Thermal Interface Material Calculator
Calculate TIM bulk thermal resistance, temperature drop, heat transfer, required conductivity, bond-line thickness, contact area, and TIM comparison results. Use it for TIM-specific bond-line thickness, effective contact area, thermal impedance interpretation, and known contact resistance checks.
Engineering tool
Thermal Interface Material Calculator
Calculate TIM bulk thermal resistance, temperature drop, heat transfer, required conductivity, bond-line thickness, contact area, and installed TIM comparisons.
Use W/(m·K), not K/W or K·cm²/W thermal impedance.
Use effective installed or compressed thickness when available.
Use effective heat-transfer contact area, not automatically the whole footprint.
Leave off for bulk-only TIM conduction. Off does not mean real contact resistance is zero.
Result console
- Total Thermal Resistance
- 0.1K/W
- Ideal Bulk Thermal Resistance
- 0.1K/W
- Area-Normalized Resistance
- 1K·cm²/W
- Area-Normalized Resistance
- 0.000100000m²·K/W
- Thermal Conductance
- 10W/K
The model is steady-state one-dimensional TIM conduction. It excludes spreading resistance unless entered separately through known contact resistance.
Formula
Formula reference
TIM one-dimensional steady-state conduction
The core formula calculates ideal bulk resistance only unless known total contact resistances are entered.
Rθ,bulk = L/(kA)Gθ = 1/Rθ = kA/LR''θ = RθA = L/kΔT = P Rθ,totalQdot = kAΔT/Lkrequired = PL/(AΔTmax)Lmax = kAΔTmax/PArequired = PL/(kΔTmax)Rθ,total = Rc1 + Rθ,bulk + Rc2Variable definitions
- k
- through-plane thermal conductivity, W/(m·K)
- L
- TIM thickness / bond-line thickness, m
- A
- effective contact area, m²
- P or Qdot
- heat flow, W
- R''θ
- area-normalized resistance
Thermal Interface Material Formula Audit
| Bulk Thermal Resistance | Rθ,bulk = L/(kA). |
|---|---|
| Thermal Conductance | Gθ = 1/Rθ = kA/L. |
| Area-Normalized Resistance | R''θ = RθA = L/k. |
| Temperature Drop | ΔT = P Rθ,total. |
| Heat Transfer | Qdot = kAΔT/L = ΔT/Rθ. |
| Required Conductivity | krequired = PL/(AΔTmax). |
| Maximum Thickness | Lmax = kAΔTmax/P. |
| Required Area | Arequired = PL/(kΔTmax). |
| Known Contact Resistance | Rtotal = Rc1 + Rbulk + Rc2. |
| Thermal Conductivity Units | W/(m·K), not K/W. |
| Thickness Units | m, mm, µm, mil converted to meters. |
| Area Units | m², cm², mm², in² converted to square meters. |
| Area-Normalized Units | m²·K/W and K·cm²/W are not total K/W. |
| Temperature Difference | 1°C difference = 1 K difference = 1.8°F difference. |
| Bulk vs Contact | L/(kA) excludes real contact resistance unless known values are entered. |
| Conductivity vs Impedance | Datasheet thermal impedance must not be entered into the k field. |
| Bond-Line Thickness | Use effective installed thickness. |
| Through-Plane k | Use through-plane conductivity for through-thickness TIM conduction. |
| Compression / Pressure | No pressure-to-resistance model is inferred. |
| Spreading / Steady State | No spreading resistance, constriction resistance, or transient thermal capacitance. |
Worked Examples
| Case | Calculation | Result |
|---|---|---|
| k=5, L=.5mm, A=1000mm² | 0.0005/(5×0.001) | Rθ=.1K/W |
| Same, P=20W | 20×0.1 | ΔT=2K |
| Same, ΔT=2K | 2/0.1 | Q=20W |
| P=20W, L=.5mm, A=1000mm², ΔT=2K | PL/(AΔT) | k=5W/mK |
| k=5, A=1000mm², P=20W, ΔT=2K | kAΔT/P | Lmax=.5mm |
| k=5, L=.5mm, P=20W, ΔT=2K | PL/(kΔT) | A=1000mm² |
| R''=L/k | .0005/5 | .0001m²K/W |
| .0001m²K/W | ×10000 | 1K·cm²/W |
| Rθ=.1K/W | 1/R | G=10W/K |
| Double thickness | L doubles | R doubles |
| Double area | A doubles | R halves |
| Double k | k doubles | R halves |
| 10W/mK 2mm vs 5W/mK .2mm | compare L/k | lower-k thinner TIM wins |
| Rc1=.02, Rbulk=.1, Rc2=.03 | sum | Rtotal=.15K/W |
| Rtotal=.15, P=20W | 20×.15 | ΔT=3K |
| 1mm | ÷1000 | .001m |
| 100µm | ÷1e6 | .0001m |
| 1mil | 25.4µm | 2.54e-5m |
| 1cm² | ×1e-4 | 1e-4m² |
| 1000mm² | ×1e-6 | .001m² |
| 1in² | 0.00064516 | m² |
| 1K/W | temperature interval | 1°C/W numerically |
| Δ10°C | same interval | 10K |
| Δ10°C | ×1.8 | 18°F |
| k=0 | invalid | rejected |
| L=0 | invalid physical TIM | rejected |
| A=0 | division by zero avoided | rejected |
| Rc<0 | invalid | rejected |
| NaN | finite check | rejected |
| Infinity | finite check | rejected |
Engineering Notes
- Ideal TIM bulk thermal resistance is L/(kA).
- Thermal conductivity is a material property; thermal resistance depends on geometry.
- Higher conductivity alone does not guarantee lower installed resistance.
- Thinner TIM and larger effective contact area reduce ideal bulk resistance.
- Thermal conductivity and thermal impedance are different quantities.
- Area-normalized thermal resistance includes area in the unit.
- Real interface resistance can include contact resistance on both surfaces.
- Bond-line thickness may differ from nominal material thickness.
- Thermal pads may compress under mounting pressure.
- Thermal grease bond-line thickness depends on assembly.
- Use through-plane conductivity for through-thickness interface conduction.
- The simple model excludes spreading resistance and transient behavior.
Common Mistakes
- Entering millimeters as meters.
- Entering mm² as m².
- Entering W/mK conductivity as K/W resistance.
- Entering thermal impedance as thermal conductivity.
- Ignoring bond-line thickness.
- Comparing only k instead of L/(kA).
- Assuming full package footprint is effective contact area.
- Using nominal instead of compressed pad thickness.
- Ignoring contact resistance.
- Using in-plane k for through-plane conduction.
- Treating R'' directly as total Rθ.
- Treating the bulk model as complete package-to-heatsink resistance.
Support reference
FAQ
What is a thermal interface material?
A thermal interface material, or TIM, is a pad, grease, paste, gap filler, film, or sheet used to improve heat transfer between two surfaces.
How do I calculate TIM thermal resistance?
For ideal one-dimensional bulk conduction, use Rθ = L/(kA), where L is bond-line thickness, k is through-plane thermal conductivity, and A is effective contact area.
What is the formula Rθ = L/(kA)?
It is the Fourier conduction result for a uniform planar material layer with constant thermal conductivity and effective area.
How does TIM thickness affect thermal resistance?
For fixed k and area, thermal resistance increases linearly with bond-line thickness.
How does contact area affect TIM thermal resistance?
For fixed k and thickness, larger effective contact area lowers bulk thermal resistance.
Does higher thermal conductivity always mean a better TIM?
No. Installed resistance depends on k, thickness, area, contact resistance, pressure, and assembly conditions.
What is bond-line thickness?
Bond-line thickness is the effective installed thickness of the TIM layer between mating surfaces.
Should I use nominal or compressed thermal-pad thickness?
Use compressed or effective installed thickness when available, because that is what appears in L/(kA).
What is the difference between thermal conductivity and thermal resistance?
Thermal conductivity is a material property in W/(m·K). Thermal resistance is a geometry-dependent result in K/W.
What is thermal impedance?
TIM datasheet thermal impedance is often area-normalized resistance-like data such as K·cm²/W, not W/(m·K).
What does K·cm²/W mean?
It is an area-normalized thermal resistance unit. To get total K/W, divide by contact area in matching area units.
How do I convert thermal impedance to thermal resistance?
If R'' is in K·cm²/W, divide R'' by area in cm² to obtain K/W.
What is area-normalized thermal resistance?
It is R''θ = Rθ × A = L/k, with area included in the unit.
What is TIM contact resistance?
Contact resistance is additional resistance at each material-surface interface due to roughness, pressure, wetting, and assembly.
Does L/(kA) include contact resistance?
No. L/(kA) only estimates ideal bulk material conduction unless known contact resistances are added.
How does mounting pressure affect TIM performance?
Pressure can change pad compression, grease bond-line thickness, voids, and real contact resistance.
What is through-plane thermal conductivity?
It is conductivity through the TIM thickness, which is the correct direction for interface conduction.
Can I use in-plane conductivity for a thermal pad calculation?
Not for through-thickness interface resistance. Use through-plane conductivity when the datasheet separates directions.
How do I calculate temperature drop across a TIM?
Calculate total TIM resistance, then use ΔT = P × Rθ.
How do I calculate required TIM conductivity?
Use krequired = P L / (A ΔTmax), after converting L to meters and A to square meters.
How do I calculate maximum TIM thickness?
Use Lmax = k A ΔTmax / P.
How is THM-007 different from the heat sink calculator?
THM-007 estimates the TIM layer contribution from material geometry, while the heat sink calculator sizes heatsink-to-ambient resistance.
Related Calculators
Thermal Resistance & Temperature Rise Calculator
AvailableUse known Rθ and heat load to calculate temperature rise.
Open calculatorHeat Sink Thermal Resistance Calculator
AvailableSize a generic heatsink thermal path once interface resistance is known.
Open calculatorThermal Resistance Network Calculator
AvailableCombine TIM, case, heatsink, chassis, and ambient thermal resistances.
Open calculatorMOSFET Thermal Resistance Calculator
AvailableEvaluate MOSFET-specific junction and heatsink thermal paths.
Open calculatorDocumentation
Related Engineering Resources
Guides and lookup references linked to this tool.
Engineering Guide
Thermal Design Workflow for Electronics: Junction, Case, Board, Heat Sink, and Enclosure
Build and verify electronics thermal paths using power loss, thermal resistance networks, junction limits, heat sinks, TIMs, conduction, convection, transient response, and enclosure constraints.
22 min · Intermediate
Engineering Guide
Power Dissipation and Component Power Ratings
Calculate actual component dissipation and choose safe power ratings using thermal resistance, derating, junction temperature, SOA, pulse conditions, PCB heat spreading, and real operating conditions.
20 min · Intermediate