PCB Thermal Conductivity: Direction and Stackup Models
PCB Thermal Conductivity Depends on Direction and Stackup
PCB thermal conductivity is not usually one number that describes every heat path. Copper patterns, insulating layers, and plated vias make the finished board conduct heat differently along its surface and through its thickness. A useful thermal model therefore specifies direction, geometry, and material assumptions before assigning an effective conductivity.
Identify where heat must travel. Lateral spreading does not necessarily solve the path to a rear heat sink, and additional vias do not remove an enclosure-to-air bottleneck.
Separate Conductivity From Thermal Resistance
Thermal conductivity, k, describes conduction within a material and uses W/(m·K). Thermal resistance describes a particular heat path and uses K/W. For an ideal uniform slab under one-dimensional steady conduction:
RTH = L / (k × A)
Here L is path length in metres and A is the area normal to heat flow in square metres. This relation does not include spreading, contact resistance, or convection. TI's PCB temperature-sensor design report, SNOA967A develops separate longitudinal, perpendicular, and via paths rather than treating a board as one uniform conductor.
Larger k does not guarantee lower assembly resistance when thickness, contact area, or interfaces also change.
Why the Direction Changes the Answer
In-plane heat flow travels approximately parallel to copper layers. Broad, connected copper can provide a spreading path, whereas slots, necks, and sparse traces interrupt that path. Through-plane flow must cross the stack thickness, where low-conductivity dielectric layers often dominate unless conductive structures provide alternate routes.
An anisotropic conductivity model can describe those directional differences. Conductivity is represented by a tensor in the general case, not by calling k a heat-flow vector. Heat flux and temperature gradient are the vectors related by that material model. COMSOL's conductivity formulation explains the relationship for layered materials.
Two Ideal Layered Models, Not One Universal Average
For continuous planar layers of equal footprint, let fi be each layer's thickness fraction, with all fractions adding to one. Assume isotropic constituent materials, constant properties, perfect interfaces, and no vias. Under those ideal conditions:
- Parallel to the layers, add their heat-carrying contributions:
kPARALLEL = Σ(fi × ki). - Perpendicular to the layers, add their series resistances:
kPERPENDICULAR = 1 / Σ(fi / ki).
These are respectively arithmetic and harmonic weighted averages. COMSOL documents the parallel-versus-perpendicular averaging distinction. Applying the arithmetic expression to every board direction can substantially overstate vertical conduction.
An Illustrative Stack Shows the Difference
Consider a deliberately simplified stack containing 10% metal and 90% dielectric by thickness. For arithmetic illustration only, assign the metal k = 400 W/(m·K) and dielectric k = 0.30 W/(m·K). These are assumed inputs, not a specification for any laminate or fabricated board.
The parallel calculation gives:
0.10 × 400 + 0.90 × 0.30 = 40.27 W/(m·K)
The perpendicular calculation gives:
1 / (0.10 / 400 + 0.90 / 0.30) ≈ 0.333 W/(m·K)
The same fractions produce different effective values because the paths connect differently. Extra plane copper can alter lateral conduction without removing the vertical dielectric bottleneck.
Real layouts depart from both idealizations. A copper percentage says little about whether islands are connected in the direction of interest. Small heat sources also create three-dimensional spreading. Use these models to check reasoning and bracket early assumptions, then represent important geometry explicitly.
Copper Planes and Vias Solve Different Parts of the Path
Plane copper spreads heat toward more board area; via barrels connect conductive regions across dielectric thickness. Neither is a heat sink by itself. TI's thermal-layout guidance for motor drives connects copper continuity, via construction, and the final heat-sink path.
A plated via conducts mainly through its metal barrel. It is not a miniature phase-change heat pipe. Its contribution depends on plating thickness, length, attachment to planes, and the number and placement of vias. “Filled” is insufficient as a thermal specification: conductive fill and nonconductive resin do not provide the same added path. Assembly requirements also constrain holes placed beneath exposed pads.
Before changing laminate, ask whether heat can reach the existing copper. A narrow connection from a hot pad may dominate even when a large plane exists nearby. Before adding more vias, ask where the opposite copper surface sends the heat next.
Turn the Model Into a Verification Plan
Start with dissipated power, not simply supply power or a universal wattage threshold. Then identify the intended sink and the permitted temperature rise. Obtain material conductivity for the relevant direction, temperature, and test method; do not infer it from glass-transition temperature alone.
For a practical review, change one assumption at a time: copper coverage, via construction, interface resistance, airflow, or enclosure temperature. If only airflow changes the predicted hot spot appreciably, further copper thickening is unlikely to be the first useful iteration. This sensitivity check is an engineering decision method, not a promise of a particular temperature reduction.
Finally, compare prototype temperatures at documented loads and ambient conditions. Surface readings are not automatically semiconductor junction readings. Record the measurement location and method, and use the component's applicable thermal guidance when estimating junction temperature.
Frequently Asked Questions
Can I use the laminate's conductivity for the whole PCB?
Only if that approximation is appropriate to the modeled path. Patterned copper and vias can make the board's effective directional behavior substantially different from the bare laminate.
Does higher PCB conductivity always improve the product?
No. It may move heat toward a sensitive sensor or temperature-dependent circuit. The objective is the intended heat path and acceptable component temperatures, not the largest possible conductivity everywhere.
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