According to "Introductory semiconductor physics" by Greg Parker (1994; Amazon.com), a photon is produced when an electron recombines with a hole across the energy gap of a direct band-gap semiconductor.

Question: How does a light-emitting diode (LED) create photons? For example, does it exploit the creation of photons that occurs when electrons drop from an excited state to a ground state?

I would understand intuitively what is the physical phenomenon of gamma production that occurs in this case.

  • $\begingroup$ Keep in mind that not all photons are gamma rays. $\endgroup$ Commented Aug 22, 2018 at 11:32
  • 1
    $\begingroup$ Note that LED acronym already has the word "diode", so "LED diode" means "light-emitting diode diode" $\endgroup$
    – Ruslan
    Commented Aug 22, 2018 at 12:07
  • 2
    $\begingroup$ You have the answer in your first paragraph. An electron and hole recombine. So, what is your real question? $\endgroup$
    – Jon Custer
    Commented Aug 22, 2018 at 12:13
  • $\begingroup$ @JonCuster, I think the OP is asking what happens when electrons and holes recombine. Perhaps, OP wants a detailed explanation of what does recombination mean. $\endgroup$
    – Nemo
    Commented Aug 22, 2018 at 12:32
  • $\begingroup$ @JonCuster My real question is that I want to know why this effect produce a photon, there is a basic physic effect like de-eccitation to explain this photons producing? $\endgroup$
    – Alex.L
    Commented Aug 22, 2018 at 19:51

1 Answer 1


High energy electrons enter the p-region from the n-region and fall from the conduction band (high energy state), though the band gap, to the valence band (low energy state), producing a photon in the process.

High energy electrons are re-supplied by the battery, so we can say that the energy of the battery is used to excite the electrons or push them to the conduction band, so that they could enter the n-region again.

  • $\begingroup$ The electrons in the n-region aren’t ‘high energy’, they are just in the conduction band at a density set by the doping level. Plus, you didn’t mention the holes flowing into the junction region to recombine either. $\endgroup$
    – Jon Custer
    Commented Aug 22, 2018 at 12:27
  • $\begingroup$ @JonCuster Thanks for your feedback. Maybe "high energy" term is ambiguous here. I just wanted to convey the idea that the electrons in the conduction band are at high potential energy level - not that they move fast. As for the holes flowing into the junction, I just treat them as electrons moving away from the junction, which they actually are. $\endgroup$
    – V.F.
    Commented Aug 22, 2018 at 12:59
  • $\begingroup$ You could have said higher energy electrons, not high energy electrons. But that's nit picky and your explanation is pretty good. $\endgroup$ Commented Aug 22, 2018 at 13:34
  • $\begingroup$ @V.F. - the problem with treating holes as electrons flowing away from the junction is that obscures the detailed balance ($np = n_{i}^{2}$) that is at the heart of understanding semiconductor physics. And it is precisely detailed balance that drives rapid recombination in the junction. As a side note, any active direct gap device is emitting photons from the junction regions - its just we choose to let them out in an LED... $\endgroup$
    – Jon Custer
    Commented Aug 22, 2018 at 14:12
  • 1
    $\begingroup$ @A.Lega Glad I could help. $\endgroup$
    – V.F.
    Commented Aug 23, 2018 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.