PCB trace width calculator

How this is calculated

IPC-2221, the generic printed-board design standard, gives the current a trace can carry for a given temperature rise as a curve fit:

I = k · ΔT^0.44 · A^0.725

with I in amps, ΔT the rise above ambient in °C, and A the cross-section in square mils. The constant is k = 0.048 for outer layers and k = 0.024 for inner ones. Solving for area and dividing by the copper thickness gives the width. Copper weight is area density — one ounce spread over a square foot is 35 µm (1.38 mil) thick.

The width sets the resistance, and the resistance sets the other half of the answer. Copper's resistivity is 1.724 × 10⁻⁸ Ω·m at 20 °C and rises 0.393 % per °C, and it is evaluated at the temperature the trace actually runs at:

R = ρ₂₀ · (1 + 0.00393 · (T − 20)) · L / (W · t)    V_drop = I · R    P = I² · R

When the drop at the IPC width exceeds your allowance, the calculator widens the trace until it does not. That is a small loop rather than a formula, because a wider trace runs cooler and cooler copper conducts better, so it is solved numerically to the exact width. The fusing current uses Onderdonk's equation for a conductor heated too quickly to lose any heat, starting from the trace's operating temperature and ending at copper's 1083 °C melting point.

An inner layer is an outer layer carrying twice the current

The inner-layer constant is exactly half the outer one. Halving k has the same effect on the required area as doubling I, so moving a trace inside the board costs exactly what doubling its current would — and because area grows as I^1/0.725 = I^1.38, that is not twice the width but 2.60 times, at every current and every copper weight. A 1 A trace that is 0.30 mm on the top layer is 0.78 mm on layer 2.

The same exponent explains three other things the curve implies and nobody states:

• Doubling the current costs 2.60× the width, not 2×. A 10 A trace is not ten 1 A traces side by side; it is 24 of them.
• Splitting a current between two parallel traces saves copper. Each carries half, each needs 1/2.60 of the width, so the pair uses 77 % of the copper one trace would. Running a supply on both outer layers and stitching them with vias is the practical form of this.
• Allowing a hotter trace buys less than it looks. Doubling the permitted rise from 10 °C to 20 °C only narrows the trace by 34 %, because the rise enters at the 0.44 power. Doubling the copper weight, by contrast, halves the width exactly.

Heat or voltage drop: which one sets the width

The IPC curves answer one question — how hot does the trace get — and every width calculator stops there. But a trace can be perfectly cool and still wrong, because the other thing it does is drop voltage, and on a low-voltage rail that is often the binding constraint.

Take 2 A on a 3.3 V rail, 100 mm long, in 1 oz outer copper. For a 10 °C rise IPC-2221 allows 0.78 mm. That trace has 67 mΩ of resistance and loses 134 mV — 4 % of the rail, before the connector, the return path or the regulator's own tolerance are counted. Holding the drop to 1 % needs 3.06 mm, nearly four times wider, and at that width the trace barely warms at all: 1.1 °C. The thermal answer would have passed every rule check and browned out a microcontroller.

The general pattern: heat decides short, high-current traces, and voltage drop decides long runs on low rails. The page runs both checks and names the one that wins. Remember that the return current flows through copper too — if the return is a trace rather than a solid plane, the loop drops twice what one trace does. The rail, the trace resistance and the load form a voltage divider whose top leg you drew by accident, and it obeys the same arithmetic.

How far to trust the curves

IPC-2221's chart descends from National Bureau of Standards measurements made in the 1950s on traces in free air. It was superseded for detailed work by IPC-2152, whose far larger data set showed that a real trace's temperature depends heavily on the board around it: its thickness, the copper planes nearby that spread heat, and the neighbouring traces that add their own. Two consequences follow.

First, treat the result as a sizing rule, not a temperature prediction. For a single outer-layer trace on an ordinary board it is generally conservative. It is not conservative for a bundle of parallel high-current traces heating each other, or for a board sealed in an enclosure whose ambient is well above the room — set the ambient to what the inside of the box will be, not the lab.

Second, the inner-layer factor is the part most often pessimistic: IPC-2152's measurements found inner traces near copper planes frequently run cooler than the old curve predicts, because the planes carry heat away. Using the 2.60× figure is safe; it may just cost board space you did not need. When the width really matters — high current, dense layout, no margin — use a thermal simulation or measure a prototype with a thermocouple or thermal camera.

What melts it

Fusing current matters for faults, not for normal operation: a trace has to survive the current that flows before the fuse or the protection circuit opens. Onderdonk's equation assumes no heat escapes during the event, which is accurate for milliseconds and increasingly pessimistic for seconds, since the board does draw heat away — so the figure shown is a floor rather than a ceiling. A trace sized for 1 A at a 10 °C rise melts at about 3 A held for one second, 9.5 A for 100 ms, and 30 A for 10 ms. If the upstream fuse is slower than that, the trace is the fuse.

FAQ

How wide should a trace be for 1 A?

0.30 mm (11.8 mil) on an outer layer in 1 oz copper for a 10 °C rise, or 0.20 mm if you can accept 20 °C. On an inner layer, 0.78 mm. Those are thermal minimums — on a long run, check the drop above before drawing it that thin.

And for 10 A?

7.2 mm on an outer 1 oz layer at a 10 °C rise, 3.6 mm in 2 oz, or 4.7 mm in 1 oz at a 20 °C rise. At currents like these, a copper pour or a pair of traces on both outer layers is usually more practical than one very wide trace, and splitting saves copper for the reason given above.

Does trace length change the required width?

Not thermally — the IPC curve describes a long trace's steady-state temperature, which does not depend on how long it is. Length enters through resistance, so it changes the voltage drop and the power lost, and past some length the drop becomes what sets the width. A very short trace between two large pads can run cooler than the curve says, because the pads sink its heat.

What does 1 oz copper mean?

One ounce of copper spread evenly over one square foot, which comes out at about 35 µm thick. 0.5 oz is common on inner layers of dense boards, 1 oz is the default outer weight at most fabricators, and 2 oz and heavier are specified for power boards. Plating adds thickness to outer layers, so a finished "1 oz" outer layer is often thicker than 35 µm — which makes the calculation slightly conservative.

What is the sheet resistance of 1 oz copper?

About 0.49 mΩ per square at 20 °C. A trace's resistance is that figure times its length divided by its width — a 1 mm wide, 100 mm long trace is 100 squares, 49 mΩ. It is the quickest mental check on any drop this page reports.

Does the calculator cover vias?

No. A via's plated barrel is a much thinner copper wall than the trace feeding it, and it is often the real bottleneck in a power path. Use several vias in parallel where a current changes layer, and size them separately.

How does this relate to the other pages here?

The resistivity and temperature coefficient are the ones the speaker cable calculator uses for wire, and the loss reasoning is the same — only the conductor's shape differs. The circuit at the end of the trace is often sized elsewhere on the site: a string of power LEDs from the LED resistor calculator draws its full current through exactly this copper, and a sense divider tapped off a rail sees the rail minus the drop reported here.