4 ms·

(Disclaimer: I have some background in physics, but I'm not an expert.) I think the heat is transferred from the "computer" to the LED itself. So the LED itsel

(Disclaimer: I have some background in physics, but I'm not an expert.) I think the heat is transferred from the "computer" to the LED itself. So the LED itself heats up. To go into a bit more detail: All physical objects emit thermal radiation (a.k.a. black-body radiation [edit, not quite correct]). The hotter the object is, the more radiation it emits and the shorter the wavelength of the radiation. For "red hot" objects, the radiation is in the visible part of the spectrum; for room-temperature objects, it's in the infrared, so you can't see it, but it's still being emitted. If you put two objects of different temperatures next to each other, they'll both emit radiation and absorb each others' emitted radiation. The hotter one will emit more radiation than the cooler one, so there will be a net transfer of energy from the hotter one to the cooler one (normally). But in this experiment, when the researchers run the LED in reverse, this somehow suppresses the LED's normal thermal radiation emissions. The "computer" continues to emit thermal radiation, which is absorbed by the LED just like normal. But the LED doesn't emit thermal radiation back to the computer (at least, not as much as it normally would). So even though the LED is hotter than the computer, the LED acts like it's colder for thermal radiation purposes. So there's a net transfer of energy from the cooler "computer" to the hotter LED. Of course, this isn't a complete solution; you still have to do something about the heat once you've transferred it to the LED. The article doesn't address this part of the problem. But I speculate you could just attach a heat-sink to the LED. Heat sinks work better at higher temperatures, so the computer+LED+heat-sink solution would be more effective than a computer+heat-sink alone.

the8472 8y ago
> All physical objects emit thermal radiation (a.k.a. black-body radiation) Not all materials are black body emitters. The black body spectrum is in fact an idealized emission spectrum which only few materials approach. An LED most definitely is not a black body emitter. You can engineer materials that absorb/emit a specific part of a spectrum and reflect/don't emit in other parts. This has uses in thermal engineering - e.g. you can shed heat by making things "white" in most of the spectrum but "black" in the infrared windows of the atmosphere. In this case they tune the behavior electrically.
- timmaxw 8y ago
  True, but the distinction is irrelevant in this case.
  - the8472 8y ago
    Far from it. If all physical objects were black bodies then the LED couldn't do anything here since emission and absorption are the same process.
    - timmaxw 8y ago
      The article notes that the LED "acts as a very low temperature object". I interpret this to mean "the LED radiates as though its temperature is lower", rather than "the LED radiates as though its emissivity is lower". In other words, AFAICT the effect can still exist even if the LED has 100% emissivity in both absorption and emission, which would make it a black-body. So the fact that the real LED's emissivity is not 100%, is not relevant to the effect. But we're splitting hairs here.
      - the8472 8y ago
        The paper is talking about effective temperature for specific photon energies. Additionally black body emission refers to thermal equilibrium conditions which the LED clearly is not. And then there's the near-field coupling they're using: > Recent experimental advances in near-field radiation have shown that heat-transfer rates on the nanoscale can exceed the blackbody limit by several orders of magnitude14–17, owing to contributions from evanescent and surface modes18,19. As a result, energy conversion rates can be greatly enhanced on this scale20. So this whole system relies on not being a black body on several levels.
        baybal2 8y ago
        Even that will only work if they don't produce more heat doing that, than they pump away. Same issue as with anti-Stokes scattering that was once held promise of "self-cooling LEDs" or using gas dynamic lasers for spaceship cooling. There is no trick against thermodynamics after all.
yummypaint 8y ago
I agree with you up to the last paragraph. It sounds to me like the led is reverse biased like a silicon particle detector. A normal led works by combining electrons and holes to make photons, in this case hole-electron pairs are made by light. The reverse bias causes them to drift apart, resulting in a small current opposing the bias. For this reason i would interpret the power supply producing the bias as the load, and the led as the power supply. This is how energy is removed from the system.

the8472 8y ago

> All physical objects emit thermal radiation (a.k.a. black-body radiation) Not all materials are black body emitters. The black body spectrum is in fact an idealized emission spectrum which only few materials approach. An LED most definitely is not a black body emitter. You can engineer materials that absorb/emit a specific part of a spectrum and reflect/don't emit in other parts. This has uses in thermal engineering - e.g. you can shed heat by making things "white" in most of the spectrum but "black" in the infrared windows of the atmosphere. In this case they tune the behavior electrically.

timmaxw 8y ago
True, but the distinction is irrelevant in this case.
- the8472 8y ago
  Far from it. If all physical objects were black bodies then the LED couldn't do anything here since emission and absorption are the same process.
  - timmaxw 8y ago
    The article notes that the LED "acts as a very low temperature object". I interpret this to mean "the LED radiates as though its temperature is lower", rather than "the LED radiates as though its emissivity is lower". In other words, AFAICT the effect can still exist even if the LED has 100% emissivity in both absorption and emission, which would make it a black-body. So the fact that the real LED's emissivity is not 100%, is not relevant to the effect. But we're splitting hairs here.
    - the8472 8y ago
      The paper is talking about effective temperature for specific photon energies. Additionally black body emission refers to thermal equilibrium conditions which the LED clearly is not. And then there's the near-field coupling they're using: > Recent experimental advances in near-field radiation have shown that heat-transfer rates on the nanoscale can exceed the blackbody limit by several orders of magnitude14–17, owing to contributions from evanescent and surface modes18,19. As a result, energy conversion rates can be greatly enhanced on this scale20. So this whole system relies on not being a black body on several levels.
      - baybal2 8y ago
        Even that will only work if they don't produce more heat doing that, than they pump away. Same issue as with anti-Stokes scattering that was once held promise of "self-cooling LEDs" or using gas dynamic lasers for spaceship cooling. There is no trick against thermodynamics after all.

yummypaint 8y ago

I agree with you up to the last paragraph. It sounds to me like the led is reverse biased like a silicon particle detector. A normal led works by combining electrons and holes to make photons, in this case hole-electron pairs are made by light. The reverse bias causes them to drift apart, resulting in a small current opposing the bias. For this reason i would interpret the power supply producing the bias as the load, and the led as the power supply. This is how energy is removed from the system.