PCB Trace Width and Current Calculator (IPC-2221)

Trace width for a current and temperature rise, or the current a trace carries, plus trace resistance and voltage drop.

Inputs

Results

IPC-2221: A[mil²] = (I / (k · ΔT^0.44))^(1/0.725), with k = 0.048 for external and 0.024 for internal layers; width = A / thickness (1 oz/ft² = 1.378 mil = 35 µm). Resistance uses copper at ambient + ΔT. Real boards depend on copper pour, vias, airflow and neighbouring traces — use it for a first estimate and prefer a thermal simulation for high currents.

Frequently asked questions

How wide must a PCB trace be for a given current?

IPC-2221 gives the cross-section as A[mil²] = (I / (k·ΔT^0.44))^(1/0.725), with k = 0.048 for outer and 0.024 for inner layers; width is area divided by copper thickness (1 oz/ft² ≈ 35 µm). The result is a first estimate: copper pours, vias and airflow change the real temperature rise.

Why are inner layers rated for less current than outer layers?

An inner trace is surrounded by dielectric and cannot shed heat into the air, so IPC-2221 uses a factor half as large for it (k = 0.024 instead of 0.048), which calls for a much wider trace at the same current and temperature rise.

For preliminary assessment only — the final decision must follow the manufacturer’s datasheet and the applicable NEC / IEC / UL standards. Results are calculated from the standard formulas stated on this page and are for design estimates only. They are not a rating of any specific part — always confirm current, voltage, temperature and environment limits against the manufacturer’s datasheet for the part you intend to use.

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