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I do understand that the wavelength of the lights depends on the energy gap of the semiconductor, but why does it consist of a narrow range instead of a fixed value? is it because during recombination the photons have random phases?

Also, how does the spectral linewidth change if I increase the temperature? If I increase the temperature, the electron distribution in the conduction band will increase as more electrons jump from VB to CB. How does that exactly change my spectral linewidth?

Meep
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  • For one thing, if it was a single well-defined wavelength, that would be an infinitely precise energy, and the Heisenberg uncertainty principle means that that would imply an infinitely *imprecise* momentum. – Hearth Mar 26 '21 at 15:49
  • The full-width half-max widens with increasing temperature. That should be rather obvious. Just as it should be fairly obvious that the band-filling effects would tend towards saturation with increasing current. Beyond that, a lot of particulars may vary with construction, intentional and unintentional impurities, and defects (vacancies, interstitial, and antisites, for example.) Look up "TSOP" (temperature sensitive optical parameters" for LEDs, for example, to help find research, – jonk Mar 26 '21 at 18:36
  • Maybe https://physics.stackexchange.com will be able to provide more helpful answers :) – Jakob Halskov Mar 27 '21 at 07:20

2 Answers2

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Each photon emitted from an LED is the product of an electron-hole pair recombination. Charge carriers (electrons/holes) in a semiconductor have an energy distribution which is a function of dopant concentration, density of states of the semiconductor and the Fermi distribution. At elevated temperature the Fermi distribution is stretched out and results in a wider range of electron and hole energies. These are all still concentrated near the bandgap, but increases the probability of a slightly more energetic electron combining with a slightly more energetic hole to emit a photon of higher energy (shorter wavelength). At higher temperatures there are also more phonons available that enable transitions that would otherwise be forbidden. This also broadens the emitted spectrum and allow for slightly sub-bandgap emission.

But temperature also affects the bandgap energy. As the temperature increases, the bandgap energy decreases. This will shift your peak to longer wavelengths at higher temperature.

Matt
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  • At low temperatures where phonon-assisted recombination is negligible (or insignificant compared to probability of occupancy past the lower edge of the conduction band), your answer seems to imply that the emission spectrum would look asymmetric, with sharp rolloff towards higher wavelengths and a slower rolloff towards lower wavelengths. Some admittedly brief Google searching suggests this is not really the case. What makes more sense is that the photon emission is a spontaneous emission process, and thus we have the sum of many photons of random phase causing spectral leakage. Any thoughts? – LetterSized Apr 01 '23 at 17:24
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I would also like to add on to the fact that why there is light emitted at higher wavelength. I think it could also be the recombination of the holes and electrons at areas that is close to the top of valence band and the bottom of the conduction band. There are also energy states and holes present there, just not in huge concentration. Recombination can happen there either in the [100] direction or the [111] direction. This will have a higher recombination band gap, thus based on the Plancks relation, the wavelength of the light emitted will have a longer wavelength.

user336554
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Apr 01 '23 at 14:57
  • What does [100] and [111] direction men? – Bruce Abbott Apr 01 '23 at 19:26