Thermal resistance is the number that actually determines whether your chip survives. Conductivity gets the marketing attention, but resistance is what you add up, budget, and measure. Here is how to calculate it properly — including the contact terms that most first-pass calculations miss.
The one-dimensional starting point
For heat flowing straight through a flat layer of thickness L and area A:
Worked example: a 0.5 mm pad, 180 W/m·K, covering a 20 × 20 mm die (A = 4×10⁻⁴ m²).
Why engineers use °C·cm²/W
Because it normalises out the area, letting you compare materials directly. Convert by multiplying °C/W by area in cm²:
The part most calculations miss: contact resistance
The formula above describes bulk conduction through the material. A real joint has two additional resistances where the pad meets imperfectly flat surfaces. Air trapped in surface valleys conducts at roughly 0.026 W/m·K — about 7,000 times worse than the graphene pad.
Building the full thermal budget
Stack every resistance between junction and ambient, then check the resulting junction temperature:
Junction to case (Rjc)
From the semiconductor datasheet. Not controllable by you.
Case to TIM (contact)
Depends on surface flatness, pressure and the TIM's conformability.
Through the TIM (bulk)
L / (k × A). The part you control with material and thickness choice.
TIM to heat sink (contact)
Usually similar in magnitude to the first contact term.
Heat sink to coolant or air (Rsa)
From the sink or cold plate supplier's flow and pressure-drop data.
Sensitivity: where to spend effort
Consider a 200 W device where the interface is 0.5 mm. Compare reducing the interface thickness versus improving conductivity:
| Change | New bulk R (°C·cm²/W) | Improvement |
|---|---|---|
| Baseline: 6 W/m·K pad, 0.5 mm | 0.083 | — |
| Halve thickness to 0.25 mm | 0.042 | 2.0× |
| Switch to 180 W/m·K at 0.5 mm | 0.0028 | 30× |
| Both: 180 W/m·K at 0.25 mm | 0.0014 | 60× |
The lesson: when a large gap is unavoidable — which is the norm in industrial assemblies with tolerance stack-up — raising conductivity gains far more than trying to machine the gap down. Halving a 0.5 mm gap usually means tighter flatness specs and higher machining cost on every heat sink.
Measuring it in the real assembly
Calculations give you a starting point. Validate with measurement:
- ASTM D5470 — the standard steady-state method for TIM thermal resistance under controlled pressure and thickness. This is how datasheet values are generated.
- Transient methods (such as laser flash) — good for diffusivity of the bulk material, but they miss contact resistance entirely.
- In-situ thermocouples — place one as close to the die as practical and one on the sink, measure ΔT at known power, back out the interface resistance.
Common mistakes
- Comparing W/m·K across materials with different thicknesses. Only compare resistances at the same BLT and pressure.
- Ignoring the contact terms. They can dominate at low pressure.
- Using nominal thickness instead of worst-case. Design for the largest gap your tolerance stack allows.
- Forgetting spreading resistance. When the die is much smaller than the heat sink base, heat spreads out and the effective area is larger than the die — this actually helps, but only if the TIM is thin enough to let spreading occur.
- Assuming the datasheet number is achieved in your assembly. It probably is not.
