What the PCB Trace Width Calculator does
This calculator sizes a PCB track using the IPC-2221 formula: the width needed to carry a current for a chosen temperature rise, on an internal or external layer, in whatever copper weight your board uses. It also gives the resistance, volt drop and power of the actual run so you can see whether the track is a heating problem, a voltage problem, or neither.
IPC-2221's equation is a curve fit to measurements made on bare boards in still air, and it is deliberately conservative. Its successor IPC-2152 takes board thickness, nearby copper planes and laminate conductivity into account and generally allows narrower tracks. This page reports the IPC-2221 figure and says so - it is a starting point, not a thermal simulation.
How to use it
- Enter the current the track carries continuously. Peak currents in a switching converter do not size a track; RMS heating does.
- Choose the temperature rise you will accept. 10 C is the conservative default; 20 to 30 C is common on boards with airflow or a low ambient.
- Choose the layer. An internal track has no air above or below it, so IPC-2221 halves its k factor and the width roughly doubles.
- Set the copper weight from your stack-up - 1 oz is 35 um and the usual default, 2 oz is common on power boards.
- Enter the run length to see the resistance and volt drop, which on a low-voltage rail often matters more than the temperature does.
Reading the results
The width returned is a minimum for heating. On a power rail the volt drop usually forces something wider: 35 mV lost on a 1.2 V core supply is 3% of the budget before the regulator's own tolerance.
Temperature rise is above ambient, and it assumes the rest of the board is not also hot. Several high-current tracks side by side heat each other, and the rise adds.
Internal layers need roughly twice the width for the same current, which is why high-current nets are routed on outer layers or, better, on a plane.
Worked example: 2 A on an outer layer of 1 oz copper
At a 10 C rise, A = (2 / (0.048 x 10^0.44))^(1/0.725) = 42.4 mil^2. One ounce of copper is 1.378 mil thick, so the width is 42.4 / 1.378 = 30.8 mil, which is 0.78 mm.
Move the same track to an internal layer and k drops from 0.048 to 0.024, so the area becomes 110.3 mil^2 and the width 80 mil - 2.03 mm, two and a half times as wide for the same current.
Over a 100 mm run the external track's resistance at 30 C is about 66 milliohms, so it drops 131 mV and dissipates 262 mW. On a 12 V rail that is irrelevant; on a 1.2 V rail it is unacceptable, and the track would be widened for the volt drop long before heating became the constraint.
Formulas and scoring rules
- IPC-2221 cross-section
A[mil^2] = (I / (k x dT^0.44))^(1/0.725)k = 0.048 external, 0.024 internal. I in amps, dT in degrees Celsius.- Width from area
width[mil] = A / thickness[mil]1 oz/ft2 of copper is 1.378 mil, which is 34.8 um.- Current from a width
I = k x dT^0.44 x A^0.725The same equation rearranged - used above to check a width you have chosen.- Track resistance
R = rho(T) x L / Arho(copper) = 1.724e-8 ohm.m at 20 C, corrected by (1 + 0.00393 x (T - 20)).- Volt drop and power
V = I R, P = I^2 R- Unit conversions
1 mil = 0.0254 mm; 1 oz copper = 1.378 mil = 34.8 um
IPC-2221 against IPC-2152
IPC-2221's charts descend from measurements published in the 1950s, taken on bare boards in still air with no adjacent copper. They ignore almost everything that actually carries heat away from a modern track: the thickness of the laminate, the planes above and below, the board's in-plane thermal conductivity and any airflow.
IPC-2152, published in 2009, replaced them with a much larger measured data set and charts that account for those factors. Its results are frequently less conservative - sometimes by a factor of two - which is why a design sized by IPC-2221 will pass an IPC-2152 assessment but not the other way round. For a production design in a thermally demanding product, use IPC-2152 or a thermal simulation; for a quick answer, the number on this page errs on the safe side.
When volt drop, not heat, sets the width
On any low-voltage, high-current rail the volt drop reaches an unacceptable level while the track is still comfortably cool. Work out the budget first: for a 1.2 V rail with 3% total tolerance, and the regulator taking most of it, a few tens of millivolts in the track is all you have.
The remedies are the same ones as for temperature - wider tracks, heavier copper, both outer layers in parallel, or a plane - but the target is different, and the calculation above shows both so you can see which one binds. Where the load is sensitive, remote sensing at the load removes the track drop from the regulation loop entirely.
Limitations: what the result does not prove
- IPC-2221 is a conservative generic rule from measurements on bare boards, not a thermal model of your board. It knows nothing about planes, airflow, laminate or nearby heat sources.
- The published charts cover up to 35 A, tracks to 400 mil, rises to 100 C and 0.5 to 3 oz copper. Outside that the curve fit is an extrapolation and the calculator says so.
- Temperature rise is above ambient, for one isolated track. Several hot tracks together, or a hot component nearby, add to it.
- This is indicative only and not a compliance statement. A qualified engineer must verify the design against IPC-2152 or a thermal analysis, and against the applicable standards for the product, before it is manufactured.
Privacy: where your data goes
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Standards and sources
- IPC-2221 - Generic standard on printed board design - checked 19 Sep 2026
- IPC-2152 - Standard for determining current-carrying capacity - checked 19 Sep 2026
- IEC 60228 - Conductors of insulated cables (copper resistivity reference)
Frequently asked questions
How wide should a PCB trace be for 2 A?
On an external layer in 1 oz copper with a 10 C rise, 0.78 mm (30.8 mil) by IPC-2221. On an internal layer it is 2.03 mm. Allowing a 20 C rise cuts the external figure to about 0.51 mm. Volt drop on a low-voltage rail may demand more than any of these.
Why do internal traces need to be wider?
They have no exposed surface to convect or radiate heat from, so IPC-2221 halves the constant k from 0.048 to 0.024. That works out at roughly twice the width for the same current and temperature rise.
What temperature rise should I allow?
10 C is the usual conservative choice and the one most fabricators assume. 20-30 C is reasonable where the ambient is low and the board has airflow. What matters is the final copper temperature against the laminate's rating and any components sitting on the track.
How much current can a 1 mm trace carry?
In 1 oz copper on an external layer at a 10 C rise, about 2.5 A by IPC-2221; internally, about 1.3 A. Double the copper weight and both roughly double. Enter your own figures above rather than relying on a single remembered number.
What is the difference between IPC-2221 and IPC-2152?
IPC-2221 uses old charts from bare-board measurements and is conservative. IPC-2152, from 2009, is based on a much larger data set and includes board thickness, copper planes and laminate conductivity, so it usually permits narrower tracks. IPC-2152 is the current standard; this page gives the IPC-2221 figure.
Does copper weight change the width I need?
Yes, in direct proportion for a given cross-section. The formula gives a required area; thicker copper achieves it in a narrower track. Going from 1 oz to 2 oz halves the width needed, at the cost of coarser etching tolerances and more expensive fabrication.
Should I use a plane instead of a wide track?
For anything above a few amps, usually yes. A plane has far more cross-section, spreads heat over the whole board and gives the return current a low-impedance path directly under the outgoing one, which is as important for noise as it is for heating.
Does this apply to pulsed currents?
Not directly. Copper's thermal mass averages short pulses, so a track can carry a much higher peak than its continuous rating. Size it on the RMS current for heating, but check the peak against fusing current if the pulse is large and long.
Last reviewed by the A2Z.Tools team against the sources listed above.
