Real laboratory tests of thermal pastes on igor’sLAB – Part 1 – Measurement setup and methodology

1 - Introduction and basics 2 - Thermal conductivity and disruptive factors 3 - From thermal resistance to thermal conductivity 4 - How the TIMA5 works with ASTM D 5470-17

From thermal resistance to thermal conductivity

Of course, you also need to know what thermal resistance is and how to calculate thermal conductivity from it. And that’s why I have to torture you for a moment. In order to convert the thermal resistance _Rth into the thermal conductivity _λeff, you need to know some parameters of the material and the geometry. The thermal resistance and thermal conductivity are linked by the thickness of the material and the area through which the heat flows.

• The thermal conductivity (_λeff) is given in watts per meter and Kelvin (W/(m-K))
• The thermal resistance _Rth) is given in Kelvin per Watt (K/W)
• Thickness of the material (d) in meters (m) (also referred to as BLT or bond line thickness)
• Area (A) through which the heat flows in square meters (m²)

The thermal resistance for a flat material is easy to calculate:

If the thermal resistance _Rth, the thickness d and the area A are known, as in my measurements, then you can rearrange this formula to calculate effective thermal conductivity _λeff:

Now we can do a real example calculation for an exemplary phase change pad made of a pure polymer, where the thermal resistance is extremely dependent on the test settings (see next paragraph):

Then the calculation of the thermal conductivity is _λeff:

_λeff= 0.00000686542 m / (0.06162 K/W * 0.0001 m² )

In this example, the effective thermal conductivity is 1.11423 W/(m-K).

Contact and interface resistance

The manufacturer sells the polymer pad used as an example above with 8.5 W/(m-K) and specifies ASTM D5470 as the test method (i.e. exactly what I do). A pressure of 60 PSI (41 N) is used and unfortunately the area, thickness and temperature are not mentioned. But where does the large discrepancy between the two values come from? Let’s remember my introduction on this page and the mention of all the factors with a negative influence. Here, the value of the so-called interface or contact resistance as the sum of all negative factors is already higher than the thermal resistance of the ultra-thin pad surface! If this interface resistance were now set at a good 5 mm²K/W, we would almost be back to the advertised 8 to 8.5 W/(m-K) with a lot of winking and a nice salute to marketing. The only problem is that under other, more realistic conditions, it is possible to fall far short of this value and in practice it can usually never be achieved!

I’ll take a quick look at the next page and illustrate what I actually mean in cross-section. We can see that the effective thermal resistance affects both the material and the two contact surfaces. Yes, there are very sophisticated methods, including pulsed lasers, which can also evaluate the pure bulk value very accurately, but in practice we ALWAYS have contact surfaces. I use reference bodies with a standardized (low) roughness for the measurements so that I can also draw conclusions from these in practice. In the end, I have two values, the effective thermal conductivity and a value averaged over all measuring points of the different layer thicknesses BLT minus the extrapolated contact resistance.

And now comes the real problem: Since the thermal resistance should be linear depending on thickness, pressure and temperature, as long as the material does not dissolve (which would of course also be an important finding), it is possible to give a single, generalized bulk thermal conductivity for all measuring points, but this ONLY applies to the one measured test series and its conditions. Conversely, the conclusion is that without information on surface area, temperature and reference bodies, such values are useless because they are not comparable! Therefore, the values that I can determine with the TIMA5 for each material under the same conditions are at least very comparable among themselves, even if they are often much lower than the manufacturer’s specifications. Let’s take a look at the curves for _Rth and _λeff for an inconspicuous, industrial silicone paste in the form of DOWSIL 340 with a stated thermal conductivity of at least 0.67 W/(m-K):

This product is honest, as the value matches the specifications. The extrapolation of the bulk value at exactly 70 °C sample temperature is 0.841 W/(m-K), which is even higher than the specified minimum value, which is honest and good. Of course, you can always do more, but please don’t do less! Of course, I know that buyers love big numbers. That’s why I ordered various industrial pastes with guaranteed and, above all, known thermal conductivity as comparison samples, whereby I took a series measurement for each paste with different layer thicknesses and one with different forces.

For the classification and evaluation of the pastes and pads, however, I will, like the industry, prefer the cleanly measured effective thermal conductivity and my own bulk estimations and not the fair-weather bulk value from the data sheets. Because I must also mention one thing here: the figures that I can now determine are used to design and continuously develop such materials. And it remains comparable! You will see how I measure and arrive at the effective thermal conductivity on the next page, where I explain the measuring method and the technology of the TIMA5.