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PCB Trace Width Calculator

Work out pcb trace width instantly with clear inputs, formula shown and shareable results.

About PCB Trace Width Calculator

IPC-2221 gives cross-sectional area from current and temperature rise: A = (I / (k x dT^0.44))^(1/0.725), with k = 0.048 for external traces and 0.024 for internal ones. Dividing by copper thickness gives the width, which is why 1 oz copper needs roughly 1 mm per 3 A for a 20 degree rise.

Formula

IPC-2221 trace width

A(mil^2) = (I / (k x dT^0.44))^(1/0.725); width = A / copper thickness

Worked example

With the default values below, PCB Trace Width Calculator returns Trace width required of 0.90. Change any input above and every figure here is recalculated in your browser.

Inputs used

Trace current
3 A
Permitted temperature rise
20 °C
Copper thickness
35 µm
Layerone of External (outer layer), Internal layer
External (outer layer)

Results

Trace width required
0.90 mm
Trace width
35.34 mil
Cross-sectional area
48.70 mil²
Resistance per centimetre
5.47 mΩ/cm

Step by step

  1. 1Values used
  2. 2IPC-2221 trace width
  3. 3Trace width required= 0.90 mm
  4. 4Trace width= 35.34 mil
  5. 5Cross-sectional area= 48.70 mil²
  6. 6Resistance per centimetre= 5.47 mΩ/cm

Assumptions and limits

  • The result is derived only from the 4 inputs listed above. Anything not among them is not part of the calculation.
  • In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
  • Figures are rounded for display to 2 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.
  • Electrical results are planning estimates only. Verify against the applicable wiring rules (IEC 60364, BS 7671, NEC or local code) and have all installation work designed and checked by a qualified electrician.

How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.

Electrical results are planning estimates only. Verify against the applicable wiring rules (IEC 60364, BS 7671, NEC or local code) and have all installation work designed and checked by a qualified electrician.

Frequently Asked Questions

Why are internal traces half as capable?

They have no convective or radiative cooling path, only conduction through the laminate, so they need twice the area for the same rise.

Is temperature rise the only limit?

No. Voltage drop along a long trace and the current-handling of vias and connectors often govern before the trace itself does.

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