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PCB trace width calculator

Current
A
Allowed temperature rise
°C
Copper weight
oz
Layer
Trace width 30.8 mil
IPC-2221: A = (I ÷ (k × ΔT^0.44))^(1/0.725)
In millimetres 0.781 mm
Cross-section 42.4 mil²
Layer type external
Copper weight 1 oz
IPC-2221 · external k = 0.048, internal 0.024

An internal layer is buried in FR4, which is a poor conductor of heat, so it cannot shed the same power as a trace on the surface exposed to air. IPC-2221 handles that with a different constant: 0.048 for external traces against 0.024 for internal. In practice an internal trace needs roughly twice the width of an external one for the same current and temperature rise.

IPC-2221 is a conservative general guideline derived from measurements on isolated traces. Dense boards, high ambient temperatures and long traces all warrant more margin, and IPC-2152 supersedes it for serious thermal work.

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IPC-2221 sizes a trace from cross-sectional area: A = (I ÷ (k × ΔT^0.44))^(1/0.725), where k is 0.048 external and 0.024 internal. Two amps on an external 1 oz layer at a 10 °C rise needs about 33 mil, or 0.84 mm.

How to size a PCB trace

1 Enter the current the trace will carry.
2 Choose an acceptable temperature rise — 10 °C is conservative, 20 is common.
3 Set the copper weight; 1 oz is standard.
4 Choose external or internal — internal traces need roughly double the width.

The temperature rise is the design choice, and it is more forgiving than instinct suggests. A trace that runs 20 °C above ambient is entirely normal and is not damaging anything on its own — the limit is the board glass transition temperature and whatever heat-sensitive part sits nearby. What the formula does not capture is context: a trace next to a hot regulator starts from a higher ambient, a trace over a ground plane sheds heat better, and a via carrying the same current is a far more constrained path than the trace either side of it. For power designs, the vias usually fail before the traces.

Questions

For 2 A on external 1 oz copper at a 10 °C rise, about 0.84 mm. Doubling the copper weight roughly halves the width needed.

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