Thermal Design on the Board
Copper area and thermal vias, how hot a part gets, and keeping heat from the parts that dislike it.
The heat path
A part that dissipates power P heats up until the heat leaving it equals P. The heat travels from the silicon (the junction) through the package to the board, along the copper and into the air. Each step has a thermal resistance in kelvin per watt, and the temperatures add up like voltages in a series circuit.
Tj = Ta + P × θJA
The data sheet gives θJA for a standard test board (JEDEC, 76 × 114 mm, usually with a copper plane): your board will differ, mostly by the copper area, the number of layers and whether there is airflow. The one thing the layout designer controls is how well the heat gets from the package into copper and out of the board.
Copper area: the cheapest heat sink
Spreading the heat into a large copper area (a pour on the top layer, connected on to planes) lowers θJA. A surface in still air loses heat at about 10 W per square meter per kelvin (natural convection plus radiation); two sides of a board double this.
Rcopper ≈ 1 ÷ (h × A), h ≈ 10 W/(m²·K)
Worked example: copper area under a SOT-223 regulator
- The package on a minimum pad has θJA of about 120 K/W (typical data-sheet value). A pour of 10 cm² loses heat at R = 1 ÷ (10 × 10×10−4) = 100 K/W; with the 15 K/W from the package to the copper that path is 115 K/W, in parallel with the package’s own 120 K/W gives 59 K/W.
- At the 0.85 W the regulator dissipates (below), the rise is 50 K instead of 102 K on a minimum pad.
| Copper area (one side) | Pour resistance | Total θJA | Temperature rise at 0.85 W |
|---|---|---|---|
| 1 cm² | 1 000 K/W | 107 K/W | 91 K |
| 2 cm² | 500 K/W | 97 K/W | 83 K |
| 5 cm² | 200 K/W | 77 K/W | 65 K |
| 10 cm² | 100 K/W | 59 K/W | 50 K |
| 20 cm² | 50 K/W | 42 K/W | 36 K |
Estimates for still air; real values depend on the board, the airflow and the other parts. The gain flattens above about 10 cm²: spreading resistance and the finite board limit it.
- Connect the thermal pad to a large pour with solid connections (no thermal relief spokes under a thermal pad).
- Use both sides and join them with vias.
- Use a heavier copper weight: 2 oz copper spreads heat twice as well (the spreading conductance k t is 13.48 mW/K per square for 1 oz and 26.95 for 2 oz).
- Leave air space and room above the board: a board in a sealed box or against a wall runs far hotter than one in free air.
Thermal vias
A plated via is a small copper tube. Placed under the thermal pad of a package, an array of them carries heat to the planes and to the bottom side, where it spreads into a large copper area. Each barrel (0.3 mm drill, 25 µm plating) in a 1.6 mm board is 163 K/W; the barrels act in parallel.
| Number of vias | Total resistance | Rise at 0.85 W |
|---|---|---|
| 1 | 163 K/W | 138 K at 0.85 W |
| 4 | 41 K/W | 35 K at 0.85 W |
| 9 | 18 K/W | 15 K at 0.85 W |
| 16 | 10 K/W | 9 K at 0.85 W |
Worked example: vias under a QFN pad
- A 3.2 × 3.2 mm exposed pad fits floor(3.2 ÷ 1.0)² = 9 vias at a 1.0 mm pitch (a typical grid for 0.3 mm vias).
- Barrel resistance with 9 vias: 163 ÷ 9 = 18 K/W. At 2.0 W that is a 36 K drop through the board (the solder layer adds only 0.20 K).
- Thicker plating and a smaller via pitch reduce this further; a pad on the bottom side that is soldered to a heat sink reduces it more.
- Filled or tented vias: open vias under a pad can wick the solder away from the joint and leave a void (see Joint Quality, Inspection and Failure). Use 0.3 mm vias with solder-mask tenting on the back, small vias (0.2 mm) or filled and capped vias, as the assembler recommends.
- Do not use vias too small for the plating: the barrel resistance goes up and the vias may not plate well.
- Spread the vias over the pad and out into the pour as well, not just under the part.
Worked example: a linear regulator
Worked example: 5 V to 3.3 V at 0.5 A
- A linear regulator dissipates P = (Vin − Vout) × I = (5.0 − 3.3) × 0.5 = 0.85 W (efficiency 66 %).
- With the 10 cm² pour (θJA 59 K/W): the junction is 50 K above ambient: 75 °C in a 25 °C room and 110 °C at 60 °C inside a case: below the 125 °C limit, but with less margin (the maximum power at 60 °C is 1.11 W).
- A 90 % efficient buck converter delivering the same 1.65 W of output loses only 0.18 W (the input power would be 1.83 W against 2.50 W): 5 times less heat.
| Ambient | Maximum power for Tj = 125 °C |
|---|---|
| 25 °C | 1.70 W |
| 40 °C | 1.45 W |
| 60 °C | 1.11 W |
| 85 °C | 0.68 W |
Rule: when the power lost in a linear regulator exceeds about 0.5 W, think of a switching regulator (see Power, Ground and Decoupling).
Power parts: QFN and aluminum boards
For LEDs and power transistors, the board’s own dielectric is the barrier. FR-4 conducts heat poorly (about 0.3 W/m·K); a metal-core (aluminum) board has a thin dielectric of 1 to 3 W/m·K between the copper and the aluminum base.
Worked example: an LED on FR-4 and on an aluminum core
- An LED with a 10 mm² thermal pad (area 10 mm²) dissipates 3.0 W.
- Through 1.5 mm of FR-4 (k = 0.3 W/m·K): R = t ÷ (k A) = 1.5 mm ÷ (0.3 × 10 mm²) = 500 K/W (the heat spreads laterally, so the real figure is lower, but still tens of K/W).
- Through a 0.1 mm aluminum-core dielectric (k = 2.0 W/m·K): 5.0 K/W. At 3.0 W: 15 K against 1 500 K: the metal-core board is the practical choice for LEDs above about 1 W.
Keeping heat away from sensitive parts
- Distance: put warm parts at one edge, with air above them, and keep the sensors, crystals, batteries and electrolytic capacitors away (see Component Placement and Floorplanning).
- Thermal isolation: cut slots or use narrow copper connections between a hot area and a precision sensor, and use the sensor’s own ground and power through a small relief.
- Temperature sensors: put them where the temperature to be measured is, with good thermal contact and thermal isolation from the board’s own heat.
- Orient parts so that the cooling air or natural convection (rising upward) passes the hot ones last.
Measuring and checking
- Calculate first, then measure: the numbers in this page are a first estimate.
- Thermocouple on the package or the pad (a fine-wire type K, taped or glued): accurate to about 1 to 2 K if the wire is thin; a thick wire conducts heat away and reads low.
- Thermal camera: good for finding hot spots; set the emissivity (about 0.95 for the package, 0.1 or less for bare copper, so paint or tape the spot).
- Test in the enclosure at the highest ambient temperature and the worst load for long enough to reach steady state (tens of minutes).
- Remember the junction is hotter than the case: Tj = Tcase + P × θJC.
Troubleshooting
| Symptom | Likely cause | What to do |
|---|---|---|
| Regulator shuts down or gets too hot | Too much power, too little copper | More copper and vias; lower the input voltage; use a switching regulator |
| Thermal pad is cold-soldered or has voids | Open vias drain the solder; too much paste | Fill or tent the vias; window-pane stencil |
| Hot spot around a power resistor | Resistor dissipation concentrated | Bigger package, more resistors in parallel, copper under it |
| Electrolytic capacitor fails early | Next to a hot part | Move away; use a 105 °C or 125 °C type |
| Board discolored or delaminated | Overheating beyond Tg | Reduce power; high-Tg material; check solder profile |
| Temperature varies with orientation | Natural convection depends on position | Test in the final position; design for the worst case |
Related
- PCB Design and Layout: the overview of this guide.
- Component Placement and Floorplanning: where the warm parts go.
- Power, Ground and Decoupling: regulators and efficiency.
- System Integration: enclosure temperature rise.
FAQ
How much copper do I need for a regulator?
For about 1 W, a pour of 10 cm² connected to the thermal pad gives a rise of about 60 K; for more power use a bigger pour, vias and a better regulator type.
How many thermal vias should I use?
As many as fit under the pad at about a 1 mm pitch (9 for a 3 × 3 mm pad); each carries about 160 K/W, so nine give about 18 K/W.
What is θJA?
The thermal resistance from the junction to the ambient air in K/W, given in the data sheet for a standard test board; yours depends on your copper and airflow.
Should I fill thermal vias?
Open vias in a pad can draw the solder away and cause voids. Use tented, filled or smaller vias, or have the vias outside the soldered area, as your assembler advises.
When do I need an aluminum-core board?
For LEDs and power parts above about 1 W with a small pad, where the FR-4 dielectric would make the pad too hot.
References
- JEDEC JESD51 series (thermal measurement and test boards); IPC-2152.
- Incropera, F. P. et al., Fundamentals of Heat and Mass Transfer, Wiley (convection and conduction).
- Texas Instruments application note Semiconductor and IC Package Thermal Metrics (SPRA953); regulator and QFN data sheets.
Last reviewed: October 2026. Thermal resistances and convection coefficients are typical: measure the finished board.