How PCB trace width is calculated
Current heats a trace through its own resistance, and the trace settles at the temperature where the heat it sheds into the board and air balances that loss. IPC-2221 fits test data to relate current, temperature rise and copper cross-section. Inner layers can only shed heat through the laminate, so for the same current they need about 2.6 times the cross-section of an outer trace.
- IPC-2221:
I = k·ΔT^0.44·A^0.725(A in mil², k = 0.048 outer, 0.024 inner) - Width:
w = A / copper thickness - Resistance:
R = ρ·L / (w·t)
Worked example: 3 A on an outer layer
3 A on 1 oz (35 µm) outer-layer copper with a 10 °C rise needs 74 mil² of copper, a trace 1.37 mm (54 mil) wide. The same current on an inner layer needs 3.56 mm.
Good to know
- Doubling the copper weight halves the width. 1 oz is 35 µm; 2 oz is 70 µm.
- A 10 °C rise is a conservative default. Power traces are often designed for 20–30 °C, provided the trace stays below the laminate's temperature rating.
- IPC-2152 replaced the IPC-2221 chart and accounts for board thickness and nearby planes; it usually allows narrower inner traces.
- On long runs, voltage drop often limits the width before temperature does. Pro calculates resistance, drop, power loss and fusing current.