Heatsink cooling

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.

say

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

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

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 resistance.

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.

John

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.

Graham

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.

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.

Graham

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.

energy absorbed. You

It shows

airflow.

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 comments.

Found this nice article:

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

energy absorbed. You

It shows

airflow.

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

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?

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

Graham

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

ones that are hopefully

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 (welder).

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

Yes.

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.

ones that are hopefully

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

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?

John

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. Thanks!

Ed