PCB Trace Width Calculator
Size a PCB trace for a target current and temperature rise with IPC-2221, then read its resistance, voltage drop and power loss.
Resistive results (optional)
- Minimum trace width
- 30.76 mil 0.781 mm
- Cross-section
- 42.4 mil² 0.0274 mm²
- Copper thickness
- 35 µm 0.035 mm
IPC-2221 is conservative
How PCB Trace Width Calculator works
A PCB trace is a thin copper conductor whose resistance heats it as current flows. Sizing a trace means choosing a cross-section large enough that the resulting temperature rise stays within what the board and its neighbours tolerate. IPC-2221 gives the standard empirical relationship, derived from measurement rather than theory.
That relationship expresses the required cross-sectional area in terms of current and allowed temperature rise: A = (I / (k × ΔT^0.44))^(1/0.725), in square mils. The constant k differs sharply by placement — 0.024 for internal layers and 0.048 for external ones — because an outer trace sheds heat to the surrounding air while an inner one is insulated by laminate on both sides. An internal trace therefore needs roughly twice the width of an external one for the same current.
Copper thickness is quoted in ounces per square foot, a historical unit describing how much copper was deposited. One ounce corresponds to about 1.4 mils, or 35 µm. Since width equals the required area divided by the thickness, doubling to two-ounce copper halves the width you need — often the cheaper answer when board area is tight.
Heating is only half the design. The trace also has resistance, and on a long run that resistance drops voltage and dissipates power. A trace can sit comfortably within its thermal limit while dropping enough voltage to matter for a sensitive supply rail, which is why both the temperature rise and the voltage drop are worth checking.
Reference
- IPC-2221: A = (I / (k × ΔT^0.44))^(1/0.725) [A in mil², ΔT in °C]
- k = 0.048 for external layers, 0.024 for internal layers
- Width (mil) = A / (thickness in mil); 1 oz copper ≈ 1.37 mil ≈ 35 µm
- Trace resistance R = ρ × L / (w × t), copper ρ ≈ 1.72 × 10⁻⁸ Ω·m
- Voltage drop = I × R; power dissipated = I² × R
How to use this calculator
Enter the current
Use the maximum continuous current the trace must carry, not the average.
Set the temperature rise
10 °C is conservative and common; 20 °C is acceptable where nothing heat-sensitive is nearby.
Choose layer and copper weight
External or internal, and the copper weight your fabricator will use — both change the answer substantially.
Check the electrical result
Enter the trace length to see resistance, voltage drop, and dissipated power alongside the width.
Worked examples
A 2 A external trace
- Given
- 2 A, 10 °C rise, 1 oz copper, outer layer
- Result
- About 32 mil (0.81 mm) wide
A routine width for a power rail on a two-layer board, and comfortably routable.
The same trace on an inner layer
- Given
- 2 A, 10 °C rise, 1 oz copper, internal layer
- Result
- About 86 mil (2.2 mm) wide
Nearly triple the width, because an internal trace has no air path to shed heat. Run high-current traces on outer layers where you can.
Voltage drop on a long run
- Given
- 2 A through a 32 mil, 1 oz trace, 100 mm long
- Result
- About 38 mΩ, dropping 76 mV and dissipating 0.15 W
Thermally fine, but 76 mV matters on a 1.2 V core rail. Widen it or use remote sensing.
When to use it
- Sizing power distribution traces on a new board layout.
- Checking whether an existing trace can carry a current you are increasing.
- Deciding between one-ounce and two-ounce copper for a power-dense design.
- Estimating voltage drop on a long supply trace before committing to a layout.
- Working out whether a high-current path needs to move to an outer layer.
Things to watch out for
- IPC-2221 is a conservative empirical fit, not an exact physical model. It assumes still air, a single isolated trace, and steady-state current.
- Nearby heat sources, poor airflow, and adjacent current-carrying traces all raise the real temperature above the prediction. Add margin in dense designs.
- Vias carry current too and are often the actual bottleneck. Use several in parallel on any high-current path between layers.
- For truly large currents, consider a copper pour or bus bar rather than a wide trace, and remember that solder mask over a trace slightly reduces its ability to shed heat.
Frequently asked questions
How wide should a PCB trace be for a given current?
It depends on the current, the temperature rise you accept, the copper weight, and whether the trace is internal or external. For 1 A at 10 °C rise on 1 oz external copper, roughly 12 mil suffices; 2 A needs about 32 mil. Internal traces need two to three times more.
Why do internal traces need to be so much wider?
Because they are surrounded by laminate, which is a poor thermal conductor, and have no air path to convect heat away. IPC-2221 reflects this with a k constant of 0.024 against 0.048 for external traces — half the current-carrying constant, and therefore a substantially larger required area.
What temperature rise should I design for?
10 °C is the common conservative default and appropriate near heat-sensitive components. Up to 20 °C is acceptable in a well-ventilated design with nothing fragile nearby. Higher rises accelerate laminate ageing and push adjacent parts toward their limits.
What does 1 oz copper mean?
One ounce of copper spread over one square foot, which works out to about 1.37 mil or 35 µm thick. It is the standard default. Two-ounce copper doubles the thickness and halves the width you need for the same current, at some extra board cost.
Should I check voltage drop as well as width?
Yes. Thermal sizing and electrical performance are independent constraints. A trace can be thermally comfortable while dropping enough voltage to put a low-voltage rail out of tolerance — long, narrow supply traces routinely fail on drop rather than heat.
How many vias do I need on a high-current path?
More than one. A typical 0.3 mm via handles roughly 0.5–1 A comfortably, so a 3 A path wants at least four to six in parallel. Vias are frequently the real thermal bottleneck in a design whose traces were sized correctly.
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