PCB Trace Width Calculator
Minimum trace width for a current at a chosen temperature rise, by the IPC-2221 method, with the resistance and voltage drop of the finished run.
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The formula
area (mil²) = (I ÷ (k × ΔT^0.44))^(1/0.725) width = area ÷ (copper weight × 1.378) k = 0.048 external, 0.024 internal
This is IPC-2221’s curve fit to measured data rather than a derivation from first principles, which is why the exponents are the odd numbers they are. The two values of k are the whole difference between an external trace, which can shed heat into the air, and an internal one, which cannot.
What to watch out for
The result is a thermal limit, not an electrical one. It is the width at which the trace stops getting hotter than you allowed — the trace will happily carry more current, it will just run hot.
- Internal traces need about twice the width. Buried copper can only lose heat through the laminate, which is a poor conductor.
- Check the voltage drop separately. On low-voltage, high-current rails the drop often demands a wider trace than the thermal rule does. A 3.3 V rail cannot spare 200 mV.
- IPC-2221 is conservative and old. IPC-2152 supersedes it with better data and usually permits narrower traces, but it needs board thickness and plane information and does not reduce to a single equation.
Frequently Asked Questions
What trace width do I need for 1 amp?
On an external 1 oz layer with a 10 °C rise, about 0.3 mm (12 mil). The same current on an internal layer needs roughly 0.75 mm, because buried copper cannot shed heat into the air.
Why do internal traces need to be wider?
They are surrounded by laminate, which insulates. An external trace loses heat to the air as well as the board, so it runs cooler at the same width — which is why IPC halves the k constant for internal layers.
How thick is 1 oz copper?
About 35 µm, or 1.378 mils. The "ounce" refers to the weight of copper spread over one square foot, which is a historical way of specifying thickness.