Thermal Impedance - Puzzled

[RustyH]

Sorry to trouble peaple for such a basic question, but Im trying to get my head around thermal Impedance.

Say you know Junction to Case Impedance of a IC chip, and say is was 5°C/w. The IC was producing 1w of power. Then you knew the Case Temperature (say 60°C). Would the Junction temperature simply be Tcase - (Power * Impedance) = 60 - (1 * 5) = 55°C?

 

[IRstuff]

That would be correct per the model.

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[RustyH]

Great, thank you, starting to make sense now, although one comment, above I put: Tjunction = Tcase - (Power * Impedance)

Should the - (subtract) actually be a + (addition) as the junction temperature would be hotter than the case right?

 

[IRstuff]

Yes, sorry

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[RustyH]

Great, thank you kindly for your help

I guess the same rings true for different powers too, so in the above, if the power was 2w, then it would be 60+(2*5)=70°C?

Also, does this junction to case resistance stay the same in all enviroments, I.E - what if there was a cooling system on top of the case, would the junction to case thermal resistance remain the same 5°C/w?

 

[IRstuff]

Things that are solid or combinations of solids would not change. The case temperature to junction temperature is a particularly two-faced kind of number in that it's great for the manufacturer to verify because it's a solid, unvarying, and repeatable number, but is not that useful for the user, because there is often an ambient to case impedance that would significantly depend on the amount of air flowing over the part.

So, in a normal thermal analysis, you would need to take the impedance from the air temperature outside of the chassis, go through the chassis, go through the air layer inside the chassis, go through the package/heatsink, and on to the junction. A 5°/W impedance is pretty reasonable when you consider natural convection from a 1-in square device would have something on the order of 150 °C/W impedance from the air to the surface.

TTFN
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