Calculating Trace Width for Current and Temperature Rise

Trace width is decided by two questions that pull in opposite directions. The trace has to carry its current without getting too hot and without dropping too much voltage, which pushes the width up; and the routing has to fit into the area available, which pushes it down. Where a controlled impedance is also required, the width is fixed by the stack-up instead, and the current question becomes a check rather than a choice.

What the Width Has to Satisfy

Three requirements apply to any conductor, and they are usually considered in this order.

Current capacity is the requirement that the conductor survive its own heating. A trace is a resistor, the heat it produces rises with the square of the current, and the temperature it reaches is the balance between that heat and the rate at which the board removes it. Beyond a certain point the laminate degrades, the copper oxidises and the mechanical strength of the joint to the board is lost.

Temperature rise is the way that limit is normally expressed. A permitted rise is chosen for the design — a modest figure for a product that will run continuously, a higher one for a part that is only loaded briefly — and the width is calculated to stay inside it.

Voltage drop is the third. A long trace has measurable resistance, and the current flowing through it produces a loss and a voltage difference between its ends. On a supply rail feeding a device at low voltage, that difference can be a substantial fraction of the voltage itself.

The Rule of Thumb and Its Limits

The most quoted approximation is that an ounce of copper — about thirty-five micrometres thick — carrying an ampere needs roughly a millimetre of width. It is a useful sanity check and a poor design method, because it says nothing about the temperature rise that is acceptable, the length of the trace, the layer it is on or the ambient temperature.

The most important thing it omits is the layer. A trace on an outer layer is exposed to air on one side and can dissipate heat reasonably well. A trace on an inner layer is sandwiched in laminate, which conducts heat away slowly, so for the same current and the same permitted temperature rise an inner trace has to be significantly wider. Designs that assume the two are equivalent produce inner-layer traces that run hotter than intended.

The Standard Calculation

The accepted method is the one published in IPC-2221, the industry standard for printed board design, and it works from the cross-sectional area rather than the width directly.

The first step determines the required conductor area from the current and the permitted temperature rise, using an empirical relationship in which the area rises with the current and falls as the permitted rise increases. The relationship is not linear in either: raising the current by a factor requires more than a proportional increase in area, which is a consequence of the fact that the heat generated rises with the square of the current while the ability to dissipate it rises only with the surface area.

The second step converts the area into a width by dividing by the copper thickness. Since copper is specified in ounces per square foot, the thickness in the units the calculation uses is obtained from that figure, and the result is a width that satisfies the current and temperature requirement for that copper weight.

The published method gives different constants for inner and outer layers, which is where the layer effect described above enters the calculation. Using the outer-layer constant for an inner-layer trace is one of the most common errors in these calculations, and it produces a trace that is too narrow by a wide margin.

power traces sized for current on a PCB layer

What the Formula Does Not Cover

The calculation produces a width for a straight isolated trace in a stated environment, and a real board is not that. Several factors shift the answer.

The copper thickness is nominal. Plating adds to it, etching removes from it, and the finished cross section differs from the drawing in both directions. On a very narrow trace the finished width can be materially less than the drawn one, which is why minimum feature limits are set by the fabricator rather than by the calculation.

The ambient temperature matters as much as the permitted rise, since a trace in a hot enclosure has less headroom before it reaches a damaging temperature. The permitted rise is a rise above ambient, not an absolute temperature.

Cooling conditions matter: still air, moving air and a board mounted against a metal chassis all remove heat at different rates, and the difference is large enough that the same trace can be comfortable in one product and marginal in another.

And neighbouring copper matters in both directions. A trace surrounded by copper, particularly copper connected to a plane, conducts heat away more effectively and can be narrower; a trace on an isolated layer has to be wider. The same applies to the total area of the board, which acts as a heatsink.

Voltage Drop and Efficiency

Where a conductor carries current over a distance, the loss is worth calculating separately, because it can dominate the decision.

The resistance of a trace rises with length and falls with cross-sectional area, so a long supply path to a distant load can waste a meaningful fraction of the delivered power even when the width is generous enough for the current alone. On a low-voltage rail the same resistance consumes a larger fraction of the voltage, and the device at the far end may be operating below the minimum its specification allows.

The remedies are the same in both cases: shorten the path, widen it, or use the plane layers to carry the current instead of a trace, which both lowers the resistance and spreads the heat. Where the current is large enough, the copper weight itself becomes a design variable, as described in the article on heavy copper in HDI boards.

When Impedance Decides Instead

On a controlled-impedance net the width is not a free variable. It is one of the parameters that produce the required impedance for the given stack-up, and changing it to satisfy a current calculation would break the impedance the design exists to achieve.

The sequence in that case is reversed. The width is taken from the impedance calculation, and the current requirement is checked against it. Where the trace cannot carry the current at that width, the answers are to route the net on a layer with thicker copper, to use several parallel traces or vias to split the current, or to reconsider whether the net needs both properties at once. What a reflection at an impedance discontinuity does to a signal is described in the note on the reflection coefficient and impedance.

Checking the Result

Because the calculation carries several assumptions, the finished board is worth measuring. A thermocouple or a thermal camera on the copper that carries the heaviest current, at full load and in the worst ambient the product will see, shows whether the rise is what the design predicted. Where the measurement disagrees with the calculation, the disagreement is usually explained by the thermal environment rather than by the arithmetic — which is exactly the factor the calculation is least able to account for.

The power stage of a converter is where these widths matter most, and the routing decisions that accompany them are covered in the article on half-bridge converter layout, while the requirements for controlled impedance are set out in our guide to impedance control.

copper width measurement on an inner layer trace

FAQ

Is one ampere per millimetre a reliable rule? As a starting estimate only. It says nothing about the permitted temperature rise, the layer, the length or the ambient temperature, all of which change the answer materially.

Why does an inner layer need a wider trace? Because the laminate around it removes heat far more slowly than air does. The same current at the same permitted rise needs a larger cross section inside the board.

What if the impedance requirement conflicts with the current requirement? The impedance sets the width and the current is checked against it. Where the trace cannot carry the current, the fix is usually a thicker copper layer or several parallel conductors rather than a wider trace.

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