PCB Trace Width Calculator (IPC-2221, Current and Temperature Rise)

What this tool does

IPC-2221 gives the cross-section a trace needs as A = [I ÷ (k ΔT^0.44)]^(1/0.725) in square mils, with k = 0.048 for outer layers, where air cools the copper, and 0.024 for inner ones. The width follows from the copper weight: 1 oz/ft² is about 35 µm (1.378 mil) thick, so 3 A with a 10 °C rise needs roughly 1.4 mm on an outer layer of 1 oz copper. The curves behind it come from old measurements, valid up to about 35 A, and IPC-2152 later showed that nearby planes and board thickness change the answer; treat the result as a good starting point, and add margin for hot enclosures.

Formula

A = [I ÷ (k ΔT^0.44)]^(1 ÷ 0.725) in mil² · width = A ÷ (1.378 mil × oz) · k = 0.048 outer, 0.024 inner (IPC-2221)

Variables

Symbol	Meaning	Unit
`i`	Current the trace must carry	A
`dt`	Temperature rise you accept	°C
`oz`	Copper weight	oz
`ly`	Where the trace is	—
`lg`	Length of the trace	mm
`WM`	Trace width needed	mm
`WI`	Trace width needed (mil)	mil
`AR`	Copper cross-section	mm²
`RS`	Resistance of that length at 20 °C	mΩ
`VD`	Voltage drop along it	mV
`PL`	Power lost in it	mW

Worked example

Current the trace must carry3 A
Temperature rise you accept10 °C
Copper weight1 oz
Where the trace is1
Length of the trace50 mm

Trace width needed1.37 mm
Trace width needed (mil)53.8 mil
Copper cross-section0.048 mm²
Resistance of that length at 20 °C17.97 mΩ
Voltage drop along it53.9 mV
Power lost in it161.8 mW

Limitations

For work that must comply with a standard or be signed off, check the result against the applicable code and have it reviewed by a qualified engineer.
The result is an estimate based only on the values you type. Real situations often include factors this calculator does not know about.