PCB Trace Width Calculator

Enter the current and the temperature rise you can tolerate. The calculator returns the minimum conductor width to IPC-2221, plus resistance, voltage drop, and power loss for your trace length.

IPC-2221 Internal / external mil + mm Voltage drop included
InputIPC-2221 conductor sizing
DC or RMS. For pulsed loads use the RMS value.
Rise above ambient, not the absolute temperature.
1 oz = 1.378 mil = 35 µm nominal.
Inner layers run hotter and need more copper.
Used for resistance and the absolute trace temperature.
Used for resistance, drop, and power loss.
Applied to the recommended width, not the minimum.
— — COPPER / SUBSTRATE CROSS-SECTION RECOMMENDED ENVELOPE SHOWN ABOVE
Width scales with the result—

How the IPC-2221 trace width formula works

The calculator solves the IPC-2221 conductor sizing equation. It starts from the current you need to carry and the temperature rise you are willing to accept, then converts the required cross-sectional area into a trace width for the copper weight you selected.

A = ( I / (k × ΔT^0.44) )^(1 / 0.725) A = required cross-section (mil²) I = current (A) ΔT = allowed temperature rise above ambient (°C) k = 0.048 external, 0.024 internal

The constant k is the whole argument for layer-aware design. External copper can radiate and convect heat into the air, so IPC-2221 allows more current per unit area. Internal copper is buried in laminate; heat has to conduct through the board before it escapes. The 2:1 difference in k means an inner trace needs roughly 2.2× the cross-section of an outer trace for the same current and rise.

Once the area is known, width follows from the copper thickness: W = A / t, where t is the copper weight in mil. One ounce of finished copper is approximately 1.378 mil (35 µm). That is a nominal number — plating, etch, and laminate tolerances move it. A “1 oz” inner layer is often a little thinner after lamination, while an outer layer gains thickness during plating.

Copper weight, in numbers

CopperNominal thicknessµmTypical use
0.5 oz0.689 mil17.5Fine-line inner layers
1 oz1.378 mil35Default for most boards
2 oz2.756 mil70Power, motor drivers
3 oz4.134 mil105High current, thermal
4 oz+5.512 mil+140+Heavy copper / bus bars

What the calculator does not model

A trace width calculator is a starting point, not a thermal simulation. These are the effects that most often make a calculated width optimistic in practice:

  • AC and skin effect. Above roughly 10 MHz, current crowds toward the copper surface. The DC cross-section overstates the useful copper. At 100 MHz in 1 oz copper, the skin depth is about 6.6 µm — a fraction of the 35 µm copper thickness.
  • Copper roughness and etch taper. High-frequency laminates use low-profile copper for a reason. Etch undercut makes a “4 mil” trace narrower at the base than at the top, and the effective resistance rises.
  • Adjacent copper and planes. A trace surrounded by copper pours runs cooler than an isolated trace. IPC-2221 does not account for this; IPC-2152 does, which is one reason the newer standard exists.
  • Vias, connectors, and neck-downs. The trace is rarely the weakest link. A 0.3 mm via with 25 µm plating has roughly the copper cross-section of a 26 mil wide 1 oz trace. Two vias in parallel are not automatically twice the current capacity if they share a poor plane connection.
  • Pulsed and fault currents. Use RMS current for thermal sizing. A short fault pulse is a different problem — that is a fusing/energy calculation, not an ampacity table.
IPC-2221 vs IPC-2152: IPC-2152 is the more recent industry method and is generally preferred for high-current or thermally critical designs. IPC-2221 is still widely used as a design rule and is conservative in many ranges, but it does not capture copper planes or airflow. If the board is safety-critical, use IPC-2152 or a thermal simulation and confirm with your fabricator.

Choosing a temperature rise

The temperature rise is the single most consequential input. It is not the ambient temperature; it is how much hotter the copper is allowed to get above the surrounding board.

RiseWhen it is reasonableTrade-off
10 °CPower rails, dense boards, long-life productsWider traces, more copper, higher cost
20 °CGeneral consumer and industrial boardsThe usual compromise
30 °CCost-constrained, short duty, airflow availableHotter board, shorter laminate life
40 °C+Rare; needs thermal test evidenceAggressive; not a default

Also check the absolute temperature: ambient plus rise. A board in a 70 °C enclosure with a 20 °C rise reaches 90 °C. Standard FR-4 has a glass transition around 130–140 °C, and long-term reliability degrades well before the Tg. If the board is in a sealed box, the “ambient” inside the box is the number that matters.

Voltage drop often governs the design

For a 12 V motor line, ampacity usually decides the width. For a 1.0 V core rail, voltage drop usually decides it first. The calculator reports resistance from the actual trace length, then voltage drop and power loss. A 3 A load on a 50 mm trace is a very different problem at 1.0 V than at 24 V.

Use the drop result as a sanity check: keep it below a few percent of the rail, and remember that the return path counts too. A wide power trace with a thin ground return is still a thin circuit.

Practical design margin

The minimum width is the point where the trace is allowed to reach the chosen temperature rise. It is not a comfortable design point. Most experienced designers apply 1.5× to 2× margin for production boards, and more where the board is enclosed, the current is pulsed, or the product must survive a fault.

  • Use 1.5× for normal industrial and consumer boards.
  • Use 2× for sealed enclosures, high ambient, or long service life.
  • Keep the width continuous through pads and neck-downs; a wide trace that pinches to 6 mil at a connector is a 6 mil trace.
  • Use copper pours or polygons for high current instead of very wide traces — they use the available copper more efficiently and spread heat.

Worked example: 3 A on 1 oz copper

Take 3 A, a 10 °C rise, 1 oz external copper, and a 50 mm run at 25 °C ambient. The IPC-2221 area is about 74 mil², which gives a minimum width of roughly 54 mil (1.37 mm). With a 1.5× margin the recommendation is about 81 mil (2.05 mm). The 50 mm trace has roughly 18 mΩ of resistance, a 54 mV drop, and 0.16 W of loss.

Switch the same trace to an internal layer and the required width rises by about 2.6×. Switch to 2 oz copper and the width roughly halves for the same current. That is the entire trade: copper thickness and layer choice versus board area and cost.

Frequently asked questions

How much current can 1 oz copper carry?

It depends on the allowed temperature rise and the layer. For a 10 °C rise, IPC-2221 gives roughly 1 A for a 10 mil external trace and about half that on an internal layer. A 20 °C rise allows about 36% more current for the same width. Always leave margin; the table is a starting point, not a guarantee.

Why does an internal layer need a wider trace?

Inner copper is laminated between dielectric layers, so heat has nowhere to go except through the board. IPC-2221 uses k = 0.024 for internal layers versus 0.048 for external, which means an internal trace needs roughly 2.6× the cross-section for the same current and temperature rise.

Is IPC-2221 still the right standard, or should I use IPC-2152?

IPC-2152 is the newer, more comprehensive method and is generally preferred for high-current or thermally critical designs. IPC-2221 remains the common design-rule approximation and is conservative in many ranges, but it does not model copper planes, airflow, or adjacent copper well. For safety-critical work, use IPC-2152 or a thermal simulation and confirm with the fabricator.

Does ambient temperature change the trace width result?

IPC-2221 sizes the trace from the allowed temperature rise above ambient, so ambient does not change the minimum width directly. It does change the copper resistance and the absolute trace temperature. A board running at 70 °C ambient with a 20 °C rise reaches 90 °C, which matters for long-term laminate life.

How do I calculate the current capacity of a via?

Treat the via barrel as a short conductor: cross-section = π × diameter × plating thickness. A 0.3 mm via with 25 µm plating has about 0.0236 mm² of copper, roughly equivalent to a 26 mil wide 1 oz trace. Use multiple vias in parallel for high current and derate for via-in-pad and filled vias.

Should I design for a 10 °C or 20 °C temperature rise?

Ten degrees is conservative and common for power rails, dense boards, and long-life products. Twenty degrees is a common compromise for consumer hardware with airflow. Thirty or forty degrees is aggressive and should be reserved for short-duty or cost-constrained designs with thermal testing.

My calculated width is below the fab's minimum. What now?

Use the fab minimum as a floor, not the calculator result. If the current needs less than the minimum, you still have to draw the minimum. If the current needs more, widen the trace, add copper, split the current across layers, or use a bus bar. Confirm the minimum with the fabricator before release.

Can I use this result directly for production?

Use it as a design starting point. Production review should include the stackup, copper thickness after plating, etch tolerance, connector and via current paths, thermal environment, and the relevant standard for your product. The estimate is not a substitute for a DFM review or a thermal test.

Quick reference: external 1 oz and 2 oz, 10 °C rise

The table below is generated from the same IPC-2221 equation, for a 10 °C rise on an external layer. Use it as a sanity check, not a substitute for the calculator.

CurrentCopperMinimum widthIn mm

Want the same number for your exact stackup, copper, and ambient? Use the calculator above. If the result is close to a process limit, send the board for a DFM review before you release it.

From number to board

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