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I want to run 80 Watts of power through 2 50W 1Ohm resistors:

http://www.upe-inc.com/upe-ate-resistors/wirewound-aluminum-housed-axial-resistor/

It's the RB50 series.

When trying to calculate what kind of heat sink I need it seems my values don't make sense.

I'm going to calculate with half power through 1 first:

So let's assume 30 degrees ambient, 40 watts of power. The resistor is rated at 1.9K/W, I'll be using thermal paste rated at 0.11K/W, and then I need a heat sink which is the unknown in the equation. The effective operating temperature range goes up to 250C, but I'd rather stay lower. 

40*(1.9+0.11+X)+30 < 250    
76+4.4+40X+30 < 250
110.4+40X < 250
40X < 139.6
X < 3.49

But if I want to take some margin into account

X < (maxTemp - 110.4) / 40

So for 150C maxTemp it would mean X < 0.99, which seems impossible for a small size heat sink? Considering the physical dimensions of the resistor, how can this be used if it needs a huge heat sink for it to function?

Does this mean that if the resistor receives 40 Watts of power, it would by itself emit 76C above ambient? So why all of a sudden I get 200+ degrees in my calculation by adding a heat sink and thermal paste? It seems I am missing something fundamental in my understanding of things.

Also I found a document (but it's in French) about this resistor in which they stated that the resistor needs derating unless it is attached to "a standard heat sink". No explanation about what a "standard" heat sink might be. For those who happen to understand French it's here:

https://produktinfo.conrad.com/datenblaetter/400000-424999/421545-an-01-fr-ATE_RB_50.pdf

If the values are correct, is there a way of calculating how warm the ambient temperature around the heat sink will become? Or is it trial and error? I will be making my own ventilated enclosure but I don't want the plastic to melt so it would be good to know how hot it will become on the inside.

Could someone explain me where I am wrong? I'd rather not burn up stuff by figuring this out through trial and error.

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  • \$\begingroup\$ Is this for a 3S LiPo battery load test? \$\endgroup\$
    – Tony Stewart EE75
    Jan 27, 2018 at 21:52
  • \$\begingroup\$ No it's not for that. \$\endgroup\$
    – Joris Mans
    Jan 27, 2018 at 22:02
  • \$\begingroup\$ can you get a used cpu heatsink/fan? thats what you need \$\endgroup\$
    – Tony Stewart EE75
    Jan 27, 2018 at 22:19
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    \$\begingroup\$ I can get any heat sink I want. It's just a matter of buying one. Would I would like to take a scientific approach towards knowing what type is needed instead of just slapping something "that'll do" on it. \$\endgroup\$
    – Joris Mans
    Jan 27, 2018 at 22:53
  • \$\begingroup\$ If you identify all thermal R's for every interface and get air speed over 2m/s and the thermal R's are then reduced by some k factor. The scientific approach here is to test it with proper pressure and grease. Imagine equating it to an overclocked 8 core CPU and match the heatsink area , pressure and fan speed to get a temp rise of 50'C max. Which may burn fingers but not boil water. If you measure < 70'C at the hotspot, you did a good job. RPM may be adjust with a series R. \$\endgroup\$
    – Tony Stewart EE75
    Jan 27, 2018 at 23:12

3 Answers 3

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If you download the "brochure" you will find that the "with heatsink" specifications apply to a 930 square centimeter X 1.5 mm heatsink. I would assume that means a flat aluminum plate that is approximately square or perhaps the same aspect ratio as the resistor.

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  • \$\begingroup\$ That's a piece of 30cmx30cmx1.5mm heat sink? I presume this is equivalent to a much smaller heat sink with fins with approximately the same total surface? \$\endgroup\$
    – Joris Mans
    Jan 27, 2018 at 21:42
  • \$\begingroup\$ Yes. I think you can get the thermal resistance from resistor to air from the data provided and then work out the heatsink to air thermal resistance of the heatsink. Also you might find thermal resistance for a flat plate somewhere. Then you could look for a heatsink with the same thermal resistance. \$\endgroup\$
    – user80875
    Jan 27, 2018 at 21:46
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I would apply the largest/highest heatsink possible either on the base (21mm) or on the top (16mm).... and experiment! You can do it with standard alu profile or with real heatsink. Don't use anodized alu for heatsink (they did it but it's wrong).

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  • \$\begingroup\$ What you're basically saying is to just build something and measure the surface temperature under continuous load to see where it gets me? \$\endgroup\$
    – Joris Mans
    Jan 27, 2018 at 22:54
  • \$\begingroup\$ Yes. Better than messing up with maths... ;) \$\endgroup\$
    – Fredled
    Jan 27, 2018 at 22:58
  • \$\begingroup\$ The problem is packaging. I want the packaging to be as small as possible. In that case maths might come in handy, or I would need a stockpile of heat sinks and start experimenting, but that's not very efficient. \$\endgroup\$
    – Joris Mans
    Jan 27, 2018 at 23:01
  • \$\begingroup\$ I understand. One more suggestion is to fix the base on the chassi (as they explained in the datasheet) and a heatsink on the top of the resistor so that it dissipate from both sides. If your chassi is metalic of course. \$\endgroup\$
    – Fredled
    Jan 27, 2018 at 23:07
  • \$\begingroup\$ It won't be. I will be able to mount the resistor base "in the air" as I will use an enclosure that can withstand heat to about 90 degrees C, which is not enough to mount this resistor directly on top. Also I will add a fan to this entire system, but I presume calculating this stuff taking the fan into account will be an even bigger nightmare. I would consider the fan as a bonus. \$\endgroup\$
    – Joris Mans
    Jan 27, 2018 at 23:16
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Lets do some thermal design. Assume the resistor is 2cm by 6cm. Install that resistor in middle of 6 cm by (2 + 4*6) = 6cm by 26cm copper plate, thickness 0.06 inches (1.5mm). Assume the airflow is adequate to keep the plate at 30 degrees Centirgrade. How hot is base of the resistor?

schematic

simulate this circuit – Schematic created using CircuitLab

We have lower than 2 degree C per watt. But assume that. Thus base of resistor is 100 degree C. Unless you can remove the heat with other than a simple flat plate.

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  • \$\begingroup\$ But then you take the assumption that you can keep the plate at 30 degrees. So you shift the problem to "what do I need to keep the plate at 30 degrees?". \$\endgroup\$
    – Joris Mans
    Jan 28, 2018 at 9:37
  • \$\begingroup\$ You'll have to solve that sort of problem whatever heat sink you use. Heatsinks are just the means to transfer heat from devices to the air, they don't absorb it. \$\endgroup\$
    – Finbarr
    Jan 28, 2018 at 9:53
  • \$\begingroup\$ Ok, but then we're running in circles here. So far my conclusion on all this is trial and error, more than do some maths and find out what you should do. \$\endgroup\$
    – Joris Mans
    Jan 28, 2018 at 10:27
  • \$\begingroup\$ To "keep the plate at 30 degrees", ventilation holes will help a lot. ... now, don't tell me it should be waterproof. \$\endgroup\$
    – Fredled
    Jan 28, 2018 at 15:04
  • \$\begingroup\$ No it shouldn't :) \$\endgroup\$
    – Joris Mans
    Jan 30, 2018 at 10:42

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