PCB trace width calculator (IPC-2221)
The minimum trace width for a current, or the maximum current for a width, on outer and inner layers, with the trace's resistance, voltage drop and power loss. Results update as you type.
Indicative calculation; verify before use.
Board
1 oz/ft² = 35 µm = 1.378 mil
Current → minimum width
- Width
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- Cross-section
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- Resistance
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- Voltage drop
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- Power loss
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- Width
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- Cross-section
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- Resistance
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- Voltage drop
- –
- Power loss
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Width → maximum current
- Voltage drop
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- Power loss
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- Voltage drop
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- Power loss
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Limits of IPC-2221
- The formula is a curve fit to the conductor charts in IPC-2221 (formerly IPC-D-275). The charts only cover up to 35 A, up to 400 mil (10.2 mm) of trace width, a temperature rise of 10 to 100 °C and 0.5 to 3 oz/ft² copper. Outside that range the formula only extrapolates, and this page warns you.
- It is meant for long traces without special heat sinking, not for short necks or thermal-relief spokes. In vacuum or in a potted assembly, use the inner-layer result for outer layers too.
- IPC-2152 (2009) is the newer standard for current-carrying capacity. It is based on new measurements, takes board thickness, board material, copper planes and vias into account, and gives different results; for example, it finds that inner traces can carry more current than IPC-2221 suggests, close to outer traces. This calculator implements the IPC-2221 curve fit, not IPC-2152.
- We have not read IPC-2221 or IPC-2152 themselves; they are paid standards. The formula and constants are as published by Advanced Circuits and DigiKey (sources below), and the results match both of their calculators.
- Resistance is the DC resistance of a uniform rectangular trace at ambient + ΔT. Vias, pads, extra plating and skin effect at high frequencies are not included.
How it works
IPC-2221 relates the current a trace can carry to its cross-section A and how much warmer than the surroundings you let it get (ΔT). An inner trace is surrounded by board material and gets rid of heat less easily, so it gets half the k: for the same current it needs 2^(1/0.725) ≈ 2.6 times the cross-section, which at the same copper thickness is 2.6 times the width. Doubling the copper weight halves the width needed.
The trace's resistance is worked out at its operating temperature, ambient + ΔT, because copper's resistance rises by about 0.4 % per °C. A mil is 0.001 inch (25.4 µm). Copper weight is the mass of copper on one square foot; 1 oz/ft² is taken as 35 µm.
Worked example: 2 A on an outer layer with 1 oz copper and a 10 °C rise. k × ΔT^0.44 = 0.048 × 2.754 = 0.1322, so A = (2 / 0.1322)^(1/0.725) = 42.39 mil² and W = 42.39 / 1.378 = 30.77 mil (0.781 mm). A 50 mm long trace at 25 + 10 = 35 °C has R = 1.724 × 10⁻⁸ × (1 + 0.00393 × 15) × 0.05 / (0.781 mm × 35 µm) = 33.4 mΩ, so it drops 66.8 mV and turns 0.134 W into heat. On an inner layer the same current needs 80.03 mil (2.03 mm).
Sources (checked 2026-11-01):
- IPC-2221 formula, k = 0.048 / 0.024, b = 0.44, c = 0.725, 1.378 mil per oz, the 35 A / 400 mil / 10–100 °C / 0.5–3 oz range and the notes on heat sinking and vacuum (Advanced Circuits): https://www.advancedpcb.com/en-us/tools/trace-width-calculator/
- The same constants, "resulting from curve fitting to the IPC-2221 curves" (DigiKey): https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-pcb-trace-width
- IPC-2152 (2009) and how it differs from IPC-2221 (Sierra Circuits): https://www.protoexpress.com/blog/how-to-optimize-your-pcb-trace-using-ipc-2152-standard/
- Resistivity of annealed copper at 20 °C (1.72 × 10⁻⁸ Ω·m), its temperature coefficient (3.93 × 10⁻³ /K) and ρ(T) = ρ0 [1 + α (T − T0)]: https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity
- Annealed copper standard, 58 × 10⁶ S/m at 20 °C (1 / 58 × 10⁶ = 1.724 × 10⁻⁸ Ω·m): https://en.wikipedia.org/wiki/International_Annealed_Copper_Standard