Sizing heatsink area

Discussion in 'LED and other Lighting' started by Pedro Mello, Aug 12, 2018.

    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

    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..
    Pedro Mello likes this.

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