A printed circuit board trace is a thin strip of copper that carries current from one component to another, and its width is a genuine engineering decision, not a cosmetic one. A trace that is too narrow for its current heats up, because the copper has resistance, and the electrical power dissipated as heat rises with the square of the current. Enough heating discolours the board, degrades the adhesive holding the copper, and in extreme cases delaminates the trace or damages nearby components. A trace that is wider than necessary, on the other hand, consumes board area, blocks routing, and costs money on dense designs. The standard method for sizing traces is the IPC-2221 formula, and understanding it will let you check any calculator output rather than trusting it blindly.
The IPC-2221 formula works in two steps. First it computes the required cross-sectional area of the trace from the current and the allowed temperature rise. The formula is: area = (I / (k × ΔT^b))^(1/c), where I is the current in amperes, ΔT is the allowed temperature rise in degrees Celsius, and k, b, and c are constants that differ for external and internal layers. For external layers, k is 0.048, b is 0.44, and c is 0.725. For internal layers, k is 0.024, with the same b and c. The smaller k for internal layers reflects the fact that a trace buried inside the board sheds heat less effectively than one exposed to air, so it needs more copper for the same current. Once the area is known, the width follows from the copper thickness: width = area / (thickness × 1.378), where thickness is in ounces of copper and the result is in mils, thousandths of an inch. One ounce copper is about 1.378 mils thick, which is where that constant comes from.
Consider a worked example. An external trace must carry 5 amperes continuously, and the designer allows a temperature rise of 10 degrees Celsius above ambient. First compute ΔT^b, which is 10^0.44. Ten to the power 0.44 is approximately 2.754. Multiply by k for external layers, 0.048 × 2.754, which is approximately 0.1322. Divide the current by that: 5 / 0.1322, which is approximately 37.82. Now raise that to the power 1/c, which is 1 / 0.725, or about 1.3793. The value 37.82 raised to 1.3793 is approximately 149.6 square mils. Finally, for one-ounce copper, divide by 1.378: 149.6 / 1.378 gives approximately 108.5 mils, which rounds to about 109 mils. Converting to metric, 109 mils multiplied by 0.0254 millimetres per mil gives about 2.77 millimetres. So a 5-ampere external trace on one-ounce copper with a 10-degree rise needs a trace roughly 2.8 millimetres wide. Every step can be checked with a scientific calculator, and checking at least once by hand is worthwhile, because unit errors in this formula are common and consequential.
The same example shows how sensitive the result is to the inputs. If the allowed temperature rise were 20 degrees instead of 10, the trace could be narrower, because more heating is tolerated; if it were only 5 degrees, the trace would need to be considerably wider. If the same 5 amperes ran on an internal layer instead, k would be 0.024 instead of 0.048, which doubles the base of the power calculation and multiplies the area by 2 raised to 1.3793, about 2.60, so the width would come out near 283 mils, or about 7.2 millimetres. Copper thickness helps directly: two-ounce copper halves the required width for the same area, which is why power boards often use heavier copper rather than enormously wide traces. Ambient temperature matters too, because the formula limits the rise above ambient, not the absolute temperature: a board in a hot enclosure has less headroom before the copper and the board material reach their limits, so the allowed rise must be set with the worst-case ambient in mind, not room temperature on a bench.
Several practical points qualify the formula. It assumes a steady, continuous current. Short pulses, such as the inrush to a capacitor or a brief motor start, heat the trace far less than a continuous current of the same peak value, because the trace has time to cool between pulses, though repeated pulses can still accumulate heat. It also assumes a single isolated trace on a board of typical material; a trace surrounded by other hot traces, or one running over a ground plane that acts as a heat sink, behaves differently. Very short traces connected to large copper pours or component pads lose heat into those structures and can be narrower than the formula suggests, while traces in the middle of a dense, hot board may need to be wider. The formula is a well-tested guideline, not a simulation, and critical power designs are verified with thermal measurement or simulation rather than by formula alone.
Voltage is a separate concern that trace width does not address. Width controls heating from current; clearance, the spacing between conductors, controls insulation and breakdown under voltage. A wide trace carrying high voltage still needs adequate spacing from neighbouring conductors, and standards specify minimum clearances that grow with voltage, altitude, and pollution conditions. Conflating the two, sizing a trace for current and assuming voltage is handled, is a classic beginner error. Both must be checked, against their own rules.
A sensible workflow for sizing a power trace runs as follows. Determine the maximum continuous current, including a margin for fault or overload conditions if the design requires it. Choose the allowed temperature rise based on the board material and the worst-case ambient temperature, commonly 10 degrees for conservative designs and up to 20 or 30 where the application permits. Decide the layer and the copper weight, then compute the width, and round up to a convenient value that the board fabricator can reliably produce. Check the result against the manufacturer's minimum width and clearance capabilities, and confirm that the trace actually fits the routing space; if it does not, consider heavier copper, a different layer, a copper pour instead of a trace, or splitting the current across multiple paths. Finally, on the first prototype, measure the temperature of the trace under full load and compare it with the prediction, because a measurement closes the loop that a formula alone cannot.
This guide is educational guidance for low-voltage board design. It does not replace the IPC standards, the board manufacturer's capabilities, or qualified engineering review, and it does not cover mains-voltage or safety-critical design. Electrical work should follow local codes, and designs that carry significant power or that affect safety should be checked by a qualified engineer or electrician.
Try it next: use the PCB Trace Width Calculator with your own current and rise, then check via current with the PCB Via Current Calculator and the Voltage Divider Calculator. For background reading, see LED Series Resistor Sizing.