Free calculator
PCB Trace Width Calculator
The IPC-2221 formula tells you how wide a copper track must be to carry a current without heating more than you allow. Enter the current, the temperature rise, the copper weight and whether the track is on an outer or inner layer.
Result
- Width in mm
- 0.303 mm
- Width in mil
- 11.9 mil
Round up to the next width your board maker allows. Longer traces also lose voltage along the way.
The formula
What each letter means
- Current (A · ampere) : How much charge flows each second.
- IPC-2221 constant k (no unit) : A number from the PCB standard: 0.048 for outer layers, 0.024 for inner ones.
- Change in Δ (delta) (no unit) : Δ in front of a letter means "the change in": ΔT is a temperature rise and ΔI_L is how much a coil current swings up and down. It has the unit of the letter after it.
- Temperature (°C · degree Celsius) : How hot something is.
- Copper cross-section (mil²) : Trace width times copper thickness, in square mils (1 mil = 0.001 inch = 0.0254 mm), the unit the IPC-2221 formula uses.
- Trace width (mil) : How wide the copper trace is, in mils (1 mil = 0.0254 mm).
- Copper thickness (mil) : How thick the copper layer is. 1 oz copper is about 1.37 mil (35 µm) thick.
How it works
Current heats copper, and the more copper there is in the cross-section (width times thickness), the cooler it stays. IPC-2221 is a curve fitted to real measurements: it turns the current and the temperature rise you allow into the cross-section you need, and the copper thickness then gives the width. Inner layers are buried in board material that holds heat in, so they use a constant k of 0.024 instead of 0.048 and need a trace about 2.6 times wider.
The fit is only trusted for currents up to about 35 A, widths up to 400 mil and temperature rises from 10 °C to 100 °C; outside those limits the calculator still answers but warns you. It looks at heating only, not the voltage lost along a long trace and not short bursts of current. Copper weight is the mass of copper per square foot: 1 oz is about 35 µm thick (1.37 mil). A mil is one thousandth of an inch, 0.0254 mm.
Worked example
- 1 A, a 10 °C rise, 1 oz copper on an outer layer (k = 0.048): the track must be at least 0.303 mm wide, which is 11.9 mil.
- The same track on an inner layer (k = 0.024) needs 0.788 mm, about 2.6 times wider.
- At 3 A the outer-layer width grows to 1.379 mm. Three times the current needs about 4.6 times the width, because the current a trace can carry grows more slowly than its copper area (the exponent is 0.725).
Common questions
Does this include the voltage lost along the trace?
No. It only limits heating. A long, thin trace can pass the heating test and still lose enough voltage to matter, so for long power traces check the resistance too. Lesson 6.2 shows both.
Why does an inner layer need a wider trace?
Outer copper loses heat into the air, while inner copper is buried in board material that holds the heat in. The IPC-2221 constant for inner layers is half as big (0.024 instead of 0.048), so the same current needs more copper.
Is IPC-2221 still the right standard?
It is a simple, widely used rule of thumb, and it is easy to calculate by hand. A newer standard, IPC-2152, is based on more measurements and can give different answers. For a critical design add a safety margin or follow your board maker’s guidance.