PCB trace width: the IPC-2221 formula, outer versus inner layers, copper weight and a worked example
A copper trace is a resistor. Push current through it and it warms; too narrow and it warms enough to lift from the board or cook the parts beside it. IPC-2221 gives the time-honoured rule relating the current, the copper cross-section and the temperature rise. This calculator turns it round: from the current you need to carry and the rise you can accept, on your copper weight and layer, it gives the width in millimetres and mils, the copper's resistance at that temperature with the drop and loss over a length, and what any standard width would carry instead, with the trace drawn to scale on the board and in cross-section.
How to use the trace width calculator
- Type the current the trace must carry continuously and the temperature rise you can accept above the board, 10 °C unless you know better.
- Pick the layer and the copper weight in the menu row. The drawing shows the outer and inner widths side by side.
- Add a length for the resistance, the voltage drop and the loss, and type a width to check if you already have one. The Standard widths tab lists what each common width carries.
The IPC-2221 formula
I = k · ΔT0.44 · A0.725 w = A / t
I is the current in amps, ΔT the rise in °C, A the copper cross-section in square mils (a mil is a thousandth of an inch) and k is 0.048 for outer layers and 0.024 for inner ones. The exponents are a fit to the curves in the old IPC-D-275 standard, drawn from measurements made in the 1950s on bare boards; IPC-2221 reprinted them and the formula is what every trace calculator uses. Inverted, A = (I / (k ΔT0.44))1/0.725, and the width is the area over the copper thickness. The newer IPC-2152 measured modern boards and gives less conservative widths, especially for inner layers near copper planes, but IPC-2221 remains the safe default.
Outer and inner layers
An outer trace sheds heat to the air by convection and radiation; an inner trace can only conduct through the laminate. IPC-2221 halves k for inner layers, which means 21/0.725 = 2.6 times the copper area, so the same 1 A that needs a 0.30 mm outer trace needs 0.78 mm inside. In practice an inner trace next to a ground plane runs cooler than that, which is what IPC-2152 documented; if you route power on inner layers routinely, it is worth reading.
Copper weight
Copper foil is specified by weight per square foot: 1 oz is 35 µm (1.378 mil) thick, ½ oz 17 µm, 2 oz 70 µm. Doubling the weight halves the width for the same current. Most boards use 1 oz on the outer layers; power electronics goes to 2 or 3 oz; fine-pitch designs use ½ oz inner layers because thick copper cannot be etched into narrow gaps. Plating adds to outer layers on a finished board, so the real outer thickness is often a little more than ordered.
Choosing the temperature rise
10 °C is the usual conservative figure and keeps the trace near the board's own temperature. 20 °C is common where there is space and airflow. Beyond 30 °C the width shrinks slowly (it falls as ΔT−0.61) while the copper, the solder joints and the parts alongside all run hotter, and the laminate's glass transition, often 130–150 °C, is approached in a warm enclosure. Remember the rise is above the board, so add the enclosure's temperature when judging it.
Your trace, step by step
- Area: (1 A / (0.048 × 100.44))1.379 = 16.3 mil² = 0.0105 mm².
- Width: 16.3 / 1.38 mil = 11.8 mil = 0.30 mm on the outer layer; 0.78 mm on the other.
- Copper at 35 °C: 1.73 Ω per metre; 100 mm is 173 mΩ, dropping 173 mV and dissipating 173 mW.
Worked example: 1 A with a 10 °C rise on 1 oz copper
Outer layer, k = 0.048. 100.44 = 2.754, so k ΔT0.44 = 0.1322 and A = (1 / 0.1322)1.379 = 16.3 mil². On 1 oz copper (1.378 mil) the width is 16.3 / 1.378 = 11.8 mil, 0.30 mm; most designers would lay out 0.30 mm. The same current on an inner layer needs 0.78 mm. At 35 °C that trace has 1.73 Ω per metre, so a 100 mm run is 173 mΩ and drops 173 mV, which is why sense lines and low-voltage rails are often made wider than the heating alone demands. At 10 A with a 20 °C rise on 2 oz copper the formula gives 2.36 mm: a motor driver's trace.
Questions
Is this the minimum width or the recommended width?
It is the width at which the trace reaches the allowed rise. Round up to a width your layout tool and fab hold easily, and go wider where the voltage drop matters.
What about short pulses?
The formula is for continuous current. A brief pulse heats the copper far less, and a motor's stall current for a second can run on a much narrower trace than its steady value would need, though a fuse and some margin are wise.
Can I use a copper pour instead?
Yes, a pour or a plane is just a very wide trace. Watch the neck where it joins a pad or a via: the narrowest point sets the limit.
Why does my fab quote a minimum width?
Etching cannot hold a trace thinner than about 0.1–0.15 mm on 1 oz copper, whatever the current. Below that figure the fab's rule wins, not this calculation.