PCB Trace Width Calculator (IPC-2221)

What actually limits a trace’s current

When to use this: sizing a power rail on a new board, checking whether the default 10 mil track your router laid down can really carry the regulator’s output, deciding whether a rail needs 2 oz copper, or answering the design-review question “where did that width come from?”.

The thing that stops a trace carrying more current is not the copper running out of ability to conduct. Copper conducts essentially the same whether you ask 100 mA or 10 A of it. What changes is the heat. A trace has resistance, that resistance turns current into heat at P = I²R, and the copper only has so much surface area to shed that heat from. The trace warms until the rate it loses heat to the air and the laminate matches the rate it is making it, and then it sits there. Push more current and it settles at a higher temperature.

So “how many amps can this trace take?” is not really a question about amps. It is a question about how hot you are willing to let the copper get, and it has no answer until you say. That is why every trace-width figure you will ever see is quoted with a temperature rise attached — 1 A at a 10 °C rise, 1.5 A at 20 °C, and so on. The same conductor carries all of those; it just runs at different temperatures. This is exactly the same logic that makes a wire’s ampacity a property of the installation rather than of the wire, and for exactly the same reason: heat has to go somewhere.

Failure, when it comes, is rarely dramatic. A trace does not usually fuse — it needs to reach several hundred degrees to do that, which takes many times the design current. What happens long before is that the laminate around it degrades, the adhesion between copper and FR-4 weakens, nearby components run above their ratings, and the board discolours. Standard FR-4 has a glass transition somewhere around 130–140 °C, and once you are near it the substrate, not the conductor, is the part that fails.

The IPC-2221 formula

IPC-2221B reduces the whole problem to one empirical curve fit. It gives you the cross-sectional area of copper needed, in square mils:

A = (I / (k · ΔT^0.44))^(1/0.725)
W = A / (thickness_oz × 1.378)

• A — required copper cross-section, in square mils (thousandths of an inch, squared). 1 mil² = 645.2 µm²; multiply by 0.000645 to get mm².
• I — the steady-state current the trace carries, in amps. Not the fuse rating, not the inrush peak.
• ΔT — the temperature rise you will permit above ambient, in °C or kelvin; the two are interchangeable for a difference.
• k — 0.048 for external conductors on an outer layer, 0.024 for internal ones buried in the stackup. This constant carries the entire assumption about how the heat escapes.
• W — the finished trace width in mils, found by dividing the area by the copper thickness, because a trace is simply a rectangle of copper.
• thickness_oz — the copper weight in oz/ft². The 1.378 converts it to mils: 1 oz/ft² of copper spread over a square foot comes out 1.378 mils, or 35 µm, thick.

Working the default case through by hand: 1 A at a 10 °C rise on an outer layer gives ΔT^0.44 = 2.754, k·ΔT^0.44 = 0.1322, I divided by that is 7.564, and 7.564^1.379 = 16.30 mils² of copper. On 1 oz copper that is 16.30 / 1.378 = 11.8 mils, or 0.30 mm. That single number is worth committing to memory — it anchors everything else, because the relationship is a smooth power law with no discontinuities in it.

Two exponents are doing all the work, and both are worth internalising. Width scales with current as I^1.379, so doubling the current needs 2.6× the copper, not 2×. Width scales with temperature rise as ΔT^(−0.607), so doubling the permitted rise from 10 °C to 20 °C buys you a width reduction of only about 34%. Neither relationship is linear, and guessing by proportion will mislead you in both directions.

Current against width: the numbers you will actually use

1 oz external copper, the default stackup on almost every board. Widths are the IPC-2221 minimum — round up, never down.

Two things fall out of that table. At the bottom end the standard is irrelevant: a 6 mil trace, which is the narrowest most fabricators quote at standard pricing, already carries 0.61 A at a 10 °C rise. Nearly every signal trace on your board is limited by what the fab can etch, not by what the copper can carry. At the top end the widths stop being routable — a 10 A rail on 1 oz copper wants 7 mm of width, which is not a track any more but a pour. The useful range of this calculation is roughly 0.5 A to 5 A, and that is where most of the arguments happen.

IPC-2221 or IPC-2152?

This calculator implements IPC-2221, and it is worth being straight about what that means. IPC-2221’s current-capacity charts descend from measurements made in the 1950s, published as a conservative curve drawn under the scatter of the data. IPC-2152, released in 2009, replaced them with a modern measured data set and explicitly supersedes IPC-2221 for current-carrying capacity. It generally allows narrower traces — often substantially narrower — because it accounts for the things 2221 ignores.

Plenty of trace-width tools present the 2221 number as though it were current best practice and say nothing about 2152. That is misleading, so: 2152 is the better standard, and if you are fighting for density on a stackup you actually know the parameters of, running the 2152 charts will usually let you take copper back. The reason this calculator still gives you 2221 is practical rather than technical. It is one equation rather than a chart lookup, it needs only the inputs you certainly have, it errs generously, and it remains what most fabricators, internal design checklists and reviewers expect to see. Use it as your default and treat 2152 as the thing you reach for when 2221’s answer will not fit.

Why internal traces come out so much wider

Move the same trace to an inner layer and IPC-2221 halves k from 0.048 to 0.024. Because the area appears under the 1/0.725 exponent, halving k does not double the copper — it multiplies it by 2^(1/0.725) = 2.601. Your 1 A trace goes from 11.8 mils to 30.8 mils, and a 2 A rail from 0.78 mm to 2.03 mm. That is a serious amount of board area to give up, and it catches people out when they move a power rail off a congested outer layer.

The stated reasoning is convection. An outer-layer trace has one face open to the air and can shed heat by convection and radiation; a buried trace is sealed in laminate and, the argument goes, cannot. It is an intuitive story and it is wrong. When IPC-2152 measured it directly, internal traces in still air ran cooler than external ones at the same current, because FR-4 — mediocre as a thermal conductor — is still a far better path for heat than stagnant air is. The 2221 internal penalty is a known conservatism carried forward from an assumption nobody had checked, not a physical law. Follow it if your process requires 2221 compliance; do not lose sleep over an internal rail that is a shade under the number.

Copper weight, and what buying more of it gets you

Copper weight is finished thickness expressed as ounces of copper per square foot. Because the standard sizes copper by area, and area is width times thickness, weight and width trade off exactly inversely: 2 oz copper needs half the width of 1 oz for the same current, and 0.5 oz needs twice as much.

Heavy copper is not free. It costs more per panel, and it etches with less precision: the etchant attacks sideways as well as down, so thick copper needs wider spaces and gives you sloped trace edges. Fine-pitch parts and 2 oz outer copper fight each other. The usual compromise on a mixed board is 1 oz outers for the signal work with a generous pour for the power, or a stackup that puts heavy copper only on the layers that need it.

Heat is only half the answer — check the drop too

Sizing for temperature rise tells you the trace will survive. It says nothing about whether the load at the far end gets the voltage it needs, and on low-voltage rails that is usually the binding constraint. Copper’s resistivity is 1.7241 × 10⁻⁸ Ω·m at 20 °C and rises about 0.393% per kelvin, the same temperature coefficient that governs cable runs. A trace sized for a 10 °C rise in a 25 °C enclosure sits at 35 °C, where its resistance is already about 6% above the room-temperature figure — which is why this calculator quotes resistance at the hot temperature rather than at 20 °C.

Take the canonical 1 A trace: 0.30 mm wide on 1 oz copper is roughly 1.74 Ω per metre at 35 °C. Over a 100 mm run that is 174 mV. On a 12 V rail nobody will notice. On a 3.3 V rail it is 5% and starting to matter. On a 1.2 V core rail it is 14% of the supply, far outside any tolerance the part will accept, and the copper is not even warm. Whenever a rail is low voltage, long, or feeding something with a tight tolerance, size for the drop first and confirm the thermal number second — the same both-constraints discipline that governs cable sizing, where ampacity protects the conductor and voltage drop protects the load.

Pours, planes and thermal vias

Once a calculated width climbs past about 200 mils you are no longer routing a trace; you are drawing a copper region, and the sensible move is to stop thinking in tracks. A filled polygon or a dedicated plane region carries the current with margin, spreads heat over a much larger area than the IPC formula credits, and is easier to route around than an 8 mm track.

A pour is only as good as its connections, though. Current entering through a single via neck concentrates exactly where the copper is narrowest, and that neck — not the pour — sets the real limit. Use several vias in parallel wherever a pour meets a plane or crosses layers. Resistance in parallel divides, and so does the heat each via has to carry. As a rough guide, a 0.3 mm finished via with 25 µm of plating handles about 1 A comfortably, so a 5 A transition wants six or more, not one.

Thermal vias under a hot part work the same way and are worth stitching generously — they are nearly free at fabrication, and a grid of them under a regulator tab turns the opposite layer into a heatsink. The corollary is that any trace running near a plane, a pour or a via field is running cooler than IPC-2221 assumes, because 2221 does not know those exist. That is the margin you are paying for by using the older standard.

Why 10 °C is the default rise

Nothing in the standard mandates 10 °C. It has become the default because it is a sensible place for several separate margins to meet. It keeps the copper far below the laminate’s glass transition even in a warm enclosure; it leaves headroom for the ambient inside a sealed box being 20–30 °C above the room; it keeps the resistance rise small enough not to disturb a voltage-drop budget; and it is the number a reviewer expects to see, so choosing it costs you no explanation.

20 °C is a perfectly reasonable choice on a power board that is already warm and where the copper cost of the extra width is real. 30 °C is defensible for short runs on heavy copper if you have checked what else is nearby. Above that you are trading board life and component derating for area, and every part, connector and solder joint the trace passes needs checking against the temperature you have chosen — the trace does not get hot in isolation. Beyond a 100 °C rise you are outside the IPC-2221 charts entirely and the formula is extrapolating, as it is above 35 A.

From the calculated number to a routed board

1. Use the steady-state current. Inrush and startup peaks are absorbed by the copper’s thermal mass; a trace takes seconds to reach equilibrium. Continuous load sizes the trace.
2. Round up to something buildable. The result is a minimum. Snap it to a width your fab quotes comfortably and your design rules already use — extra copper is the cheapest thermal margin available.
3. Respect the fab minimum. 6 mil trace and space at standard pricing, 4 mil at a premium. If the calculation says 3 mil, draw 6 and take the margin.
4. Check the voltage drop. On anything below about 5 V, or over about 100 mm, the drop usually asks for more copper than the heat does.
5. Mind the necks. Widths are meaningless if the current has to squeeze through a single via, a thermal relief on a pad, or a gap between two connectors. The narrowest point is the trace.
6. Watch the neighbours. Several warm traces side by side heat each other, exactly as bundled conductors in a conduit do. Give power traces space, or give them a plane.