LEDs get less bright as they heat up. It's in every LED datasheet. Why does that happen? What is the mechanism here?

I know the bandgap decreases causing the wavelength shift. I also guess that they become less efficient at higher temperatures (that is just a guess, perhaps due to more energy in the system increasing the chances of carrier recombination). But I'm not sure if either of those are directly related to the decrease in brightness.

LED Output Decreasing with Temperature Datasheet Graph

  • \$\begingroup\$ The factors you stated, plus the decrease in efficiency of some phosphors at higher temperatures. The metastable elevated state may be destroyed by heat before it can emit a photon. \$\endgroup\$ Commented Mar 30, 2022 at 0:10
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    \$\begingroup\$ You may find this paper helpful in finding a few mechanisms worth a moment. \$\endgroup\$
    – jonk
    Commented Mar 30, 2022 at 6:08

2 Answers 2


The shift in energy levels of valence and conductive band does not explain why the amount of light (i.e. the generated photon flux) decreases. From a quantum mechanical perspective, the emitted wavelength is determined by the forward drop (band gap) and as the forward voltage decreases, the emitted wavelength increases.

The photon flux on the other hand is strictly proportional to the electrical current in the diode - each electron has a certain chance to cause the emission of one photon. The "certain chance", commonly called quantum yield, decreases with higher temperatures, which is the effect you are describing. The major reasons for a decreasing quantum yield are

  • the dramatically changing lifetimes of radiative and non-radiative recombinations, which leads to non-radiative recombinations taking over (see e.g. Shockley-Read-Hall and Auger recombination)

  • phonons (crystal oscillation), which can interact with electrons and interrupt a radiative recombination. As those oscillations increase with temperature, fewer electrons participate in light generation

  • more recent research (2019 - at least after I studied) seems to suggest that transport mechanisms are the dominant contributor to thermal droop (paper I have not read myself)

So, there are two things happening when the temperature increases, which both cause the emitted power to get smaller:

  1. Fewer electrons participate in light generation and
  2. The electrons participating in light generation generate photons with less energy (this one does not decrease the LED's efficiency because the consumed power decreases respectively).

It is important to differentiate those two effects, that's one of the key points why we had to come up with quantum mechanics.

I recommend this article and this paper which cover the topic.


Why do LEDs emit less light as they get hotter?

They only do so when the current is constant. That's why I hate those graphs, because the constant current constraint is all-important here. Without it, the graph makes no sense and can't be interpreted at all.

You can regulate the current against the forward voltage drop to completely compensate this effect, although if the LED has not enough cooling and you don't limit maximum current adequately, it'll happily thermally run away. Typically the regulator will have a temperature cut-off (temp feedback via forward voltage), so this won't happen. But simple breadboard contraptions of this kind a excellent led killers. Don't try them without a series resistor in place to limit damage :)

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    \$\begingroup\$ Those diagrams typically show the relative intensity at a defined test current, the 100% intensity (at test current) is given elsewhere in the datasheet. So the CC constraint is fulfilled. The graph contains the temperature-efficiency dependency indirectly. It can be calculated if also the I-U curve is taken into account. \$\endgroup\$
    – Sim Son
    Commented Apr 1, 2022 at 20:47
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    \$\begingroup\$ @SimSon The datasheet writer mentality is still in the times when these things had a per-page cost. They could long afford putting a sentence in there "at I_const=...". The existence of this very question affirms such need. I know they are taken at constant current. I hate that I have to know it, because those who don't are at a disadvantage. Yes, this stuff is a kind of general knowledge, but if someone wanted to learn it, there's no single source... The world is imperfect, alas. \$\endgroup\$ Commented Apr 2, 2022 at 23:20
  • \$\begingroup\$ I aggree, it would be nice to simply have a graph that shows the over-all efficiency versus current... \$\endgroup\$
    – Sim Son
    Commented Jul 14, 2022 at 16:22

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