I am looking to find an appropriate range of heatsink selections to maintain my resistor from damage due to heat. I am using a flat plate (SS) resistor from ARCOL (TFBR900). It is rated for a 900 watts (w/ heatsink). Will be applying 600 volts, to a 440 ohm value. Current will be 1.36A. I have researched and dug up many articles and most are for transistors but should be able to apply to my case. I am looking for the thermal resistance of the resistor but no luck. I have the datasheet which I will attached a link for. I contacted the manufacturer and vendor but they did not help me out at all. The only relevant information I get from the datasheet is operating temp, temp coefficient, and max surface temp. I have never dealt with heatsink selection so any information would be greatly appreciated. Is there anything I am over looking? Can I somehow manipulate the temp coefficient to get any thermal resistance values? TFBR datasheet

I also plan on adding some type of forced air system to help with the cooling.

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    \$\begingroup\$ Wow ! What are you trying to achieve with this ? On a semi-facetious note, maybe a water heater will work better for you (with the water, of course). \$\endgroup\$ Oct 4, 2021 at 20:12
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    \$\begingroup\$ The resistive device(s) inside any given power resistor, may or may not be able to tolerate 600V for some period of time. This one is spec'd to withstand 1kV, so should be ok. And "900W" is probably implying an infinite heatsink. My guess that it should be ok on a modest heatsink if used intermittently, for say, a motor braking application. But it might die if left on for too long (that heatsink can't take away enough thermal energy.) Also consider nichrome resistance wire - wind 440Ω of wire into coil(s) and enclose in a metal box with adequate ventilation holes. \$\endgroup\$
    – rdtsc
    Oct 4, 2021 at 21:25
  • \$\begingroup\$ @rdtsc Yes the manufacturer states only 60% of the power should be used (540W). I am just calculating to worst case scenario. The box design will have perforated holes and fans to help with the cooling. Also can't have a very large heatsink as there will be several of these in the box. worst case is left on for an hour. \$\endgroup\$
    – a_valle1
    Oct 4, 2021 at 21:33
  • \$\begingroup\$ over designing and adjusting as we move on \$\endgroup\$
    – a_valle1
    Oct 4, 2021 at 21:33
  • \$\begingroup\$ I'll just throw this in because nobody has yet. The true test will be to run it while measuring the operating temperature. I might just eyeball it with the heatsink, test it, and go from there. \$\endgroup\$
    – Drew
    Oct 4, 2021 at 23:28

3 Answers 3


Some back-of-the-envelope calculations: your power, assuming you are connecting the resistor between 600V and ground, is 818W. This means your average heat flux is about 58 W/cm2 (the peak flux will be higher, since the packaging and mounting areas of the resistor will not be producing heat), which is on par with a computer processor, but the total amount of heat you're dissipating is much higher.

I think the problem you're going to run into is getting heat away from the resistor and into the dissipating areas of a heatsink. The derating curve for your resistor starts to roll off at ~130C so you're looking at a maximum temperature rise of about 100C, which means that your heatsink needs to have a maximum thermal resistance of 0.12 C/W. On top of that, you need to deal with their ambiguous datasheet. They mention a maximum "operating temperature" of 175C. Is that internal element temperature? Baseplate temperature? This Riedon datasheet gives a maximum flange temperature and the thermal resistance between the flange and heatsink of 0.1 C/W which seems high; similar resistors have thermal resistances of 0.025 C/W, which means that your heatsink needs to be lower than 0.095 C/W at the very least (and if your thermal resistance to the heatsink is 0.1 C/W, it needs to be better than 0.02 C/W).

0.095 C/W in air is certainly achievable; a Wakefield-Vette 512-12M heatsink can do 0.045 C/W in a 100 lfm stream and if you really blast it, I'm sure you could do better. But like I said, getting that heat out of a 188mm long package and into a 300mm long heatsink is the difficult part; you'll have a thermal gradient from the center to the ends, which will reduce the effectiveness of the heatsink and you're already on the edge. You could machine grooves into the base and press-fit some flat heat pipes in there, but I'm not bullish on that.

Where does that leave you? Going back to the Riedon datasheet, you'll see a note further down that states "liquid cooling highly recommended". 818W is about 2800 Btu/hr. A modest-sized (230mm square) liquid-air heat exchanger can deal with that with room to spare. It will cost a pretty penny but it's your best option. You could also, now that I think about it, split up the resistance across several smaller resistors that you could then easily cool with smaller heatsinks. If the cost of multiple resistors was a factor before, knowing how much it will cost to cool a single resistor and how much work it will take might tip the scales.


I am looking for the thermal resistance of the resistor but no luck

I believe we can estimate a crude thermal resistance from the datasheet to get you started. The free air rating for this part is 120W. While the max temperature is listed as 200C, the derating chart show the part can sustain 100% full load up to about 130C. We can assume the thermal rise (change from room temperature) as 130C - 20C = 110C. This lead us to 110C/120W or 0.917 C/W. We are making some assumptions that makes this estimate on the cautious side, if we allow thermal rise to 200C, that would yield a lower thermal resistance.

As for picking a heatsink, if we assume all the heat goes to the heatsink then you are looking for a 110C thermal rise at 900W. This give us a resistance of 110/900=0.122 C/W. Digikey does list heatsinks rated for that and better in both convection and forced air. Now you have to decide how much air you can move across this heatsink.

Here is a straight forward explanation for picking a heatsink that might be useful: https://www.cuidevices.com/blog/how-to-select-a-heat-sink


I strongly prefer to use large resistors that don't need heatsinking. I suggest you buy a resistor that is actually rated for the power you need to dissipate without a heatsink. This may even be cheaper than buying a heatsink for the resistor you have now.

Here is one example: TE Passive Products part number TE1500B470RJ

This is a 1500 Watt, 470 Ohm resistor, however, if you read the datasheet you will see that it can dissipate 1500 Watts with no heatsink when the ambient temperature is 70 C. If you put the resistor inside an enclosure, you will need to keep the enclosure temperature below 70 C for 1500 Watts, or maybe 100 C for 900 Watts. The temperature rise at the surface of the resistor is about 375 C over ambient under full load. So the resistor will be hot enough to sear flesh (make sure nobody can touch it).

I use these to discharge lithium ion battery packs and also as brake resistors on motor controllers that have to operate in regenerative mode for an extended period. I always use resistors rated for 1.5x to 2x the power I need to dissipate so that they don't get too hot.

Here is an excerpt from the datasheet:


If you really need 440 Ohms, you can add another much larger resistor in parallel (it will not need such a high power rating).

Oh, also, the resistor will be very hot, so you need to make sure nobody touches it. It is also a shock hazard when it is energized. So that is another reason to make sure nobody can touch it when it is in use.

  • \$\begingroup\$ I wish it were possible to purchase other resistors but this has already been decided and the resistors have been purchased. But, thank you for the information \$\endgroup\$
    – a_valle1
    Oct 7, 2021 at 18:32
  • \$\begingroup\$ I do not think there is any practical way for you to make it work with the existing resistors. \$\endgroup\$
    – user57037
    Oct 8, 2021 at 18:24

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