LED light becomes extremely hot

I recently took apart a cheap ($7) LED light to use in a project. It was attached to a heat-sink that I removed and discarded, thinking nothing of it. Later when I turned on the light, I noticed that the circular piece of metal that had about 11 tiny LEDs on it was getting extremely hot! If LEDs are suppose to be super energy-saving, why are they giving off so much heat? - Why would you remove a heat-sink and think nothing of it? Why would a company spend money on incorporating components into their products that have no real use? Waste of money. Everything is there (or missing) for a reason (cost)... – dext0rb Jun 11 '12 at 17:16 Ever put your tongue on a 100 watt incandescent bulb? And it has a lot of surface air to dissipate the heat. – kenny Jun 11 '12 at 18:38 You also need to cool LEDs because they tolerate much lower temperatures than incandescent lights. – starblue Jun 12 '12 at 7:04 "Super energy saving" means that instead of making large amounts of waste heat, they only produce small amounts of waste heat. But, that is not the same as "no waste heat"! – gbarry Aug 17 '12 at 20:16 add comment 3 Answers Russell commented in a recent answer that efficiency(*) up to 20 % is possible. That's 4 or 5 times the efficiency of an incandescent bulb. So we're doing fine here. But of course, a 20 W LED lamp will still produce 16 W in heat (that bulb would have 19 W in heat). And if it comes with a heat sink, better leave it as it is; it's there for a reason. (*) I'm using the word "efficiency" here because that's the word used for it in everyday language. Szymon correctly points out that "efficacy" would be better, expressed in lumen per watt. Actual efficiency as Szymon uses the word is always 100 %, because that's the ratio energy out/energy in, due to the laws of thermodynamics. - Incandescent bulbs are actually much more efficient than LEDs. See my answer below. – Szymon Bęczkowski Aug 13 '12 at 22:16 @Szymon - By your reasoning they should be both 100 % efficient, since no energy is lost, thanks to the laws of thermodynamics. When we talk about a bulb's efficiency it's about the useful output, what we need it for. That's visible light. I don't need IR radiation on a summer day. You're the only one who includes IR in the efficiency. – stevenvh Aug 14 '12 at 4:48 But he also uses wrong metric to describe efficiency. If we take a white LED with no heat loss, all electrical power is converted into light, its luminous efficiency will not be 100%! It should max out around 50–60% as far as I can remember. There is no physical possibility of going better. If you ar interested in useful output to power ratio then use efficacy (lm/W). Efficiency people are talking about is a colloquial term for efficacy. – Szymon Bęczkowski Aug 14 '12 at 6:55 @Szymon - lm/W would indeed be better. But remember that the whole efficiency/efficacy debate was to convince Joe the plumber that LEDs or CCFL are better than incandescent. Most people will understand the word efficiency, especially when used comparatively, but have no idea what a lumen is, and don't want to hear about it, because too technical. (Though lm/W also gives comparative numbers.) – stevenvh Aug 14 '12 at 7:06 @Szymon - there's no sharp limit, but since the invention of integrals that's not needed. There's no sharp limit in the radiated lumens, either. If you want to talk about scientific, please note that the 5 % is given with 1 significant digit, not as 5.00 %. – stevenvh Aug 14 '12 at 8:51 show 3 more comments Your actual question is If LEDs are suppose to be super energy-saving, why are they giving off so much heat to which the simple answer is • "The very best LEDs that modern physics can make well enough for them to be able to be commercially used will dissipate 80% or so of their drive energy as heat. LEDs used in a cheap light can reasonably be expected to turn 90% to 95% of their drive energy into heat. " As I noted recently - the BEST LEDs are around 20% efficient in converting electrical energy to light. 20% is extremely good for an LED - just not extremely good in absolute terms. Most are less efficient. It was attached to a heat-sink that I removed and discarded, thinking nothing of it. The ability of a heatsink to work depends on it being in thermal contact with the heat. Alas, no matter how little you think of them they are still needed just as much if they were required in the first place. Removing them stops them form working as intended in all known cases to date. I recently took apart a cheap ($7) LED light to use in a project

A $7 light probably cost under$1.50 at a factory door somewhere in China. Makers of products that cost that little tend to include nothing that is not utterly essential. If it is there it will probably not work well if you remove it, whatever it is.

11 tiny LED

If the LEDs are white they are probably rated at 20 mA each and drop about 3V.
At full rated output power in = V x I = 3V x 20 mA x 11 LEDs = 660 mW. A circular piece of metal large enough to accomodate 11 x LEDs would probably get hot but not very hot i still air so the LEDs are probably being run at above rated current - this is very usual and would cause them to die rather quickly in normal use. They will die much much much more quickly without a heatsink.

LED efficiency:

• In some systems of units, luminous flux has the same units as radiant flux. The luminous efficacy of radiation is then dimensionless. In this case, it is often instead called the luminous efficiency, and may be expressed as a percentage.

A common choice is to choose units such that the maximum possible efficacy, 683 lm/W, corresponds to an efficiency of 100%.

The distinction between efficacy and efficiency is not always carefully maintained in published sources, so it is not uncommon to see "efficiencies" expressed in lumens per watt, or "efficacies" expressed as a percentage.

The luminous coefficient is luminous efficiency expressed as a value between zero and one, with one corresponding to an efficacy of 683 lm/W.

On that basis, the 20% efficient figure that I mentioned would be
683 x 20% = 137 lumen/Watt.

I have used large quantities of the utterly marvellous but relatively low power (50 mA max) Nichia NSPWR70CSS-K1 "Raijin" LEDs which have been on the market for about 3 years. These were well ahead of any other commercial offering when they first came out and even now are extremely good.

At "full power" of 50 mA they are rated at slightly over 120 l/W.
At 30 mA they are typically specified at 165 l/W (depending on binning).
165/683 ~= 24%.
At this point they are outputting about 20 mW of actual light energy = 20 x that of a Class 1 LASER diode. The light is however vastly more diffuse (80 degree radiation angle).
According to tests which (to my surprise) Nichia were kind enough to do for me, they actually constitute a formal minor eye hazard at the blue end of the spectrum, but you would have to be beyond perverse to look into the LED from close enough for long enough to acquire eye damage from a 80 degree 150 mW white LED.

The 24% efficiency is exceptionally high even by current standards - but whereas the Raijin is rated at only 90 mW input at 165 l/w and 150 mW in at 120 l/W, current 3W to 6W LEDs are quoting l/W figures in the 120-140 l/W range at significant percentages of their full power and over 170 l/W at low % of full power in some cases.

Such results are attainable only from top bins of top performance parts and my suggested 20% efficient = 136 l/W upper guideline is still in excess of what you would expect from the large majority of devices currently available. No name parts from Asia (as are liable to be used in the light in question) will almost always be well below this level.

Note that the lumen is defined with the human eye spectral response as part of the specification. Thus red LEDs are easier to produce with high l/W ratings than are white LEDs and deep blue and near UV LEDs have such low l/W ratings that they are specified instead in mW output.

The chart below (said to be from the U.S. Department of Energy) which shows a lumen per watt comparison of common light sources and white LEDs as well as the projection for the future.purports to show past and future efficincies of various light sources.

Note their "White LED lamp" line, which is about correct so far. These are handicapped by needing to operate from AC mains and provide power factor correction and in many cases dimmer compatibility. Energy conversion efficiency from mains to LED input is often under 80% and they usually use diffusers to 'improve' user perception of the light source so, if a diffuser reduces output by 10%, a delivered 80 l/w means the LEd will be 80/80%/90% =~ 110 l/W.

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Nice answer! Can you add a reference for the efficiency? –  clabacchio Jun 12 '12 at 9:37
@clabacchio - Added. No rest for the wicked, eh ? :-) –  Russell McMahon Jun 12 '12 at 13:55
it's just because I didn't expect such efficiency, it's very interesting to have it referenced. But the way it's defined it doesn't represent the amount of power dissipated in heat. –  clabacchio Jun 12 '12 at 13:56
Très artistique, that graph :-) –  stevenvh Jun 12 '12 at 13:56
@SzymonBęczkowski - There MAY be a type (or numero?) there (or not) but I do not believe I have mixed up the calculations. Note that (using the cited Wikipedia equivalents that I posted) I have explictly mentioned both efficincy and efficacy AND provided a means of conversion between the two for practical purposes. The aim here is to achieve practical usefulness and to avoid pedantic "correctness" which may leave people confused. | –  Russell McMahon Aug 14 '12 at 13:31

This is a common misconception about LEDs.

A lightbulb will convert 100W of electrical power into ~80W of electromagnetic radiation (20W lost in contacts, etc.). Relationship between em. radiation to electrical power is called efficiency. It is equal to 80% (!). However, most of this 80W is in infrared. The visible part of the spectrum will yield only 1700lm giving in total 17lm/W. Lightbulbs do not need heatsinks because they convert most of its input power to em. radiation.

A LED of 10W input power will generate only ~3W of blue photon radiation (30% efficiency, much less than in normal lightbulb). Remaining 7W are dissipated as heat. LEDs, contrary to incandescent, will not generate any infrared or UV. White LEDs have a blue chip inside and a phosphor cover, that converts some of the blue photons to yellow photons (loosing some more power in the process). Blue and yellow light mixed together gives white light. This yellow part of the diode spectrum lies almost in the maximum of the human eye perception. This is why tiny amount of radiant power can be seen as a bright light. Because some of the power was lost for converting blue photons to yellow photons, the final em. radiation will have only 2.5W of power. This is, however, translated to e.g. 1000lm of light, yielding 100lm/W. LEDs need heatsinks, because of their size and their temperature dependency. Typical high power LED chip is only 1–4mm². It is really hard to remove few watts of power through such a small structure. LEDs need to be grown on semiconductor structures that add thermal resistance to the heat flow path. In this example 7W needs to be dissipated through the structure. If the junction to ambient thermal resistance is equal to 10K/W we get a junction temperature rise of 70K. If the ambient temperature is 25°C we have 95°C. LEDs can only operate safely to around 125–150°C. Furthermore the higher the junction temperature the lower the light output.

So when you ask about efficiency, you actually mean efficacy (lm/W). LEDs are not very efficient but they have really high efficacies.

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-1 While your answer is technically correct, it is not useful. You're just being pedantic in your answer. Technically, Energy in always equals energy out-- regardless of what the system is. It doesn't matter if we're talking an LED, incandescent light, or a presidential candidate. What matters is the useful energy (or work) out; which in my examples is High, Low, and almost non-existent in that order. –  user3624 Aug 14 '12 at 0:30
@DavidKessner, It seems like a useful distinction. If an incandescent bulb's waste heat is radiated away as infrared photons, then no heat sink is needed to cool the bulb. If an LED lamp's waste heat is produced in the form of conducted heat in the chip, a substantial heatsink is needed to cool the LED. And the LED potentially is hotter to the touch than the incandescent bulb even though it's more efficient at generating visible light from electrical input energy. –  The Photon Aug 14 '12 at 0:58
@ThePhoton But that answer makes no reference to heat sinking. Further, that is not the reason why incandescent bulbs don't need heat sinks. Incandescent bulbs work better when hotter, while LEDs work better when cooler. Cool an incandescent bulb to the same temperature as an LED and it won't emit any visible light. Heat an LED to the same temp as an incandescent filament and it also won't emit any visible light. It is also not fair to say that the "bulbs waste heat is radiated away as IR photons". With an incandescent bulb there is much more "conducted heat energy" than an equiv. LED. –  user3624 Aug 14 '12 at 2:26
I have added heatink to my answer. You say incandescent bulbs work better when hotter but I would rather say incandescents work because they are hot. I am not trying to be pedantic in my answer but just scientifically correct. –  Szymon Bęczkowski Aug 14 '12 at 11:40
Your comments on light bulbs MAY be correct if they were thought of as an incandescent bulb BUT a "light bulb"'s job is to make visible light and that's the metric that matters. You allow an incandescent bulb's IR output to count in efficiency but discount heat produced by LEDs. | ALL electrical devices that do not send energy outside a boundary are 100% efficient heat producers within that boundary. LED, CFL, incandescent - all make heat at the end of the day (or night). . –  Russell McMahon Aug 14 '12 at 13:34
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