There is no such thing as a trace's current rating
A copper trace does not fail at a particular current. It simply gets hotter as current rises, and you decide how hot is acceptable. Everything else follows from that choice. Invert the IPC relationship for a 10-mil outer-layer trace in 1 oz copper and it carries 1 A at about a 13 °C rise, 1.5 A at about 33 °C, and 2 A at roughly 64 °C — the last of which is approaching the glass transition temperature of standard FR-4. All three are the same trace.
What makes the choice non-obvious is that the heat has to leave. On an outer layer it leaves by convection and radiation into the air and by conduction into the laminate. On an inner layer only conduction is available, and the laminate is a poor conductor — roughly 0.3 W/m·K through the thickness against copper's 400. IPC-2221 handles this with a blunt instrument: the constant k is 0.048 for external conductors and 0.024 for internal ones, so an internal trace needs about 2.6 times the width of an external one at the same current and rise.
The second thing the width buys you is a lower resistance, and that matters independently of temperature. A 50 mm run of 20-mil 1 oz trace is about 51 milliohms. At 3 A that is 154 mV of drop and 0.46 W of heat — enough to matter in a 3.3 V rail and enough to be the dominant loss in a small buck converter's output path. Sizing for temperature and sizing for regulation are different problems that happen to have the same knob.
Where the IPC-2221 numbers come from
IPC-2221 does not publish a physical model. It publishes charts, drawn from measurements made in the 1950s, and the relationship everyone uses is the curve fit to those charts:
A = (I ÷ (k · ΔT0.44))1/0.725, with A in square mils, I in amperes and ΔT in degrees Celsius.
Read the exponents as behaviour, not as physics. The 0.44 on temperature says that doubling the permitted rise buys you about a 36% increase in current for the same trace (20.44 = 1.36). The 1/0.725 = 1.379 on the outer bracket says current capacity scales as area0.725, which is sublinear — doubling the area gives you only 1.65 times the current, because a wider trace is also a slightly worse radiator per unit area. Both numbers are empirical.
Converting area to width needs the copper thickness. Copper weight is quoted in ounces per square foot, and one ounce spread over a square foot is 1.378 mils (35 µm) thick. So width = area ÷ (oz × 1.378). Doubling the copper weight halves the width for the same area, which is the cheapest way to fit a high-current trace through a congested board — though it costs you etch resolution everywhere else on that layer, because thicker copper undercuts more and forces wider minimum spacing.
Resistance uses ordinary copper physics. R = ρL/A, with ρ = 1.724 × 10−8 Ω·m at 20 °C and a temperature coefficient of about 0.00393 per degree. This page evaluates ρ at ambient plus the temperature rise, since that is where the trace actually operates. The correction is not trivial: a trace at 85 °C has 25% more resistance than the same trace at 20 °C, and that extra resistance produces extra heat, which raises the resistance further. For high-current designs the runaway is worth checking explicitly.
One subtlety about the area. The IPC relationship assumes a rectangular cross-section of the full nominal thickness. Etching produces a trapezoid, and the finished copper on an outer layer includes plating that may exceed the base weight. The result is normally conservative for outer layers and close to nominal for inner ones.
Worked example: 3 A on an outer layer in 1 oz copper
A buck converter output rail carries 3 A over 50 mm of outer-layer trace. The board sits at 25 °C ambient and you will allow a 10 °C rise.
- Temperature term. ΔT0.44 = 100.44 = 2.754229.
- Denominator. k · ΔT0.44 = 0.048 × 2.754229 = 0.132203.
- Cross-sectional area. (3 ÷ 0.132203)1.37931 = 22.69241.37931 = 74.161 mil².
- Width. 1 oz copper is 1.378 mils thick, so 74.161 ÷ 1.378 = 53.818 mil, which is 1.367 mm.
- Operating temperature. 25 + 10 = 35 °C, so ρ = 1.724 × 10−8 × [1 + 0.00393 × (35 − 20)] = 1.724 × 10−8 × 1.05895 = 1.82563 × 10−8 Ω·m.
- Area in SI. One square mil is (25.4 µm)² = 6.4516 × 10−10 m², so 74.161 mil² = 4.78458 × 10−8 m².
- Resistance. 1.82563 × 10−8 × 0.05 ÷ 4.78458 × 10−8 = 0.019078 Ω.
- Drop and dissipation. 3 × 0.019078 = 0.0572 V, and 3² × 0.019078 = 0.172 W.
Now suppose the routing channel only allows 20 mils. That trace has an area of 20 × 1.378 = 27.56 mil², or 1.77806 × 10−8 m², so its resistance rises to 1.82563 × 10−8 × 0.05 ÷ 1.77806 × 10−8 = 0.05134 Ω — 2.69 times the 0.019078 Ω of the full-width trace, exactly the ratio of the two areas, 74.161/27.56 = 2.69. The drop becomes 0.154 V and the dissipation 0.462 W, and the temperature rise will exceed 10 °C because the trace no longer meets the IPC area for 3 A. On a 3.3 V rail, 154 mV is 4.7% of the supply, which is why power traces get routed first.
Choosing a temperature rise, and knowing when width stops helping
10 °C is the default for a reason. It leaves room for everything the calculation ignores: neighbouring traces that are also warm, a component body radiating onto the copper, and an enclosure that is hotter inside than the room. Use 20 °C where the board is well ventilated and the trace is short. Use 30 °C only when you have measured the assembly or the trace is a short stub between a connector and a plane.
Diminishing returns arrive quickly. Because capacity scales as area0.725, going from 50 mil to 100 mil at 1 oz raises the allowable current by only 65%. Past roughly 200 mils a trace is a poor use of board area, and the right answers become heavier copper, splitting the current between an outer trace and an inner one stitched with vias, or a copper pour rather than a trace.
Vias are the constraint people forget. A single 0.3 mm finished via with 25 µm of barrel plating has far less copper cross-section than the trace it connects. Current that a 100-mil trace carries comfortably needs several vias in parallel, and via current capacity follows a different relationship in IPC-2152.
Voltage drop and temperature rise disagree about what matters. A 3 A trace 10 mm long meets a 10 °C rise at 53.8 mils and drops only 11 mV; the same width over 300 mm drops 343 mV while the temperature rise is unchanged, because rise depends on current density and drop depends on total length. Check both. On low-voltage rails — a 1.0 V core supply at 10 A — the drop is almost always the binding constraint, and remote sensing or a plane rather than a trace is the real fix.
Pulses are a different question. The IPC relationship is a steady-state result. A trace that would fail thermally at 20 A continuous can carry a 20 A pulse of a few milliseconds without noticeable heating, because copper's thermal mass absorbs it. Fusing current — where the trace vaporises — is a separate calculation entirely and is far above anything you would design to.
Required external trace width in mils, IPC-2221
| Current | 1 oz, ΔT 10 °C | 2 oz, ΔT 10 °C | 1 oz, ΔT 20 °C | 2 oz, ΔT 20 °C |
|---|---|---|---|---|
| 0.5 A | 4.55 | 2.27 | 2.99 | 1.49 |
| 1 A | 11.83 | 5.91 | 7.77 | 3.88 |
| 2 A | 30.76 | 15.38 | 20.20 | 10.10 |
| 3 A | 53.82 | 26.91 | 35.34 | 17.67 |
| 5 A | 108.87 | 54.44 | 71.49 | 35.74 |
| 10 A | 283.23 | 141.61 | 185.97 | 92.98 |
Every value is A = (I/(0.048·ΔT^0.44))^(1/0.725) divided by the copper thickness — 1.378 mil for 1 oz and 2.756 mil for 2 oz. Doubling the copper weight exactly halves the width, because the required area does not change.
Assumptions this calculation makes, and where they break
- It is steady state. Pulsed and intermittent currents heat the trace far less than the same peak sustained, and the correct treatment is a thermal-mass calculation rather than this one.
- It ignores adjacent copper. A trace running alongside a ground plane or between two other warm traces behaves differently from an isolated conductor. Planes help by spreading heat; parallel current-carrying traces hurt.
- It ignores board thickness and laminate. IPC-2152 introduced exactly these variables and generally allows narrower traces than IPC-2221 for the same rise on ordinary boards.
- It assumes a rectangular cross-section. Etching produces a trapezoid, and plating on outer layers adds copper beyond the base weight, so the true area differs from the nominal one in both directions.
- Connectors, vias and pads are not included. The narrowest copper in the path governs, and it is often a via barrel or a pad neck rather than the trace.
- Solder mask changes the thermal path slightly and is not modelled; a bare or solder-coated trace runs cooler than a masked one at the same current.
- The resistance figure uses the trace only. Add the resistance of vias, connector contacts and any series sense resistor before comparing a total drop against a regulation budget.
IPC-2221 against IPC-2152
IPC-2221B is the generic printed board design standard and carries the conductor charts this calculator implements. IPC-2152, published in 2009, superseded those charts for current-carrying capacity: it is based on modern measurements and accounts for board thickness, laminate thermal conductivity, the presence of adjacent copper planes, and whether the board is in still air or a vacuum. Where IPC-2152 applies, it typically permits narrower traces than IPC-2221 for the same rise, because the older charts were derived without an internal plane present. IPC-2221 remains what most design reviews, tools and reference tables use, and its conservatism is the reason. Use this page for a first cut and consult IPC-2152 or a thermal simulation when the trace is carrying serious current or the board is unusual.
Where trace sizing sits in a power design
Trace width is one of three copper decisions on a power path, and it is usually the least important of them. Plane resistance comes first: on any board with more than two layers, a solid pour on an inner layer beats any trace you could route, and a split in that pour is the failure mode to look for. Via count comes second, because the barrels are where cross-section collapses. Trace width comes third and is the one every calculator addresses.
Downstream of the copper, the same current has to be delivered to the load without violating its supply tolerance. Add the trace drop from this page to connector contact resistance and any sense element, then compare against the regulation window. The Ohm's law calculator and the electrical power calculator handle the arithmetic of that budget, and the voltage drop calculator covers the same problem at the cable scale.
On the signal side, width means something different again: at frequencies where the trace behaves as a transmission line, width and dielectric height set characteristic impedance, and the current-carrying question is irrelevant because the current is microamps. A 50 Ω microstrip on 1 oz copper over a 0.2 mm dielectric is around 14 mils wide whatever current it carries. Do not confuse the two problems — a signal trace that is widened for current handling will no longer be 50 Ω.
For small-signal parts of the same board, the LED resistor calculator and the op-amp gain calculator cover the component-level sizing that goes alongside the layout work.
Key terms
- Copper weight
- Mass of copper per square foot of board, quoted in ounces. One ounce corresponds to 1.378 mils or 35 micrometres of thickness; the term describes a thickness even though its unit is a mass.
- Mil
- One thousandth of an inch, 25.4 micrometres. PCB dimensions are quoted in mils in North America and millimetres elsewhere; 10 mil is 0.254 mm.
- Temperature rise
- How much hotter the trace runs than the board around it under load. It is a design choice, not a property of the trace, and it is what the IPC relationship solves for.
- External and internal layer
- External conductors sit on an outer surface and lose heat to the air; internal ones are laminated between dielectric layers and lose heat only by conduction. IPC-2221 halves the constant k for the internal case.
- Temperature coefficient of resistance
- Copper's resistance rises about 0.393% per degree Celsius. Over the range from 20 °C to 85 °C that is a 25% increase, which feeds back into the heat the trace generates.