PCB Trace Width Calculator
Find the minimum PCB trace width for a given current and temperature rise using the IPC-2221 formula, for external or internal copper layers.
How it works
A = (I ÷ (k · ΔT^0.44))^(1/0.725) · width = A ÷ (thickness in mils) · k = 0.048 external, 0.024 internal
A trace heats up as current flows through its resistance, and the IPC-2221 standard relates the allowable temperature rise to the copper cross-section needed to carry a given current. Because inner layers are buried and shed heat poorly, their constant k is half that of outer layers, so they need wider traces for the same current. Once you have the required cross-section, dividing by the copper thickness (0.5, 1 or 2 oz) gives the width — thicker copper lets you run the same current in a narrower track.
Worked example
For 2 A on an external 1 oz layer with a 10 °C rise: A = (2 ÷ (0.048 × 10^0.44))^(1/0.725) ≈ 42.4 mil². One-ounce copper is 1.378 mil thick, so the trace must be at least 42.4 ÷ 1.378 ≈ 30.8 mil, or about 0.78 mm wide.
Sources: IPC-2221 trace-width formula
Frequently asked questions
Is this the trace width or a safety-margin width?
It is the minimum width for the stated temperature rise. Good practice adds margin — widen high-current traces beyond the minimum, and treat a 10 °C rise as a comfortable target rather than 30–40 °C.
Why do inner layers need wider traces?
Outer traces are exposed to air and the board surface and cool relatively well, while inner traces are sandwiched in the laminate. IPC-2221 captures this with a smaller constant for internal layers, which roughly doubles the required cross-section.
Does copper weight change the answer?
Yes. The formula gives a required cross-sectional area; splitting that between width and thickness means 2 oz copper needs only about half the width of 1 oz copper for the same current.
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This calculator is for educational purposes. Double-check important results before acting on them.