PCB Trace Width Calculator (IPC-2221)

Find how wide a PCB trace must be to carry a given current without heating up more than you allow.

Minimum width
In mils
Cross-section

Runs entirely in your browser. Nothing you enter is sent anywhere.

How it works

This uses the IPC-2221 design chart equation, which links current, the trace's cross-sectional area and its temperature rise. Internal layers are surrounded by laminate and dissipate heat less well, so they need roughly twice the area of external traces.

Copper weight sets the thickness: 1 oz/ft² copper is about 35 µm (1.378 mil) thick. The result is a minimum; widen power traces where space allows, and check voltage drop on long runs.

Area (mil²) = (I / (k × ΔT^0.44))^(1/0.725)
Width (mil) = Area / (thickness_oz × 1.378)
k = 0.048 external, 0.024 internal

Worked example

2 A on an external 1 oz layer with a 10 °C rise needs an area of (2 / (0.048 × 10^0.44))^(1/0.725) = 42.4 mil², a width of 42.4 / 1.378 = 30.8 mil (0.78 mm).

Frequently asked questions

What temperature rise should I allow?

10 °C is a common, conservative choice. Some designs allow 20–30 °C, but warmer traces raise the board temperature and the resistance of the copper.

Is IPC-2221 still accurate?

It's widely used and conservative for most designs. The newer IPC-2152 standard models heat flow in more detail and often allows narrower traces, especially on boards with large copper planes.

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