The PCB trace width calculator above applies the IPC-2221 conductor sizing relationship: given a current and the temperature rise you are willing to tolerate, it returns the copper cross-section required and converts that into a width for the copper weight you have specified. It then reports what that trace will actually do electrically — its resistance at the working temperature, the volts lost along it, and the heat it produces.
Arb Digital builds free calculators that show their working. This one is explicit that the IPC-2221 curve is an empirical fit to measured data with a defined range of validity, and it tells you when your inputs have wandered outside that range instead of extrapolating quietly.
What This PCB Trace Width Calculator Does
The headline number is the minimum trace width in millimetres, with the same figure in mils underneath, since board houses and design rules still speak in mils. That width is what the IPC-2221 relationship requires for your current at your chosen temperature rise.
The grid adds four numbers the width alone does not tell you. Cross-sectional area is the quantity the formula really solves for, and it is worth seeing directly, because width and copper weight are interchangeable ways of reaching it. Resistance is computed at the trace's own working temperature, meaning ambient plus the rise you allowed, using the temperature coefficient of copper. Voltage drop is that resistance times your current, and power dissipated is the heat the trace sheds into the board.
The last two matter more often than designers expect. A 3.3 V rail that loses 150 mV along a trace has given away a meaningful slice of its regulation budget, and the IPC width calculation will not warn you about that, because it was never about voltage.
How to Use It
- Enter the continuous current, not the peak. Thermal rise depends on sustained heating. A motor that draws 20 A for 40 milliseconds at startup and 3 A thereafter is a 3 A thermal problem.
- Choose a temperature rise you can defend. Ten degrees is conservative and common. Larger rises give narrower traces, but the rise adds to whatever the board is already sitting at.
- Set the layer correctly. An internal trace is buried in laminate with no air on either face, and IPC-2221 halves its constant to reflect that. Getting this wrong understates the width by roughly a factor of two.
- Pick the copper weight your fabricator will actually supply. Two-ounce copper needs half the width of one-ounce for the same area, but it costs more and etches with wider tolerances.
- Read the voltage drop before you accept the width. On a low-voltage rail, drop rather than heat is usually the binding constraint, and it will push you wider than IPC-2221 does.
The Formula: How IPC-2221 Sizes a Conductor
IPC-2221 expresses current capacity as I = k × ΔT0.44 × A0.725, with the current in amperes, the temperature rise in degrees Celsius and the cross-section in square mils. The constant k is 0.048 for external conductors and 0.024 for internal ones. Rearranged to give the area you need: A = (I ÷ (k ΔT0.44))1/0.725. Dividing that area by the copper thickness gives the width. The relationship comes from IPC-2221, Generic Standard on Printed Board Design, published by IPC.
Work the defaults through. Two amps, ten degrees rise, external layer, one-ounce copper. ΔT0.44 = 100.44 = 2.754. Multiply by k = 0.048 to get 0.1322. Divide the current: 2 ÷ 0.1322 = 15.13. Raise that to the power 1 ÷ 0.725 = 1.379 and you get 42.4 square mils. One-ounce copper is 1.378 mils thick, so the width is 42.4 ÷ 1.378 = 30.8 mils, which is 0.78 mm.
The resistance follows from the geometry. Copper's resistivity is 1.68 × 10−8 Ω·m at 20 °C with a temperature coefficient of 0.0039 per degree, as tabulated in OpenStax University Physics Volume 2, section 9.3 on resistivity and resistance. At a working temperature of 35 °C, 50 mm of that 42.4 square mil trace comes to about 32.5 mΩ, which drops 65 mV at two amps and dissipates 130 mW.
Where the IPC-2221 Curve Comes From, and Its Limits
The exponents in that equation are not derived from first principles. They are a curve fit to a set of measurements made decades ago on bare, isolated conductors, and every part of the test arrangement is baked into the answer. The charts were produced for conductors up to about 400 mils wide and cross-sections up to roughly 700 square mils, with temperature rises between about 10 and 100 degrees. Push far outside that envelope and the fit stops describing anything real.
Three effects the formula ignores are large in modern boards. A trace sitting directly over a copper plane runs considerably cooler than the isolated conductor that was tested, because the plane spreads heat. A trace on a thick board with poor thermal conductivity runs hotter. And a trace passing through vias meets extra resistance the width calculation never sees.
IPC published IPC-2152, Standard for Determining Current Carrying Capacity in Printed Board Design, precisely because of these gaps. It accounts for board thickness, adjacent planes and dielectric conductivity, and its results differ substantially from IPC-2221 in both directions depending on the construction. IPC-2221 remains widely used because it needs one line of arithmetic, and it is usually conservative for external traces on ordinary boards. Treat it as a first estimate, not a verdict.
Why Internal Traces Need Roughly Twice the Width
Halving the constant does not halve the current; it changes the width you need for a given current. Because area appears with an exponent of 0.725, doubling the required current capacity term does not double the area. Work it through: cutting k in half doubles the bracketed term, and two raised to the power 1.379 is 2.6. So an internal trace needs about 2.6 times the cross-section, not twice it.
The physical reason is straightforward. An external trace loses heat to moving air on one whole face and radiates from it. An internal trace is encased in laminate whose thermal conductivity is poor, perhaps a hundredth that of copper, so its heat has to travel sideways through copper or slowly outward through resin before it reaches anything that can shed it. It is not a small difference, and it is the single most common oversight in multilayer power routing.
Heat Is Not the Only Constraint
Four other limits regularly decide a trace width before temperature rise does, and none of them appears in the IPC-2221 formula.
Voltage drop comes first on low-voltage rails. Point-of-load regulators may tolerate only tens of millivolts of static drop before their sense point misreads the load. Use the drop figure in the grid, and if it is uncomfortable, widen the trace or move the sense point rather than accepting the thermal minimum. Our voltage drop calculator handles the same question for cable runs.
Conductor spacing comes second, and it is governed by voltage rather than current. IPC-2221 has separate spacing tables driven by peak voltage between conductors, coating and altitude, and on a mains-referenced board those clearances are safety-critical and non-negotiable.
Fusing current comes third. A trace that carries three amps happily will still vaporise at some much larger current, and if it is the only thing standing between a fault and a fire, the design is wrong. Protection belongs in a rated device.
Manufacturing tolerance comes fourth. Etching removes copper from the sides as well as the top, so a finished trace is narrower than drawn, and the effect grows with copper weight. Heavy copper commonly needs several mils of allowance.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Using peak current instead of continuous — thermal rise is driven by sustained heating, so a short inrush spike does not size the trace.
- Leaving the layer set to external for a buried trace — an internal conductor needs around 2.6 times the cross-section, so the error is not marginal.
- Adding the temperature rise to nothing — the rise is above the local board temperature, which may already be well above room temperature inside an enclosure.
- Ignoring voltage drop — on a 1.8 V or 3.3 V rail, drop usually forces a wider trace than heat does, and this formula never mentions it.
- Treating a trace as a fuse — overcurrent protection belongs in a rated device selected by a qualified engineer, never in the geometry of a conductor.
Related Free Tools From Arb Digital
Work out the current from a supply voltage and a load with the Ohm's law calculator, then check the loss along a cable feeding the board with the voltage drop calculator. The heat a resistive conductor produces is covered in more depth by the Joule heating calculator, and component dissipation by the power dissipation calculator. For reading the resistors around that trace, use the resistor color code calculator. Anything on the mains side of the supply is a different discipline entirely and starts with the breaker size calculator. To move between mils, millimetres and inches, use the length converter. The full catalogue is at the free online tools hub.
Frequently Asked Questions
Ten degrees is a common conservative starting point and twenty is widely used where board temperature is well controlled. The rise is above the local ambient, so a board already running at 60 degrees inside an enclosure reaches 80 degrees with a 20 degree rise, and that has to sit within the laminate and component ratings.
IPC-2221 halves the constant for internal conductors because they are encased in laminate with poor thermal conductivity and cannot shed heat to air. Because cross-section carries an exponent of 0.725, halving the constant raises the required area by a factor of about 2.6 rather than two.
It is an empirical fit to old measurements on bare isolated conductors, and it ignores adjacent copper planes, board thickness and dielectric conductivity, all of which change the real temperature substantially. IPC-2152 was published to address exactly those gaps. Treat the IPC-2221 result as a quick first estimate.
No. The width comes from cross-section, which depends on current and allowable temperature rise only. Length changes the resistance, the voltage drop and the total heat produced, which is why those three figures are shown separately and why a long trace often ends up wider than the thermal minimum.
It should not be. A conductor will eventually vaporise at some large current, but the current at which it does so is poorly controlled and depends on board construction, adjacent copper and airflow. Overcurrent protection must come from a rated protective device chosen by a qualified engineer.
Etching attacks the sides as well as the top, so the finished conductor is narrower than the artwork, and the effect grows with copper weight. Heavy copper commonly needs several mils of allowance. Ask your fabricator for their etch factor rather than assuming the drawn width is what you get.
It is a thickness stated as the weight of copper spread over one square foot. One ounce corresponds to about 1.378 mils or 35 micrometres. Two-ounce copper is twice as thick, so it reaches the same cross-section in half the width, which is often the cheaper way to carry a heavy current on a crowded board.
This tool is provided for educational and preliminary design use. Conductor sizing, spacing and overcurrent protection on any board connected to mains electricity or handling hazardous energy must be reviewed against the applicable standards by a qualified engineer, and this page publishes no ampacity table or clearance table of its own.
