Given the surface area, volume, and temperature of an IC, can you calculate how much power is dissipated?

Assume the temperature of the IC is uniform and constant and the surface area and volume are known.

Can you calculate from these values the power dissipated by the IC through heat? If not, what other variables do you need to do this calculation? Can you make a fairly accurate estimation with just surface area, temperature, and volume?

• Look up "Theta-J" the thermal resistance which is an important factor in the process how an IC losses heat. (I started to answer but it is late and I am going to bed. It would take me too long to straighten out statements like "power dissipated ... through heat" ) Feb 3, 2020 at 22:24
• No. Because typically most heat is conducted to the PCB by the IC's pins (and thermal pads) Feb 3, 2020 at 22:37

Sometimes the $$\R_{\theta JA}\$$ and $$\R_{\theta JC}\$$ are separately specified so you could subtract the two to get $$\R_{\theta CA}\$$, but wow, that's going go be a very inaccurate guess. I wouldn't depend on it. The $$\R_{\theta JA}\$$ is highly variable with the PCB layout and specifications, air flow and all sorts of others things, particularly for SMT parts. Usually the manufacturer has some (perhaps obscure) document keyed to the package code number, not the part, where they describe the test conditions and maybe provide a PCB layout. A difference such as the use of a copper lead frame rather than steel in an apparently identical-looking package can make a huge difference.

It's fairly hard to get an accurate reading of the case temperature as well. Thermocouples draw heat away, RTDs are too big, thermistors don't couple well and IR often gives a fairly inaccurate reading for various reasons.

$$\R_{\theta JA}\$$ = thermal resistance junction to ambient (in °C/W)

$$\R_{\theta JC}\$$ = thermal resistance junction to case (in °C/W)

You could also model the entire chip, the PCB and surrounding conditions and use a finite element software package but I think that's a lot of work unless you do that sort of thing all the time. The software tends to be very expensive.

Alternately, make the chip (or a suitable proxy of similar mechanical characteristics) into heater, drive the current and thus the temperature carefully up to the measured temperature, and measure the input power from volts and amps. Most chips you could just reverse the power supply polarity and get a diode, and drive it with a constant-current source. Of course you won't want to use the chip for much after that. Maybe a tie clip.

Obviously, the best and easiest method in most cases is to measure it electrically in the first place, but that's not always possible. Most digital oscillocopes will let you multiply two input channels together, say voltage and current, and the product is the instantaneous power. The average of that is the average power on that supply rail.

• There is sometimes a third R theta for junction to copper pad (for parts designed to dump heat into the PCB). Feb 4, 2020 at 0:06

That's a good question. Quick answer no, it depends on the conductor surface area and not the insulator area (Epoxy). But let's say there is no board heatsink.

Using datasheet specs    Package     θja      Pdis
sides & top area  Trise=125°C for Ta= +25°C,Tj= +150°C


I multiplied the area by 1k and divided by the max power rating. So the correlation is pretty weak but exists to some extent.