Here is something I never quite understood; look at the temperature drop for a typical power, area, thickness and thermal conductivity
View attachment 360503
P = 100W
A = 3 cm^2
k = 20 W/(m.K)
Dx = 1/20 mm
How then can thermal pastes differ by say 2 to 3
°C degrees?
I'm not sure I fully parse your question, so apologies in advance for any misunderstanding or if I say too much you already know. But that equation (a
model for thermal conduction) is
not one for Δ
T between CPU and ambient, nor is it one for comparing two TIMs. Instead it tells you about conduction through a single TIM application. Your Δ
T is the temperature difference between the CPU-end and cooler-end
of the TIM itself. (Are you asking: if that number is so tiny, how can there be so much variation among different ones leading to single-digit CPU temp changes? If so, completely valid, I think I just didn't read you write the first time. Regardless...)
Seperately, how is your Dx so small? 5% of one millimeter? Where is that from? (This will show up later as well.)
In any case, I got curious, so here's an attempt to model the CPU temperature difference between two TIM applications. Lots of simplifications and bla bla bla I don't expect extreme accuracy or anything, but I burned an evening with this in my head so here we are.
Modeling CPU Temp. Difference Between TIMs
Here's a visual sketch of our two imaginary setups.
I'll call the TIM applicaiton thicknesses by
d rather than Δ
x to save some space. The two cross-sectional areas are denoted by
A. Temperatures are all denoted
T, and a note about each one:
- T1 is the temperature at the CPU/TIM 1 interface. T2 is the temperature at the CPU/TIM 2 interface. The quantity we're after is T1−T2, the difference in CPU temps given two different TIM applications and all else being equal.
- Tb is the temperature at the TIM/cooler base plate interface. For reasons I'll outline in the assumptions, this will be the same for both TIM applications.
- Tambient is the ambient air temperature. We assume this is the same for both TIM applications.
Addtl. Assumptions I'm Making
- The CPUs are identical, operating in steady state, and dissipating power P in both cases. This means the power through each TIM and through each cooler must be P: otherwise the TIM (or cooler) would itself be accumulating or losing internal energy over time, neither of which are stable scenarios.
- Perfect (read: I will completely ignore) thermal contact conductance between the CPU/TIM and TIM/cooler interfaces. You're welcome to follow the process here and relax this assumption on your own to see how it turns out!
- I'm ignoring any inhmogeneity in temperature across the surface of the chip, so I'm not considering "hot spots."
- Identical coolers. This makes a big difference, because it allows me to get away with not modeling the cooler at all. The point is that the cooler thermally connects the TIM contact with the ambient temperature—perhaps with heatpipes and fins and a fan; perhaps through liquid cooling and a radiator and fans; perhaps something more exotic. But as long as they conduct the same power P, and as long as that's linear in ΔT between b and ambient (as it would be for a convective model), that forces the two Tbs to equality, regardless of what other dependencies there are in the cooler's operation (as long as those conditions are equal for both TIM applications). This lets us ignore the cooler.
Working it out
Start with TIM 1. We have a relationship between the heat conducted and temperature change along the thickness
d1 (the heat conduction equation you cited):
,
where
k1 is the thermal conductivity of TIM 1. We have
the same equation for TIM 2, with the subscripts changed. These are both "true" equations for conduction along each of the TIMs independently. But the fact that they must be conducting the same power and the equality of their
Tbs connects them.
Now we'll
take the difference:
,
and voilà! I'm leaving this here to highlight the fact that not only the conductances but also the thickness and cross-sectional areas of the applied TIMs will make a difference here. But that's difficult to control and/or measure, so I'll make one further simplificaiton before plugging in numbers:
suppose the two TIMs are applied with the exact same thickness over the exact same area. (This is obviously not how it goes irl: different properties of the TIM and variance in user application will lead to different numbers there.)
Then we have:
.
We can quickly see this passes a sanity check: if
k1 >
k2, then
T1 <
T2, so a higher conductance TIM indeed gives a lower CPU temperature. Plugging in your numbers
P = 100 W,
A = 9 cm^2 (assuming you meant 3 cm × 3 cm area),
d = 0.05 mm (though this seems
insanely thin, where'd you get this from?) and comparing, say, Conductonaut at 73 W/m K vs TL-A40 at 15 W/m K (both from
here) we get a temperature difference of about 0.3 K. But I'll draw attention again to that
d: the temp. difference is linear in that, so adjusting to half a millimeter of thickness brings us to ~3 K temperature difference. We can also see this scales linearly with conducted power: so the difference will get bigger with more power draw. It does
not scale linearly with the difference in conductances, though, which is why liquid metal isn't like an order of magnitude better than the competition.