Because the LED convert the electricity directly to the light, this conversion by itself could be indeed 100% efficient (and I would say it nearly is). And this is the reason, why it would be for other light sources known today hard to compete. And if any other light source would ever "overtake" them, it would have to use the direct conversion as well. And it can not rely to given temperature, as maintaining any accurate temperature mean quite high energy demand only for that, while such energy demand mean losses for the thing as a light source. So discharges are out...
But that does not mean the complete LED's could be 100% efficient. There are two major things, why even the monochromatic LED is not (and never would) 100% efficient: - You have to carry the current to the place, where it could be converted to the light (the energetic barrier on the junction) - You have to carry the generated light out from the place, where it was generated.
The first require very good electrical conductor. But as the light is an electromagnetic wave, any electrical conductor would partially reflect or absorb the electromagnetic wave, so the light. And the reflection/absorbtion would be stronger, as the conductivity increase.
Going into an extreme: When reaching ideal electrical conductor, you would get an ideal mirror as well. The ideal mirror would reflect all the light back into the place, where it was "born". As the photon matches the energetic transitions of that area, it would likely be converted back to electron/hole pair, so in fact consumed. But this electron-hole pair would recombine again, forming another photon. All these reflections and conversions would repeat so many times, till the photon finally leaves the chip. But with ideal mirror, it would require realy many reflections and conversions, so even when each of them would approach 100% efficiency, as the number of the conversions "approach" infinity, the overall efficiency to the ones finally leaving approach zero.
When making a layer, what pass majority of the photons, it would have to have low electrical conductivity. But that mean high ohmic losses when carrying the electricity to the junction, again the final efficiency approaching zero.
As the layer can not be conductive and not conductive at the same time, there should be made some compromise, so the resistive losses won't be so high, yet the generated light could escape moreless freely. Some materials may feature some resonances, so they could be rather conductive for "DC", but non-conductive on frequencies corresponding to the generated light, so the "best compromise" could yield better efficiency, but still it is far from the ideal. And the higher the photon energy, more likely it could "tunnel" through the conductive layer, so more transparent the conductor is. Moreover higher voltage drop mean the relative resistive losses are lower. Both cause the blue LED's to be the most energy efficient.
And the resistance vs photon transparency is one of the main reasons, why LED's loose efficacy on higher currents: The ohmic resistance losses are proportional to current^2, while the power delivered to the converting junction are proportional just to the current. SO higher the current, higher the relative component of the ohmic losses, while still keeping the same resistances (so the efficiency of passing the generated photons to the outside).
For white LED's there is another aspect: The voltage drop across the junction is equal to the energy of the photon (at this point assume the effects mentioned above mean no losses at all)
As the voltage across the whole LED junction is everywhere the same (assume an ideal conductors), it would all correspond to the same photon energy. As we assume no losses, each electron/hole pair then create just one photon of that energy. So such creation is able to generate only a monochromatic light. And as the energy band vary with temperature, so does the wavelength of the generated light.
As we want white light source, we need multiple such monochromatic radiators, well spaced over the visible range. But all have to be positioned so, the overall color would be white.
Now as each of them generate different wavelength, it would have different voltage drop, so they can not be connected directly parallel. But you want to feed all from a single supply, so only three possibilities remain: - Feed each of them with separate source (quite complex) - Feed them in series (the relative intensity is not controllable, as the current is the same for all of the components; but they could be spaced over the spectrum so, it gives white light) But all that allow only few distinct lines to be generated, not any continuum, so the color rendering is limited.
To generate (at least some) continuum, only two possibilities remain: - Use a phosphor to convert the primary radiation into the required light - fundamentally lossy method... Vary the band gap across the chip surface. But that mean a parallel connection with ballast resistances (that is the case for green LED's with wide spectrum, or for the phosphor-less white LED's). But ballast resistors mean fundamental losses as well.
So today vast majority of the white LED's are designed as blue LED's (the most energy efficient, still visible) with the red/green generated by the phosphor
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