Electrical – What heat sink should I use for this 50W resistor


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


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:


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.

Best Answer

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.