How this instrument works
This is the companion question to "how much current can this trace carry?" — instead you already know the current you need to carry and want to know how wide to draw the trace. Both questions use the same underlying IPC-2221 empirical relationship between cross-sectional area, allowed temperature rise, and current; this calculator just solves it in the other direction, for area (and therefore width) given a target current.
The same two physical facts that govern the forward direction still apply here. More current needs more copper cross-section, since resistive heating (I squared R) rises faster than the cooling a given cross-section provides. And a trace buried internally, insulated by the surrounding laminate instead of exposed to open air, needs roughly double the cross-section of an external trace to carry the same current at the same allowed temperature rise — which is why the internal-layer option in this calculator consistently comes back with a noticeably wider required trace than the external option for the same current.
Once you have the required width in mils, remember it's a minimum, not a target to hit exactly — PCB fabricators generally work to standard trace-width increments and manufacturing tolerances, so most designers round up to the next convenient width their fab house supports rather than drawing the theoretical minimum exactly.
- Enter Required current in amps — the current this trace needs to carry.
- Enter Trace thickness in mils — 1oz copper is about 1.378 mils, 2oz about 2.756 mils.
- Enter Allowed temperature rise deltaT in degrees C — how much hotter the trace may get above ambient (10C is a common default).
- Select Layer — External (exposed to air) or Internal (buried, needs more copper for the same current).
- Read Required trace width in mils — round up to your fab house's next available trace-width increment.
Worked example — the round-trip check against 1.4635A
Feed in 1.4635A, essentially the current a 20-mil, 1oz-copper, 10C, external trace produces in the forward IPC-2221 current calculation, with the same 1.378-mil thickness, 10C, and external layer settings: A = (1.4635 / (0.048 x 10^0.44))^(1/0.725) = (1.4635/0.13220)^1.37931 = 11.0701^1.37931 = 27.56 mils squared. Width W = 27.56 / 1.378 = 20.00 mils — the calculator reconstructs the original 20-mil trace, confirming the two directions are true inverses of each other.
Ask for a higher 3A on the same external, 1oz, 10C setup: A = (3/0.13220)^1.37931 = 22.692^1.37931 = 74.16 mils squared; W = 74.16/1.378 = 53.82 mils — current only a bit more than doubled (1.4635A to 3A), but the required width grew nearly 2.7x, since area scales faster than linearly with current here (the 1/0.725 exponent is a bit above 1).
Questions
Why does the internal-layer option give a much wider required trace?
Because an internal trace, buried in the board's laminate, can't shed heat by convection into open air the way an external trace can — so it needs more copper cross-section to carry the same current at the same allowed temperature rise. IPC-2221 models this with half the empirical constant for internal traces (k=0.024 versus 0.048 external), which works out to roughly double the required width for the same current, thickness, and temperature rise.
Should I use the exact calculated width, or round up?
Round up. The formula gives a theoretical minimum; real fabrication has fixed manufacturing increments and etch tolerances, and most designers add margin rather than drawing the exact minimum. Round to your fab house's nearest supported trace width at or above the calculated value, not below it.
How does trace thickness affect the required width?
Thicker copper needs less width for the same current, since it's the cross-sectional area (width times thickness) that ultimately matters, not width alone. Doubling the copper weight from 1oz to 2oz roughly halves the required width for a given current — which is why high-current boards often specify heavier copper rather than making every trace impractically wide.
Is this the inverse of your PCB Trace Current calculator?
Yes, exactly — both tools implement the same IPC-2221 relationship between current, cross-sectional area, temperature rise, and layer type; this one just solves for the width needed to hit a target current, instead of the current a given width can carry. Feeding this calculator's width output back into the trace-current calculator, with the same thickness, deltaT, and layer, reproduces the original current.
What if my required width doesn't fit my available board space?
You have a few real options: use heavier copper (which lowers the required width for the same current), split the current across two or more parallel traces, add a copper pour connected to the trace to help spread heat, or, if the design allows it, relax the allowed temperature rise slightly. Simply drawing a narrower trace than the calculation requires isn't one of them — it directly trades away thermal margin.