Read the Whole Heat Path in a Thermal Interface Material Offer

A TIM provides a path across imperfect surfaces

Surfaces that look flat can meet only at scattered high points, leaving air-filled voids. A thermal interface material, or TIM, improves coupling by conforming between the mating surfaces and replacing poorly conducting air in that path. Adding a suitable material can therefore improve heat transfer even though it separates the solid parts.

Laird’s February 2019 Thermal Management 101 separates the resistance within the TIM from resistance at each contact. Here, the complete interface means those three contributions between the two mating surfaces. It does not mean the entire heat path from semiconductor junction to room air.

References: [TIM1]

Keep the layer and both contacts in the calculation

For a simplified steady, uniform, one-dimensional path through a layer, write Rinterface = t/(kA) + Rc1 + Rc2. Thickness t is in metres, bulk through-thickness conductivity k is in W/(m·K), area A is in m², and every resistance term is in K/W. Rc1 and Rc2 describe the two mating contacts; they need not be equal.

Area normalization multiplies each resistance by the same area. Define Zinterface = Rinterface × A and Zc1 = Rc1 × A, with the same definition for the second contact. Then Zinterface = t/k + Zc1 + Zc2, in m²·K/W. Multiplying a numerical value in m²·K/W by 10,000 converts it to K·cm²/W.

Keeping the area factor visible prevents a common mistake: t/k alone is an area-normalized bulk term, rather than a resistance in K/W. To recover Rinterface, divide Zinterface by area expressed in the matching units. These equations describe the stated uniform model; an actual assembly needs evidence that its geometry and contacts fit the comparison.

References: [TIM1]

Identify what the conductivity field includes

A bulk conductivity describes conduction inside the material. It does not, by itself, describe entry through one mating contact and exit through the other. Laird’s technical-terms guide defines apparent or effective conductivity as a measured conductivity that takes contact resistance into account.

Consequently, two fields carrying W/(m·K) may have different measurement meanings. Ask whether the reported value represents bulk material behaviour or an effective result for a characterized interface. Calculating t/k from an effective value and then adding contacts already included in it would count those effects twice.

Bondline is the material thickness in the application. Identify whether a quoted thickness describes that bondline or the nominal supplied material. The same guide links flatness and conformity to contact resistance, which explains why the two mating surfaces belong in the evidence package.

References: [TIM2]

TI-44: the smaller bulk term still gives the larger total

TI-44 is an original fictional comparison of two uniform interfaces, each with an area of exactly 1 cm². Each bondline belongs to its own stipulated configuration. Assume steady heat flow only through the stipulated layer and its two contacts. All conductivities, bondlines and contact terms below are invented explanatory inputs, rather than measurements, product data or approved assembly choices.

Offer A has bulk k = 10 W/(m·K), a 1.00 mm bondline and a combined area-normalized contact term of 0.20 K·cm²/W. Convert 1.00 mm to 0.00100 m. Its bulk term is t/k = 0.00100/10 = 0.000100 m²·K/W = 1.00 K·cm²/W. Adding both contacts gives Zinterface,A = 1.00 + 0.20 = 1.20 K·cm²/W.

Offer B has bulk k = 5 W/(m·K), a 0.25 mm bondline and a combined contact term of 0.80 K·cm²/W. Its thickness is 0.00025 m, so t/k = 0.00025/5 = 0.000050 m²·K/W = 0.50 K·cm²/W. Its complete result is Zinterface,B = 0.50 + 0.80 = 1.30 K·cm²/W.

A has twice B’s conductivity. B nevertheless has the smaller area-normalized bulk term because its layer is only one quarter as thick: its t/k is half A’s. Ranking the bulk terms would favor B. But B’s contacts add 0.60 K·cm²/W more than A’s, outweighing its 0.50 K·cm²/W bulk advantage. A therefore has the lower complete interface resistance by 0.10 K·cm²/W.

Ranking k alone would miss B’s bulk advantage. Ranking t/k alone would miss the contact penalty that reverses it. A’s larger k happens to agree with the final ordering here, but that agreement does not make conductivity alone sufficient evidence.

At exactly 1 cm², dividing the totals by that area gives 1.20 K/W for A and 1.30 K/W for B. The numerical equality depends on the chosen area; K/W and K·cm²/W remain different units. For another area, the uniform-model conversion is R = Z/A with matching units, and the assumed area-normalized contact behaviour must still be applicable.

A historical announcement shows why both fields matter

Henkel’s announcement dated April 6, 2022 identifies BERGQUIST LIQUI FORM TLF 10000 and separately reports conductivity of 10.0 W/m-K and thermal impedance of 0.45 K·cm²/W at a 0.5 mm bondline. The units and thickness qualification show that these fields answer different questions.

The release illustrates how a supplier can report both quantities. It does not supply TI-44’s inputs or enough characterization detail to reconstruct separate contact terms. Retain the product, date and stated bondline with the figures; the historical announcement establishes neither present availability nor suitability for a particular GPU assembly.

References: [TIM3]

Ask which supplied assembly the evidence describes

Begin the comparison with the interface the offer actually has to bridge. The evidence should identify the material and supplied form, the relevant bondline or supported thickness range, the two surface materials and their finish or flatness, and the supported mechanical conditions. Stated assembly pressure or loading is a condition to match in the supplier’s evidence, rather than a setting to choose from a general article.

Next, identify the reported quantity, measurement method, temperature basis and whether the value is typical or a stated limit. Ask the supplier to connect the complete resistance or impedance result to those conditions. If contact effects are already included, retain that definition. If only bulk k is supplied, the complete interface comparison remains unresolved.

TI-44 gives a useful way to read an incomplete offer. A package reporting only A’s 10 W/(m·K) cannot establish its stipulated 1.20 K·cm²/W total. Adding its bondline establishes the 1.00 bulk term under the model, but still leaves the contacts open. The decisive information is the applicable complete result, or a justified decomposition that supplies those missing terms.

The thinner B layer is a premise of the example, not an available adjustment for every product. A real gap and supported mechanical arrangement constrain what can be supplied. Neither a smaller arbitrary bondline nor a higher catalog k closes that assembly question. Material selection and any change to an existing assembly belong with its appointed product and thermal specialists.

Keep other thermal quantities at their own boundaries. Package theta or psi metrics and a device’s transient thermal-impedance curve do not provide the two surface contacts and TIM layer required here. The related thermal-metrics and pulse-response guides explain those separate comparisons.

References: [TIM1][TIM2]

Carry the identified system into a physical installation discussion

After the equipment owner establishes the complete system and its requirements, the physical installation is a separate project decision. Leviathan’s existing rack-and-stack service is the relevant route for discussing agreed equipment mounting, IT cabling, labels and physical handoff records. That discussion concerns the identified equipment and site; it makes no thermal-material selection or rework offer.

Use the linked service and project inquiry to share the exact equipment models and quantities, delivered assembly state, rack positions, external connection plans and requested physical work. Those inputs allow a discussion of a separately agreed installation scope, requirements and availability.