Cooling a DPAK2 Resistor?

[tekati]

I have a BOURNS Power Resistor in a DPAK2 case.  It shows it is capable of 35 watts.  The problem is cooling.  If I try to pass that much current through it not even close to 35 watts it basically starts thermally running away to the point it will melt down.  I was hoping for maybe 25 watts.  How do you keep it from thermally melting down?  I was thinking some sort of heat sink but I have no real experience with thermal dynamics and this sort of thing.  When they list it being good for 35W are they only claiming the device will withstand that but you must find your own way of cooling the device?  I completely admit I have 0 clue when it comes to thermal dynamics and this sort of thing so please any advice would be very welcome.


http://www.bourns.com/data/global/pdfs/PWR263S-35.pdf

[SeanB]

35 W is mounted on a big heatsink where you keep the heatsink at 25C, while the resistor soldered to the copper heatsink will be derated depending on how big a copper pad the PCB has. If you have a 100mm by 100mm pad on both sides, and a good number of thermal vias under the device, it might be able to handle 7W of dissipation without unsoldering itself. To get the full 35W you will need a big aluminium heatsink with a copper inlaid slug that you solder the device tab to. Think of a CPU/GPU heatsink with a copper thermal pad on it.

[Bored@Work]

Long ago Dave made a video about heatsink dimensioning. The principles are the same. You might want to dig through Dave's videos to find it.

[codeboy2k]

basically, it's like any other themal problem ..

From that datasheet , it says:
    Power Rating @ 25  ?C Case Temperature = 35 W

So you can get 35W from it only if the case stays at 25  ?C.. But that won't happen without getting rid of the heat generated, using a heatsink.

The datasheet also says the thermal resistance is Rthjc 3.7  ?C/W, and the maximum temperature is 155  ?C, although it's not clear if that is case temperature or junction temperature. Lets assume junction temp and be safer.

so the 25 W that you want will make the thermal difference across the junction-case be 25 W *3.7  ?C/W = 92.5  ?C. That's going to be
the temperature rise above ambient temperature at 25W.  So if the ambient air temperature is 25C, the case temperature will be 117  ?C at 25W.

That's hot, but still within spec. ... you would think .... however, there's a gotcha.. there's a derating curve.

At 117  ?C the maximum power has to be derated by 40% , so it can't really handle 25W at 117  ?C, it can only handle max*40% = 35*40% = 14W max at that temperature.  So it's clear from the datasheet that you can't operate at 25W without a heatsink.

So, what is the maximum temperature at 25W? figure it out using the derating curve => 25W is what percentage of 35W? about 72%
Now look at 75% derating, find the temperature, that's the maximum temperature you can allow at 25W.  It's about 85 ?C. So shoot for 80 ?C and you'll be safe.

So without a heatsink, you'll be at 117 ?C, but you need the case to be at 80 ?C.  You need to get rid of 117-80 = 37 ?C at 25W.

 37 ?C/25W = 1.48 ?C/W .  So you need a heatsink for the dpak that can at least achieve 1.5 ?C/W.  That will keep your case temperature around 80 ?C.

Also keep in mind the TCR value. any temperature rise on the device above 25C means it's value will change, +-100PPM /   ?C

If that's important you need to consider the change in resistance due to this heating. If you need less change in resistance at that power level, you need a really big heatsink, to get rid of 117-25 = 92 ?C ....  92 ?C/25W = 3.7 ?C/W..  not surprising, this is the Rthjc from the datasheet.  If you have a heatsink that can get rid of that temperature rise, your junction stays at 25C and the resistance is stable.

[tekati]

basically, it's like any other themal problem ..

Wow that was above and beyond and greatly appreciated.  I learned a lot in that little tutorial you just wrote.  It makes me want to learn more as well.  Looks like I have my research project for the (USA and others) Labor Day weekend.


Thank you again.