LEDs waste 75% as heat

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MikeAusC

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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

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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!
 

MikeAusC

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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

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. . . .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.
 

jtr1962

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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.
 

beerwax

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

MikeAusC

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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.
 

Oznog

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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.
 

beerwax

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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.
 

MikeAusC

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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.
 

Oznog

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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.
 

onetrickpony

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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.:ohgeez:
 
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onetrickpony

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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.
 

bbb74

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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).

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...
 

jirik_cz

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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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