| Assumed non-ceramic R | Original total R | AlN-scenario total R | Total R reduction | Temperature reduction at 40 W |
|---|---|---|---|---|
| 0.10 K/W | 0.2042 K/W | 0.1147 K/W | 43.8% | 3.58 K |
| 0.90 K/W | 1.0042 K/W | 0.9147 K/W | 8.9% | 3.58 K |
The useful question is the temperature reduction
Aluminum nitride improves a ceramic PCB thermal path most when resistance through the ceramic contributes substantially to the temperature rise being controlled. High conductivity alone does not establish the value of a material change. Define the required reduction at a stated heat load and cooling condition, then estimate how much of the existing resistance the substitution can remove. The AlN substrate guide covers material and process selection; this article develops a practical screening calculation and an experiment for deciding whether an upgrade is worthwhile.
Choose one thermal boundary and keep it fixed
Write the path from the heat source to a named reference: for example, die to a controlled cold-plate surface. Include die attachment, metallization, ceramic, substrate attachment and any baseplate or thermal interface within that boundary. A device-to-ambient result includes the cooler and air conditions; a device-to-case result does not. Never compare those two resistance values as though they describe the same system. Our thermal-management design guide explains the complete path. For this screening exercise, assume steady state, a single effective heat path, constant properties and unchanged dissipated power. Real modules may require a distributed model.
Calculate the ceramic share of the original resistance
Let R₀ = Rother + Rc, where Rc is the original ceramic resistance and Rother represents every other contribution inside the chosen boundary. For one-dimensional conduction, Rc = t/(kA). If thickness t and effective area A stay fixed, replacing conductivity k₀ with k₁ gives Rc,new = Rc × k₀/k₁. The fractional reduction in total resistance is therefore f × (1 − k₀/k₁), where f = Rc/R₀. This is a derived screening relation, not a manufacturer performance claim. Even an ideal ceramic with infinite conductivity cannot reduce total resistance by more than f under these assumptions. If the ceramic accounts for only 10% of the original path, a 30% system improvement is impossible from that layer alone.
Use a grade-specific value, then test its sensitivity
CeramTec's Alunit AlN HP announcement reports thermal conductivity of 170 W/m·K. That named product value is useful context, but a purchase specification still needs the current grade data, measurement conditions and tolerances. In the worked scenario we use k₁ = 170 W/m·K and an assumed baseline k₀ = 24 W/m·K; these are fixed model inputs, not guaranteed operating-temperature properties. The alumina versus AlN comparison gives broader selection trade-offs. Research by Xu and colleagues on crystalline AlN shows that temperature and defects affect conductivity; its crystalline specimens are not interchangeable with commercial sintered circuit substrates. Request the actual grade's k(T) data instead of substituting a research maximum.
Worked scenario: identical ceramic, different surrounding paths
Assume 0.50 mm ceramic thickness, a uniform 200 mm² heat-flow area and 40 W passing through that area. The baseline ceramic resistance is 0.1042 K/W; the AlN scenario gives 0.0147 K/W. The reduction is 0.0895 K/W, corresponding to 3.58 K at the assumed power. If Rother is 0.10 K/W, total resistance falls from 0.2042 to 0.1147 K/W, a 43.8% reduction. If Rother is 0.90 K/W, it falls from 1.0042 to 0.9147 K/W, only 8.9%. The absolute temperature reduction remains 3.58 K in this fixed-power model. A small percentage can still matter near a temperature limit; a large percentage can be commercially irrelevant if the absolute benefit is small.
Turn the required benefit into a screening threshold
For the same assumed conductivities, 1 − 24/170 = 0.8588. To achieve a 20% reduction in the modeled total resistance, the original ceramic must contribute at least 0.20/0.8588 = 23.3% of that resistance. Alternatively, a required 5 K reduction at 40 W needs a resistance saving of 0.125 K/W. Our 0.0895 K/W scenario cannot meet that target through this material substitution alone. Run different thicknesses and areas in the thermal-resistance calculator, but identify which assumptions you changed. If thickness also changes, use Rc,new/Rc,old = (tnew/told) × (kold/knew) × (Aold/Anew), and recheck electrical and mechanical suitability separately.
Check whether interface variation could hide the improvement
Infineon's TIM selection guidance highlights surface conformity, uniform application, thickness and assembly pressure, alongside material conductivity. For an illustrative interface with thickness 50 µm, k = 3 W/m·K and area 200 mm², ideal bulk resistance is 0.0833 K/W. Doubling that thickness adds another 0.0833 K/W before contact resistance is considered, nearly erasing the 0.0895 K/W ceramic saving in our scenario. These interface numbers are hypothetical, not an assembly instruction. Do not reduce a real interface below the thickness needed for reliable contact or isolation. Control flatness, application and mounting, and include contact effects in the test rather than treating the TIM's advertised conductivity as sufficient.
Compare matched assemblies and quantify repeatability
Build a comparison in which die layout, conductor geometry, finish, attachment process, mounting method and cooling conditions are held constant as far as practical. Record actual dissipated power, reference temperature, sensor location and stabilization criteria. If comparing junction temperatures, use a calibrated method appropriate to the device; a surface measurement is not automatically junction temperature. Repeat the mounting operation or test multiple assemblies to expose interface variability. Report the observed spread and measurement uncertainty beside the mean improvement. If a predicted 3.58 K benefit is comparable with the experiment's variation, improve the experiment before treating the result as a material ranking. Do not invent sample counts or confidence bounds without test data.
Know when the simple model stops answering the question
The formula is a screening tool for unchanged power and geometry. Closely spaced heat sources, lateral copper spreading, small hot spots and parallel paths can change the effective area. A short pulse requires transient thermal impedance rather than a steady-state resistance. Device losses may themselves change with temperature, creating feedback absent from the fixed-power example. Different metallization or attachment may alter Rother as well as Rc. In these cases, compare complete candidate assemblies using suitable simulation and measurement. Our 800 V inverter stack-up example illustrates why thickness and interfaces must remain explicit when comparing practical constructions.
Release a supported material decision
Document the target temperature reduction, the modeled ceramic share and the evidence from the matched comparison. Include the cost of the finished assembly and any process or qualification changes, not just the price of the bare ceramic. Thermal improvement is one requirement: assembly yield, insulation and service reliability still need their own acceptance criteria. Use the thermal-cycling guide when the material change affects the strain path. Select AlN when the complete design gains useful margin and that gain survives realistic assembly variation. If the cooler or interface dominates, address that bottleneck or evaluate both changes together.
Engineering example
All table values are hypothetical screening results. With t = 0.00050 m and A = 0.000200 m², Rc,old = t/(24A) = 0.10417 K/W and Rc,new = t/(170A) = 0.01471 K/W. Their difference times 40 W is 3.578 K. The two table rows change only Rother; neither represents a tested product or a junction-temperature guarantee.
Before you release the design
- State the controlled thermal boundary, heat load and required absolute temperature reduction.
- Use current grade data and evaluate conductivity at the intended operating temperature.
- Calculate the original ceramic share and check whether the requested improvement is physically achievable in the screening model.
- Control interface thickness, contact conditions and mounting repeatability in comparative tests.
- Verify insulation, manufacturing and cycling requirements independently of thermal improvement.
Sources and further technical reading
Manufacturer references support the material and process context. Worked examples and checklists are engineering guidance; they are not test results or supplier guarantees.