm not sure I would be prepared to testify under oath, my guess is that it will reduce the current but not by very much. The main effect will be to heat up the LED.


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