Track width for a current
The narrowest copper track that carries a current without heating more than you allow, by the IPC-2221 curves.
Inputs
Use a dot or a comma for decimals.
Results
- Minimum width
- 0.78mm
- Minimum width in mil
- 30.8mil
- Copper thickness
- 35µm
- Cross-section
- 42.4mil²
- Resistance at 20 °C
- 31.52mΩ
- Resistance at working temperature
- 33.37mΩ
- Voltage drop
- 66.7mV
- Power lost in the track
- 133.5mW
An estimate from a published formula. Add margin.
The formula
I = k × ΔT^0.44 × A^0.725
I is the current in amperes, ΔT the temperature rise in °C, A the cross-section in square mils (width × copper thickness), and k is 0.048 for an outer layer or 0.024 for an inner layer. The page solves for A, divides by the copper thickness (1 oz is 35 µm, 1.378 mil) and converts to millimetres. Resistance is ρ × length ÷ (width × thickness) with ρ = 1.724e-8 Ω·m at 20 °C, raised by 0.393 % per °C for the working temperature (ambient plus the rise).
Worked example: 2 A, 1 oz, 10 °C rise gives 0.78 mm on an outer layer and about 2.0 mm on an inner layer.
Where the formula holds
The IPC-2221 curves cover 0.1 to 35 A, a rise of 10 to 100 °C and 0.5 to 3 oz copper; the page refuses inputs outside them. IPC-2221 is conservative for inner layers, and its newer companion IPC-2152 is considered the better source; treat the width as a minimum and widen the track where you can. Check the narrowest point of the route: a pad neck, a via, a connector pin.
Source: IPC-2221, generic standard on printed board design (section 6.2, conductor current capacity). Calculated in your browser. Nothing is sent to a server.