Light source efficiency math?

LEDs are (thermally) maybe producing 90% heat. So, if you feed it

1W, plan on heatsinking... one watt.

If they've got so much more efficient that the current number is really 85% heat, you're still good to go with a 1W heatsink.

-----------------

** Not the important question.

** OK, count on the heatsink having to remove all the power with a temp rise not more than 25C above ambient - assuming the ambient is never more than 50C.

The crucial thing is to keep the LED chip cool, never more than 75C.

Calculations are impossible, you must do real tests on a real set up.

I hope you have a K-type thermopcouple handy.

..... Phil

mike wrote on 11/13/2017 12:00 AM:

I don't think it is that complex. If an incandescent bulb is only 5% efficient and an LED bulb uses about 4 times less power for the same light... that means the LED is producing about a quarter of the waste heat as the incandescent bulb. Isn't the rest in the noise? Do you really need to know to some small percentage?

--

Rick C 

Viewed the eclipse at Wintercrest Farms, 
on the centerline of totality since 1998

For monochromatic light, determine the eye sensitivity for a specific wavelength, such as 683 lm(W @ 555 nm (yellowish green).

If a green LED produces 170 lm/W, the efficiency is 25 %

For white sources, depending of your definition of "white" light, the

100 % efficiency is about 200-300 lm/W.

So, if you input 1 W of electricity, 33 % to 50 % is visible light, the rest is heat. thus, the heatsink needs to dissipate 500 -700 mW.

For long life time, run the LED at Imax/3, so for "1 W" (Imax=350 ma) LED run it at100-150 mA and you do not need any accurate measurements with any reasonable sized heatsinks.

An infinite heat sink will prevent the heat sink temperature from ever rising above ambient. You can cram an infinite number of watts into an infinite heat sink and not see any temperature rise.

"Carry Away" is not a thermodynamic option for moving heat. Your choices are conduction, radiation, and convection.

100 lumens per watt is a measure of luminous efficacy: Notice that the lumens are just the visible part of the spectrum and does not include UV (leakage from the UV LED), NearIR (near infrared), and FarIR (heat). Also notice that the definition of total power is rather vague and offers 3 different variations (see first paragraph).

Convert lumens/watt to what?

When you take a 1 watt white LED, and shove enough current into it so that the voltage across the LED times the current through the LED equals 1 watt, exactly 1 watt of "radiation" is produced. Some of this radiation is in the visible light range, while other parts are wasted mostly in the IR region. The luminous efficacy rating for the LED does not define the ratio of useful visible light to useless heat and invisible light.

What I suspect you need to do is determine what part of the spectrum produced by the LED is visible and consider the rest to be heat. I don't consider this a very useful calculation as it would probably suffice to simply assume that all the power going into the LED becomes heat and use that for heat sink calculations. The real numbers are less, but unless you're working with a tight thermal budget, it's not really worth calculating. For example, some NearIR and UV are blocked by the plastic lens, which converts those wavelengths to heat. You would need to add those to your heat sink budget.

Drivel: LED efficiency can exceed 100%: "Ultra-efficient LED puts out more power than is pumped in"

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

About 80% of the input power. The word you want is efficacy.

An optimum luminous efficacy would be about 700 lumens/watt. See

The highest efficacy so far is up around 200 just behind the low pressure sodium lamp and tetchy sulphur lamps.

More like 98%

Ballpark is that good LEDs are 10x more power efficient at generating light than an incandescent filament (which are lousy at it).

0.85 (but as low grade thermal heat) 0.14

I know the answer is complex and the problem is vague.

--
Regards, 
Martin Brown

Well, he already knows that his LED has a luminous efficacy of 100 lm/W.

What he wants to know is the efficiency, or, how many Watts are 100 lm of light (so he can calculate the efficiency).

How is that calculated? It appears it requires some estimates. Of course, the LED will output light and some other radiation, plus it will conduct heat to the heatsink. Apparently from the 100 lm/W efficacy it cannot really be told how much it actually radiates and how much (the remainder) it will have to get rid of by conduction. Because the number of lm has been calculated taking eye response into account (which isn't relevant for thermal calculation).

Not sure that's "possible", basically because it's hard to define what "ambient" is for an infinite heat sink. Using a 2D metal plate as an example, either the energy source is uniformly distributed, or it isn't. If it's uniformly distributed over the plate and you consider ambient to be the temperature "at infinity" then the plate can't be at a uniform temperature over its entire area because the heat equation doesn't admit steady-state solutions with discontinuities on the boundary.

If the energy being put in isn't uniformly distributed then the plate also can't be at a uniform temperature over its entire area because e.g. any circle you can draw around a point heat source on such a plate will have finite circumference, so only finitely much heat can be conducted across the boundary per unit time.

When you put your hand in front of say a 1 watt blue LED running at full power do the blue photons coming off it burn a hole thru your hand? No? Must mean that a negligible amount of the input power is being converted directly to visible radiation.

If it's a commercial LED that you can mount to a heatsink it should have thermal resistance specs and derating curves n stuff, just do the calculations as you would for any other semiconductor power device

Now, that's interesting. No indication that it can scale to useful levels, but interesting. Unless it's a pons/fleischmann folly.

You're overcomplicating things. Assume that the temperature at the LED is maintained at 25C by the action of the "heatsink". The whole point was to eliminate temperature from the equation and stick to the ratio of light output to power input.

Thanks. At least, those numbers correlate.

An infinite heat sink sure is gonna be complicated!

I think you're over-complicating things, though. The "wall-plug" efficacy of an average LED isn't stupendously greater than that of an average halogen bulb; LED bulbs consume less power for the same useful light output because their luminous efficiency in lumens/watt is stupendously greater. That is they both waste gobs of power as heat but the LED consumes much less power overall because it requires less overall power to function.

Sort of like how there are "0 calorie" sodas but artificial sweeteners don't usually have any fewer calories than regular sugar, they're just ~200 times and sweet so they can use 1/200th as much.

At high currents the diode's bulk ESR will completely dominate and that will be dissipated as heat in the heatsink, so design the heatsink similar to how you would design it for any other power semiconductor device

I've you'll accept a very large specific heat or heat capacity as an acceptable substitute for an infinite heat sink, it's easy enough to demonstrate that it's possible. Place a resistor that dissipates 1 watt into an Olympic size swimming pool. Measure the temperature of the water before and after applying 1 watts of heat. Unless you have an extremely sensitive thermometer, you won't see any change.

An infinite heat sink is assumed to also have an infinite thermal conductivity. However, even a very large heat sink, such as the Olympic size swimming pool will eventually show a temperature gradient if you're willing to wait a lifetime for it to be measurable.

Huh? Just assume that the test medium has a sufficiently low thermal conductivity so that it doesn't have a huge effect on the temperature rise of the infinite heat sink. A vacuum will do nicely. Dry air, less so, but quite effective. Still atmospheric air is probably good enough. However, that just takes care of thermal conductivity, and doesn't do much to affect radiation and convection. Incidentally, initial ambient temperature is the temperature of the test medium (air).

Very true if you want to quibble over boundary conditions between the heat source and the heat sink.

Anyway, throwing in an infinite heat sink into the original question on LED efficiency and efficacy is not going to produce a usable answer or calculation, nor answer the original question.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

Should be "luminous efficacy", rather

Yes, as you say the answer to the original question is "If you wanted to cool a glowing hot resistor you'd treat it as a hot resistor, if you want to cool a hot semiconductor diode you treat it as a hot semiconductor diode"

Not negligible any more unlike the early LEDs. Surface brightness of a modern high efficiency LED is broadly comparable with the photosphere of the sun at its specific emitting wavelength. They come with warnings not to stare directly at them on the higher power devices.

Some of the newer UV ones might give you a nasty UV burn as might the purple diode pumped frequency doubled laser at 400nm.

--
Regards, 
Martin Brown

That which is not easily understood, it most easily described as magic.

These should help: which explains that the LED was acting as a horribly inefficient Seebeck Effect thermoelectric generator partly powered by the room temperature heat.

I once tried to do something similar with a tiny LED and a large copper plate soldered to each lead. Two glasses of water, one steaming hot and one full of ice cubes, are placed next to each other with some foam poster board in between as thermal insulation. The LED and copper plates straddled the insulation. My hope was that the respective temperatures would conduct up the LED leads and produce some visible light. It did somewhat function, but required a microscope to see the very dim light. The future for perpetual power LED flashlights, powered by room temperature heat, is not looking good.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558