How to use it
- Type the Current (A) the trace must carry and the Allowed temperature rise (°C).
- Choose the Copper weight (oz), usually 1 oz, and whether the trace is on an Outer layer or an Inner layer.
- Type the Trace length (mm) if you want the resistance and the voltage drop along it.
- Read Trace width (minimum). Use the next wider width your board maker offers, and check the result with the designer's rules.
The IPC-2221 formula
IPC-2221 relates the current a conductor carries to the temperature rise it causes and its cross-section. The standard's curve-fit constants, as quoted by DigiKey's trace-width calculator, are 0.048 for outer layers and 0.024 for inner layers, with the exponents 0.44 and 0.725. The area is in square mils and the width follows from the copper thickness.
I = k × ΔT^0.44 × A^0.725
A = (I ÷ (k × ΔT^0.44))^(1/0.725)
width = A ÷ (oz × 1.378 mil)
k = 0.048 outer layer, 0.024 inner layer
ΔT temperature rise (°C)
A cross-section (mil²)
1 oz copper = 35 µm = 1.378 mil- Copper thickness: Wikipedia gives 35 µm for 1 oz/ft², while some manufacturers quote 34 µm. The calculator uses 35 µm.
- The resistance uses the resistivity of annealed copper at 20 °C; a hot trace has a little more resistance.
| Layer | k | Width for 2 A, 10 °C, 1 oz |
|---|---|---|
| Outer | 0.048 | 0.781 mm (30.76 mil) |
| Inner | 0.024 | about 2.6 times the outer area |
What this calculator does not cover
- Planes and wide copper areas, where IPC-2152 gives different and usually more useful figures.
- Thermal vias, nearby copper, forced air and enclosed boards, which change the rise a trace reaches.
- Pulses and high frequencies, where the current crowds to the surface of the copper; a steady current is assumed.
Worked example: 2 A, 10 °C rise, 1 oz, outer layer
The area is A = (2 ÷ (0.048 × 10^0.44))^(1/0.725) = 42.39 mil². With 1 oz (1.378 mil) the width is 42.39 ÷ 1.378 = 30.76 mil, which is 0.781 mm.
Over 50 mm the trace resistance is 1.72 × 10⁻⁸ Ω·m × 0.05 m ÷ (0.781 mm × 35 µm) = 31.4 mΩ, so the drop at 2 A is 62.9 mV.
Frequently asked questions
What width do I need for 1 A on an outer layer?
With a 10 °C rise and 1 oz copper, the calculator gives the width for any current you type. Enter 1 and read the result: a narrower trace than for 2 A, because the area grows with the current in the formula.
Why does an inner layer need a wider trace?
An inner trace is enclosed by the board, so it cannot lose heat to the air as easily. The formula uses k = 0.024 for inner layers, which gives a trace about 2.6 times the outer area for the same rise.
What is 1 oz copper in millimetres?
1 oz/ft² is about 35 µm thick, which is 1.378 mil. Some manufacturers quote 34 µm; the difference changes the width by about 3 %.
Is a 10 °C rise safe?
It is a common target for a board that runs at room temperature. The safe rise depends on the components and the board's maximum temperature, so check those too.
Can I use this for power planes?
No. Planes and wide copper areas spread heat differently, and IPC-2152 is the better guide for them.
Sources
- DigiKey: PCB trace width conversion calculator (IPC-2221 formula and constants)
- Wikipedia: Printed circuit board (copper weight 1 oz/ft² is 35 µm)
- Wikipedia: Electrical resistivity and conductivity (copper at 20 °C)
Figures and text last checked on Tuesday, November 10, 2026.