About the PCB Trace Current Calculator
This tool estimates the maximum current a copper trace on a printed circuit board can carry continuously without exceeding a chosen temperature rise, using the widely used IPC-2221 empirical formula. It also reports the trace's cross-sectional area, resistance over a given length, and the voltage drop at that current — the numbers PCB designers check when sizing power and high-current traces.
The IPC-2221 current-capacity formula
The standard formula, derived from curves originally published in IPC-2221 (the successor to MIL-STD-275), is:
I = k × ΔT0.44 × A0.725
- I — maximum current in amps
- ΔT — allowed temperature rise above ambient, in °C
- A — trace cross-sectional area in square mils (width in mils × copper thickness in mils)
- k — a constant that depends on where the trace sits: 0.048 for external traces (on an outer layer, exposed to air) and 0.024 for internal traces (buried inside the board)
Internal traces get half the constant of external traces because they can only shed heat by conduction into the surrounding laminate, while external traces also lose heat by convection and radiation from the exposed copper surface. For the same width, copper weight, and temperature rise, an internal trace therefore carries noticeably less current than an external one.
Copper weight and cross-sectional area
PCB copper is specified by weight per square foot rather than thickness directly. One ounce of copper rolled out over one square foot of board is about 1.378 mils thick (roughly 35 micrometers). So 0.5 oz copper is about 0.689 mils thick, 2 oz copper is about 2.756 mils thick, and 3 oz copper is about 4.134 mils thick. Cross-sectional area is simply trace width multiplied by that thickness, both in mils, giving an area in square mils.
Trace resistance and voltage drop
Once the maximum current is known, it is useful to see what that current costs in resistive loss. Copper trace resistance can be estimated with sheet resistance: 1 oz copper has a sheet resistance of roughly 0.49 milliohms per square at room temperature, and sheet resistance scales inversely with copper weight (2 oz copper is about half that, 0.5 oz is about double). The number of "squares" in a trace is its length divided by its width, in the same units. Multiplying sheet resistance by the number of squares gives total resistance, and multiplying that resistance by the calculated maximum current gives the voltage drop across the trace.
Practical notes
- This is a steady-state, continuous-current estimate for still air at typical FR-4 board conditions — it does not model pulsed currents, forced airflow, or unusual board stack-ups.
- Vias, connectors, and copper pours in the current path are not modeled; a trace that meets the target may still be limited by a narrow via or connector pin elsewhere.
- For high-reliability or safety-critical designs, use the result as a starting point and confirm against your fabricator's copper tolerances and any applicable design standard.