PCB Trace Width Calculator
Enter current, copper weight, and the maximum temperature rise you can accept. Get the minimum trace width for external and internal PCB layers per IPC-2221, plus resistance and voltage drop for any trace length. Nothing uploaded.
Set inputs on the Trace Width tab, then add rows here to compare multiple traces side by side.
IPC-2221 Trace Width Formula
The IPC-2221 standard defines the relationship between trace current capacity, conductor cross-sectional area, and temperature rise using empirical curve-fit equations:
External layers: I = 0.048 × ΔT0.44 × A0.725
Internal layers: I = 0.024 × ΔT0.44 × A0.725
Where I is current in amps, ΔT is temperature rise in °C above ambient, and A is the copper cross-sectional area in mil².
Solving for minimum area and then width:
A (mil²) = ( I / (k × ΔT0.44) )1/0.725
W (mil) = A / T where T = oz × 1.378
Copper thickness: 1 oz/ft² copper = 1.378 mil (approximately 35 µm) thick.
Internal layers run hotter because heat cannot dissipate as easily - hence the lower k factor (0.024 vs 0.048 for external). Always use the internal formula for buried and blind traces.
Safety recommendation: Add at least 50% margin over the calculated minimum width. Use 2x for primary power distribution traces. IPC-2221 minimums represent the absolute edge of safe operation.
Resistance formula: R/m = rho / (W × T) where rho for copper = 1.724×10-8 Ω·m. Temperature increases resistance (about 0.4%/°C above 20°C) - the calculator uses room-temperature resistivity.
Learn more: PCB trace width and current capacity
The formula the tool uses
The tool solves the IPC-2221 equation, I = k x dT^0.44 x A^0.725, for the cross-section area A in square mils. Here I is the current in amps, dT is the allowed temperature rise in degrees C, and k is 0.048 for external layers and 0.024 for internal ones. It converts the copper weight to a thickness (1 oz is 1.378 mil) and divides the area by the thickness to get a width.
For 2 A with a 10 C rise on 1 oz copper, the area is 42.4 square mils, which is 30.8 mil (0.78 mm) wide on an external layer. The internal setting uses half the k value, so the same job needs 80.0 mil (2.03 mm). Our post on IPC-2221 and IPC-2152 trace width numbers looks at where those two constants came from.
Copper weight and temperature rise
Width is area divided by thickness, so doubling the copper from 1 oz to 2 oz halves the width. The same 2 A and 10 C rise on 2 oz copper needs 15.4 mil (0.39 mm). Allowing a bigger rise also shrinks the width, at the price of a hotter trace. The rise is measured above ambient, not an absolute temperature.
Rounding and the safety rating
The formula's exact width is seldom a round number, so the tool rounds up to the next 0.1 mm and the next 5 mil. For 30.8 mil that gives 0.8 mm and 35 mil.
When you enter the width you plan to use, the tool compares it to the calculated minimum. A ratio of 1.5 or more is rated safe, from 1.0 up to 1.5 marginal and under 1.0 unsafe. Those bands are the tool's own and not part of IPC-2221. A margin helps with etching tolerance and copper thickness differences between board makers.
Resistance, voltage drop and vias
A trace wide enough to stay cool can still drop too much voltage over a long run. The tool uses copper's resistivity at 20 C, 1.724e-8 ohm metres. A 0.8 mm trace on 1 oz copper has 0.0063 ohms per cm, so 100 mm is about 0.063 ohms. At 2 A that is a drop of about 0.12 V, which matters on a 3.3 V rail.
The via figure applies the same equation to the plated barrel, with area equal to pi times the mean barrel diameter times the plating thickness. The pad does not carry current, so pad size does not change the result. A 0.3 mm drill with 1 oz plating comes out at about 2.5 A for a 10 C rise.
FAQ
What does temperature rise mean?
It is how many degrees above ambient the trace may heat when it carries the current. The tool defaults to a conservative 10 C. A 20 or 30 C rise allows a narrower trace, which then runs hotter.
Does copper weight change the width that much?
Yes. Current capacity follows cross-section area, and area is width times thickness. Going from 1 oz to 2 oz halves the width for the same current in this tool, from 30.8 mil to 15.4 mil in the 2 A example.
Why show resistance and voltage drop as well?
They catch a separate failure. A trace can pass the heating limit and still drop too much voltage over a long run, especially on a low-voltage rail. Check both numbers.