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Thread: Led cooling question

  1. #1
    Flashaholic
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    Default Led cooling question

    Hello, lets say i want to make this cooling in copper, will it cool enough for lets say a xpg2?
    Or maybe a xhp35 hi?


  2. #2
    Flashaholic* FRITZHID's Avatar
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    Default Re: Led cooling question

    At what drive current?
    I Got tired of looking for the light at the end of the tunnel so i lit that bitch up myself! Convoy s2 365nm, Maxa-Beam Gen II, 55w hid/100w incan Vector Twin, Amondotech n30, vss-3A, Reylight Ti Lan v3, Helius Sigma 9, astrolux s41 219, Shadow JM35, BLF GT,

  3. #3

    Default Re: Led cooling question

    fed from a single 18650, probably. just know that 100% copper is not fun to lathe, it's too gummy. There are other alloys that would work, like regular old 6061 aluminum.
    Have torch, will travel.

  4. #4
    Flashaholic* FRITZHID's Avatar
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    Default Re: Led cooling question

    Probably won't move the heat to the base fast enough at those currents, could try and put a heat pipe into the center of that post down to the thermal mass.
    I Got tired of looking for the light at the end of the tunnel so i lit that bitch up myself! Convoy s2 365nm, Maxa-Beam Gen II, 55w hid/100w incan Vector Twin, Amondotech n30, vss-3A, Reylight Ti Lan v3, Helius Sigma 9, astrolux s41 219, Shadow JM35, BLF GT,

  5. #5
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    Default Re: Led cooling question

    You could always make the tall smaller diameter portion out of copper and press it into a base of either copper or aluminum. Better than removing all of that material.

    Copper cuts just fine if you have the right tooling and setup.
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    http://www.candlepowerforums.com/vb/...ead.php?383660

  6. #6
    Flashaholic* DIWdiver's Avatar
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    Default Re: Led cooling question

    It's commonly claimed that copper has a thermal conductivity twice that of aluminum (commonly quoted as 400 vs 200 W/m-K). However, small amounts of alloying materials and even heat-treating can have a HUGE impact on the thermal conductivity. Beryllium copper, for example is well below most aluminum alloys.

    I did some research into this a while ago, and IIRC, most alloys that are called "copper" had numbers around 385, while most brasses and bronzes were below 200, some way below. Also, most aluminum alloys are well below 200.

    According to my Reynolds Aluminum Data Book (which is older than I am, and that's saying something), here are some alloys and their thermal conductivities (W/m-K):
    EC: 239
    2024-T0: 188
    2024-T3, T4, T36: 121
    6061-T0: 172
    6061-T4, T6: 155
    6063-T5, T6: 201
    7075-T6:121

    EC is electrical conductor grade, and is the highest listed. 6063 was about the highest of any hardened alloy.
    Last edited by DIWdiver; 12-06-2018 at 08:59 PM.

  7. #7
    Flashaholic* DIWdiver's Avatar
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    Default Re: Led cooling question

    I'll take a crack at actually answering the OP's question, or at least getting us into the ballpark.

    If we replace the word 'big' with the word 'infinite', then the part of the diagram that's relevant is the rod. What we need to know is what will be the temperature drop across the length of the rod given some power (heat) input. Then the temperature at the tip of the rod is the heatsink temperature plus the drop across the rod. Conversely, we could state the maximum tip temperature, then calculate the maximum power input (probably more useful here).

    The formula we need is dT = P/k * (l/A), where
    dT is the temperature differential
    P is the power input, watts
    l is the length of the rod, meters
    A is the area of the rod, m^2
    k it the thermal conductivity of the material, W/m-K

    IF YOU CAN GROK THIS EQUATION, YOU WILL BECOME A THERMAL GENIUS!

    Note the distinction between upper and lower case: K, for Kelvin (equivalent to one °C) and k, for thermal conductivity.

    For our purposes, I rearrange the formula to
    P = dT*k * A/l

    If we want a max tip temp of 75°C, and the heatsink temp is 25°C, then dT is 50°C or 50K
    If we're using some relatively pure copper alloy, k=385 W/m-K
    A= (0.01m)^2
    l = 0.06m

    P = 50K * 385W/m-K * 0.0001m^2 / 0.06m = 32W.

    Noting that this is only the heat component, not the full electrical power input, clearly this is sufficient for almost any LED (but not large COBs). If you need a lower tip temperature, have a higher heatsink temperature, and a lower k value (and it's quite possible you could have all of these), the results could be quite different.

    Knowing the acceptable tip temperature is another discussion, though extremely similar to this one. It uses the same equation several times. Grok it!
    Last edited by DIWdiver; 12-06-2018 at 09:02 PM.

  8. #8
    Flashaholic* DIWdiver's Avatar
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    Default Re: Led cooling question

    Okay, I used Google (I'm not a TOTAL Luddite). The thermal conductivity of beryllium copper is 66 W/m-K.

    Can you tell I'm not busy this evening?

  9. #9
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    Default Re: Led cooling question

    I wish to drive the led at its max potential.
    Problem is indeed : is the heat going to be fast enought be absorbt to the big base?

    Problem is that i have only have 11mm hole in my reflector.
    Maybe i can make reflector hole bigger? Its a mbs410 MAXABEAM (total defective ofcoarse) that i want to convert to led (and keeping the zoom function).

  10. #10

    Default Re: Led cooling question

    The XPG2's maximum appears to be 1.5 amps at 3.15 volts. That's not much.

    https://www.cree.com/led-components/...te/xlamp-xp-g2

    Assuming, of course, that I grok the spec sheet properly
    Have torch, will travel.

  11. #11

    Default Re: Led cooling question

    Quote Originally Posted by bonhomme View Post
    I wish to drive the led at its max potential.
    Problem is indeed : is the heat going to be fast enought be absorbt to the big base?

    Problem is that i have only have 11mm hole in my reflector.
    Maybe i can make reflector hole bigger? Its a mbs410 MAXABEAM (total defective ofcoarse) that i want to convert to led (and keeping the zoom function).


    i do not see why it should not work, assuming this heat sink is connected to a aluminum body, or a radiator, heat sinks do not cool, they move heat away, in some cases when heat sink is big and current is low, it may not require anything to transfer heat into.

    i moded this light 7 years ago, xpg led sits on aluminum rod, (exacto knife handle) and some heatsink. the light is sealed, so except that rod and heatsink, there is nothing else to dissipate the heat but current is 1A. so it is enough, the light is used often, and still works just like the day i build it. btw it has one of the few reflectors that are made for bulb, but work great with led. very narrow beam, but funny spill due to faceted reflector
    http://forum.fonarevka.ru/attachment...4&d=1297697070
    http://forum.fonarevka.ru/attachment...5&d=1297697070
    http://forum.fonarevka.ru/attachment...4&d=1297975618

    here is the thread about it at fonarevka.ru http://forum.fonarevka.ru/showthread.php?t=2954
    Last edited by alpg88; 12-07-2018 at 02:06 PM.

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