How to use
- Enter the continuous current through the trace in amperes.
- Enter the temperature rise above ambient you accept, for example 10 °C.
- Choose the copper weight and whether the trace is on an outer (external) or inner (internal) layer.
- Read the minimum width in millimetres and mils. Round up to your design grid.
Formula
IPC-2221 gives the current a trace can carry:
I = k × ΔT^0.44 × A^0.725
- I — current, A; ΔT — temperature rise, °C; A — cross-section, mil²;
- k = 0.048 for external layers, k = 0.024 for internal layers.
Solved for the cross-section: A = (I / (k × ΔT^0.44))^(1 / 0.725). The width is A divided by the copper thickness (1 oz = 1.378 mil), and 1 mil = 0.0254 mm.
Worked example
1 A on an outer layer of 1 oz copper with a 10 °C rise (the default values):
- 10^0.44 = 2.754; 1 / (0.048 × 2.754) = 7.564;
- A = 7.564^(1/0.725) = 16.3 mil²;
- width = 16.3 / 1.378 = 11.8 mil, or 0.3 mm.
Twice the current, 2 A, on an inner layer of 1 oz copper needs 110.3 mil², a width of 2.033 mm.
Tips and limits
- The formula comes from measured curves for fairly ordinary boards. Treat the result as a minimum and use wider traces where there is room.
- Short spikes matter less than the continuous (RMS) current: size the trace for the current it carries for minutes, not microseconds.
- For high currents, pour copper areas, use thicker copper or route the current on several layers in parallel.
- Wide traces also lower voltage drop: the resistance of a trace falls in proportion to its cross-section.