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PCB Trace Width Calculator

Calculate the required PCB trace width from current, temperature rise and copper weight per IPC-2221.

Trace width
0.3004 mm
Width (mils)
11.83 mils
Cross-section
16.3 mils²

Per IPC-2221. Add margin for long traces, vias and dense boards.

Disclaimer: This calculator is provided for general informational and educational purposes only, on an “as is” basis and without any warranty of accuracy or fitness for a particular purpose. Results may contain errors — always verify independently before relying on them in real designs. PartAndStock accepts no liability for any loss or damage arising from use of this tool, including when embedded on third-party sites.

How to use

  1. 1Enter the current the trace will carry, in amperes.
  2. 2Enter the allowable temperature rise (typically 10 °C).
  3. 3Select the copper weight (1 oz ≈ 35 µm is most common).
  4. 4Select the layer (external/internal); the required width is shown in mm and mils.

How it works

Find the minimum width a PCB trace needs to safely carry a given current, based on the IPC-2221 standard.

A = ( I / (k · ΔT^0.44) )^(1/0.725) · Width = A / (copper[oz] × 1.378) · k = 0.048 external, 0.024 internal

How IPC-2221 works

When current flows through a trace, the copper's resistance heats it up. IPC-2221 gives, with an empirical formula, the current a trace can carry without exceeding a given temperature rise: it first computes the required cross-sectional area, then divides by the copper thickness to get the width. External (surface) layers shed heat to air better, so they need a thinner trace than internal layers (the k coefficient is twice as high).

Limits and margin

This formula only accounts for heating; it does not cover voltage drop along long traces. For long power rails, dense multilayer boards, via transitions and vibration/reliability, add margin to the calculated width. At high currents it is often more practical to use thicker copper (2–3 oz) or parallel traces / copper pours instead of a single wide trace.

Worked examples

  • 1 A, ΔT 10 °C, 1 oz, external → ≈ 0.30 mm (11.8 mils)
  • 3 A, ΔT 10 °C, 1 oz, external → ≈ 1.37 mm (53.8 mils)
  • 1 A, ΔT 10 °C, 1 oz, internal → ≈ 0.79 mm (internal ~2× wider)

Current → Trace Width (1 oz · external · ΔT 10 °C)

CurrentWidthmils
0.5 A0.12 mm4.6
1 A0.30 mm11.8
2 A0.78 mm30.8
3 A1.37 mm53.8
5 A2.77 mm108.9
10 A7.20 mm283.5

Frequently Asked Questions

How is PCB trace width calculated?+
Per IPC-2221 you first find the cross-section A = (I/(k·ΔT^0.44))^(1/0.725), then divide by the copper thickness to get width. k is 0.048 for external and 0.024 for internal layers.
What is the difference between external and internal layers?+
Internal layers can't shed heat to air, so they run hotter; the same current needs roughly twice the width (the k coefficient halves).
What does copper weight (oz) mean?+
It is the weight of copper spread over one square foot. 1 oz ≈ 35 µm ≈ 1.378 mils thick; 2 oz is twice as thick and carries more current.
What temperature rise should I choose?+
A typical, safe value is 10 °C. 20 °C allows a thinner trace but heats it more; for critical/sensitive boards prefer 10 °C.
Does this account for voltage drop?+
No — it only gives width based on heating. For voltage drop on long traces you also need to compute the trace resistance (length/cross-section).

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