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PCB trace width calculator

The minimum width of a copper trace for a current and temperature rise, the current a trace can carry, or how hot it gets, from the IPC-2221 formula for external and internal layers. It also gives the trace’s resistance, voltage drop and power loss.

Solve for
A
°C
Layer
°C

Minimum width –

Cross-section
–
Copper thickness
–
Trace temperature
–
Resistance
–
Voltage drop
–
Power loss
–

How it works

IPC-2221 relates the current a trace carries to its cross-section and its temperature rise above ambient with a curve fit to the standard’s conductor chart:

I = k · ΔT0.44 · A0.725

Solving for width rearranges the formula to A = (I / (k · ΔT0.44))1/0.725; solving for the rise gives ΔT = (I / (k · A0.725))1/0.44.

Resistance, voltage drop and power loss

For a trace of length L, R = ρ(T) · L / (w · t), with ρ(T) = 1.7241 × 10−8 Ω·m · (1 + 0.00393 · (T − 20 °C)), the resistivity and temperature coefficient of annealed copper (IEC 60028), at the trace temperature T = ambient + rise. The voltage drop is I·R and the power loss I2·R.

Source

IPC-2221, Generic Standard on Printed Board Design (formerly IPC-D-275). The constants are those widely reproduced from its conductor chart, for example in the notes of Advanced Circuits’ trace width calculator and on NinjaCalc.

Validity range

The chart behind the formula covers currents up to 35 A, widths up to 400 mil (10.16 mm), temperature rises of 10 to 100 °C and copper of 0.5 to 3 oz/ft². Outside that range the formula extrapolates, and the calculator says so.

Limits

Worked examples

  1. Width for 1 A at a 10 °C rise, 1 oz copper. External: A = (1 / (0.048 · 100.44))1/0.725 = 16.3 mil², so w = 16.3 / 1.378 = 11.8 mil (0.300 mm). Internal: 30.8 mil (0.781 mm). Both match published IPC-2221 tables.

  2. Width for 5 A at a 20 °C rise, 2 oz external, 50 mm long, 40 °C ambient. The width is 35.7 mil (0.908 mm). At 60 °C the trace measures 15.7 mΩ, so it drops 78.5 mV and dissipates 0.392 W.

  3. Current for a 20 mil (0.508 mm) trace, 1 oz external, 10 °C rise. A = 20 · 1.378 = 27.6 mil², so I = 0.048 · 100.44 · 27.60.725 = 1.46 A. Over 100 mm at 35 °C: 103 mΩ, 0.150 V and 0.220 W.

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