Author Topic: holonyak's remarks on led technology being 100% efficient  (Read 1996 times)
Silverliner
Administrator
Member
*****
Offline

Gender: Male
View Posts
View Gallery

Rare white reflector


GoL
holonyak's remarks on led technology being 100% efficient « on: March 10, 2013, 08:53:00 PM » Author: Silverliner
heres an interesting discussion with holonyak who invented the first red led. he said leds can be 100% efficient, because there is no incandescence, fluorescence, chemical reaction etc involved in leds to create light, it is only generated by the electrical current itself. he does admit there are still ohmic losses in the wires and contacts though. read the link then feel free to share what you think.

http://www.npr.org/2012/10/12/162790782/fifty-years-ago-a-bright-idea
Logged

Administrator of Lighting-Gallery.net. Need help? PM me.

Member of L-G since 2005.

Collector of vintage bulbs, street lights and fluorescent fixtures.

Electrician.

Also a fan of cars, travelling, working out, food, hanging out.

Power company: Southern California Edison.

Medved
Member
*****
Offline

Gender: Male
View Posts
View Gallery

Re: holonyak's remarks on led technology being 100% efficient « Reply #1 on: March 11, 2013, 05:25:26 PM » Author: Medved
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
Logged

No more selfballasted c***

James
Member
*****
Offline

Gender: Male
View Posts
View Gallery


WWW
Re: holonyak's remarks on led technology being 100% efficient « Reply #2 on: March 13, 2013, 05:22:43 PM » Author: James
Today the quantum efficiency of the best LED die is already at about 85% of the theoretical maximum.  However the light extraction efficiency (getting the light out of the die) is only at about 60% efficiency.  There is strong potential for further improvement in this area.  Additionally the conversion efficiency of the phosphor (for white LEDs) is not so high.  Typically this is about 60% today.  For warm white LEDs having good colour rendering, the conversion efficiency may be as low as 50%.

When all of these losses are added up, and we combine them with the thermal droop effects which mean that LEDs are less efficient at normal operating temperatures vs the cold temperatures that manufacturers tend to quote data for, the overall light conversion efficiency of good commercial white LEDs is only about 30%.  The very best blue LEDs can have a conversion efficiency just over 50%.  For white this is of course about 5-6 times better than halogen, but only the same as fluorescent which also has about 30% efficiency for converting electrical energy into visible light.

So we still have a lot of exciting potential for future improvements over the next few decades, as we try to get closer to the theoretical efficiency limits!
« Last Edit: March 13, 2013, 05:32:01 PM by James » Logged
James
Member
*****
Offline

Gender: Male
View Posts
View Gallery


WWW
Re: holonyak's remarks on led technology being 100% efficient « Reply #3 on: March 13, 2013, 05:30:28 PM » Author: James
.
Logged
Medved
Member
*****
Offline

Gender: Male
View Posts
View Gallery

Re: holonyak's remarks on led technology being 100% efficient « Reply #4 on: March 13, 2013, 05:43:40 PM » Author: Medved
What I expect to improve is the quantum efficiency (semiconductor purity, crystal defect density), maybe a bit of the light "extraction" efficiency (more "selective" connection layer)
For white LERD's it would be quantum efficiency of the phosphors (principal materials, but mainly the manufacturing quality), but as it have to convert the blue to red-green range, for 100% quantum efficiency (I guess the theoretical maximum) it indeed could not be above the 60..70% (depend on till how deep red it would have to generate)
and better optimisation between CRI vs efficacy (less or shorter wavelength red mean more lumens per radiated watt, so more lm/W per the same energy efficiency) - the CRI95+ is on most places not needed...

For efficiency, with all of the losses I do not believe the white LED would ultimately go above 40..50%, what mean the ~130..150lm/W, while today, with the 30% (so ~100lm/W) we are already not that far away, so I really think there is no room for any significant efficiency improvement anymore, it would improve only very slowly.
What could improve is the temperature stability (so they would better maintain the efficiency at elevated temperatures) and general reliability (simpler, cheaper, but better performing packages, module design,...).

The LED's have the similar 30% (so about the 100lm/W for CRI~80) as fluorescents, but LED's maintain this even with smaller lumen packages (600lm and below), what make them more practical for home use (or more efficient compare to the <1000lm fluorescents).
And the LED's emit the whole light already into a half-sphere, so practical fixture designs would yield better optical efficiency (less light losses when redirecting it towards the illuminated area).
And today, most major brand LED manufacturers rate their illumination LED's at Tj between 70..100degC, so on realistic temperatures for real practical fixture.

But the largest changes would be in driving the manufacturing costs down: Larger wafers, lower defect density so scrap rate, higher volumes, generally more experience with the manufacturing. Basically what happened with silicon semiconductors since they were first manufactured till today...
« Last Edit: March 13, 2013, 06:08:58 PM by Medved » Logged

No more selfballasted c***

Print 
© 2005-2024 Lighting-Gallery.net | SMF 2.0.19 | SMF © 2021, Simple Machines | Terms and Policies