# LEDs waste 75% as heat



## MikeAusC

The ideal LED would convert all the electrical energy into light, but it's clear that a lot gets converted to heat.

Unlike filament bulbs, most of the heat needs to be conducted out of the back of the LED to keep the LED cool - the hotter it gets, the less efficient it is.

I've seen a few attempts on CPF at working out how much heat an LED puts out, but I think the only reliable method is substitution - find out how much heat will raise the same heatsink to the same temperature. 100% of the electrical energy that goes into a resistor gets converted to heat - it's very hard to make an inefficient electrical heater !

I found two identical unadonised aluminium finned heatsink - 10 x 5 x 3.5 cm.

On one, I mounted an XM-L T6 on a 20mm star from www.ledsales.com.au. On the other, I mounted two 10 watt 1 ohm resistors in aluminium extruded housings with two screw mounts.


*LED 8.42 watts*
- efficiency was 18% - heat output from resistors was 6.86 watts = 82%
- heatsink temperature = 56.7 degC, ambient = 25.5 degC
- LED 3.12v, 2.70 amp
- Resistor 3.71v, 1.85amp 


*LED 2.95 watts*
- efficiency was 31% - heat output from resistors was 2.04 watts = 69%
- heatsink temperature = 35.5 degC, ambient = 23.7 degC
- LED 2.95v, 1.00 amp
- Resistor 2.02v, 1.01 amp 


So over a normal operating range, modern high power LEDs waste around 75% of the power going in to them.

Of course this testing method ignores radiant heat loss from the LED - when I hold my hand 1cm in front of the LED it gets a lot warmer than holding it 1cm in front of a resistor putting out the same heat. EDIT- as pointed out by jtr1962, the warmth I feel would be due to the light being converted to heat when it strikes my skin - the energy has to go somewhere.


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## jtr1962

MikeAusC said:


> Of course this testing method ignores radiant heat loss from the LED - when I hold my hand 1cm in front of the LED it gets a lot warmer than holding it 1cm in front of a resistor putting out the same heat.


Good work but don't confuse the phenomenon you're describing with radiant heat loss. In order to have enough radiant heat loss to feel it with your hand 1 cm away, the LED die would need to be a few thousand degrees. What you're feeling is the light energy from the LED absorbed by your hand, and turned into heat. I first noticed this with a Rebel where if I put black electrical tape very close to the LED dome, it would become hot enough to start smoking. Even my lighter colored finger quickly gets too hot to hold above the dome. It's an interesting phenomenon.

Incidentally, the 10 watt resistors end up slight increasing the surface area from which heat is dissipated compared to when the LED is mounted, so your experiment might be overestimating the amount of power required for a given heat sink temperature rise. At 1 amp, a T6 XM-L should be outputting about 370 lumens. This might be equivalent to around 1.1 watts of light energy, so the heat would be 1.85 watts instead of 2.04 watts. A slight refinement then might be to put insulation over the resistor body so that there is as little heat dissipation there as practical. Other than that, great experiment and interesting results!


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## MikeAusC

I had thought of using the Dichroic Reflector that's used with 50mm Halogen Bipin lamps - they're designed to reflect the light forward, but let the heat pass throught the reflector, to avoid setting fire to the object being lit up. If you look at the bulb filament from the back of the reflector, you can see it only lets a small amount of the light through.

So you could arrange the LED so the radiant heat which passes through dichroic reflector also heats up the heatsink, but the reflected light radiates into space.


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## MikeAusC

jtr1962 said:


> . . . .What you're feeling is the light energy from the LED absorbed by your hand, and turned into heat. I first noticed this with a Rebel where if I put black electrical tape very close to the LED dome, it would become hot enough to start smoking. Even my lighter colored finger quickly gets too hot to hold above the dome. . . .


 
Good point - I'd forgotten that even light will get converted to heat. If you have a light inside an opaque container, the only way the energy can escape, is as heat from the outer surface of the container.


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## srfreddy

MikeAusC said:


> Ito avoid setting fire to the object being lit up.


 
 We wouldn't want that now would we? An "ideal" LED could never convert all electricity into light.


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## jtr1962

MikeAusC said:


> I had thought of using the Dichroic Reflector that's used with 50mm Halogen Bipin lamps - they're designed to reflect the light forward, but let the heat pass throught the reflector, to avoid setting fire to the object being lit up. If you look at the bulb filament from the back of the reflector, you can see it only lets a small amount of the light through.
> 
> So you could arrange the LED so the radiant heat which passes through dichroic reflector also heats up the heatsink, but the reflected light radiates into space.


Interesting idea. I'm reasonably sure though the amount of energy leaving an LED as radiant heat can be measured in microwatts, and is thus entirely negligible for the purposes of this experiment. Remember that radiant heat is proportional to absolute temperature to the fourth power. If the LED die were at incandescent lamp temperatures, it might radiate a couple of watts. However, at perhaps 50°C, it'll only radiate about 1/10000th the power.

In any case, your assessment that only about 25% of the input power is converted to visible light seems quite correct and reasonable for an LED operated at medium to high currents. You can approach or even exceed 50% at very low currents, but at the expense of using more LEDs. I expect by the end of the decade we'll be pushing efficiencies of 75% to 80%. Such efficiencies have already been reached in the lab.


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## beerwax

what sort of thermal path was there from the resistors to the aluminium extrusion housing the resistors ?


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## MikeAusC

These were commercial high-power resistors and I assume they are cemented into the aluminium housing. They have a flat base so I used Arctic Silver compound, just like under the star.


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## Oznog

This is hard to measure, and your methods are somewhat shot-in-the-dark for accuracy.

Just look it up. Lumens are a measure of total light power, rather than a light intensity that increases with focusing. Unfortunately, rather than a fixed relationship between lumens and power, lumens is a scale compensated for human eye response. 5mW of green laser appears 10x brighter than 5mW of red laser. A green LED putting out 100mW of green light will score about 10x the lumens of a red LED putting out 100mW of red light.


Given a particular wavelength or white color temp, you can look up how many watts of light energy = 1 lumen. So look up the spec sheet's "typical" for current, voltage, and lumens and you can calc the efficiency straight-out.


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## beerwax

i tried to figure a margin of error for mikeausc s work, but figured it didnt matter or affect the conclusion.
i think he clearly demonstrated that while it seems leds are quite efficient (energy to light 25 percent seems a long way from a candle or a heated wire) theres still plenty of room for improvement. so we are probably not at the peak just yet. 

theres no substitution for practical experiment.


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## MikeAusC

The point of this experiment was to answer the frequently-asked question "I'm feeding x watts to my LED, how much heat will I have to remove."

If anyone has a more accurate basis for answering this, please enlighten me.


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## Oznog

20%-25% is about right for LEDs. Maybe a bit better now.

There are theoretical limits which would keep LEDs far from being able to 100% efficiency no matter how perfect they are, at least with AlInGap and InGaN technology.


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## Lego995743

it is impossible for an led to be 100% light because light is heat energy.


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## Walterk

Very clarifying. I dont longer have to think


> I'm feeding x watts to my LED, how much heat will I have to remove.


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## Oznog

Lego995743 said:


> it is impossible for an led to be 100% light because light is heat energy.


 That... what??


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## onetrickpony

Lego995743 said:


> it is impossible for an led to be 100% light because light is heat energy.


 
Yeah, um, no. It's not. It's light energy. Photons, which according to quantum mechanics, leave and arrive as particles, but travel as waves. Yada, yada, yada.

EDIT: And, in fact, if you want to SEE heat energy, you can throw on some infrared goggles.


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## bbb74

What you see as light ... what you feel as radiated heat - its all the same thing. Just different frequencies.


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## onetrickpony

bbb74 said:


> What you see as light ... what you feel as radiated heat - its all the same thing. Just different frequencies.


 
Yeah, but I feel that is oversimplifying things. That's like saying that a candle and a plasma torch are the same thing.

Light and heat may fall under the electromagnetic umbrella, but they are definitely different as far as how humans perceive them. Aka, you can see heat, and you can feel light, but you're much more likely to be able to perceive them vice versa.

Light IS a form of energy. Various sources of both will emit the other, but let's be clear, heat is NOT light energy.


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## bbb74

onetrickpony said:


> That's like saying that a candle and a plasma torch are the same thing.


Getting OT here .. but thats not what I mean, although I think we are agreeing with each other though  A plasma torch will radiate infrared heat which you can feel, but it also has a cutting flame of hot plasma, which yes is obviously a very different kettle of fish. The plasma does the cutting but it also loses some energy as radiant (IR) heat which you can feel. Applying the torch to your hand results in a different method of energy transfer (and would be a bit painful).



onetrickpony said:


> Light IS a form of energy. Various sources of both will emit the other, but let's be clear, heat is NOT light energy.


I agree but can I reword it as "Electromagnetic radiation within certain ranges of frequencies can be detected by the human body as either 'light' or 'heat' with varying sensitivities". 

Not sure why I'm making this post now...


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## jirik_cz

Oznog said:


> 20%-25% is about right for LEDs. Maybe a bit better now.
> 
> There are theoretical limits which would keep LEDs far from being able to 100% efficiency no matter how perfect they are, at least with AlInGap and InGaN technology.



Theoretical maximum for white light is somewhere around 300lm/W. So Cree XM-L driven at 0.35A will have 50% efficiency. Of course it goes down with increasing current.


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## dellayao

it is "ideal",so it is hard to achieve. we cann't avoid the power to convert into heat.:duh2:


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## CKOD

Lotta confusion going on in here with numbers and what is and isnt heat. 



> Theoretical maximum for white light is somewhere around 300lm/W. *So Cree XM-L driven at 0.35A will have 50% efficiency*. Of course it goes down with increasing current.


Incorrect statement, if it achieves 150 lm/W and the theoretical max is 300 lm/W, then it achieves 50% of theoretical maximum efficiency, not an efficiency of 50% ( watts out / watts in ) I wish I could find a good table of wavelength vs lumens per watt, Lumens are defined by the response to the eye compared to 1W of light energy. In daytime vision 555nm is ~600 lumens per watt IIRC, and it its lower for all the other frequencies, tapering off towards infrared and UV obviously. 



> I agree but can I reword it as "Electromagnetic radiation within certain ranges of frequencies can be detected by the human body as either 'light' or 'heat' with varying sensitivities".


close but still confusing, possibly misleading. The only sense organ in the body that directly detects electromagnetic frequencies is the eyes, 'heat' as felt by the skin, is an indirect sense from the heating of the skin itself. I.E. a thermometer doesnt sense electromagnetic frequencies, but it you shine a 1W laser on it, its still gonna go up. 
Various frequencies of the elctromagnetic spectrum are absorbed by the body at differing rates, so heating varies. Everything from UV down to a few GHZ will be absorbed at least partially by the skin, causing local heating, felt as warmth. UV and above is ionizing and bad and starts to just zing right though you, breaking DNA along the way and causing cancer, but not inducing much heating, stuff below a few GHz penetrates the body in varying depths, causing internal heating, but thats not felt as there isnt heat sense organs inside your body, so by time you do feel it, damage via heating can be done. And lower frequencies have wavelengths larger then you and pass though without a care in the world. 


Yes, longwave IR cameras are traditionally called "heat vision", but they just see the IR emissions from the warm object. Kind of like how incan lamps emit light, you emit IR light also, just in the 310K spectrum. IR just falls into a sort of niche where we cant see it, we can feel the heating effects due to it being absorbed by the skin very quickly, its too high frequency for us to do direct emissions via an antenna (Getting there! THz transmitters arent stuff of fiction anymore) 

I would post a certain picture of jackie chan right now, but I know the moderators on here wouldnt be too fond of it :nana:


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## slebans

CKOD said:


> Lotta confusion going on in here with numbers and what is and isnt heat.
> 
> 
> Incorrect statement, if it achieves 150 lm/W and the theoretical max is 300 lm/W, then it achieves 50% of theoretical maximum efficiency, not an efficiency of 50% ( watts out / watts in ) I wish I could find a good table of wavelength vs lumens per watt, Lumens are defined by the response to the eye compared to 1W of light energy. In daytime vision 555nm is ~600 lumens per watt IIRC, and it its lower for all the other frequencies, tapering off towards infrared and UV obviously.
> :nana:


 
Not sure where you are coming from but if the Luminous Efficacy of Radiation(LER) value for the XM-L at 350ma is 300 lumens and the LED emits 150 lumens then the electrical efficiency of the LED is 50%(watts in to watts out). 

Stephen Lebans


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## Potato42

MikeAusC said:


> The point of this experiment was to answer the frequently-asked question "I'm feeding x watts to my LED, how much heat will I have to remove."


 
excellent experiment MikeAusC. It's nice to have a rough guideline for this sort of thing. Do you have any idea how similar the results would be with other LED's? Would results correspond to the relative efficiencies of the LED tested vs the XM-L T6?


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## CKOD

slebans said:


> Not sure where you are coming from but if the Luminous Efficacy of Radiation(LER) value for the XM-L at 350ma is 300 lumens and the LED emits 150 lumens then the electrical efficiency of the LED is 50%(watts in to watts out).
> 
> Stephen Lebans


Ahh ok, I thought you were referencing a theoretical ideal LED vs the actual led, not the lm/w value for its emisison spectrum vs its actual output. That makes more sense and is correct then, though there would be some error introduced by the color temperature, but thats not as significant.


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## jirik_cz

CKOD said:


> Incorrect statement, if it achieves 150 lm/W and the theoretical max is 300 lm/W, then it achieves 50% of theoretical maximum efficiency, not an efficiency of 50% ( watts out / watts in ) I wish I could find a good table of wavelength vs lumens per watt, Lumens are defined by the response to the eye compared to 1W of light energy. In daytime vision 555nm is ~600 lumens per watt IIRC, and it its lower for all the other frequencies, tapering off towards infrared and UV obviously.



I'm afraid that I'm not the one who is confused here  

300lm/W is roughly theoretical maximum for white light (100% energy converts to white light). So white LED with 150lm/W emits 50% energy as white light and 50% goes to waste heat.


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## slebans

jirik_cz said:


> I'm afraid that I'm not the one who is confused here
> 
> 300lm/W is roughly theoretical maximum for white light (100% energy converts to white light). So white LED with 150lm/W emits 50% energy as white light and 50% goes to waste heat.


 
It is not accurate to state that "300 lm/w is the theoretical maximum for white light. For a detailed explanation go to the DOE website and read the current LED Roadmap report. Or for a specific(but somewhat dated) reference see:
Color Rendering and Luminous Efficacy of White LED Spectra

Stephen Lebans


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## jirik_cz

Stephen do you mean this NIST document? (your link doesn't work)

Most of the current white LEDs are actually blue LEDs with YAG phosphor. According to this document theoretical maximum for LED with YAG phosphor, 6800K CCT and CRI 81 is *294 lm/W*.


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## jtr1962

That link is dead. Here's a direct link to the .pdf you're referring to:

http://lib.semi.ac.cn:8080/tsh/dzzy/wsqk/SPIE/vol5530/5530-88.pdf


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## slebans

jtr1962 said:


> That link is dead. Here's a direct link to the .pdf you're referring to:
> 
> http://lib.semi.ac.cn:8080/tsh/dzzy/wsqk/SPIE/vol5530/5530-88.pdf


 
THe link was fine but it seems to be the "Add Link" feature within the posting interface that did not correctly parse the URL I entered:
http://lib.semi.ac.cn:8080/tsh/dzzy/wsqk/SPIE/vol5530/5530-88.pdf

I am still relatively new to this interface so perhaps I am doing something wrong. In the future I will test the links within the Preview window to ensure they are parsed correctly.

Thank you for posting the correction.

Stephen Lebans


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## beerwax

so , if we had a new led, type W, that converted 90 % of the input electrical energy to radiation of a non visible wavelength , it would generate little heat and would be rated 'very inefficient'. see no reason not to like mikes concept.


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## onetrickpony

Someone needs to come up with a driver that regeneratively converts heat back into electricity. I hope if someone reads this, then does it, that I AT LEAST get some credit, if not a nice fat check in the mail. Thanking you in advance....


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## jtr1962

onetrickpony said:


> Someone needs to come up with a driver that regeneratively converts heat back into electricity. I hope if someone reads this, then does it, that I AT LEAST get some credit, if not a nice fat check in the mail. Thanking you in advance....


There is an entire body of research devoted to converting low-level waste heat into electricity via improved thermoelectrics. The laws of thermodynamics say that even best case you won't be able to convert much waste heat from an LED because conversion efficiency increases with source temperature. An LED by definition can't run at a high enough temperature to make converting waste heat worthwhile. On the flip side, if someone were to invent relatively efficient thermoelectric convertors which could survive being placed in close proximity to a lamp filament, then you could potentally recover a large percentage of waste heat in theory. In practice this wouldn't make much sense. Even with waste heat recovery, overall the lamp would still be less efficient than LED or fluorescent sources. The main application for waste heat recovery is power plants where even a 1% efficiency increase translates into millions of dollars.


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## onetrickpony

jtr1962 said:


> There is an entire body of research devoted to converting low-level waste heat into electricity via improved thermoelectrics. The laws of thermodynamics say that even best case you won't be able to convert much waste heat from an LED because conversion efficiency increases with source temperature. An LED by definition can't run at a high enough temperature to make converting waste heat worthwhile. On the flip side, if someone were to invent relatively efficient thermoelectric convertors which could survive being placed in close proximity to a lamp filament, then you could potentally recover a large percentage of waste heat in theory. In practice this wouldn't make much sense. Even with waste heat recovery, overall the lamp would still be less efficient than LED or fluorescent sources. The main application for waste heat recovery is power plants where even a 1% efficiency increase translates into millions of dollars.


 
I'm pretty sure that the efficiency of the converter is less relevant than the cost of the converter relative to the savings it provides. In other words, if someone comes out with a miniature converter that costs $3 and saves 100 ma per hour on a 10 watt led, you've got progress.


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## jtr1962

onetrickpony said:


> I'm pretty sure that the efficiency of the converter is less relevant than the cost of the converter relative to the savings it provides. In other words, if someone comes out with a miniature converter that costs $3 and saves 100 ma per hour on a 10 watt led, you've got progress.


The theoretical maximum efficiency of a heat engine is 1 - Tc/Th, where Tc is the cold side temperature and Th is the hot side temperature. LEDs generally need to keep the die at less than 100*°C ( 373K) for decent life. Assuming you use some fancy setup where the cold side can be close to room temperature ( ~290K ), then the maximum _possible_ efficiency of a thermoelectric is ~22%. Now I've been studying the development of thermoelectrics for a long time as a personal interest of mine, mostly for their refrigeration capabilities as opposed to use for heat recovery (although the same device can do both). Most studies I read point to achieving 50 to 60% of Carnot efficiency as "ambitious and probably unlikely". Assuming we reach this lofty goal (and people have been trying for the last 50 years), then you're talking about recovering ~13% of the waste heat at the upper limit of the LED's operating temperature. Since the goal here is more efficient conversion of power to light, you can achieve about the same increase by simply bringing the die temperature close to room temperature via a better thermal path. In fact, generally power savings is mainly a concern for general lighting where the LEDs must operate at lower die temperatures for long life. These lower die temperatures are even less conducive to heat recovery. For example, at a 60°C maximum die temperature, which is often a goal for long-life general lighting, maximum possible Carnot efficiency drops to not much over 10%. A practical device might not recover more than 5%. OK, 5% might still make a difference, but only if the cost of the device exceeds the power savings over the life of the LED. If we have a 100 watt LED, then we save 5 watts. Over the LED's 100,000 hour life that's 500 kW-hr, or about $50 at today's average electric rate of 10 cents per kilowatt-hour. Can we make an efficient thermoelectric capable of recovering 5 watts at such a low temperature differential, along with the associated heat sinks/pipes/fans to get the cold side as close to ambient as possible, all for $50 or less? I doubt it. Might as well just use that same heatsinking setup on the LED itself. If you can drop the die temperature by that same ~25°C temperature difference over which the thermoelectric is recovering heat, you increase output (i.e. efficiency) by about 6 or 7 percent. In short, it makes more sense to just bring die temps as close to ambient as possible. In fact, regardless of the die temperature, it makes more sense to just cool the LED better than to try and recover waste heat. Don't forget that waste heat recovery systems by definition use huge heat sinks to get the cold side as close to ambient as possible.

On another note, thermoelectrics which operate at a good fraction of the Carnot efficient _might_ make sense to use to cool the LED die. As a refrigerator, Carnot efficiency is Tc/Th-Tc. Note how this can be much larger than 1 if Th-Tc is small. For example, if we bring the LED die down from 50°C to 25°C, we'll obtain the aforemented 6-7 percent increase in output. Maximum Carnot efficiency doing this would be 11.92 which is ~12. Assume the 100 watt LED is 50% efficient at converting power to light, so the heat load would be 50 watts. Operating at half Carnot efficiency we would need 50/6 = 8.33 watts. This is about 8% more power in order to obtain a 6-7% increase in light. It's almost worthwhile. If you can approach Carnot efficiency then you actually end up using less power to cool the LED compared to the increase in output. Still probably not worthwhile from an economics standpoint, but at least here you stand to actually increase overall LED efficiency by using a thermoelectric (in theory anyway). And a while back I did an experiment which vividly demonstrates the effects of cooling LEDs. Even with the much more temperature sensitive output of an amber LED, I concluded that cooling LEDs with today's thermoelectrics has no practical value.


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## beerwax

i followed your link . thx for sharing.
i have some questions please. 
does that sort of efficiency curve hold for xpg and for temperatures from 25 deg to 60deg ? 
and does the lost output all become heat ? 

are we going to see thermoelectrics in motor vehicles.


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## monkeyboy

What about radiant heat loss from the resistor? Is this enough to throw off the results?

If we know the temperature of the resistor and the exposed area, we could use the black body approximation with the stefan-boltzmann law to calculate an order of magnitude...

OK I've lost interest already


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## jtr1962

beerwax said:


> i followed your link . thx for sharing.
> i have some questions please.
> does that sort of efficiency curve hold for xpg and for temperatures from 25 deg to 60deg ?
> and does the lost output all become heat ?


No, actually the output of the amber Luxeon I tested varies MUCH more with temperature than an XP-G would. An XP-G would only increase output by perhaps 8-9% if die temperature were reduced from 60°C to 25°C.



> are we going to see thermoelectrics in motor vehicles.


If we can get Carnot efficiency up, then there is talk of replacing the A/C compressor with thermoelectrics. There is also some talk of increasing mpg by converting waste heat from the engine into electricity. I'm personally dubious of this second possibility because the car would need an electric motor to make use full of this generated power, and in my opinion internal combustion engines will be obsolete for ground transport within a decade anyway due to improved batteries for EVs. Nevertheless, using thermoelectrics for A/C will remain viable regardless of the vehicle's source of motive power, and I think we'll see this.

On another note, I've been waiting since the early 1990s for improved thermoelectrics. What's available commercially now isn't a whole lot better than what was available then. I've read some papers, such as this one which talk of great improvements over today's devices, but I've yet to see any reach production. Page 2 for example shows that a single stage cooler based on their approach could reach 130K. I would love to get my hands on a thermoelectric with that kind of performance. With what exists nowadays, I'm lucky to approach 200K (and that's with a two-stage setup, water cooling, and very little heat load). The best I've done with bulk cooling is to get my temperature chamber down to -58°F ( -50°C = 223K ).


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## cloverhsu7

I wonder which is the best way to improve the efficiency


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## Bright+

If you were to express how efficient LEDs are in VISIBLE light in PERCENT, you will have to measure how many watts of radiant energy is coming out within the visible light spectrum. Anything outside of this band is a waste, just like infrared from incandescent lamps.


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## monkeyboy

Bright+ said:


> If you were to express how efficient LEDs are in VISIBLE light in PERCENT, you will have to measure how many watts of radiant energy is coming out within the visible light spectrum. Anything outside of this band is a waste, just like infrared from incandescent lamps.


 
exactly, and even within the visible band of light, the human eye sensitivity varies greatly with wavelength. e.g. 1W radiant output of 750nm red would still be visible but appear several orders of magnitude less bright than 1W 588nm green. The 750nm would be a dim glow and the 588nm would appear extremely bright even if both were produced by a 100% efficient source. Therefore IMO, it doesn't make sense at all to quote efficiency as a percentage which is why we quote things in lm/w.

I guess the OP was distinguishing conducted and radiated energy, so in this case it does make sense.


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## zzonbi

This only means the real efficiency is even less (he didn't measure so little heat to worry about invisible spectrum, on the contrary).
Unless the 294lm/W phosphor efficacy is overstated, or there's something quite wrong with the measurement (unsinked resistor heat) it seems like that led has 2/3 of the expected efficiency (t6 should be about 46% efficient at 1A, but is at most 31%)


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## Oznog

jtr1962 said:


> If we can get Carnot efficiency up, then there is talk of replacing the A/C compressor with thermoelectrics. There is also some talk of increasing mpg by converting waste heat from the engine into electricity. I'm personally dubious of this second possibility because the car would need an electric motor to make use full of this generated power, and in my opinion internal combustion engines will be obsolete for ground transport within a decade anyway due to improved batteries for EVs. Nevertheless, using thermoelectrics for A/C will remain viable regardless of the vehicle's source of motive power, and I think we'll see this.
> 
> On another note, I've been waiting since the early 1990s for improved thermoelectrics. What's available commercially now isn't a whole lot better than what was available then. I've read some papers, such as this one which talk of great improvements over today's devices, but I've yet to see any reach production. Page 2 for example shows that a single stage cooler based on their approach could reach 130K. I would love to get my hands on a thermoelectric with that kind of performance. With what exists nowadays, I'm lucky to approach 200K (and that's with a two-stage setup, water cooling, and very little heat load). The best I've done with bulk cooling is to get my temperature chamber down to -58°F ( -50°C = 223K ).


 
Currently essentially ALL thermoelectrics are bismuth-tellurium junctions. This type is VERY limited, and are about 1/10th the cooling efficiency of phase change (freon) systems used in air conditioning and refrigeration. And that's about the extent of bismuth-tellurium technology and it's not expected to get better. It will not replace them.

The quantum well superlattice form of thermoelectric device IS being worked on, and DOES seem to have potential for amazing cooling and power generation. If it does work, LEDs are hardly the "big" application. They can potentially generate almost as much energy out of a car's exhaust system waste heat as the engine itself already generated! At that point one might wonder why bother with the engine at all, instead of just running a naked flame into a thermoelectric generator. And that situation could come up. And there's plenty of sources of temperature differentials that could be exploited, IF the devices are cheap enough. And quantum well devices are primarily silicon, so they could be cheap, small, and durable.


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## Kinnza

The way of knowing out the heat to be dissipated just on paper before installing anything requires to know the exact LER (Luminous Efficacy of Radiation) of the LED. Only way to know it accurately is with an spectrometer, although a digital camera and some software might serve aswell with somewhat lover accuracy. Anyway, the heat load for the heatsink dont need to be excessively accurate, so an approximation is usually enough. For that purpose, using 300 lm/W (optical watt, energy of emitted light) is usually good enough. 

I have calculated the LER of several white LEDs and the typical ones (CRI 65-80) usually are around 300lm/W, while high CRI ones have LERs lower. Cool tones often are below 300lm/W, most are 270-290lm/W, so using 280lm/W often works fine. While warm whites often are over 300lm/W, up to 330lm/W, similar to neutrals. Generally, LEDs increases a CCT (K rating of light tone) increases (up to 2700K or so, when it often goes down again, but it depends more of the exact spectrum which achieves that CCT than with other tones). Very cool whites (7500K and higher) can have way lower LER (250lm/W). 

But in general, it is possible to use 300lm/W for all and get a decent approximation to the heat load. If you want somewhat more accurate results, use 280lm/W for coolwhites and 310lm/W for warmer tones.

After that, just simply calculate or measure lm output, divide it for watts burned (If*Vf) and divide the obtained figure for the LER, so you get the amount of input energy emitted as light, the rest up the unity being heat.

So for example, a coolwhite LED that emits 420lm and burns 3W (say an XP-G for example, simplifying 3V at 1A), it emits at 420/3= 140lm/W. Using a LER of 280lm/W, 140/280=0.5. So 50% of input energy is converted as light, and the remaining 50% (1.5W) is the heat load.

Or a warm white emitting 300lm with 3W of power: 300/3= 100lm/W. Applying a LER of 310lm/W, 100/310=0.32, thus 32% of input light is converted as light and 68% (2.04W) is wasted as heat.

Other threads cover the topic of knowing the lm emitted, either measuring (integrating sphere) or calculating it (with datasheet, first getting the light output for a given current and the derating it for the LED actual temperature).


----------



## MikeAusC

Kinnza said:


> . . . . . So 50% of input energy is converted as light, and the remaining 50% (1.5W) is the heat load. . . . .



So how do you explain the big difference between theoretical calculations and real-world tests ?


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## David_Campen

Kinnza said:


> So for example, a coolwhite LED that emits 420lm and burns 3W (say an XP-G for example, simplifying 3V at 1A), it emits at 420/3= 140lm/W. Using a LER of 280lm/W, 140/280=0.5. So 50% of input energy is converted as light, and the remaining 50% (1.5W) is the heat load.


The amount of power converted to white light is much less than that.

A white LED is a blue emitting LED that is surrounded by a mixture of phosphors that convert the blue photons into other colors to give a white appearing light.

For a moment, ignore the efficiency of the phosphor and look only at the efficiency of the blue LED. For these blue LEDs manufacturers actually publish the output of blue light in watts. For example, a 460 nm LED from LEDengin:
http://www.ledengin.com/files/products/LZ1/LZ1-00DB00.pdf
Driven at 1000ma it has a forward voltage of 3.6 for a power dissipation of 3.6 watts and produces 900mw of blue photons for an efficiency of 25%.

Now consider an inefficiency of the phosphor in that it is taking a 460nm photon and converting it into a photon with a longer wavelength. This results in a loss because the emitted photon is less energetic than the absorbed photon. If, for example, we assumed that the photons emitted by the phosphor were at 555nm then this conversion efficiency would be 460/555 = 83%.

Multiplying these 2 efficiencies we get an overall efficiency of .25X.83 = 21%


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## MikeAusC

That's what drove me to do these tests using the substitution method to work out how much heat an LED puts out.

There are way too many assumptions needed in theoretical predictions - although it's good to see that some agree closely with the real-world tests.


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## slebans

David_Campen said:


> The amount of power converted to white light is much less than that.


 
Given the values stated by Kinzza - his calculations are correct. I do not understand your logic where you then use a less efficient LED Engin product to compare with the XP-G used by Kinzza. That does not seem fair or logical. 

The underlying XP-G Royal Blue LED has an efficiency higher than 60% and that a significant percentage of the photons produced by said Royal Blue LED are emitted directly and not subject to phosphor conversion losses?

Just my $.02

Stephen Lebans


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## David_Campen

slebans said:


> Given the values stated by Kinzza - his calculations are correct. I do not understand your logic where you then use a less efficient LED Engin product to compare with the XP-G used by Kinzza. That does not seem fair or logical.
> 
> The underlying XP-G Royal Blue LED has an efficiency higher than 60% and that a significant percentage of the photons produced by said Royal Blue LED are emitted directly and not subject to phosphor conversion losses?
> 
> Just my $.02
> 
> Stephen Lebans



The post I was responding too was discussing the efficiency of white LEDs so, yes, there is the phosphor conversion inefficiency to be considered.

Yes, I should have used values from a Cree datasheet. I couldn't find a datasheet for a royal blue XP-G but the data sheet for a Cree XT-E royal blue driven at its rated capacity shows that you get an efficiency of about 40%; multiply that by the phosphor efficiency of 83% and you get about 32% overall.
http://www.cree.com/products/pdf/XLampXT-E_ROY.pdf

By driving the Cree XT-E royal blue at 1/3 its rated power then you can get an efficiency of 50%.


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## Kinnza

I just used some easy figures in order to illustrate the calculation with an example. 

It was not intended as calculation of a real sample, except if you get a true output of 420lm with a power of 3W for a coolwhite. Actually, I don't think any real LED had reached such performance yet, but latest ones are close.

Main difference between theoretical calculations and empirical ones result for underestimate the decreasing of light emission due to heating. That's why I didn't enter into the topic of calculating lm output theoretically with the datasheet. Actually, it is way better to measure the emission with an IS and BTW measure the actual power used, the exact forward voltage and the current.

But if you have accurate figures of true emission and power burned, the method of calculating the heat load by crosschecking with LER is very accurate. More accurate as finest the figures, of course, but thermal load don't need an extra accurate figure, so it works fine. If you are interested on knowing the actual energetic efficiency of a given LED for comparison with other lights, then initial figures needs to be as accurate as possible.

Given the large differences that heatsink positioning, amount of air flow, etc, plays determining the temperature of a heatsink for a given heat load, I would say that trying to measure the heat load by comparing heatsink temperature is not very accurate either. So actually, if you have accurate figures for lm emission and power burned, this theoretical method works probably better. Crosschecking both methods using same LED would be very interesting.


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## Kinnza

David_Campen said:


> The post I was responding too was discussing the efficiency of white LEDs so, yes, there is the phosphor conversion inefficiency to be considered.
> 
> Yes, I should have used values from a Cree datasheet. I couldn't find a datasheet for a royal blue XP-G but the data sheet for a Cree XT-E royal blue driven at its rated capacity shows that you get an efficiency of about 40%; multiply that by the phosphor efficiency of 83% and you get about 32% overall.
> http://www.cree.com/products/pdf/XLampXT-E_ROY.pdf
> 
> By driving the Cree XT-E royal blue at 1/3 its rated power then you can get an efficiency of 50%.


 
It is very difficult to calculate the final efficiency of a white LED still when you know accurately the efficiency of the blue LED inside. It is not so easy as derating the emission for the quantum efficiency of the phosphor, as photons converted have different effect on luminosity, thus LER changes (from 40-75lm/W of a deep blue LED to 300lm/W for a white). Phosphor layer affects too the light extraction efficiency of the package, and different concentration of phosphor lead to very different light scattering. 

Probably with remote phosphor technology all those factors are minimized, but I would say such calculation is anyway subjected to too many parameters to give off any accurate figure.

But what is sure that white LEDs with efficiencies of 50% are waiting on the near future and the XT-E is close enough to do it with two bins higher or so.

Top brands top bins of coolwhites are all well over 30%, best ones over 40% when running soft. Due to heating, is very different to compare LEDs working at 350mA or 2.5A. Junction temps varies a lot, thus actual efficiencies does it aswell. In order to see efficiencies of 50% with LEDs running at 1A or over, probably major advances reducing current density droop and improving thermal resistance are required


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## MikeAusC

Kinnza said:


> . . . . Given the large differences that heatsink positioning, amount of air flow, etc, plays determining the temperature of a heatsink for a given heat load, I would say that trying to measure the heat load by comparing heatsink temperature is not very accurate either. . . . . .


 
That's why the Substitution Method uses IDENTICAL heatsinks operating at IDENTICAL temperature in IDENTICAL air flow and temperature.


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## MikeAusC

What if you used an Ultraviolet LED to excite the phosphor.

What's the Luminous Efficiency of an Ultraviolet LED ?


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## slebans

MikeAusC said:


> What if you used an Ultraviolet LED to excite the phosphor.
> 
> What's the Luminous Efficiency of an Ultraviolet LED ?


 
At 400nm the LER value would be .270 lumens for a 100% efficient emitter at 1 watt of power. The luminosity function reflects the fact a human's retina is not sensitive to UV wavelengths. 

Toshiba - and other companies - were/are planning to produce white LEDs based on UV instead of Blue source LEDS using phospor conversion. I honestly do not know if anything has come of their research.

Stephen Lebans


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## Kinnza

Zero, of course.

Thats what I was meaning when I said that LER changes when you apply the phosphor, and that knowing the blue LED efficiency dont allows to calculate final outcome of the white LED.

Im not critiquing your method, Mike. Somebody asked how to know the efficiency for other means so I chimed in and explain the method using measured lumen emission. As it allows to calculate the emitted energy accurately, it gives you the heat load, as energy emitted as light cant go as heat on the heatsink and you cant get more energy output than energy input. I think your way is good to crosscheck results. If any, it could be somewhat improved as others suggested, by isolating well the resistors, placing them glued with thermal adhesive and measure heatsinks temperature well ahead of the heating elements.


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## MikeAusC

I don't see the point of knowing the LUMEN efficiency of the Blue LED - we don't see most of the LED's output.

We need to know the Phosphor-sensitive-range efficiency of the LED.


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## SemiMan

----------


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## Kinnza

Instead of debating so much about whites, why not testing the resistor setup with a royal blue? It would be a straightforward way of checking the accuracy of the method without messing with the conversion factors of the whites. Once done, you can be sure your testing with whites is accurate. Anyway, I think it is relatively accurate actually, but my experience says thermal energy is tricky to measure


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## MikeAusC

SemiMan said:


> . . . . Not sure why you used what seemed to be such a different resistor compared to the CREE part. Why not use a good size SMT resistor which would have better equated the packaging? . . . .


 
Compared with the heatsink, the resistor or LED+star have very small surface area, so they'll have little impact on heat removal. In fact the surface area of the star and the resistor aren't much different. 

At these temperatures, there'll be minimal heat removal from radiation.


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## MikeAusC

Kinnza said:


> Instead of debating so much about whites, why not testing the resistor setup with a royal blue? It would be a straightforward way of checking the accuracy of the method without messing with the conversion factors of the whites. Once done, you can be sure your testing with whites is accurate. Anyway, I think it is relatively accurate actually, but my experience says thermal energy is tricky to measure


 
I don't see how this would improve accuracy. The Substitution Method measures the HEAT that ANY object puts out, by substituting an item that puts out an accurately measurable amount of HEAT. 

I did the tests with a couple of white LEDs, so it tells me how much heat white LEDs put out. 

I'm not interested in the heat output from Blue LEDs.


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## MikeAusC

Kinnza said:


> . . . . So for example, a coolwhite LED that emits 420lm and burns 3W (say an XP-G for example, simplifying 3V at 1A), it emits at 420/3= 140lm/W. Using a LER of 280lm/W, 140/280=0.5. So 50% of input energy is converted as light, and the remaining 50% (1.5W) is the heat load. . . .


 
Why do Cree only claim 100lm/W in the XM-L datasheet ?

[Edit - the 100lm/W is NOT LER. It's Luminous Efficacy - the ratio between the total luminous flux emitted by a device and the total amount of input power (electrical, etc.) it consumes - which this thread has always been about.]


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## MikeAusC

Here's a theoretical calculation that isn't dependant on too many assumptions.

One Lumen is DEFINED as 1/683 Watt at 555nm. As a first approximation, let's average this to 1/340 watt over the visible spectrum.

Cree state that the XM-L produces 100 Lumen per watt of input.

So 10 watts input produces 10x100/340 watts of light = 2.92 watts which leaves 7.08 watts exiting as heat i.e 71% wasted as heat.


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## Kinnza

MikeAusC said:


> Here's a theoretical calculation that isn't dependant on too many assumptions.
> 
> One Lumen is DEFINED as 1/683 Watt at 555nm. As a first approximation, let's average this to 1/340 watt over the visible spectrum.
> 
> Cree state that the XM-L produces 100 Lumen per watt of input.
> 
> So 10 watts input produces 10x100/340 watts of light = 2.92 watts which leaves 7.08 watts exiting as heat i.e 71% wasted as heat.



The efficacy of any LED depends strongly of the current level and the junction temperature. The quoted 100lm/W for the XM-L is for it emitting 1000lm. Actually, its not realistic, Cree hasnt released any XM-L emitting 1000lm on real conditions:

A T6 bin emits a minimum of 280lm (max 300lm) when running at 700mA (max 300lm). At 3A, maximum power rated for this LED, it emits 3.25x the emission at 700mA. So a T6 running at 3A emits between 910 and 975lm, with a typical power of 10W. But all that is refered to a Tj of 25ºC, impossible on normal conditions, as the absolute lower junction temp (Tj)at 3A is going to be about 27ºC over heatsink temperature, likely 30-35ºC still for a good thermal path. Rounding to 30ºC, for a heatsink at 40ºC, it means a Tj=70ºC, and for a heatsink at 60ºC, Tj=90ºC. 

So taking a good thermal path and a good heatsink running at 40ºC, Tj is 70ºC at 3A. At that Tj, as best emission is 90% of that at Tj=25ºC.

So the actual emission of a T6 at 3A with very good conditions (tj=70ºC) is about 850lm, for a power of 9.64W (using average forward voltage of the model), getting 88lm/W.

While when running at 700mA, a T6 on real conditions emits about 266lm (minimum, for a heatsink about 40ºC) for a power of 1.99W, getting 134 lm/W.

Meaning that at 3A, the XM-L is 34% less efficient than at 700mA (of course, for a sample, it may vary a lot depending of the thermal path, this is for a best case scenario).

These are calculations based on the datasheet, of course the most accurate data of the efficiency would be obtained from actual measurement of samples on a IS. I just made this calculation to have realistic figures for the T6 performance to work with.

If such T6 is a coolwhite (actually, no such bin on other tones yet), at a LER of 280lm/W, it mean that at 3A and 88lm/W, it has an efficiency of 88/280=31.4% (68.8% heat). While at 700mA, its 134/280=47.8%.

The way of mounting the LED for the trial affects the lm emission, specially at high currents. The higher the thermal resistance of your way of mounting, the lower the light emission, thus higher the heat load (lower efficiency).

So lets analyze your results at 1A, where thermal effects are less noticiable. At 1A, XM-L emission is 1.375X of nominal (700mA). So 385lm (Tj=25ºC). Increase of temp due the own LED package is 2.5ºC/W, so for a power of 2.95W, its 7.4ºC over LED base temp. As heatsink was at 37.5ºC, it mean Tj was 45ºC plus thermal resistance added by the interface LED-Heatsink. At Tj=45ºC, emission is about 95% of that at 25ºC, thus 366lm. 366lm/2.95W=124lm/W without taking into account LED-heatsink interface resistance.

At a LER of 280lm/W, 124/280=44% efficiency. Say your thermal interface was not very good and added 10ºC to Tj. Emission would be about 354lm (120lm/W). And suppose your color bin had an inusual high LER for a coolwhite of 310 lm/W. Still efficiency of a T6 wouldnt be below 120/310=38.7%.

You got 31%. Too large difference, meaning that either thermal resistance of your setup was rather high and/or, your comparison method over estimates thermal load.


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## Kinnza

And lets clear something.

Luminous flux of any light source is calculated or measured by integrating the energy emitted in each wavelength into the photopic curve. Please read any reference text about that.

By definition, Luminous Efficacy of Radiation is the Luminous Flux (in lm) divided by the Radiometric flux (in Watts). When normalized to a radiometric Flux of 1W, it says how many lm produces a given spectrum (when emitting 1W).

Thus, when you have the luminous efficacy of a light source (lm/W), which is the division of lm emitted by power input, by LER, you have the ratio between emitted power and input power. By DEFINITION, the energetic efficiency of a light source:

lm/W(input) divided by

lm/W(output) (LER, by DEFINITION)

=W(output)/W(input)

This is indisputable. All you need to get an accurate figure of efficiency is to have accurate figures of lm/W (input) and lm/W(output).

I have measured many LER of white LEDs. But it you dont trust on my figures, calculate them yourself by integrating spectrum in the photopic curve. But on many sites you can find the average LER of white LEDs being around 300lm/W. While your assumption of white LED being 1/340 is a mere trial.

While this system is as accurate as your lm emission figure, yours is dependent on the measurement of thermal energy, a very little accurate science. As you could check if you perform some measurements with a royal blue of known efficiency.

I would say your system over estimates heat load due resistors dissipate a little more heat by radiation, more heat directly by convection at the own resistors and probably you measure heatsink too close of the heating elements with the result of different deltas affecting your measurement.


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## slebans

Excellent post, Kinnza.
Thank you very much for taking the time to do this.

Stephen Lebans


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## MikeAusC

Kinnza said:


> . . . . While this system is as accurate as your lm emission figure, yours is dependent on the measurement of thermal energy, a very little accurate science. As you could check if you perform some measurements with a royal blue of known efficiency. . .



No, the subsitution method means I don't have to measure thermal energy - I equalise them.




Kinnza said:


> . . . . I would say your system over estimates heat load due resistors dissipate a little more heat by radiation, more heat directly by convection at the own resistors and probably you measure heatsink too close of the heating elements with the result of different deltas affecting your measurement.



At these temperatures, there will be minimal heat loss by radiation. The LED or resistor are a small fraction of the heatsink surface, so any difference will have minimal impact.


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## MikeAusC

Kinnza said:


> . . . . I have measured many LER of white LEDs. But it you dont trust on my figures, calculate them yourself by integrating spectrum in the photopic curve. But on many sites you can find the average LER of white LEDs being around 300lm/W. While your assumption of white LED being 1/340 is a mere trial. . . . . .


 
"So 10 watts input produces 10x100/340 watts of light = 2.92 watts which leaves 7.08 watts exiting as heat i.e 71% wasted as heat."

OK, lets use replace my assumption of 340 lm/w with your 300 - 
So 10 watts input produces 10x100/300 watts of light = 3.30 watts which leaves 6.7 watts exiting as heat i.e 67% wasted as heat.


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## MikeAusC

Kinnza said:


> . . . .By definition, Luminous Efficacy of Radiation is the Luminous Flux (in lm) divided by the Radiometric flux (in Watts). When normalized to a radiometric Flux of 1W, it says how many lm produces a given spectrum (when emitting 1W). . . . .



And that's where the discussion falls apart.

The standard definition of Luminous Efficacy of Radiation relates ONLY to the conversions of radiated watts (NOT electrical input watts) to visible-light. 

LED light designers are only interested in the measuring how much ELECTRICAL INPUT WATTS are converted to light - as the remainder is converted to heat.

LER tells us how much of the RADIATED energy (always less then the electrical input energy) is actually producing visible light. A 10 watt filament may radiate 9 watts electromagnetic radiation (1 watt gets conducted away ) but if the 8 of these 9 watts are infra-red heat, then we only have 1 watt as visible light.

The LER of a filament at 2800K is 15 lumens per radiated watt.

The LER of the star Betelguese is 30 lumens per radiated watt.


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## MikeAusC

Quote from http://www.led-professional.com/downloads/LpR_13_468932.pdf page 53

"Thermal management of sophisticated LED solutions - Dr Michel Kazempoor - Perkin Elmer"

"Despite the high efficiency of LEDs still 70-80% of the electrical input power is converted to heat"


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## Kinnza

MikeAusC said:


> And that's where the discussion falls apart.
> 
> The standard definition of Luminous Efficacy of Radiation relates ONLY to the conversions of radiated watts (NOT electrical input watts) to visible-light.
> 
> LED light designers are only interested in the measuring how much ELECTRICAL INPUT WATTS are converted to light - as the remainder is converted to heat.
> 
> LER tells us how much of the RADIATED energy (always less then the electrical input energy) is actually producing visible light. A 10 watt filament may radiate 9 watts electromagnetic radiation (1 watt gets conducted away ) but if the 8 of these 9 watts are infra-red heat, then we only have 1 watt as visible light.
> 
> The LER of a filament at 2800K is 15 lumens per radiated watt.
> 
> The LER of the star Betelguese is 30 lumens per radiated watt.



It is being somewhat boring to discuss against a closed mind. Maybe you could learn something and be able to improve your knowledge and experiments if you concentrate on understanding my arguments, instead of just looking the way of negate them (before you have understood them).

Please read again post #64 and try to understand it. Method explained obtain the quotient of the light output (in watts) by the power input (in watts). BY DEFINITION, the energetic efficiency of a device.

LER dont say how much of the radiated energy is actually producing light. It says how many lm (human perceived light) produces a radiated watt into the visible range.

LER is the ratio between photometric flux and radiometric flux (lm/W emitted). In lighting applications, that ratio is calculated between 400 and 700nm, the most meaningful and most often used (the one I'm giving), or more extensively, along the full visible range (380-780nm). Astromers may use a LER covering the full electromagnetic spectrum, but that is relevant for them, not for us.

It is important to note that when LER is calculated, wavebands used for photometric flux and radiometric flux are obviously the same. Due LEDs emits very little energy outside the band 400-700nm, this is the range most often used. If we use a wider bandgap, for those LEDs emitting some of the energy beyond 700nm, energy emitted result on a negligible increase on lm emission, thus calculated LER is lower than that calculated for 400-700nm.

When calculating the energy efficiency, a lower LER results on a higher radiometric efficacy, because we are taking into account the energy emitted in other bands but that dont add to lm production. Thus, more energy leaving the package as electromagnetic energy and lower thermal load.

I decided to be conservative and use the range 400-700nm, that result in a slightly higher thermal load, as it consider negligible the energy emitted beyond 700nm so sum it to the thermal load. With LEDs, this is not a problem due the low far red emission. With other lighting technologies, it's different, but in those cases thermal load is dissipated either by radiation or by convection, very different to the thermal load of LEDs which is dissipated almost fully through the heatsink.

Values cited in the quote use wider wavebands, 380-780nm (or still 360-830nm of the latest CIE photopic curve).


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## Kinnza

MikeAusC said:


> Quote from http://www.led-professional.com/downloads/LpR_13_468932.pdf page 53
> 
> "Thermal management of sophisticated LED solutions - Dr Michel Kazempoor - Perkin Elmer"
> 
> "Despite the high efficiency of LEDs still 70-80% of the electrical input power is converted to heat"



Its difficult to believe you are a flashaholic aware of how fast are improving efficiencies of LEDs, when you cited a May 2009 article about LED's efficiency. At that time, best coolwhite LEDs where about 100lm/W (for Tj=25C, not operating conditions). Cree just announced the huge leap on efficiency of the XPG but it was not released commercially until near end of 2009.

The same article says "the efficiency of these chips is up to 20-30% *and more in practice*, depending on the current, chip material, cooling, etc..". The reason of after that choosing of the lower range of efficiency, of not best LEDs on good conditions, is to push on his own product, which by that time offered a low thermal resistance, which offer more advantages as higher the thermal load.

From that time, we have jumped from just 100lm/W to 160lm/W on same tones. 30% efficiency of the best LEDs of that time have improved on 60%: from 30 to 48%.

Once you have put so much effort trying to negate the reality, you could use it reading some other articles on the same magazine. Actually, 2 cited the method I explained as the method most accurate for the measurement of the thermal load, because of how tricky can be thermal measurements (page 27). He cites the CIE standards for luminous flux and radiometric flux measurements (0.7% error margin) and two papers about determining the heat load using this method, first is "THERMAL INVESTIGATION OF HIGH POWER OPTICAL DEVICES BY
TRANSIENT TESTING" (pdf) by Farkas et al . Check how engineers experts on thermal measuring and modeling used luminous flux corrected by photopic curve to obtain the radiometric flux and the thermal load. Basically, because it is the accepted accurate method of doing it. 

Who knows why they didnt used resistors.


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## SemiMan

Kinnza, I was wondering how long it would take you to lose it. You showed admirable patience.

MikeAusC, you have refused to actually consider the arguments put forth by others more knowledgeable.

LER is important ... otherwise there is no ability to convert between emitted lumens (of a visible LED) and the radiated power of that LED.

Without the radiated power, you have no knowledge of power out versus power in and hence efficiency and waste heat.

It was offered to you that thermal experiments are difficult yet you refuse to consider that your experiment may be flawed. 

I don't know what else to say other than Kinnza, I have enjoyed your posts!

Semiman


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## MikeAusC

Kinnza said:


> . . . . when you cited a May 2009 article about LED's efficiency. . . . . . .



. . . and then you reference an EIGHT YEAR OLD article that only addresses phosphorless red LEDs and presents the conclusion "Measuring power LEDs we found that their thermal behaviour depends on the applied power level. We defined for optical devices an effective thermal resistance for calculating the thermal stress and a residual thermal resistance for characterising package quality" ????


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## MikeAusC

Kinnza said:


> Its difficult to believe you are a flashaholic aware of how fast are improving efficiencies of LEDs, when you cited a May 2009 article about LED's efficiency. At that time, best coolwhite LEDs where about 100lm/W (for Tj=25C, not operating conditions). Cree just announced the huge leap on efficiency of the XPG but it was not released commercially until near end of 2009. . . .


 
100lm/W is EXACTLY the figure Cree are publishing TODAY in their XM-L datasheet - "XLamp XM-L is 20% more efficient than XLamp XP-G at the same current, and can deliver 1000 lumens with 100 lumens per Watt efficacy." - but then maybe you have some better information than Cree has.


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## THE_dAY

I think it means when its pumping out 1000 lumens it drops to 100 lumens per Watt.


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## Bullzeyebill

Folks please maintain decorum here.. Be respectful to each other particularly when there are disagreements.

Bill


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## LukeA

THE_dAY said:


> I think it means when its pumping out 1000 lumens it drops to 100 lumens per Watt.


 
Exactly. That efficacy is impressive because it's a typical drive level, not the peak value and it's a production LED, not some lab baby.


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## bbb74

Ok I'm confused, but following the thread with interest 



Kinnza said:


> But on many sites you can find the average LER of white LEDs being around 300lm/W. While your assumption of white LED being 1/340 is a mere trial.


I don't get how a LER could be 300lm/W, nor how it has a metric of lm/W. Isn't it a dimensionless ratio, of luminous flux/radiant flux. Eg (with "example" data): LER = 134lm/W / 200lm/W = 0.67? If the LER was 300lm/W for a white LED, with the maximum possible being around ~400lm/W, doesn't this mean you are only losing 25% as heat? That sounds too low to me so something is wrong with me or the 300lm/W.



Kinnza said:


> If such T6 is a coolwhite (actually, no such bin on other tones yet), at a LER of 280lm/W, it mean that at 3A and 88lm/W, it has an efficiency of 88/280=31.4% (68.8% heat). While at 700mA, its 134/280=47.8%.



From where did you get the LER of a XML T6 as 280lm/W, I can't find this anywhere? Do you mean its radiant flux is 280lm/W? I've read the cree spec sheet and can't find it. To get this you would need to look at the spectral curve and calculate it, is that what you have done or something?




Kinnza said:


> Thus, when you have the luminous efficacy of a light source (lm/W), which is the division of lm emitted by power input, by LER, you have the ratio between emitted power and input power.



Quote from wikipedia:


> The main difference between the luminous efficacy of radiation and the luminous efficacy of a source is that the latter accounts for input energy that is lost as heat or otherwise exits the source as something other than electromagnetic radiation. Luminous efficacy of radiation is a property of the radiation emitted by a source. Luminous efficacy of a source is a property of the source as a whole.



According to wikipedia the theoretical maximum LED luminous efficiency is about 38.1–43.9% (260-300lm/watt). If a white LED was "perfect" and only emitted "white" light and a) lost no light b) converted none of its input energy to heat etc, it would get around ~400lm/watt but obviously this isn't achievable.

I was going to write more but I have a cracking headache now.... thanks...


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## MikeAusC

bbb74 said:


> Ok I'm confused, but following the thread with interest
> I don't get how a LER could be 300lm/W, nor how it has a metric of lm/W. Isn't it a dimensionless ratio, of luminous flux/radiant flux. Eg (with "example" data): LER = 134lm/W / 200lm/W = 0.67? If the LER was 300lm/W for a white LED, with the maximum possible being around ~400lm/W, doesn't this mean you are only losing 25% as heat? That sounds too low to me so something is wrong with me or the 300lm/W.
> . . . . .



That's the first opportunity for confusion.

LER can be used for - 

Luminous Efficacy of Radiation - the ratio of luminous flux (lumens) to radiant (watts). - of limited use here, since we don't know the radiated watts of LEDs - and the white LED is the result of direct leakage through the phosphor and wavelength conversion by the phosphor. Unfortunately confusion will start if people are uncertain about whether the watts being radiant output or electrical input - when building LED lights, we're really only interested in electrical input.

Luminous Efficiency Ratio - if you measure the radiant power in watts and luminous power in watts (using 683 lumen/watt at 555nm) you can come up with a dimensionless ratio - as you seem to be using.


----------



## shao.fu.tzer

Well, scientists have been working on a way to transform waste heat into sound, and then back into electricty. Maybe this technology can find its way into flashlights in... say... 20 years... 

Shao


----------



## slebans

MikeAusC said:


> That's the first opportunity for confusion.
> 
> LER can be used for -
> 
> Luminous Efficacy of Radiation - the ratio of luminous flux (lumens) to radiant (watts). - of limited use here, since we don't know the radiated watts of LEDs - and the white LED is the result of direct leakage through the phosphor and wavelength conversion by the phosphor. Unfortunately confusion will start if people are uncertain about whether the watts being radiant output or electrical input - when building LED lights, we're really only interested in electrical input.


 
Back in April I posted a link to a document explaining all of the terms being bandied about within this thread. Here are the relevant bits:


The main driving force for solid-state lighting is the potential of huge energy savings on the national or global scale
6.
Thus, when considering spectra of light sources for general illumination, another important aspect to consider is
luminous efficacy (lumens per watt). The term ​​
_luminous efficacy _is normally used as the conversion efficiency from the
input electrical power (watt) to the output luminous flux (lumen). The luminous efficacy of a source is determined by
two factors: the conversion efficiency from electrical power to optical power (called ​​
_radiant efficiency _or _external_
_quantum efficiciency_​​
7) and the conversion factor from optical power (watt) to luminous flux (lumen). The latter is called
​_luminous efficacy of radiation _​_
_​_
_(lumen/watt) and is hereinafter denoted as LER. Since LER and color rendering are
determined solely by the spectrum of the source, white LED spectra should be optimized for both of these aspects.​
I have to post this next part as an Image due to the use of symbols/formatting not supported via this forum. I had trouble getting this image to show so I hope it is visible to everyone in this thread.







Mike, if we know the luminous efficacy of the LED and the LER value for this specific LED at a specific color temperature over a fixed spectral width then we can derive the radiant efficiency. For example if we drive the cool white XM-L at a power level that produces 100 lumens per watt, with a LER value of 300, then the radiant efficiency is 33%. For every watt of energy fed to this LED only 1/3 is emitted as light and the rest is heat.

Stephen Lebans​


----------



## slebans

bbb74 said:


> Ok I'm confused, but following the thread with interest
> 
> Quote from wikipedia:
> 
> According to wikipedia the theoretical maximum LED luminous efficiency is about 38.1–43.9% (260-300lm/watt). If a white LED was "perfect" and only emitted "white" light and a) lost no light b) converted none of its input energy to heat etc, it would get around ~400lm/watt but obviously this isn't achievable.
> 
> I was going to write more but I have a cracking headache now.... thanks...


 
That Wikipedia article is inaccurate in that it uses a LER value derived from monochromatic Green at 555nm and then calculates efficiencies of other light sources relative to this LER value of 683 lumens per watt.

Within this thread, we have been discussing Cree cool white LEDS and the spectra produced by these LED based on a Blue LED with YAG phosphor. The LER value will depend on the exact Bin being used and the power level the LED is being driven at. A range of LER values from 280 - 370 lumens per watt have been posted. The values within the documents I have read range from 300-350 lumens per watt but it depends on the spectral range used in the calculation. 

LER values exceeding 400 lumens per watt are possible but are based on RGB or RGBA configurations.

Back in April, in this thread, I suggested readers should visit the DOE website and download the Multi-Year Roadmap plan and the Manufacturing Roadmap documents. They helped me to understand the issues we are discussing here.

Stephen Lebans


----------



## Kinnza

I hope Stephen's post had clarified this topic. I going to try to go deeper on the explanation, in order to cover concepts and definitions that is difficult to obtain well defined (specially on Wikipedia).

Efficacy of light sources refers to quotients of light flux by power input. The typical *Luminous Efficacy* uses photometric flux divided by input power, thus divide lm by W, obtaining lm/W

*Wall plug Luminous Efficacy* is the same, but including all power after the wall plug (ballast,drivers, etc). Generally its applied tho a whole luminary instead of a bare lamp, so it consider aswell the optical losses (secondary optics, reflectors, etc). So it's always lower than bare Luminous Efficacy. But unit used is the same lm/W. (Note: in efficacy figures, W refers to input Watts).

*Luminous Efficacy of Radiation* (LER) is the quotient of Luminous Flux (lm) by Radiometric Flux (emitted light, in Watts, radiometric, optical ones). Thus its unit is lm/W, but here W are output watts in the form of light (officially, 360-830nm; I already explained before why using just 400-700 for LEDs gives a very close figure, only very slightly higher). This figure is a direct consequence of a given Spectral Power Distribution (SPD): only SPD and luminous coefficients (photopic curve) are required to calculate it. Each SPD has its own corresponding LER figure. It is a pure spectral characteristic, thus sometimes it is called Spectral Luminous Efficacy.

In other contexts, the whole electromagnetic spectrum is used, but not in lighting, as it don't gives any useful information.

On the other hand, figures of Efficiency are often dimensionless as they refers to how good is a device performing work compared to energy used to do it. As light is a form of energy, it can be expressed as Watts. In general, efficiency figures compares output/input on same units.

So when we compare the light emitted with the energy required to do it, we obtain the *gross* *Efficiency* of the light source: output Watts/input Watts. W/W=1, so its a dimensionless figure. As max efficiency is 1, it often is expressed as percentage.

*Net Efficiency* or wall plug efficiency is the same, but as Wall Plug Efficacy, obtained including all the input power and only the true light emission.

In the entry of Wikipedia, defines Overall Luminous Efficacy of Radiation and Overall Efficiency Of Radiation, which uses the full electromagnetic spectrum output. As it includes infrared energy, for lighting they are unmeaningful concepts and unuseful figures, especially for LED lighting. The latter calculates efficiency by comparing the Luminous Efficacy (lm/W) with the maximum possible efficacy of 683 lm/W (555nm monochromatic source of 100% efficiency).



bbb74 said:


> Ok I'm confused, but following the thread with interest
> 
> 
> I don't get how a LER could be 300lm/W, nor how it has a metric of lm/W. Isn't it a dimensionless ratio, of luminous flux/radiant flux. Eg (with "example" data): LER = 134lm/W / 200lm/W = 0.67? If the LER was 300lm/W for a white LED, with the maximum possible being around ~400lm/W, doesn't this mean you are only losing 25% as heat? That sounds too low to me so something is wrong with me or the 300lm/W.



LER is measured in lm/W (emitted, output, optical Watts).

Quotient of Luminous Efficacy (lm/W) by LER (lm/W) gets a figure of Efficiency, dimensionless. 0.67=67% would be the Efficiency of a light source with those figures.

Notice LER only depends of the SPD of the LED. It says how many lm produces an optical watt of such spectrum. You just need to multiply it by the actual power output (optical watts) to obtain the total flux emitted (lm).

On the reverse, if you have the lm emission and the LER, you can get the light emitted in Watts. This is the procedure used to calculate the Efficiency of a given LED and thus, its thermal load.

The fact that white sources has typically LERs between 250 and 400 lm/W don't says itself nothing about heat dissipated. LER refers to optical energy, light emitted. It says how many lm produces a unit (1W) of light emitted. It only depends of the SPD weighted by the photopic curve. The maximum possible is for a pure monochromatic light of 555nm, 683lm/W. As to get a white light you need emission on other colors, which have luminous coefficients below 1 (just 555nm is 1), any spectrum of white light results on a LER below 400lm/W. But its only due blue and red light produces in our eyes a lower brightness sensation than green.





bbb74 said:


> From where did you get the LER of a XML T6 as 280lm/W, I can't find this anywhere? Do you mean its radiant flux is 280lm/W? I've read the cree spec sheet and can't find it. To get this you would need to look at the spectral curve and calculate it, is that what you have done or something?



Yes, exactly. I calculated it weighting its SPD by photopic coefficients, in 1nm wavebands steps.

Actually, not a XML T6. The LER 280lm/W for a coolwhite is just a good average for typical coolwhite spectrums (I have measured many). It actually depends of the color bin. You can find coolwhite LEDs with higher LER (I have measured up to 310lm/W). But in most cases is lower (LER tends to decrease as CCT, K temperature, raises). If you prefer to be conservative and you dont have the means of measuring the SPD of a given LED, use a 300lm/W for LER. It will probably under estimates slightly the efficiency and over estimates thermal load, but for little.

Without having an spectrometer, the only you can do is calculate the LER based on the typical SPD in the LED's datasheet. Resize graph to fit graph width to nm in the scale, multiply each value by its photopic coefficient, sum them all and multiply by 683.

Actually, LER varies with operating temperature change and current level. Fortunately for us, the way it does for phosphor based white LEDs is very small, almost negligible for the purpose of calculating efficiency and thermal load.




bbb74 said:


> Quote from wikipedia:
> 
> 
> According to wikipedia the theoretical maximum LED luminous efficiency is about 38.1–43.9% (260-300lm/watt). If a white LED was "perfect" and only emitted "white" light and a) lost no light b) converted none of its input energy to heat etc, it would get around ~400lm/watt but obviously this isn't achievable.
> 
> I was going to write more but I have a cracking headache now.... thanks...



Actually, maximum LER for a white source of decent chromatic rendering is about 400lm/W. For actual technology of phosphor based white LEDs, about 350lm/W (due Stokes losses). 

So 100% efficient LEDs could produce as max an efficacy 400lm/W. 

In the practice, with actual phosphor technology, true 250lm/W looks like the best achievable on the medium term, as it implies efficiencies over 70%.


----------



## Kinnza

MikeAusC said:


> . . . and then you reference an EIGHT YEAR OLD article that only addresses phosphorless red LEDs and presents the conclusion "Measuring power LEDs we found that their thermal behaviour depends on the applied power level. We defined for optical devices an effective thermal resistance for calculating the thermal stress and a residual thermal resistance for characterising package quality" ????



I only linked it so you can see how people who knows about LEDs and thermal energy measures efficiency by measuring lm emission and LER. I hoped it does you think why.



> 100lm/W is EXACTLY the figure Cree are publishing TODAY in their XM-L datasheet - "XLamp XM-L is 20% more efficient than XLamp XP-G at the same current, and can deliver 1000 lumens with 100 lumens per Watt efficacy." - but then maybe you have some better information than Cree has.



In post #63, I show you with Cree datasheet in hand that actually that Cree statement is pure marketing, as it is only possible with the best T6 working at Tj=25ºC, impossible in real conditions. I calculated that in probably best case, 850lm at 88lm/W is what a XM-L T6 can perform, give or take.

But when working at 700mA, it easily achieves true 134lm/W


----------



## uk_caver

Kinnza said:


> The LER 280lm/W for a coolwhite is just a good average for typical coolwhite spectrums (I have measured many). It actually depends of the color bin. You can find coolwhite LEDs with higher LER (I have measured up to 310lm/W). But in most cases is lower (LER tends to decrease as CCT, K temperature, raises). If you prefer to be conservative and you dont have the means of measuring the SPD of a given LED, use a 300lm/W for LER. It will probably under estimates slightly the efficiency and over estimates thermal load, but for little.


Thanks for your explanations.

Just to make sure I'm not misunderstanding anything, am I right in thinking that what you're saying is that if a light source of _any_ technological construction had the same spectral output as a 'standard' phosphor-based white LED of 300lm/W LER, then a source like that which was producing 150lm and consuming 1W, would necessarily be wasting 50% of its input power compared to a perfect device with the same output, typically with much of that waste ending up as heat?

And regarding any particular kind of design, all the inefficiencies (such as Stokes losses) would end up collected together in the 'waste' category and don't have to be considered individually when calculating from the lumen output and LER to work out the likely heat production - the LED can be treated as a 'black box' in that respect?

And effectively, even if the maximum theoretical luminous efficacy of a phosphor-based white LED might be ~350lm/W, such a device would have to have a 'more subjectively efficient' spectral output than current LEDs do, and so the 350lm/W figure is not useful in calculating the gross efficiency of a current LED, but we should probably be using a figure of ~300lm/W?


----------



## MikeAusC

uk_caver said:


> . . . . . .white LED of 300lm/W LER, then a source like that which was producing 150lm and consuming 1W, would necessarily be wasting 50% of its input power compared to a perfect device with the same output . . . .


 
No, LER is Luminous Efficacy of Radiation. In 300lm/W, the W is optical watts, not electrical input watts ("consuming 1 watt"). It's a measure of how much of the radiated power is visible, not a measure of how well electrical input gets converted to visible radiation.

http://en.wikipedia.org/wiki/Luminous_efficacy
"Luminous efficacy of radiation measures the fraction of electromagnetic power which is useful for lighting. It is obtained by dividing the luminous flux by the radiant flux."
"In SI, luminous efficacy has units of lumens per watt (lm/W). Photopic luminous efficacy of radiation has a maximum possible value of 683 lm/W, for the case of monochromatic light at a wavelength of 555 nm (green)"


----------



## Kinnza

uk_caver said:


> Thanks for your explanations.
> 
> Just to make sure I'm not misunderstanding anything, am I right in thinking that what you're saying is that if a light source of _any_ technological construction had the same spectral output as a 'standard' phosphor-based white LED of 300lm/W LER, then a source like that which was producing 150lm and consuming 1W, would necessarily be wasting 50% of its input power compared to a perfect device with the same output, typically with much of that waste ending up as heat?



You understood perfectly.

If a device with a LER of 300lm/W emits 150lm,it means optical power emitted is 0.5W. So if it consumes 1W ,efficiency is 50%.

For the calculation, you just need the efficacy of the light source, in this case (150lm consuming 1W), 150lm/W. If you divide a figure in lm/W (input) by lm/W(output), you get W(output)/W(input).

The rest of the energy is dissipated as heat.



uk_caver said:


> And regarding any particular kind of design, all the inefficiencies (such as Stokes losses) would end up collected together in the 'waste' category and don't have to be considered individually when calculating from the lumen output and LER to work out the likely heat production - the LED can be treated as a 'black box' in that respect?



Exactly. LER is derived directly from the final Spectral Power Distribution, by weighting it for the photopic curve (official way of doing it is integrating it along all the waveband 360-830nm, as shown in Stephen's excellent post).

How the device obtain that emission falls apart of LER.

As Stokes losses are inherent to phosphor conversion, they limit the max efficiency achievable by this technology. As phosphor convert short wavelength photons to lower energy longer wavelength photons, the remaining energy is dissipated as heat in the phosphor. But this relates to how the LED achieves its light emission, so it don't affect LER, which is determined solely by the final SPD: phosphor conversion limits the max efficiency of the LED. If you use phosphors, you cant get an efficiency of 100% converting input energy to output light.

So all the others causes that reduces efficiency, won't show in the LER figure, but in the lm emission.



uk_caver said:


> And effectively, even if the maximum theoretical luminous efficacy of a phosphor-based white LED might be ~350lm/W, such a device would have to have a 'more subjectively efficient' spectral output than current LEDs do, and so the 350lm/W figure is not useful in calculating the gross efficiency of a current LED, but we should probably be using a figure of ~300lm/W?



I have measured some warm white LEDs with LER of 340lm/W. But obviously they correspond to greenish tones and a worse chromatic rendering than tones with lower LERs. Warm whites without excess red can achieve a LER of 350lm/W, maybe more yet. Maybe with not good color rendering, but say, for example, a CRI of 60-65 (BTW, I don't like CRI metric very much, but lets take it as a general figure for chromatic rendering). Useful for many applications, unacceptable for others.

On the other hand, the blue peak of white LEDs is usually at 450-460nm. It is a range with a low photopic coefficient (32-41 lm/W), so the larger it is, the lower the overall LER. Thus, coolwhite tones usually have lower LERs than warm whites. Extreme case are violet (instead of blue) chip phosphor based LEDs. For example, the GE VioLED of 3000K has a LER of 267,38lm/W (380-780nm) (267,30lm/W for 400-700nm) due the peak is at 404nm (which only produces 2.89lm per optical watt).

But in general, for good quality color rendering LEDs, the range 270-330 lm/W is the appropriate. I agree 300lm/W is a good average, although for warm tones it can be higher and still have good color quality.

For no phosphor based LEDs (for once the green-yellow gap has been solved), theoretically is possible to obtain LER in excess of 400lm/W with 4 band LED. Actually, attending to just theory, it would be possible to build "white" sources to near 500lm/W of LER (but forgiving decent color rendering).


----------



## Kinnza

MikeAusC said:


> No, LER is Luminous Efficacy of Radiation. In 300lm/W, the W is optical watts, not electrical input watts ("consuming 1 watt"). It's a measure of how much of the radiated power is visible, not a measure of how well electrical input gets converted to visible radiation.
> 
> http://en.wikipedia.org/wiki/Luminous_efficacy
> "Luminous efficacy of radiation measures the fraction of electromagnetic power which is useful for lighting. It is obtained by dividing the luminous flux by the radiant flux."
> "In SI, luminous efficacy has units of lumens per watt (lm/W). Photopic luminous efficacy of radiation has a maximum possible value of 683 lm/W, for the case of monochromatic light at a wavelength of 555 nm (green)"



First, Wikipedia have many errors. Specifically, that entry is clearly written by someone who knows about photometry, but not about radiometry.

If LER is the quotient of Luminous Flux by Radiant Flux, how could it says anything about the fraction of electromagnetic power useful for lighting? No sense, man. If you divide lm emitted by watts emitted (radiant flux), you obviously get how many lm per emitted watt. Meaning, the efficacy of the spectrum to produce light (as humans sense it, lumens). Hence its name, Luminous Efficacy of Radiation.

Thus effectively, if you have a SPD of pure 555nm light, which produces 683lm for each optical watt, you get a LER of 683lm/W. And that refers exclusively to the SPD of the source.

As you can see in Stephen's quote, the *official* CIE procedure to integrating LER is between 360 and 830nm. So hardly LER can inform about the fraction of electromagnetic power used for lighting when its officially calculated for just the visible range.

The first paragraph of the entry is more clear and accurate:

"The luminous efficacy of radiation describes how well a given quantity of electromagnetic radiation from a source produces visible light: the ratio of luminous flux to radiant flux.[5] Not all wavelengths of light are equally visible, or equally effective at stimulating human vision, due to the spectral sensitivity of the human eye; radiation in the infrared and ultraviolet parts of the spectrum is useless for illumination. The overall luminous efficacy of a source is the product of how well it converts energy to electromagnetic radiation, and how well the emitted radiation is detected by the human eye."

It is the defined as "Overall Luminous Efficacy" that includes all the electromagnetic spectrum. It is not useful outside Astronomy. Actually, when it is required to know it, it is descomposed on 2 useful terms: LER multiplied by the quotient of visible light energy (360-830nm) by total radiant flux (0-infinite).

But for the purpose of the thread, to know the energetic efficiency of LEDs, only LER, as it is defined by CIE, is useful.


----------



## slebans

Kinnza said:


> On the other hand, the blue peak of white LEDs is usually at 450-460nm. It is a range with a low photopic coefficient (32-41 lm/W), so the larger it is, the lower the overall LER.


 
Hi Kinnza,
Last year I built a spreadsheet to help me visualize the luminosity function. My calculated values do not agree with your post. At 450nm I get a value of 25.954 lumens. I've only checked a handful of values with other sites and so far my tables have been accurate. Can you have a look and let me know if my numbers agree with yours.

I've attached an image below my sig of the relevant data.

Thanks!

Stephen Lebans


----------



## uk_caver

Kinnza said:


> I have measured some warm white LEDs with LER of 340lm/W. But obviously they correspond to greenish tones and a worse chromatic rendering than tones with lower LERs. Warm whites without excess red can achieve a LER of 350lm/W, maybe more yet. Maybe with not good color rendering, but say, for example, a CRI of 60-65 (BTW, I don't like CRI metric very much, but lets take it as a general figure for chromatic rendering). Useful for many applications, unacceptable for others.


 I'm curious as to why it often seems that for a given LED type, LED vendors have the higher lumen bins available at the cooler end of the spectrum.

[edit]
Is it just that even though the LER is higher, that's more than compensated for by other losses, giving a lower overall efficacy.
If so, is that unavoidable in practice, or something that's theoretically alterable?
[/edit]


----------



## MikeAusC

"the theoretical ηL limit of white LEDs (260–300 lmW−1)."

"White light emitting diodes with super-high luminous efficacy"
Nitride Semiconductor Research Laboratory, Nichia Corporation, 491 Oka, Kaminaka, Anan, Tokushima 774-8601, Japan

http://iopscience.iop.org/0022-3727/43/35/354002/

So if an LED was producing 300 lm/Watt of electrical input, it would be working at the maximum theoretical limit of white LEDs !!!!


----------



## uk_caver

MikeAusC said:


> So if an LED was producing 300 lm/Watt of electrical input, it would be working at the maximum theoretical limit of white LEDs !!!!


I suppose it's worth clarifying whether 'white LED' covers _any_ white LED, or just blue+phosphor ones.

And also even if the theoretical maximum efficacy of blue+phosphor was ~300lm/W, that could presumably square with a potential ~350lm/W LER.


----------



## Kinnza

Hi, Stephen! Thank you for pointing me to that discrepancy. 

I checked it, and your values are correct, it is the official Photopic function of CIE, V(λ), approved on 1931 (built in 1924). It is the one used to calibrate all photo detectors, so it is the adequate to calculate LER (as lm emission are calculated according to it).

I realized I was using the Vm(λ) CIE function, approved in 1988 (built in 1978). Actually, it represent more accurately the human response, which is higher in the blue region, below 460nm, than the original function. But I have to change it in my spreadsheets, as it overestimates LER of short wavelengths (from 460nm and up, they are identical). Due all the standards were built for the older, CIE refused to change the standard and just endorsed the new function as more "physiologically accurate" for research purposes.

But your values, still being the official ones, have a problem. They are the values for the peak wavelength, not for the range until the next peak. LER value for 450nm is just for 450nm, while 451, 452, 453,454 have higher values. As you get SPDs usually in wavebands, if you use a 5nm wavebands, as 450-455nm, 455-460nm, etc, results are less accurate. For balanced spectrums, probably difference will be little, as in the blue underestimate lm and in the red, overestimate them.

This is the table with 1nm figures. If you use 5nm wavebands, it is better to average them in groups of 5. But anyway, for LEDs is good to perform calculations on 1nm. They are just the coefficient for each nm, to obtain LER you need to multiply it by 683.002 (I had V(λ) in text format, now I have to import it to Excel)

Wavelength V(λ) 

360 0.0000039170000 361 0.0000043935810 362 0.0000049296040 363 0.0000055321360 364 0.0000062082450 365 0.0000069650000 366 0.0000078132190 367 0.0000087673360 368 0.0000098398440 369 0.0000110432300 370 0.0000123900000 371 0.0000138864100 372 0.0000155572800 373 0.0000174429600 374 0.0000195837500 375 0.0000220200000 376 0.0000248396500 377 0.0000280412600 378 0.0000315310400 379 0.0000352152100 380 0.0000390000000 381 0.0000428264000 382 0.0000469146000 383 0.0000515896000 384 0.0000571764000 385 0.0000640000000 386 0.0000723442100 387 0.0000822122400 388 0.0000935081600 389 0.0001061361000 390 0.0001200000000 391 0.0001349840000 392 0.0001514920000 393 0.0001702080000 394 0.0001918160000 395 0.0002170000000 396 0.0002469067000 397 0.0002812400000 398 0.0003185200000 399 0.0003572667000 400 0.0003960000000 401 0.0004337147000 402 0.0004730240000 403 0.0005178760000 404 0.0005722187000 405 0.0006400000000 406 0.0007245600000 407 0.0008255000000 408 0.0009411600000 409 0.0010698800000 410 0.0012100000000 411 0.0013620910000 412 0.0015307520000 413 0.0017203680000 414 0.0019353230000 415 0.0021800000000 416 0.0024548000000 417 0.0027640000000 418 0.0031178000000 419 0.0035264000000 420 0.0040000000000 421 0.0045462400000 422 0.0051593200000 423 0.0058292800000 424 0.0065461600000 425 0.0073000000000 426 0.0080865070000 427 0.0089087200000 428 0.0097676800000 429 0.0106644300000 430 0.0116000000000 431 0.0125731700000 432 0.0135827200000 433 0.0146296800000 434 0.0157150900000 435 0.0168400000000 436 0.0180073600000 437 0.0192144800000 438 0.0204539200000 439 0.0217182400000 440 0.0230000000000 441 0.0242946100000 442 0.0256102400000 443 0.0269585700000 444 0.0283512500000 445 0.0298000000000 446 0.0313108300000 447 0.0328836800000 448 0.0345211200000 449 0.0362257100000 450 0.0380000000000 451 0.0398466700000 452 0.0417680000000 453 0.0437660000000 454 0.0458426700000 455 0.0480000000000 456 0.0502436800000 457 0.0525730400000 458 0.0549805600000 459 0.0574587200000 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## Kinnza

uk_caver said:


> I'm curious as to why it often seems that for a given LED type, LED vendors have the higher lumen bins available at the cooler end of the spectrum.
> 
> [edit]
> Is it just that even though the LER is higher, that's more than compensated for by other losses, giving a lower overall efficacy.
> If so, is that unavoidable in practice, or something that's theoretically alterable?
> [/edit]


 
You are right, the warmer the tones, the lower the efficiency. The loss in efficiency is higher than the gain due higher LER, thus warmer colors have lower efficacies than coolwhites.

As both uses blue chips of similar performance, the problem is with the phosphor. 

First because warmer tones means converting photons to longer wavelengths, thus higher Stokes losses. 

Second, they need to convert more blue light (check how blue peak in warm tones are way smaller than in cool ones), achieved by using a larger concentration of phosphor. This result on more losses at the phosphors (they are not 100% efficients) and more light scattering that reduces light extraction efficiency (a larger fraction of light emitted is unable to leave the package).

Manufacturers are working hard on these issues, and each day differences in efficacy between cool and warm whites are smaller.

Better phosphors, with higher quantum efficiency, reducing losses at the phosphor layer (actually, now there is little to improve here).

New ways of using the phosphor, reducing scattering and reduction in light extraction efficiency. For example, remote phosphor systems and phosphors embedded in ceramics.

Another trick which is resulting very useful is using red (or red and yellow,as Cree) LEDs along the white ones. This allows to use cooler LEDs and still get a warm tone, without reducing the efficiency of the white LED. But this trick is basically only valid at the whole luminary level. Some manufacturers (Epistar, at least, reported it for no far production) are trying to implement it into a single LED. Most of the top performance warm white luminaries in the market uses this solution.


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## Kinnza

MikeAusC said:


> So if an LED was producing 300 lm/Watt of electrical input, it would be working at the maximum theoretical limit of white LEDs !!!!


 
Great! You finally got it.

If a LED has a LER of 300lm/W, meaning that emits an spectrum able to produce 300lm per optical watt, at 100% efficiency it will emit 300lm/W.

Im glad you understand it, now we can move along and maybe look for alternative efficiency measurement systems as the one you tried. Always is good to have alternative ways of measuring things.


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## slebans

Kinnza said:


> .
> 
> This is the table with 1nm figures. If you use 5nm wavebands, it is better to average them in groups of 5. But anyway, for LEDs is good to perform calculations on 1nm. They are just the coefficient for each nm, to obtain LER you need to multiply it by 683.002 (I had V(λ) in text format, now I have to import it to Excel)


 
Thank you very much for posting the table, Kinnza. When you get it imported into Excel would you mind Emailing me a copy? 

Finally, I was using the value of 683 so with your correction I should be 100% accurate.

Stephen Lebans


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## bbb74

Wow I think I'm getting it, that didn't hurt at all  Actually there's probably a few things I don't quite get...

Lets say I invented some new type of device, that magically converts 100% of its input electrical energy into pure 555nm light. Would it produce 683lm per watt of input electricity? And its LER would be ... also 683lm/W? If the device was only 50% efficient, it would produce 683/2 lm/W but the LER would still be 683lm/W?

Lets then say I invented some new type of device, that magically converts 100% of its input electrical energy into "white" light. Would it produce almost ~400lm per watt of input electricity? I know it depends on the cct but what would its LER be roughly in this case? The same? I get stuck here...

The maximum theoretical efficacy of leds of 300lm/W I see about - what is that based on? The fact that LEDs are designed in a certain way (eg blue+phosphor) and there are certain known losses that must occur (eg Stokes) that simply cannot be avoided by improved technology?

Last question ... Kinzza you seem to be able to do some calculations regarding the junction temperature. I was wondering if you could guestimate (or guess) the junction temperature in real world led torches if you could measure the temperature of the body? Eg. Tk41, XM-L at 10 watts. Is the junction temp going to be "nice" down around say 80 degrees, or is it going to be more likely cooking up around the 150 degree mark?

Based on what you've been saying, a moderately driven XM-L has about 30% of input current wasted as heat, but at 3A it will be more like about 50% wasted as heat, right?


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## slebans

uk_caver said:


> I suppose it's worth clarifying whether 'white LED' covers _any_ white LED, or just blue+phosphor ones.
> 
> And also even if the theoretical maximum efficacy of blue+phosphor was ~300lm/W, that could presumably square with a potential ~350lm/W LER.


 
It is not accurate to state that the theoretical maximum efficacy of a white LED is 300lm/W. The statement must be qualified concerning the specific attributes of the LED and its spectral output. 

With regard to the document Mike quoted, please note the research was performed in 2009. There are also several issues with the document that are probably related to the translation of the source Japanese document and the desire to protect the company's IP. I am not questioning the science behind the document -those scientists have forgotten more than I know about LEDS.

Stephen Lebans


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## uk_caver

bbb74 said:


> Lets say I invented some new type of device, that magically converts 100% of its input electrical energy into pure 555nm light. Would it produce 683lm per watt of input electricity? And its LER would be ... also 683lm/W? If the device was only 50% efficient, it would produce 683/2 lm/W but the LER would still be 683lm/W?


Yes - the LER is effectively a 'subjective colour efficiency' which doesn't depend on how good the device is at turning electricity into radiation.



bbb74 said:


> Lets then say I invented some new type of device, that magically converts 100% of its input electrical energy into "white" light. Would it produce almost ~400lm per watt of input electricity? I know it depends on the cct but what would its LER be roughly in this case?


The LER in that case would depend only on how 'efficient' the spectrum it produced was - if you like, on the precise kind of white the device produced.

For comparison, it was stated earlier that 'kind' of white light typically produced by current cool white phosphor LEDs has an LER of ~300lm and that by warmer LEDs a bit more, and that doesn't depend on the basic electricity-light power conversion efficiency of the devices.
*Taking those figures as correct for the sake of argument*, they'd mean that if you could produce a new device that was 100% efficient at converting electricity to light *and* it produced an output spectrum the same as a cool white LED, you'd have an overall luminous efficacy (electricity-to-lumens) of ~300lm/W, or higher if your device mimicked the spectrum of current warmer white LEDs.

If you could make a device that produced an 'efficient white' spectrum rather than mimicking the spectrum of phosphor devices, potentially you could get 400lm/W.



bbb74 said:


> The maximum theoretical efficacy of leds of 300lm/W I see about - what is that based on? The fact that LEDs are designed in a certain way (eg blue+phosphor) and there are certain known losses that must occur (eg Stokes) that simply cannot be avoided by improved technology?


If someone's stating an maximum theoretical efficacy of 300lm/W for a particular device, that will be based on the unavoidable losses in that kind of device and the LER of the light such devices produce.


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## Kinnza

Any given spectrum produces more lumens (higher LER) as larger is the percentage of green light related to the full visible spectrum. On the opposite, it has lower LER as higher is the emissions in blue and red, and especially if such emissions are deeper on both ranges (shorter wavelengths for blue, longer for red).

You can check in the photopic coefficients above how dramatic is the drop the conversion of radiant energy to lm:

While 1W of 555nm (green) produces 683lm, near it, at 600nm (yellow-orange) produces 431lm and at 510nm (cyan) 344lm. And once we enter in blue and red, it drops a lot more yet. For red, 260lm (620nm), 148lm (635nm, typical peak of red LEDs) and 42lm (660nm, deep red). Worse yet for blue, 62lm (470nm), 26lm (450nm, most used peak for white LEDs) and 16lm (440nm) (figures are rounded)

So generally if an spectrum has a broad and large peak around green (plus cyan, yellow, some orange) and little on blue and red (and uses near bands of them), it may obtain "white" light (with an ugly tint, of course) at very high LER (over 450lm/W).

Current LEDs using blue chips and phosphors are somewhat limited about the wavelength range that can be used. Coupling with most phosphors is not good over 460nm, so manufacturers need to use 445-455nm peak (give or take) blue leds (longer ones often are used for warmer tones), and this limits the maximum LER achievable by this technology. But I believe is excessive to put the theoretical limit of this technology at 260-300lm/W of efficacy. But for sure that the authors of that article knows their stuff, so probably I'm unaware of more things that limits it (not related to LER, but to efficiency, probably limits on the external quantum efficiency)

But for sure that statement refers to phosphor based white LEDs. I have read papers designing 4 color bands LEDs surpassing 400lm/W of LER and CRI close to 80. Not using phosphor allows to use longer wavelength blue LEDs and shorter wavelength red LEDs. But this way of obtaining white, apart of the problem of different behavior with changing temperature (color stability problems) for the moment is limited due the low efficiency of green and yellow LEDs. Recently Ive seen proposals of building such 4 band LEDs but using a blue and red one, and two blues with phosphor fully emitting in green and yellow (actually, there is some models of yellow-amber LEDs in the market that are blue chips with phosphor, as the Rebel). But anyway, 4 bands technologies need large improvements on the efficacy of green and yellow LEDs to compete with the simpler blue+phosphor scheme (or the blue with phosphor+red, currently the solution with higher efficacy).

PS: Stephen, I'll do that Excel next week most probably. As soon as I have it completed, Ill send you it. Im thinking on improving it adding the color matching and color coordinates function, so the spreadsheet calculates x,y,z coordinates, u' and v', CCT, CRI and maybe some other color indexes.


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## slebans

Kinnza said:


> PS: Stephen, I'll do that Excel next week most probably. As soon as I have it completed, Ill send you it. Im thinking on improving it adding the color matching and color coordinates function, so the spreadsheet calculates x,y,z coordinates, u' and v', CCT, CRI and maybe some other color indexes.


 
That would be perfect, Kinnza. I worked through the math last night to convert the color coords to CCT but then got stuck trying to convert to (u,v) and onto to CRI. A couple of hours of math and color theory - my head started to ache!

Stephen Lebans


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