I would like to learn how to calculate the size (inc. thickness) of an aluminum / metal sheet that I will need given an average of a high-power COB LED chip in order to cool it.

Basically, I'm working on a DIY lighting project on an indoor open space.
I will be using a non-brand / eBay / "Alibaba" LED chip (but a "better" one with larger chips on the board).
They do not have exact specs but I believe I can roughly estimate wattage that will be dissipated.

So.... If I know the wattage, how can I get a rough idea of aluminum sheet I will need?

Eg., (please correct if I'm deeply wrong):
Chip: 75w (18-22v, 3500 mAh) red LED chip, powered at 20v, 1800 mAh == 36 w
With the advantage of driving at lower current, let's say we get max. %25 efficiency and 27w needs to be taken care of.​

If my constraint is the thickness of aluminum sheet (example 5mm or 0.2 in)..
.. How can I calculate the surface area I should have?

There can and probably will be slight breeze through to a fan passing over it.

Aluminum sheets I can buy likely to have a density of about ~2.7 gr per cm3.

For 5,m thickness I can buy either "6082 T651" or "5083 O" type.
For 3mm thickness I can buy 1050A or 5251 H22 type aluminium.

P.S. I really do not want to skip hunting on eBay or around the neighbourhood to find cheap but nice heat sinks. A sheet would work much better for my project.

Thanks!

 

There are many on-line heat sink calculators.
Read the Wiki

 

MikeML, thank you. Would you mind linking to a SINGLE WEB PAGE that helps you to calculate ALUMINUM SHEET SIZE to dissipate the heat from COBs?
The only relevant discussion I found was here where somebody mentioned in2 per C.

 

Flat plates don't make very good heat sinks.
My rule of thumb calculator is:

Temp rise in C=50/ Sq. Rt of area in cm.

 

Why square root, without taking into account volume and thickness of the material which has a huge impact?

 

Not so much. No matter the volume it will eventually reach the final temperature limited by the surface area. It does matter if the size gets out of hand and you can't conduct the heat throughout the plate.

 

Given that the sheet will be radiating out heat on its own and with the fan on top as I mentioned in the question, I highly think the volume is the key here.

Isn't here anyone who knows about mounting hot stuff on aluminum sheets?

 

Eg., (please correct if I'm deeply wrong):
Chip: 75w (18-22v, 3500 mAh) red LED chip, powered at 20v, 1800 mAh == 36 w
​

​

LEDs are driven by current and you must have a voltage source higher than the max forward voltage (Vf).. So you cannot reliably drive an LED with a 18-22Vf with only a 20V source.
Yes it will be sufficient if the Vf stays at 18V.. but if the real Vf is 22V and your source is 20V it won't turn on.
You would want a driver with a 1800ma output current and capable of greater than 22V DC output voltage.

Seeing as very few if any cheap "ebay" LEDs actually state a thermal resistance for the COB package itself there isn't much you can do to calculate accurate numbers.


And with heatsinks is all about the surface area as thats whats exposed to the air (convection happens at the surface). Yes there is conduction through the body but for this purpose exposed surface area is king.
Hence why heatsinks have fins/pins,etc... they serve to increase surface area.
There is alot more to the formulas than you would think..
BUT One can "approximate" the thermal resistance of aluminum plate using the following.. (see Ron above)
Rha=50/(sqrt A)
Where A is in cm^2

and the lower the junction temp the longer the life of the LED..
I've build numerous "high power LED" fixtures.. many with 3W LEDs and a few with 50W and 100W LEDs.
https://dl.dropboxusercontent.com/u/9726802/Reef/IMG_20121215_164729.jpg

and here is the best tip for you (and the most cost effective).. Assuming using a "high power LED" under 50W dissipation. Just use an old fan cooled duron heatsink from a computer.. It works perfect and keeps junction temps low and if you don't have one laying around the house you can pick them up for as little as $5 USD with the fan.
I've used them many times and you simply drill/tap the heatsink..slap on some thermal grease..screw the LED down and voila.. It will easily keep junction temps under 60degC in a 25deg ambient.

 

Thank you @mcgyvr for this detailed explanation. I hope you can excuse the errors I made in the example I tried to give.
I was just trying to state that the LED would be underdriven via limiting the current and voltage would be on the lower side of the range of what the LED uses.

Anyway, after reading your point I ordered 2 heatsinks. One copper piped aluminum with a fan on (found second hand cheap, prolly will replace the fan just in case it might fail on the way) and a very large, anodized black one with spaced fins (weighs about 3 pounds! also used found cheap).

The large heat sink says it has thermal resistance of about 0.5 C / W.

Does it mean a 100 w LED, ran at 60 W would increase its temp. to ambient + 30 C ?

For cheap eBay LEDs, what is really the acceptable heatsink temp. besides the junction?

PS. I ordered an IR temp. measuring device. Will it work to measure junction temp or would it be inaccurate due to LED possibly radiating more IR?

 

yes If all devices thermal resistances are know then you simply add them up and multiply by the wattage.. and that number should be "close".. but again real testing is required.
I stated above they typically don't post the junction to case ambient of the LEDs on ebay.
But you know one part so you know its going to be at least 30degC over ambient.
Here is some good basic information about that
http://sound.westhost.com/heatsinks.htm

I try to keep junction temps as low as possible.. At the very least keeping junction temps under 85deg C.. but I typically try for 65degC max for longer life. 85degC is typically where they do lifespan testing and most specify 50K hours if junctions are kept below 85C

The problem with IR temp devices is to be really accurate they require you to input the emissivity value of the surface being measured and having the wrong one can throw the numbers off (sometimes quite a bit). Thats why I always use a thermocouple for temp measurements as it doesn't come into play there. IR is really only good for approximate temps.