What sets the width
A trace carrying current dissipates I²R as heat, and it loses that heat to the board and the air around it. The temperature it settles at is where those two balance. Widening the trace does two things at once: it lowers the resistance so less heat is made, and it increases the surface shedding heat. That is why the relationship between current and width is not linear — doubling the current needs well more than double the width.
The relation used here is the IPC-2221 curve, which expresses the required cross-sectional area in square mils as a function of current and allowable temperature rise, with a different constant for external and internal layers. Width follows by dividing the area by the copper thickness: 1 oz copper is 1.378 mils thick, 2 oz is twice that. So 1 A on 1 oz external copper at a 10 °C rise wants about 12 mils, and the same current on 2 oz copper wants about 6.
Internal traces get roughly half the current for the same width, because they are surrounded by laminate rather than open to air and have no easy path to shed heat. The standard bakes that in as a constant that is half the external one.
Choosing a temperature rise
| Rise chosen | What it implies | Typical use |
|---|---|---|
| 10 °C | Trace stays close to board temperature | Default for anything you have not measured |
| 20 °C | Noticeably warm, still safe on FR-4 | Space-constrained boards running in cool air |
| 30 °C and up | Trace is a heat source in its own right | Only with deliberate thermal design and testing |
The rise is above the board around the trace, not above room temperature. A board already at 60 °C inside an enclosure with a 20 °C rise puts the copper at 80 °C, and the laminate, the solder mask and the adjacent components all live at that temperature too. This is why a design that passed on the bench fails in a sealed box: the ambient moved, and the rise stacked on top of it.
Resistance, drop and where it actually matters
One ounce copper has a sheet resistance near half a milliohm per square, so a 12 mil trace an inch long is about 83 squares, roughly 41 mΩ. At 1 A that is 41 mV and 41 mW. Trivial on a 12 V rail; not trivial on a 1.2 V core rail where 41 mV is 3.4 percent of the budget, and actively wrong on a current-sense line where the drop lands directly in the measurement.
Copper resistance also rises about 0.39 percent per degree, so a trace at 85 °C is about 25 percent more resistive than at 20 °C. That is a mild positive feedback: hotter means more resistive means more heat. It does not run away in normal designs, but it means the drop measured on a cold board understates the drop in service.
What the calculation ignores
Vias are the most common oversight. A trace sized generously for 3 A that passes through a single small via has a bottleneck the width calculation never sees; use several vias in parallel wherever a current path changes layer. Connectors, pads and the copper necking down around a component footprint are the same problem in a different place — the narrowest point sets the temperature, not the average width.
Neighbouring copper cuts both ways. A ground plane one layer down conducts heat away and makes the real rise lower than calculated, while three current-carrying traces packed side by side heat each other and make it higher. Solder mask, conformal coating, board thickness and airflow all move the answer, and none of them appear in the formula.
The honest summary is that this gives you a starting width with the industry's standard margin built in. On anything carrying more than a couple of amps, put a thermocouple on the trace or point a thermal camera at the assembled board under real load and find out what it does. See also the heat sink calculator for the components at the ends of these traces and the power supply rail calculator for the current each rail actually has to carry.
Questions people ask
How wide should a trace be for 1 amp?
About 12 mils on 1 oz external copper for a 10 °C rise, or about 6 mils on 2 oz. Internal, on 1 oz, it wants roughly 30 mils for the same rise. Those are the numbers this calculator produces from IPC-2221 and they match the rules of thumb most people carry. In practice many designers route 1 A on 20 mils or more simply because the space is there and the extra copper costs nothing, which lowers the drop as well as the temperature.
What does copper weight mean in ounces?
It is the weight of copper spread over one square foot, which translates to a thickness. One ounce is about 1.4 mils or 35 µm, half an ounce is 0.7 mil, and two ounce is 2.8 mil. Standard boards are 1 oz on outer layers and often 0.5 oz on inner layers, so an internal trace can be thinner than you assume in two ways at once — less copper and no air. Heavier copper costs more and limits how fine the etched features can be, so it is a choice for power boards rather than a default.
Why do internal traces need to be so much wider?
Because they cannot get rid of heat. An outer trace sits against air and radiates and convects from its surface. An inner trace is encased in laminate, which is a fairly poor thermal conductor, and the heat has to travel through it before it reaches anything that can shed it. IPC handles this by halving the constant in the current equation, which works out to needing roughly twice the width for the same current and rise. If your inner-layer trace also sits directly under a hot component, the real situation is worse than the formula suggests.
Is the IPC-2221 formula accurate?
It is a curve fit to measurements made decades ago on bare test coupons, and it is used industry-wide because it is conservative and simple, not because it models your board. Real results vary in both directions: a trace over a solid ground plane runs cooler than predicted, while a bundle of parallel traces in still air inside a sealed enclosure runs hotter. Thermal simulation or measurement beats it every time. Treat the output as a starting width with margin already in it, and verify on the actual assembly if the current is significant.
Does a via limit how much current a trace can carry?
Frequently, yes, and it is the most common thing people miss. The copper in a plated via barrel is thin and its cross-section is much smaller than the trace it connects, so a single via can be the hottest point in an otherwise generous power path. The usual fix is several vias in parallel at every layer change on a current path, spread out rather than clustered, so the heat has somewhere to go as well. The same logic applies to any point where the copper narrows — a pad neck, a connector footprint, a thermal relief.