PCB Trace Width Calculator
Minimum PCB trace width for a current and temperature rise from IPC-2221, for inner or outer layers and 0.5 to 3 oz copper, with resistance and voltage drop.
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About this toolWhat is the PCB Trace Width Calculator?
A copper trace on a circuit board is a small resistor, and the current through it heats it. Make a power trace too thin and it runs hot, discolours the laminate or, in the worst case, fuses open. This PCB trace width calculator uses the IPC-2221 design equation to find the narrowest trace that carries a given current with no more than the temperature rise you allow, on an outer or inner layer and for 0.5, 1, 2 or 3 oz copper.
It also works the other way, giving the current a trace of known width can carry. Enter a length and the result adds the trace resistance at its operating temperature, the voltage drop and the power lost as heat, which often matter more than the temperature on long low-voltage runs. A table compares the width needed on outer and inner layers across all four copper weights, in both mil and millimetres.
- IPC-2221 width for a current, or current for a width
- External and internal layers, 0.5 to 3 oz copper
- Width in mil and mm, cross-section in mil² and mm²
- Resistance, voltage drop and power loss at operating temperature
- Comparison table across layers and copper weights
- Warnings outside the range the IPC-2221 curves cover
How to use the PCB Trace Width Calculator
- Choose Width from current, or Current from width to check an existing trace.
- Pick External for top and bottom layers or Internal for buried layers.
- Enter the current (or the width in mil or mm) and the temperature rise you can accept. 10 °C is a common default; 20 to 30 °C is often fine for power traces.
- Select the copper weight your fabricator will use, and optionally the ambient temperature and the trace length.
- Read the minimum width, the cross-section, and for a length the resistance, voltage drop and power loss. Compare layers and weights in the table below.
Worked examples
2 A on an outer layer
Input
2 A, 10 °C rise, 1 oz, external, 50 mm long
Output
30.76 mil (0.781 mm); 33.37 mΩ, 66.75 mV drop and 133.5 mW lost over 50 mm
This matches the widely quoted figure of about 31 mil for 2 A at 10 °C on 1 oz outer copper. The same current on an inner layer needs 80.0 mil.
1 A signal-and-power trace
Input
1 A, 10 °C rise, 1 oz, external
Output
11.83 mil (0.300 mm)
A 12 mil trace is fine for an amp on the surface. Many designers still round power traces up to 20 mil or more to cut the voltage drop.
What does a 30 mil trace carry?
Input
Current from width, 30 mil, 10 °C rise, 1 oz
Output
1.964 A external, 0.982 A internal
Moving a trace to an inner layer halves its current rating for the same width and temperature rise under IPC-2221.
10 A on 2 oz copper
Input
10 A, 10 °C rise, 2 oz, external, 100 mm
Output
141.6 mil (3.60 mm); 7.25 mΩ, 72.5 mV and 0.725 W over 100 mm
Heavier copper halves the width. At these currents a copper pour or polygon is usually more practical than a single trace.
Background and insight
IPC-2221, the generic standard for printed board design, carries a chart of current against conductor cross-section for several temperature rises. The equation most calculators use is a curve fit to that chart: I = k × ΔT^0.44 × A^0.725, with I in amps, ΔT in °C and A in square mils, and k = 0.048 for external layers and 0.024 for internal ones. Solving for the area gives A = (I ÷ (k × ΔT^0.44))^(1/0.725), and the width is the area divided by the copper thickness. One ounce per square foot is taken as 1.378 mil, or 35 µm. Halving k for inner layers means an inner trace needs about 2^(1/0.725) = 2.6 times the area of an outer one, because laminate conducts heat far worse than air convects it from the surface.
The curves behind the fit only span roughly 0 to 35 A, widths up to 400 mil, rises of 10 to 100 °C and 0.5 to 3 oz copper, and the calculator warns outside that range. The data dates from tests in the 1950s on boards without planes. IPC-2152 (2009) replaced it with a much larger test programme and shows that internal traces are not nearly as bad as IPC-2221 suggests when the board has copper planes, while outer traces on thin boards can run hotter. Many engineers treat IPC-2221 as a conservative first pass for outer layers and refer to IPC-2152 charts or a thermal solver for high currents.
Resistance uses the resistivity of annealed copper, 1.724 × 10⁻⁸ Ω·m at 20 °C, raised by 0.393% per °C to the ambient plus the allowed rise. Plated copper is a few percent more resistive, and etching usually leaves a trapezoidal cross-section a little smaller than width times thickness, so treat the resistance as a slightly optimistic estimate.
Common use cases
Power distribution
Size the traces from a connector or regulator to the loads so they stay cool and the drop is acceptable.
Motor and LED drivers
Check that high-current switch paths and LED strings have enough copper on the layers you have chosen.
Design rule setup
Pick width classes for power nets before routing, for each copper weight in the stack-up.
Reviewing an existing board
Estimate the safe current of a trace you have measured, or the temperature margin of a repair.
Tips and limitations
- Short traces between large pads and vias run cooler than the formula predicts, because the copper around them spreads heat. Long isolated traces are the critical case.
- Vias need sizing too. A single 0.3 mm via carries roughly an amp; use several in parallel for power paths.
- Outer copper is often plated up beyond the base foil, so a "1 oz" outer layer may finish nearer 1.4 to 2 oz. Ask the fabricator for finished thickness.
- Voltage drop can be the tighter limit. 100 mm of 10 mil 1 oz trace is about 0.2 Ω, which drops 0.2 V at 1 A on a 3.3 V rail.
- Neighbouring hot traces and components add up. Leave margin when several power traces run side by side.
Frequently asked questions
What is the IPC-2221 trace width formula?
I = k × ΔT^0.44 × A^0.725, with A in square mils and k = 0.048 for outer layers or 0.024 for inner layers. Solve for A, then divide by the copper thickness (1.378 mil per ounce) to get the width.
How wide should a PCB trace be for 1 amp?
About 12 mil (0.30 mm) on an outer layer with 1 oz copper and a 10 °C rise, and about 31 mil (0.78 mm) on an inner layer. Many designers use 20 mil or more to reduce voltage drop.
Why do internal traces need to be wider?
Heat leaves an inner trace only by conduction through the laminate, which is a poor conductor, so IPC-2221 halves the constant k. The result is about 2.6 times the cross-section for the same current and rise.
What temperature rise should I design for?
10 °C is a conservative default. 20 to 30 °C is common for power traces if the board and nearby parts can tolerate it. Add the rise to the highest ambient inside the enclosure and keep the total well below the laminate rating.
Is IPC-2152 better than IPC-2221?
IPC-2152 is newer and based on far more test data, including the effect of planes and board thickness. IPC-2221 remains a quick conservative estimate for outer layers, but for high currents or tight inner-layer designs IPC-2152 or thermal simulation is more accurate.
How thick is 1 oz copper on a PCB?
One ounce of copper spread over one square foot is about 1.37 to 1.4 mil thick, or 35 µm. 2 oz is 70 µm and 0.5 oz is 17.5 µm.
How do I reduce voltage drop on a PCB trace?
Widen it, use heavier copper, shorten it, or run the current on a pour or plane. Resistance falls in proportion to the cross-section, so doubling the width halves the drop.
Is there an in-depth guide to the PCB Trace Width Calculator?
Yes. “PCB Trace Width for Current: IPC-2221, IPC-2152 and Judgment” covers the background, how it works, common pitfalls and worked examples. Read the full guide.
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