Sizing heatsink area

[Pedro Mello]

Hi guys,

I'm calculating the dimensions of an aluminium passive heatsink for my setup.
After some study I got that my cobs working with aprox 80w (480/6) ea would generate 20% of heat at the worst case temperature.
Then I made:

0,2*(80)=16 heat watts

and sized the dimensions of the heatsink.

I've reached that a 8'' x 8'' heatsink would handle each cob.

Is it make any sense?
Thank you.


PS: my setup is: 6 vero29 3500K B on a HLG 480w (54V) driver.

 

[nfhiggs]

After some study I got that my cobs working with aprox 80w (480/6) ea would generate 20% of heat at the worst case temperature.

A COB running at 80% efficiency? - not on your life. Realistic efficiencies are more in the range of 35-55%. At 80W I'd hazard to guess those cobs are running in the 40-45% efficiency range.

I've reached that a 8'' x 8'' heatsink would handle each cob.

By 8" x 8" - do you mean you need 64 square inches of surface area?

 

[Pedro Mello]

A COB running at 80% efficiency? - not on your life. Realistic efficiencies are more in the range of 35-55%. At 80W I'd hazard to guess those cobs are running in the 40-45% efficiency range.

I looked at the vero 29 gen 7 catalog :

Imagining 105°C, then I made 100%-85%=15% of heat generation (which I increased to 20% to include an error margin)

Which info should I search on catalog to find the correct percentage of heat generate?

By 8" x 8" - do you mean you need 64 square inches of surface area?

I was planning a 7.984'' wide extruded with 8'' lenght and 46.29'' outer perimeter.

My way to think was:
lenght * outer perimeter = surface area

and 17in² of passive dissipation area for each heat watt



Thank you for your time, man!

 

[CobKits]

After some study I got that my cobs working with aprox 80w (480/6) ea would generate 20% of heat at the worst case temperature.

thats impossible

 

[Pedro Mello]

thats impossible

I've just realized that but still don't know which info should I look for on the cob's catalog to find the real efficiency and the percentage of heat generate.
Do you know how may I find that?

Thanks!

 

[nfhiggs]

I looked at the vero 29 gen 7 catalog :
View attachment 4180046
Imagining 105°C, then I made 100%-85%=15% of heat generation (which I increased to 20% to include an error margin)

Which info should I search on catalog to find the correct percentage of heat generate?


Thank you for your time, man!

That chart just shows the reduction in light output in relation to temperature normalized at 25C.

There is a lot more math involved in calculating heat generated. But I would work from an estimate of 40/60 light/heat at 80 watts. Cobkits could probably give you a closer estimate.

The 8 x 7.984 heatsink you're looking at would probably be enough to cool a COB at 80W with a fan on it, but not passively.

 

[CobKits]

I've just realized that but still don't know which info should I look for on the cob's catalog to find the real efficiency and the percentage of heat generate.
Do you know how may I find that?

Thanks!

call it 50% and go from there for design purposes

 

[GBAUTO]

I will say that my Vero's @80w run right at 50 degC with a 140mm pin heatsink.

 

[nfhiggs]

I will say that my Vero's @80w run right at 50 degC with a 140mm pin heatsink.

The pin sinks are extremely effective at passive cooling - the 7.984 profile.... not so much.

 

[Randomblame]

You need 110cm² surface area for every 1 heat watt(50°C case temps in mind).
A 50% efficient 80w COB would produce ~40heat watt and ~40PAR/w of light. A heatsink should have at least 4400cm² to keep the COB around 50°C(cool to touch).
If you use a small oszillating fan to create a slight breeze across the heatsinks you could get away with 3000cm² but without fan the heatsinks probably reach up to 80°C. Not a problem they are binned at 85°C but there is a risk to burn your fingers, lol.

BTW,
active cooling requires 40cm²/heat watt..