Thermal Impedance - Puzzled
Thermal Impedance - Puzzled
(OP)
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?
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?





RE: Thermal Impedance - Puzzled
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RE: Thermal Impedance - Puzzled
Should the - (subtract) actually be a + (addition) as the junction temperature would be hotter than the case right?
RE: Thermal Impedance - Puzzled
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RE: Thermal Impedance - Puzzled
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?
RE: Thermal Impedance - Puzzled
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.
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