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Heatsink cooling

Discussion in 'Electronic Design' started by Ignoramus18299, Jan 4, 2006.

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  1. Let's say that I have a source of about 800-1000 watts (two dual IGBTs
    inverting 200 amps). They should run relatively cool, to stay safe say
    under 80 C. I have a heatsink that is about 6x13 inches and weighs
    perhaps 12 lbs. I am cooling it with a fan mounted right next to the
    ribs, pushing air along them, the fan is about 20 watts. Ambient temp
    could be quite hot, say 40 C (inside welding machine). Would you say
    that this amount of cooling should be enough? I will have a overheat
    switch mounted on the sink, so, hopefully, worst case would be a
    inconvenience of having to stop welding, but I would like to have some
    idea of adequacy of this setup.

  2. Guest

    What's the CFM (cubic feet of air) rating of the fan? And what's the
    surface area of the heatsink? (Material of construction of the

    1 kW of heat to dissipate... have you considered a water-cooled system?

    Is this a system you're building yourself, or is this setup on an
    off-the-shelf piece of equipment?

    Cross-posting to sci.engr.mech, they should be able to contribute more
  3. Pooh Bear

    Pooh Bear Guest

    1kW is nothing really. My amps can generate about that much ( in 2u of 19"
    rackmount ) and I get rid of it with a max heatsink temp of 95C with 2 80mm
    'boxer' style fans.

  4. Guest

    If this is a commercial heat heat sink extrusion, you should be able to
    find data on it's thermal resistance to ambient on the manufacturers
    web site. There will be figures for operation in still air and for
    various air-flow rates (mostly in feet per second).
    The fan power is strictly an indicator. You need to know how many cubic
    feet of air get pushed through the fan per minute, which depends on the
    areodynamic reistance of the whole air path on both sides of the fan.

    The crucial measureable parameter is the air-speed over the heat sink
    (which is why it comes up in the heatsink manufacturer's data sheets.
    Measure the input and output air-temperatures at your heat sink while
    dissipating a known amount of heat in the heat sinks. The heatsinks
    will see an "ambient air temperature" which will be about half way
    between, and from that you can work out how how your junctions will get
    to be.

    You can also work out how much air-flow you are getting from the
    temperature difference across the heat sink, the power being
    dissipated and the heat capacity of air
  5. That's very nice to know PB. 1 kW is a bit of an overestimation. 800
    watts would be the maximum (200 amps x 2 volts voltage drop x 2 sides
    of the bridge), and welding would not be continuous, as well.

  6. John  Larkin

    John Larkin Guest

    I did a big ugly 800-watt CAMAC power supply, with a heatsink about
    that size. A single 120 CFM fan was ducted (custom vacuum-formed
    plastic thing) to blow all its air through the fins, not *near* or
    *around* the fins as fans are wont to do.

    I got a tad over 0.04 K/W, but that was with the heat load fairly
    uniformly distributed over the heatsink baseplate. Concentrated heat
    loads would be worse because of lateral (spreading) thermal

    If you don't duct the air, impingement is almost as good: orient the
    plane of the fan parallel to the sink and blast the air directly into
    the fins at short range. Tweak distance maybe.

    Your numbers sound tough to do, but not impossible.

  7. Pooh Bear

    Pooh Bear Guest

    Back in the office I have a neat little equation for temp rise in airflow vs energy absorbed. You
    can do it from first principles but it may come in handy. I'll post it later. It shows
    importantly that for certain temp rises vs watts you need a given mimimum airflow.

  8. Measure It - you should have an idea what the losses are; since you have a
    prototype that actually runs you can even measure the difference between the
    input power and output power. Assuming 80% goes in the heat-sink is a good
    first guess IMO.

    The hard part of cooling is determining the heat-sink to air thermal
    resistance - often it is easier to use high-wattage resistors to dump power
    in the heat sink and *measure* how hot it gets in the real situation rather
    than trying to calculate it. Especially with natural convection.

    If you want to be really clever, you can mount your devices then run them as
    constant-current sinks and apply the appropriate voltage for the desired
    dissipation. Then measure the temparture close to the device and work back
    to chip temparature. *really clever* is using part of the internal structure
    to measure the chip temparature - some MOSFETS (IRF, I think) come with an
    extra pin for this purpose. If you cannot use the real device - some large
    IGBT's have electronics in them so they can only be used for switching -
    then you have to use the thermal resistance of the target device to
    calculate what it's temparature will be.
  9. Thanks! Yes, I would like indeed to do some calculations. Now that I
    am packaging stuff for real installation inside the welder, I want to
    avoid making stupid avoidable mistakes.

    I made some pictures last night, of the heatsink assembly with some

  10. Found this nice article:

    Will check it out. Turns out that my heatsink with glued in fins is
    the highest performance heatsink.


  11. I used this calculator:

    Volumetric Flow Rate m^3/s 0.06
    Number of Fins 13
    Fin Width m 0.001
    Fin Length m 0.3
    Fin Height m 0.13
    Sink Width m 0.12
    Fin Thermal Conductivity W/mK 237

    Result: Thermal Resistance C/W 7.46266e-02

    So, for a continuous 800 watt load, the temperature rise would be
    56C. That's probably acceptable. In reality, my load would be less
    than 800 watt (current lower than maximum, non-100% duty cycle in a
    welder). So, I am not worried about adequacy of my heat sink assembly.

  12. Pooh Bear

    Pooh Bear Guest

    Here's the info I had in mind. It conveniently mixes units so as to use the ones that are hopefully
    most accessible.


    Heat transfer equation from Sunon ( fan manufacturer )

    Q = Cp . W . T

    Q = Amount of heat transferred
    Cp = specific heat of air
    T = temperature rise
    W = mass flow

    putting in the relevant values I got to this..................

    Air flow required ( CFM ) = 1.76 * power / temp rise ( degrees C )

    Worked example e.g. 200W and delta T = 50C

    gives 1.76 * ( 200 / 50 ) = 7 CFM
  13. So each device is dissipating 500W. Never mind what the heat sink temp is,
    that's irrelevant except as it affects what it's mounted to (and whether it
    burns your finger); what you care about is junction temp, which probably
    wants to be below 150C, that is, 100C above ambient in round numbers. So,
    you need to have 0.2K/W total thermal resistance, from junction to ambient.

    What's the junction-to-case rating of your IGBTs? Randomly picking the
    first TO247-cased IGBT I find on Google, I see it's rated at 0.43K/W.
    Lessee, that means you need -0.2K/W in your heat sink; that's going to be
    hard to do.

    Graham, I wonder if your amps are using more than two transistors to
    dissipate that 1kW?
  14. Thanks. I also tried to find some place with a calculator and found
    something nice. Below is a copy of ,y text file with a little detail
    of these calculations. It looks like my setup is just about adequate
    for 100% duty, which it will obviously not see due to application

    Heatsink calculations:

    Size 7x12 width, 5" depth, 10 fins.

    According to

    Volumetric Flow Rate m^3/s 0.06
    Number of Fins 13
    Fin Width m 0.001
    Fin Length m 0.3
    Fin Height m 0.13
    Sink Width m 0.12
    Fin Thermal Conductivity W/mK 237

    Result: Thermal Resistance C/W 7.46266e-02


  15. According to datasheet, 0.11 K/W.
    I think that these IGBTs have a sensible rating for their power, so, I
    need to make sure that the heatsink is cool.

  16. John  Larkin

    John Larkin Guest

    That looks about right. But that assumes perfect coupling from the
    heatsink to the air stream, which doesn't happen.

    This gets complex, and I don't really understand it, but it's
    something like...

    If the fins on a heatsink are few and far between, the 7cfm will zip
    through easily and not pick up much heat. So if the sink is
    dissipating the 200 watts, the air exit temperature will certainly be
    50K above intake (by conservation of energy) but the heatsink may be a
    lot hotter. Imagine half the air being in good contact with fins, half
    zipping through unaffected, and the halves mixing at exit, averaging
    +50. The heatsink only contacts half the air, so it rises +100.

    If the fins are very dense, coupling from sink to air will be good,
    but you'll develop a lot of back pressure, and a "7 cfm" fan won't
    move 7 cfm, so again the sink temp rise will be above 50.

    My working theory is that you get the best heat transfer if the
    heatsink restricts the fan to delivering about half its rated cfm. And
    the fins should be nearly isothermal (ie, not long and skinny) because
    cool fins restrict flow without coupling heat very well.

    Anybody know more about this?

  17. ehsjr

    ehsjr Guest

    I wish more people would do what you do - post updates
    of the progrees in your project. You do a very good job
    of keeping interested people informed.

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