The IPC-2221 formula
A PCB trace carrying current heats up; the trace must be wide enough that the temperature rise stays acceptable. IPC-2221 gives the empirical relationship between current, cross-sectional area, and temperature rise:
A = (I / (k · ΔT0.44))1/0.725
with area A in square mils, current I in amps, ΔT the rise in °C, and k = 0.048 for external (surface) traces or 0.024 for internal ones. Divide the area by the copper thickness to get the width. Copper weight sets that thickness: 1 oz/ft² ≈ 35 µm (1.37 mil), ½ oz half that, 2 oz double.
Worked example
2 A on an external 1 oz trace at a 10 °C rise: the formula gives about 26 mil (0.66 mm) wide. On an internal layer (k halved) the same current needs roughly 60 mil — internal traces run hotter because they can't shed heat to the air.
Why external and internal differ so much
The IPC-2221 curves bake in that a buried trace is a worse radiator, so it needs far more copper for the same current and rise. That factor-of-two in k translates to a big width difference — always pick the layer your trace actually runs on.
Frequently asked questions
- Is IPC-2221 accurate?
- It's deliberately conservative and dates from older data; the newer IPC-2152 accounts for board material, thickness, and adjacent copper, and often permits narrower traces. For most hobby and general work IPC-2221 is a safe starting point — verify thermally for high-current or dense designs.
- What temperature rise should I choose?
- 10 °C is a common conservative default; 20 °C is often acceptable and allows narrower traces. The rise is above ambient, so a hot enclosure eats into your margin.
- What about vias?
- Vias carry current too and have their own limits; a rough rule is to treat a plated via like a short trace of similar copper area, and parallel several for high current. This tool sizes traces, not vias.
- How much does a trace drop?
- Enter a trace length to see its resistance, voltage drop, and I²R heating. For wiring rather than traces, see the voltage drop calculator.