Power LED Heatsinking.


Does anyone know a good rule of thumb for Power LED heatsinking?
(PLease, no guesswork ;) )

Basically, I would like to know the ratio of Watts-in to Watts-heat for the average modern large surface mount LED.
I can't find info anywhere, regardless of make. I'm Googled-out.

e.g. assume a rated 3W white LED . It gets hot. How do I calculate the heatsink rating? Does anyone know of a link anywhere that will give a reasonable guide of power/heat/light 'efficiency'?
I realise I could heatsink it based on Watts In = Watts heat.
Nothing wrong with belt-and-braces, but something a bit more 'accurate' would be really nice.


Senior Member
From the Philips site there is this. It is a two week trial.

And also this pdf

Haven't tried software or fully read pdf, so hope it isn'ta waste of your time.


That was a good find lanternfish.
Shame the models only cover Luxeons.

This is becoming a growing problem with LEDs. The technology is moving fast but the datasheets rarely contain sufficient information to make thermodynamic models and even if they did, the model would be obsolete by the time you bought your next LED:mad:

Short of that Dippy, I can only suggest that you take worst case and assume power in = heat dissipated. Always pays to have a little safety margin.


Thanks, all this sort of stuff is useful.

I've seen similar for other makes, but it is very specific for a specific part.

I was really after the answer to: "How much heat will an LED, say 3watt rated running at (elctrically) 3 Watts, chuck out?"

And I'm referrng to the various modern SMD power LEDs in various formats.
There are so many PCB/Heatsinking techniques.

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Good question Dippy.
How many watts of light does a 3W (in) LED put out?
I bet there is some correlation with lumens.
The datasheet will give wavelength and lumens (or maybe candles) so it should be possible to work it out but darned if I know how.


Senior Member
My understanding is that an incandescent light is about 3% efficient. LED lamps are supposed to be 10 times more efficient. That suggests 30% light and 70% heat but don't ask me for hard data.


This Canadian article suggests LED’s are up to 90% efficient.
“There are many advantages to LED lighting, the greatest being efficiency. Many LEDs offer 90 percent efficiency, compared with 5 percent for traditional lighting sources. This represents significant cost advantages to consumers and reduced greenhouse gas emissions that contribute to climate change.”

I have read that OLED’s are seeing a big jump and will soon approach 50% efficiency.

= = = = = = = = = = = = = = = = = = = =

Saw an LED based industrial light a few weeks back.
4 sub-boards each with an 8 x 8 array of 3Watt LED’s. LED’s only running at 2Watts each for long life. So that is 256 x 3W LED’s operating at 512Watts.
Total lamp face area was around 300mm x 300mm but the entire back of the light was a very impressive heat sink with a matrix of fins around 75mm in length The heat sink did not suggest only 50 Watts of heat needed to be dissipated even considering higher ambient temperatures.


Senior Member
This Canadian article suggests LED’s are up to 90% efficient.
"In January 2009, it was reported that researchers at Cambridge University had developed an LED bulb that costs £2 (about $3 U.S.), is 12 times as energy efficient as a tungsten bulb, and lasts for 100,000 hours." So I suspect that 90% efficiency is a bit high in reality.

There don't seem to be many people that will commit to how much heat will come off a 3W LED lamp. They all want to go into the number of lamps you will need to replace aver 10 years; or the efficiency of AC to DC power supplies for mains installations etc etc.

Yes, Dippy, I'm trawling (read: Googling) the internet for answers - mainly for my own interest.


I feel a figure of 90% is way, way too high. Thanks for looking.

You'll probably find the author (an ice-cream salesman during the day) is getting confused with efficiencies of LED power controllers.

I can't find any info whatsoever. ... but still searching.
It would certainly help Heatsink calcs.

And to anyone about to dabble: the heatsinking is very significant. They do NOT run cool as said by some (overpaid) marketing ninnies.
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Senior Member
Maybe 25% efficient.

You'll probably find the author (an ice-cream salesman during the day) is getting confused with efficiencies of LED power controllers.
My first thought too.

The following 2007/2008 document suggests 25% efficiency for high power LEDs, which is in the range of my original suggestion. LED Manufacturing Technologies 2008 may offer some credibility.

That suggests that 2.25W from a 3W LED will be heat. Of course, not all of that will be conducted out the back of the lamp to the heatsink, something that many people overlook.


Power LED efficiency

Well it seems that 25 to 30% efficiency is more in the realisticv realms:

From: http://www.ledinside.com/forum/viewtopic.php?f=2&t=1780

For the high power LED over 1w,the electro-optical conversion efficiency is about 30 percent at present, the balance translated into heat while the chip’s size is 1mm * 1mm ~2.5mm * 2.5mm.That is to say, the power-density of chips is very large. Different from traditional lighting devices, there is not infrared spectrum included in white light-emiting spectrum of LED. so the heat cannot be released relied on radiation. Then, how to improve ability in heat-elimination will be one of the complex problems which decide if high power LED can be put into industrialization or not.

From: http://www.electronicsweekly.com/Articles/2008/11/20/41947/LED-technology-White-LEDs.htm
Reality checking
Firstly, firms which claim 90 per cent efficiency from LED light sources are making it up. Even the latest ones convert far more electricity to heat than they do to light.
Next, according lighting industry experts, LEDs will remain expensive compared with light bulbs and florescent tubes. And whilst they will increasingly appear in homes and offices, they will almost certainly will not replace florescent tubes in office lighting.
Florescent tubes, at 100 lm/W for the best fittings, are equal in electrical efficiency to the best LEDs

from: http://www.coolon.com.au/downloads/ims.pdf

Thermal management
The tendency for LED die temperature to rise is related to the mechanisms by which energy leaves the chip. For example, even though up to ninety percent of the electrical energy supplied to a red AllnGaP LED will be converted into visible light, only part of this light energy can be ejected from the chip. The remaining energy, which cannot leave the chip, will be converted into heat.

Not quite sure how visible light gets back to being heat again (particularly since there is mention in other articles that there is no IR compnent in the light). :confused:


Nice finds Westy. Thank you.

Even though I usually baulk at 'expert' info on Forums, the quotes you show from that Forum (in your first link) look realistic.

"Firstly, firms which claim 90 per cent efficiency from LED light sources are making it up. Even the latest ones convert far more electricity to heat than they do to light."
- now you see why I spluttered at the info in your previous post :)

"Not quite sure how visible light gets back to being heat again.."
- if you had to study Quantum and Nuclear Physics, like I had to, you'd see why.

Thanks again.

Oh btw.
I emailed 5 Tech Supports from manufacturers of Power white LEDs.
I'm brave you see.
I received 5 email replies.
They're ALL out of their bleedin' offices on holiday !

I will put a Fiver on either:
1) Please see our App Notes.
2) Don't know or no reply.

If only they were all as good as Microchip and AD.


Thanks iPete.
Thread went over a page so I missed it.
So, I think I can assume 25% light 75% Heat would be realistic (currently).

Obv I wil have to rely on the package heatsink pad and arguably (for the pedants) some heatsinking via 'pins'.

I honestly think that the necessary heatsinking is undervalued in some designs. Lifetime , brightness and brightness degradation versus temp and time is very important.

Anyway, now I have some figures to work with.
Greatly appreciated.


Yes, it would make a nice school experiment.
Sadly, I haven't got the time or equipment to do a study.

Manufacturers have the time, money and incentive to produce 'efficiency' values (as opposed to 'efficacy') but seem reluctant for some reason.


Senior Member
from wikipedia

"The highest efficiency high-power white LED is claimed[27] by Philips Lumileds Lighting Co. with a luminous efficacy of 115 lm/W (350 mA)

Note that these efficiencies are for the LED chip only, held at low temperature in a lab. In a lighting application, operating at higher temperature and with drive circuit losses, efficiencies are much lower. United States Department of Energy (DOE) testing of commercial LED lamps designed to replace incandescent or CFL lamps showed that average efficacy was still about 31 lm/W in 2008"


Senior Member
I make my own heatsinks for various stuff, including power LEDs, out of scrap 4mm aluminum plate. I never calculate the Watts-per-degree or whatever, just estimate and experiment. Worked fine for me so far. As I’ve said before, my rule of thumb is ‘If it’s too hot to hold your finger on, then it needs better heatsinking’. If I were forced to guess, I’d say about 50%. So 3W LED = 1.5W heat.


Thanks boriz, but it may have worked fine for you under your conditions.
At what (watt) power are you running and for what period of time and duty?

I'm afraid guessing is only OK if you aren't worried about lifetime and light degradation vs temp&time and continuous running.

Your thumbermometer can't tell you the junction temperature (Tj) and that is crucial for long-term performance.

Thumb tests between very warm and f hot may be fine for a MOSFET heasink, but LED performance varies hugely at raised temps. Don't take my word for it, read up on it.

A number of other 'rules of thumb' have indicated >6 sq inches area of heatsink per LED input watt.

A design data sheet by Lamina , and using the BL-21D0-0130 running at 4.4Watts, have calculated a heatsink TR of 2.2oC / Watt to be used for a Tj of 60oC @ ambient 25oC.
That's a hefty heatsink.
But that is for a specific product.

If we had an 'efficiency' value, then using the Manuf's Tj and TR values we could have a general reasonable equation for all LEDs.
Yes, I realise it depends on the mounting and ambient conditions etc. - but it would take most of the guesswork out of things.

I think most people will be surprised at the heatsinking required for Power LEDs.
A few watts doesn't sound much, but when it is generated in a tiny area/volume then the temps reached can be huge. But, hey ho, no-one studies Physics anymore :)

PS. Come on ec - I know you're seaching ;)
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Senior Member
watt sauce do you want on these bacon butties?


I've been searching, as a good excuse/distraction for not getting out of bed for the last half hour
quote below from....


following the instructions given below you find lumens per watt, the rest is wasted as heat?

Calculate LED Efficiency

Corresponding component datasheets contain the necessary information to calculate for LED efficiency. Start by locating the typical luminous output stated in lumens (lm). Some datasheets may state the candela rating (mcd), in which case a conversion is required. Continue by noting the test condition stated in milli-amperes (mA), for the corresponding typical luminous output. For example, a datasheet may state the typical luminous output is 45 lm at 350 mA. In this case, the test condition is equal to 350 mA. Move on to determine the maximum forward voltage drop at test condition. This information is normally located in tabular format within the first several datasheet pages. Note that the test condition stated for voltage drop must correspond with the test condition stated for typical luminous output. Moving on, multiplying the maximum forward voltage drop by the test condition. The result is power dissipation expressed in watts. Finally, multiply the typical luminous output by the reciprocal of calculated power dissipation, to determine LED efficiency expressed as lumens per watt. Higher numbers correspond with LEDs that are more efficient.


But that still doesn't give absolute efficiency!
It gives lumens/watt which is very often quoted in the datasheet anyway.
It is only a figure that can be used to detemine relative efficiency.

The LED holy grail is this:-
If an LED was 100% efficient at converting electricity into light (and putting it out in a useful manner), how many lumens would be put out for one watt of electricity put in?


Senior Member
we've (the father in law and i) have actaully looked at using an array of these hi powered leds, and you're actaully better off using a small array (10X10) of the 5mm high intensity leds in parallel and put 3.0v through them,
on the panels we've built so far we get no measureable heat in the leds , the mosfet, and thats on a 8 X16 that runs at just under 6 watts,

even better if you can get the surface mount high intensity leds with the 120 degree viewing angle,they are a little more expensive but they aparently put out the same as around 10 to 13 5mm leds, that we are using

i'll post a photo of the panel later when i get my camera back


put 3.0v through them ??? (they're CURRENT devices)
Mosfet runs at 6W ??? (meaningless without the FET loads and specs)

Sorry, but none of that makes any sense.
Many luxeon type LEDs are single die multi-junction requiring up to 15v before conduction even starts.

5mm LEDs cannot in any shape or form be compared to the HIGH POWER serious heat generating LEDs being discussed here.
There are LEDs available now which generate up to 30W as heat at the (very small) junction. This not only requires a good understanding of thermal heat transfer but also the actual thermal resistance characteristics of the heat-sink itself. Know the basic W/oC of the heatsink resistance to air is not good enough anymore. Local heat spots on the heatsink itself come into play with regard to longevity of the LED. This is an entirely different league.


Thanks DPG, but we're talking different leagues here as outlined by BB.

Very kind, but no need for photo of panel. I think I have a shrewd idea what a load of 5mm LEDs look like :)

A photo (unretouched) of it's illumination quality would be of more interest.
How about a self-portrait, after dark, with your panel 2m away?


"Appears" is the correct word Rick.

A quick check shows it is schoolboy Ohms law and Ivy Watts.

Aha BB. You've cracked it!!