Sizing heatsink area

Discussion in 'LED and other Lighting' started by Pedro Mello, Aug 12, 2018 at 4:29 PM.

    Pedro Mello

    Pedro Mello Member

    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.
    Last edited: Aug 12, 2018 at 4:36 PM

    nfhiggs Well-Known Member

    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.

    By 8" x 8" - do you mean you need 64 square inches of surface area?
    Randomblame likes this.
    Pedro Mello

    Pedro Mello Member

    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?

    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 Well-Known Member Rollitup Advertiser

    thats impossible
    Randomblame and diyled like this.
    Pedro Mello

    Pedro Mello Member

    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?


    nfhiggs Well-Known Member

    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 Well-Known Member Rollitup Advertiser

    call it 50% and go from there for design purposes

    GBAUTO Well-Known Member

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

    nfhiggs Well-Known Member

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

    Randomblame Well-Known Member

    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.

    active cooling requires 40cm²/heat watt..

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