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PCB Trace Width Calculator (IPC-2221)

PCB trace width for a current and temperature rise, or the current a trace can carry, per IPC-2221 — with resistance, voltage drop and power loss.

Find
Minimum width
Cross-section
Resistance (hot)
Voltage drop
Power loss
Trace temperature

IPC-2221 formula with annealed-copper resistivity. Checked against the formula written out independently and the standard 1 A / 10 °C reference values.

How to use

  1. Choose whether to find the width for a current or the current for a width.
  2. Enter the current or width, the allowed temperature rise, the copper weight and whether the trace is on an outer or inner layer.
  3. Optionally give the ambient temperature and trace length to see the resistance, voltage drop and power loss at operating temperature.

About this tool

Show

IPC-2221 relates the current a copper trace carries to how much it heats up: I = k · ΔT^0.44 · A^0.725, with A the cross-section in square mils and k = 0.048 for outer layers, 0.024 for inner ones. This calculator runs it both ways — the width needed for a current, or the current a given width can take — for any copper weight (0.5 to 4 oz/ft² or a thickness in µm or mil, with 1 oz taken as 1.378 mil = 35 µm) and any allowed rise above ambient. With a trace length it also gives the resistance at the operating temperature (annealed copper, 1.7241 µΩ·cm at 20 °C, rising 0.393 % per °C), the voltage drop and the power the trace dissipates. The classic check: 1 A with a 10 °C rise on 1 oz copper needs 11.8 mil (0.30 mm) on an outer layer and 30.8 mil (0.78 mm) inside the board. Treat the answer as a starting point, not a guarantee. The formula is a curve fitted to measurements from the 1950s on isolated traces in still air, and it knows nothing about copper pours, vias, neighbouring hot parts, board thickness or airflow. It was drawn for up to 35 A, widths up to 400 mil and rises of 10 to 100 °C; beyond that the tool still answers but says so. IPC-2152, which replaced these charts in 2009, found the inner-layer factor far too pessimistic — internal traces usually run cooler than IPC-2221 predicts because the laminate spreads heat — so an inner-layer width from here errs on the safe side. Manufacturing limits such as minimum width, spacing and the plating added to outer layers are separate questions.

Frequently asked questions

How wide should a trace be for 1 A?

With a 10 °C rise on 1 oz copper, IPC-2221 gives 11.8 mil (0.30 mm) on an outer layer and 30.8 mil (0.78 mm) on an inner layer. Allowing a 20 °C rise brings the outer-layer figure down to about 7.8 mil. Most designers round up generously for anything that matters.

Why do inner layers need wider traces?

IPC-2221 halves k for internal traces because they are buried in laminate rather than exposed to air. Later measurements for IPC-2152 showed the laminate conducts heat well enough that inner traces usually run cooler than that, so the IPC-2221 inner-layer answer is conservative.

What temperature rise should I allow?

10 °C is the traditional conservative choice; 20 to 30 °C is common for power traces. The limit is set by the board's material and the parts nearby: the trace temperature is ambient plus rise, and it must stay well under the laminate's rating (often 130 °C for FR-4) and whatever the neighbouring components tolerate.

Does the resistance include the temperature rise?

Yes. It is computed at ambient plus the allowed rise, using copper's temperature coefficient of 0.393 % per °C, so a trace running 30 °C above a 25 °C ambient has about 14 % more resistance than at 20 °C.

How thick is 1 oz copper?

One ounce of copper spread over one square foot is 1.378 mil, or about 35 µm. Half-ounce is about 17.5 µm and 2 oz about 70 µm. Outer layers usually gain some plating during manufacture, which the nominal weight does not include.

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