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PCB Trace Width Calculator

Find the minimum copper trace width for a given current using the IPC-2221 standard — in mils and millimetres, for external or internal layers.

IPC-2221
mils & mm
Copper weight
Ext / int layer
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PCB trace width — Quick answer

IPC-2221 links current, temperature rise and copper cross-section. Solve for the area, then divide by the copper thickness to get the width.

I = k · ΔT0.44 · A0.725  (k = 0.048 ext, 0.024 int)  |  width = A / thickness

Worked example: 2 A, 1 oz copper, 10 °C rise, external layer → area ≈ 42 mils², thickness 1.37 mils → width ≈ 31 mils ≈ 0.78 mm. An internal trace for the same current would be roughly twice as wide.

External trace width, 1 oz copper, 10 °C rise

CurrentWidth (mils)Width (mm)
0.5 A50.13
1 A120.30
3 A541.37
5 A1002.54

Used for: power traces, motor and LED drivers, battery and charger boards, connectors and fuse links.

⚡ PCB Trace Width Calculator

Enter the current, copper weight and the temperature rise you allow. Choose external for surface traces, internal for buried layers.

Trace width
Trace width
Cross-section
Copper thickness

⚠️ IPC-2221 estimate. Add margin for vias, connectors, fusing and high ambient; for tight power designs cross-check with IPC-2152.

A PCB trace is just a thin copper conductor, so like any conductor it heats up as current flows through its resistance. The trace width must be large enough that this self-heating stays within an acceptable temperature rise. The long-standing IPC-2221 formula relates the current, the allowed temperature rise and the copper cross-sectional area; dividing the area by the copper thickness gives the width. Internal layers need wider traces because they cannot shed heat to the air as easily.

Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: IPC-2221B and IPC-2152 trace current-capacity standards.

The IPC-2221 formula

Current capacity
I = k × ΔT0.44 × A0.725

Where I is current in amps, ΔT is temperature rise in °C, A is cross-sectional area in mils², and k = 0.048 for external (outer) layers or 0.024 for internal layers. Rearranged to find the area you need:

Required area
A = ( I / (k × ΔT0.44) )1/0.725
Width from area
width = A / (copper thickness)

Copper thickness comes from the weight: 1 oz/ft² ≈ 1.37 mils (35 µm). So a 2 oz board has twice the thickness and needs about half the width for the same current.

Worked example — a 3 A power trace

Scenario: 3 A on a 1 oz external layer with a 10 °C rise.

Area
A = (3 / (0.048 × 100.44))1/0.725 = (3 / 0.1322)1.37974 mils²
Width
width = 74 / 1.37 ≈ 54 mils ≈ 1.37 mm

On a 2 oz board the same 3 A trace would be about 27 mils (0.69 mm) — half the width, because the copper is twice as thick. If the trace were on an internal layer, double the width again.

Frequently Asked Questions

How do I calculate PCB trace width for a given current?

Use IPC-2221: I = k × ΔT0.44 × A0.725 (k = 0.048 external, 0.024 internal). Solve for area A in mils², then divide by copper thickness for the width. 2 A, 1 oz, 10 °C external ≈ 0.78 mm (31 mils).

What copper thickness does 1 oz correspond to?

1 oz/ft² ≈ 35 µm (1.37 mils); 2 oz ≈ 70 µm; 0.5 oz ≈ 17.5 µm. Thicker copper carries more current for the same width.

Why are internal traces wider than external ones?

Buried traces can't shed heat to air as well, so IPC-2221 uses k = 0.024 (vs 0.048) — about twice the cross-section, and therefore roughly twice the width, for the same current and rise.

What temperature rise should I use?

10 °C is a common conservative default; 20 °C is often used for power traces. A higher allowed rise lets the trace be narrower, but keep the total board temperature within the laminate and component ratings.

Is IPC-2221 conservative compared with IPC-2152?

Yes. IPC-2221 is older, simpler and more conservative (wider traces). IPC-2152 is newer and accounts for board material and copper planes, often allowing narrower traces. IPC-2221 is a safe first estimate.

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