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

heatheatsinkresistors

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: