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PCB Trace Width Calculator

Enter the current the trace must carry, the allowed temperature rise and the copper weight. The calculator applies the IPC-2221 formula for external or internal layers.

Result

Minimum trace width
30.76 mil
Trace width in millimeters
0.781 mm
Cross-section area
42.39 sq mil
Copper thickness
1.378 mil

IPC-2221 charts cover currents up to 35 A, temperature rises from 10 to 100 °C and widths up to 400 mil. Add margin for manufacturing tolerance.

Formula

  • Area (mil²) = (I ÷ (k × ΔT^0.44))^(1 ÷ 0.725)
  • Width (mil) = area ÷ (copper weight in oz × 1.378)
  • k = 0.048 for external layers and 0.024 for internal layers

How the PCB Trace Width Calculator works

A trace heats up as it carries current. IPC-2221 gives an empirical relation between current, the cross-section of the copper and the temperature rise above ambient. External traces cool better than internal ones and can be about half the area for the same current.

Copper weight is the mass of copper per square foot of board. One ounce copper is 1.378 mil thick, which is about 35 micrometers.

Worked example

For 2 A with a 10 °C rise on an external 1 oz layer: area = (2 ÷ (0.048 × 10^0.44))^(1 ÷ 0.725) = 42.4 mil². Dividing by 1.378 mil gives a width of 30.8 mil, or 0.78 mm.

Frequently asked questions

How wide should a trace be for 1 amp?

On an external 1 oz layer with a 10 °C rise, about 11.8 mil or 0.30 mm. Internal layers need about 30.8 mil.

What is 1 oz copper thickness?

1.378 mil, which is 0.035 mm. Two ounce copper is twice as thick, so a trace can be half as wide for the same current.

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