Trace Width and Current: Calculating the Right Copper

Sizing a trace for current is one of the few layout decisions with a closed-form answer, and it is also one of the most frequently skipped. A trace that is too narrow will heat up, increase its resistance, heat up further, and in the worst case open. A trace that is too wide wastes routing space that the design may not have. Between those extremes there is a range of acceptable widths, and choosing within it requires understanding what actually limits the conductor. The limit is nearly always thermal rather than electrical, which is why the calculation starts with heat rather than with resistance.

Current Density and Cross Section

The quantity that determines heating is current density: the current carried per unit of cross-sectional area, expressed in amperes per square millimetre. The relationship is simple. The cross section of a trace is its width multiplied by its thickness, so the current density is the current divided by that product.

A worked example makes the arithmetic concrete. Taking a working current density of 20 amperes per square millimetre and a conductor thickness of 100 micrometres, which is 0.1 millimetres, the required width for a one ampere current is one divided by twenty times 0.1, which equals 0.5 millimetres. Converted to imperial units, 0.5 millimetres is about 19.7 mil, so the theoretical minimum width for one ampere under those assumptions is roughly 20 mil.

The assumptions are the important part. A current density of 20 amperes per square millimetre is a reasonable working figure for a modest temperature rise on an outer layer with good convection, but it is not a universal constant. A design constrained by a tight temperature rise, or one whose traces are buried between layers with no airflow, needs a lower current density and therefore a wider trace.

Temperature Rise Is the Real Criterion

The standard treatment of trace sizing relates the allowable temperature rise to the current, the width and the copper thickness, rather than fixing a current density. The reason is that the failure mode being prevented is thermal: the trace must be able to dissipate the heat it generates without exceeding a temperature that would damage the laminate, degrade the solder mask or affect the reliability of neighbouring components.

A copper trace on an outer layer dissipates heat by convection and radiation, so its allowable current for a given width is higher than an inner layer trace of the same dimensions, which can only conduct heat into the laminate. That asymmetry is significant in practice, and charts that give a single current for a given width without distinguishing inner from outer layers are misleading.

The presence of copper nearby also matters. A trace surrounded by a ground pour conducts heat into the pour through the dielectric, which raises its capacity; a trace isolated on a bare area of the board runs hotter. The same reasoning explains why the copper area attached to a component affects the thermal behaviour of that component far more than the nominal trace width alone, and the role of flooded copper in that process is described in this discussion of copper flooding.

Copper Thickness and Plating

Copper thickness is expressed in ounces, meaning the weight of copper in one square foot of the material. One ounce corresponds to approximately 35 micrometres, and standard boards use one ounce on the outer layers and half an ounce or one ounce on the inner layers. Heavy copper constructions increase that figure substantially, and they are chosen precisely when the current requirement cannot be met with a wider trace.

Plating adds to the outer layer thickness, but not uniformly. The barrel of a plated hole and the surface copper both receive plating, and the current density in the plating bath determines how much. Where a design depends on a specific copper thickness for its current capacity, that dependence should be stated explicitly, because the finished thickness is a process outcome rather than a specification the fabricator can hit exactly. The chemistry that controls deposition is described in this article on electroplating additives.

Voltage Drop Over Length

Current capacity is only half the calculation. A long trace has resistance proportional to its length and inversely proportional to its cross section, and the resulting voltage drop may matter more than the temperature rise. A trace carrying one ampere with a resistance of 50 milliohms drops 50 millivolts, which is negligible on a 12 volt rail and significant on a 1.2 volt core supply.

The rule that follows is that power distribution networks should use planes rather than traces wherever the space allows, because a plane offers a much larger cross section for the same footprint. Where a trace must carry a substantial current, the designer should compute the drop at the maximum current and compare it with the tolerance of the load, not with the nominal rail voltage.

Vias are part of the same calculation. Each via has a resistance determined by the wall area and the plating thickness, and a path that passes through several vias in series can have more resistance than the trace connecting them. Where a current path crosses layers, the number of vias should be chosen to keep the total resistance acceptably low, and the same dimensioning logic applies to the via barrel as to the trace, as described in this discussion of via to trace clearance.

Practical Sizing Procedure

Determine the maximum continuous current the trace will carry, including the transient margin the load requires. Decide the allowable temperature rise based on the laminate rating and the enclosure conditions. Choose the copper thickness available from the fabricator for that layer. Select a width that satisfies both the temperature rise and the voltage drop criteria, using inner-layer data for inner layers and outer-layer data for outer layers.

Then check the corners. Traces that step down in width create a local hot spot, so the smallest width on the path is the one that must satisfy the calculation rather than the nominal width. Where a trace passes through a via field or a component pad, the copper on either side of the restriction should be widened, and thermal relief connections to a plane should be sized with the same care as the trace itself.

FAQ

How wide should a trace be to carry 1 ampere? Roughly 20 mil on an outer layer with 100 micrometres of copper if the allowable current density is 20 amperes per square millimetre. That figure assumes a modest temperature rise and good convection; an inner layer trace or an enclosed design needs to be wider.

Why do inner layers carry less current than outer layers of the same width? Because an inner layer trace can only dissipate heat by conduction into the laminate, while an outer layer also loses heat by convection and radiation. The same width therefore reaches a higher temperature inside the board than on the surface.

Is temperature rise or voltage drop the more important limit? Both must be checked, and which one governs depends on the application. High-current power rails are usually limited by voltage drop, while a signal or a modest supply on a densely routed board is more often limited by temperature rise.