PCB Trace Width Calculator
Estimate an external or internal PCB trace width from current and temperature rise, then the resistance of the length you type.
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Result
About this tool
The inputs are current, an allowed temperature rise in degrees Celsius, copper thickness in ounces, a layer choice of external or internal, and a trace length. The defaults are 1 amp, a 10 degree rise, 1 ounce, external, and 50 millimetres. Current is multiplied by its unit, so the milliamp choice divides by 1000. Length is multiplied too: the millimetre choice uses 0.001, so 50 becomes 0.05 metres. Every one of those four numbers must be strictly positive. The layer constant k is 0.048 on an external layer and 0.024 on an internal layer.
Cross-section in square mils is (I / (k × ΔT^0.44)) raised to the power 1/0.725. That is the IPC-2221-style estimate the result line itself names. Thickness in mils is ounces times 1.378, so 1 ounce is 1.378 mil. Width in mils is the area divided by that thickness, and width in millimetres is the mil width times 0.0254. For the defaults the area prints as 16.296001347209 mil², the width as 0.30037622221997 mm, which is 11.825835520471 mil. The same current and rise on an internal layer, where k is half, prints a width of 0.7814106716856 mm, about 30.764199672661 mil. Internal copper has to be wider for the same rise.
Resistance uses a second model. Thickness in metres is ounces times 35e-6, so 1 ounce is 35 micrometres, and the width in metres is the millimetre width divided by 1000. Area is width times that thickness. Resistance is 1.724e-8 times length divided by that area. For the default 50 mm external trace the resistance prints as 0.081992413535767 Ω. Voltage drop is current times that resistance, 81.992413535767 mV, and power loss is current times the drop, 81.992413535767 mW. The 1.378 mil per ounce figure and the 35 micrometre figure are both in the script. They are not identical conversions, so the width used for the IPC area and the thickness used for resistance are not the same physical assumption carried to the last digit. The last line of the report says this is an IPC-2221-style estimate and that a board house should confirm it before production. The page does not apply IPC-2152, does not add a plane or a solder mask, and does not know the copper's actual temperature coefficient beyond the fixed 1.724e-8 ohm-metre resistivity. A higher allowed rise makes the trace narrower. A longer trace does not change the width. It only scales the resistance, the drop, and the loss.
Example. 1 A, a 10 °C rise, 1 oz external copper, and 50 mm of length report a width of 0.30037622221997 mm (11.825835520471 mil), resistance 0.081992413535767 Ω, drop 81.992413535767 mV, and loss 81.992413535767 mW.
Limit. Internal layers use k of 0.024 instead of 0.048. The width is an IPC-2221-style estimate, and the resistance uses 1.724e-8 Ω·m with a 35 µm per ounce thickness that is not the same conversion as 1.378 mil per ounce.
How wide is a 1 amp external 1 oz trace at 10 °C?
Those are the defaults, with the length left at 50 mm. Area is 16.296001347209 mil² and the width is 0.30037622221997 mm, or 11.825835520471 mil. k for an external layer is 0.048. The report also prints the resistance of that 50 mm run.
What changes on an internal layer?
k becomes 0.024. The same 1 A, 10 °C, and 1 oz then need 0.7814106716856 mm, about 30.76 mil. The formula is (I / (k × ΔT^0.44)) raised to 1/0.725, then divided by the ounce thickness of 1.378 mil per ounce.
Where does the resistance number come from?
Resistance is 1.724e-8 Ω·m times the length in metres, divided by the cross-section. Thickness for that step is 35e-6 metres per ounce, which is not the same conversion as the 1.378 mil used for the width. The default 50 mm external trace is 0.081992413535767 Ω, with a drop of 81.992413535767 mV at 1 A.
Is this a fabrication sign-off?
No. The report ends by calling the width an IPC-2221-style estimate and telling you to verify it with the board house. It does not run IPC-2152, and it ignores mask, nearby planes, and copper roughness.