How this calculator works
Current flowing through a copper trace generates heat because copper has resistance. If a trace is too narrow it runs hot. The IPC-2221 standard links a trace’s current capacity to its cross-sectional area and the temperature rise you allow above ambient.
- I — current in amps
- ΔT — allowable temperature rise (°C)
- k — 0.048 external, 0.024 internal
- A — cross-sectional area (mil²)
- t — copper thickness (1 oz ≈ 1.378 mil)
- W — resulting trace width (mil)
More detail
External vs. internal traces
Traces on the outer layers shed heat to the air and can be narrower. Internal traces are buried in the board with nowhere for heat to escape, so IPC-2221 halves the constant (k = 0.024) — which raises the width needed for the same current by roughly 2.6× (the ½-constant is applied through the 1/0.725 exponent, not linearly). Switch Internal above to see the difference update live.
Design tip. IPC-2221 is intentionally conservative. For dense boards, check the newer IPC-2152 data too, and add margin for vias, connectors, and long runs. When in doubt, go wider — copper is cheap, reworks are not.
Frequently asked questions
What trace width do I need for 1 A?
For a 1 oz external trace with a 10 °C rise, IPC-2221 gives roughly 0.30 mm (≈11.8 mil). Enter your exact current and copper weight above for the precise value — an internal trace needs roughly 2.6× the width.
Should I use IPC-2221 or IPC-2152?
IPC-2221 (used here) is the older, more conservative standard and is fine for most designs. IPC-2152 is based on newer thermal data and often allows narrower traces, but needs more inputs. Use IPC-2221 for a safe first pass, then check IPC-2152 for high-density or high-current boards.
Why do internal traces need to be wider?
Internal traces are buried in the board and can't shed heat to the air, so IPC-2221 halves the current constant (k = 0.024). For the same current and temperature rise, that raises the required width to roughly 2.6× that of an external trace.