I purchased this 0.5 Ohm power resistor, which comes in a TO-220 Package, with the aim of pushing 3A through it (i.e. 4.5W)

The datasheet has the following information:

enter image description here

It says it will rise 6.5 K/W, so it should rise 29.25C + 23C ambient = 52.25C. Using my thermometer probe, the temperature on the metal body is well over 150C, so I turned it off.

Have I messed up something in my calculations, or is this datasheet incorrectly specified?

  • \$\begingroup\$ To recommend/calculate for you a heatsink (value) you should tell us what is the maximum temperature you want this to run at, and what is the max the ambient temperature will get for your application. \$\endgroup\$ – Fizz Jan 6 '15 at 11:18
  • \$\begingroup\$ Stick it on a big aluminum plate and see how that works. \$\endgroup\$ – George Herold Jan 6 '15 at 14:15
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    \$\begingroup\$ My very conservative rule of thumb (for semiconductors, but this part has similar limitations) is that more than 0.6W without a heatsink in a TO-220 is time to start thinking about a heatsink in normal industrial environments. More than that, and you could burn your thumb on the device. \$\endgroup\$ – Spehro Pefhany Jan 6 '15 at 15:04

The level of heatsinking you need depends on the maximum ambient temperature you can guarantee (let's assume 50C). At an operating temperature of 150C you have a temperature gradient between the resistor's internals and the air of 100C. You want to push 5W (rounded up) through this gradient, which means that the total thermal resistance can not be more than 100C per 5W or 20C/W (C/W is how a heatsink is rated).

If the 6.5C/W is the correct internals-to-case heat resistance that leaves 13.5C/W for the heatsink. To put that in perspective: the heatsink below is rated at 11C/W. Note that for heatsinks a lower value is better (and larger, and heavier, and more expensive), so stay below the 13.5C/W figure.

enter image description here

You can do such calculation the same way you would calculate with voltage (temperature in C), current (power in W), and resistance (thermal resitance in C/W).

Note1: With a 13.5C/W heatsink that heatsink will be at (13.5/20)*100 = 67.5C above the ambient temperature. If that is too hot for your purpose you will need a (much) bigger heatsink.

Note2: The heat must be transferred to the ambient. Even th largest heatsink is useless when it is put in a thermally isolated box!

Note3: The C/W of eany heatsink can be improved (lowered) dramatically by using forced circulation (= a fan). But think about what happens when the fan fails, and the air must still go somewhere.

Note4: A a quick rule of thumb anything in a TO220 casing can dissipate up to 1W, but the case will get too hot to touch. Above 1W a heatsink is probably needed.

  • \$\begingroup\$ The part I can't get is how this part can be rated at 20W. \$\endgroup\$ – Scott Seidman Jan 6 '15 at 11:39
  • \$\begingroup\$ Got it. It's rated at 8 watts at 100 degree C \$\endgroup\$ – Scott Seidman Jan 6 '15 at 11:42
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    \$\begingroup\$ When you keep the casing at 25C and dissipate 20W the internals will be at 20W * 6.5C/W + 25 C= 155C, which is exactly the maximum operating temperature. Just like statistics datasheets don't lie, you just have to read very carefully what they say! \$\endgroup\$ – Wouter van Ooijen Jan 6 '15 at 11:43
  • \$\begingroup\$ This answer reminds me of an exam where the verbal instructions were: if you need a value that is not specified in the problem-statement, just assume one and continue. \$\endgroup\$ – Fizz Jan 6 '15 at 12:04
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    \$\begingroup\$ Although I have upvoted this (for its valiant effort at being a tutorial), there is one thing you've missed: derating. The device in question is derated to 0W(!!) at 150C (backplate temp) according to its datasheet. \$\endgroup\$ – Fizz Jan 6 '15 at 12:18

I would venture a guess that \$\mathrm{R_{thj}}\$ is not junction-to-air (well, technically, this being a resistor doesn't have a junction, but I'm using the familiar term) but rather it is to case/packaging.

EDIT: based on a Bourns app note and the corresponding data sheet for the product used as example in that appnote, \$\mathrm{R_{thj}}\$ is definitely to case, not to air.

Bootnote: Since it appears impossible to get more precise specification regarding the desired operating point from the OP/question, I'll just vaguely point toward Arrhenius' law as it applies to temperature-induced failures in electronics; the following graph is excerpted from Dorf's Electrical Engineering Handbook.

enter image description here

So, yeah, you can push components to their limit temp... as detailed in Wouter van Ooijen's answer... but it's not risk free.

  • \$\begingroup\$ So is there a way to know what thermal capacity heatsink I should use from the datasheet? \$\endgroup\$ – tgun926 Jan 6 '15 at 10:56
  • \$\begingroup\$ Well, I was looking into that. Bourns has an appnote, which (1) unforntunately uses slightly different notations than their datasheets and worse (2) in the example they give the data they show (2.1 C/W) doesn't match the actual datasheet of their product PWR220 (used as example in the appnote) which has 4.8C/W in its datasheet. \$\endgroup\$ – Fizz Jan 6 '15 at 11:00
  • \$\begingroup\$ Ah yeah, I was looking at the wrong datasheet. Bourns has both a "PWR220 S" which is the first hit in google, and a PWR220. For the latter there is indeed a "Version C & D" that has only 2.1C/W for its Rthj matching the appnote example; so that's definitely the thermal resistance to case based on the appone. \$\endgroup\$ – Fizz Jan 6 '15 at 11:07

The 6.5 degC/W is how much the "internals" will warm up compared to the plate/case on the TO-220 package. So with 4.5 watts the internals will rise about 30degC above the case.

But how does the case get rid of heat. This will likely be a much larger figure.

If you look here it tells you that for 1 watt generated, a TO-220 package will be 65degC warmer than ambient temperature. This implies, that for 1 watt, the internals will be 6.5 degC warmer.

Intenal will be at over 70 degC for 1 watt and you are wanting to dissipate 4.5 watts.

  • \$\begingroup\$ You cannot really use generic values from Wikipedia. For these TO-220 Bourns resistors, the thermal resistance to case can vary by a factor of 3 from one product to another. You need to use the datasheet of the specific product. \$\endgroup\$ – Fizz Jan 6 '15 at 11:14
  • \$\begingroup\$ @RespawnedFluff the data sheet doesn't tell you this value. I did check you know. If you have better information please supply a link. \$\endgroup\$ – Andy aka Jan 6 '15 at 11:27
  • \$\begingroup\$ @RespawnedFluff In this case Tca dominates. You get some variation with manufacturing variations but to keep the case under say 150C you need a Tja or ~ Tca of around 30 C/W max. That's about half Andy's generic value. Without diving into data sheets I'd suggest that this confirms what you'd expect - that 4.5 W in a TO220 without heatsink is "overdoing it" \$\endgroup\$ – Russell McMahon Jan 6 '15 at 11:29
  • \$\begingroup\$ You (Andy and Russell) are both quite right about this, i.e. Tja being dominated by Tca for TO-220 to the point where 65-70 C/W is a good approximation regardless of Tjc or vendor. I apologize for the red herring! \$\endgroup\$ – Fizz Jan 17 '15 at 4:36

A specific worked example : you want to safely dissipate 4.5W in that resistor.

It is rated to work at 155C internally; however derating usually has a positive effect on reliability. Let's see what happens if we aim for 120C internal temperature. At Rthj=6.5K/W, the junction will rise 6.5K*4.5W = 29.25K above the case temperature. Therefore we want to hold the case below 120-29.25 say 90C.

Now what is your ambient temperature? Say this is in an enclosure with other electronics and we can expect the air to be a little warmer than room temp, say 40C.

So our heatsink needs to have a thermal resistance of (90 - 40) C / 4.5 W = 11.11C/W (or less) - I would design for 10C/W unlesss space or budget are particularly tight.


Thermal resistance is used to tell you how fast heat travels from the resistor to the package exterior. That allows you to estimate the temperature rise from power applied.

If you think of heat as being like voltage and how well or poorly different materials allow heat to pass through them as being like resistors then you have a better idea what the spec is telling you.

  • \$\begingroup\$ If you want to do the electrical -> thermal analogy, then heat is like charge, (heat flow is like current) and temperature difference is like voltage difference. \$\endgroup\$ – George Herold Jan 6 '15 at 14:13

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