PCB Trace Width Calculator
Minimum trace width for current, copper weight, layer, and temperature rise.
What it does
A copper trace is a resistor with a small cross-section. Current through that resistance produces heat, and the heat has to escape into the board, the air, and the surrounding copper. If the trace is too narrow, the temperature rise can soften solder, damage the laminate, or shorten the life of the board. Trace width is the main geometry lever that keeps that rise under control.
The long-used IPC-2221 relationship starts with a cross-sectional area. The area depends on current, the allowed temperature rise, and a constant that changes for external and internal layers. Once the area is known, the width is the area divided by the copper thickness. A 1 oz copper layer is about 1.38 mils thick, so a 74 square mil cross-section becomes a 53.8 mil wide trace.
Layer location matters because heat leaves an external trace through the air and the board, while an internal trace is surrounded by laminate. The IPC curve fit uses a smaller constant for internal layers, which produces a wider trace for the same current and temperature rise. The default 3 A, 10 degrees Celsius, 1 oz external case needs 53.8 mils. The same case on an internal layer needs 140 mils.
The default is a useful benchmark, but modern boards often behave better than the old curve because copper planes, pours, and multiple layers spread heat. They can also behave worse when a connector, a dense bundle of traces, or a hot enclosure adds heat that the trace model does not see. IPC-2152 is the newer standard and accounts for more of these effects. Use the IPC-2221 result as a conservative minimum, then check voltage drop and thermal reality.
This page returns the minimum width in mils or millimeters, the cross-section area, the copper thickness, and the layer used. It does not choose a fuse, a connector, a via, or a plane. It also does not check high-voltage spacing, impedance, or manufacturability. Those are separate design rules with their own margins.
How to use
- Choose imperial or metric for the width and area output.
- Enter the continuous current the trace will carry. Use the worst-case load, not the idle current.
- Enter the temperature rise you can tolerate above the board ambient.
- Pick the copper weight from the stackup. If the board has 2 oz outer layers, choose 2 oz.
- Choose external or internal. A trace on an inner layer needs the internal setting.
- Read the minimum width, then round up to a standard width and add margin for etching, plating, and connector heat.
Formula
- current
- Continuous current through the trace in amperes.
- temperature rise
- Allowed rise above ambient in degrees Celsius.
- k
- Layer constant: 0.048 external and 0.024 internal in the IPC-2221 fit.
- area
- Copper cross-section in square mils.
- copper
- Copper weight in ounces per square foot; 1 oz is about 1.378 mils thick.
- width
- Minimum trace width in mils before rounding and margin.
The exponents 0.44 and 0.725 come from a curve fit to the IPC-2221 charts, not from a first-principles heat transfer model. The result is sensitive to the temperature rise: a small allowed rise forces a much wider trace. The copper thickness multiplier assumes the finished copper weight, so plating and process variation can reduce the real thickness. The formula assumes a single trace in still air, with no nearby copper plane and no additional heat from components. Traces in parallel, copper pours, airflow, and exposed surface area can lower the real temperature, while bundles, enclosures, and connectors can raise it. IPC-2152 gives a more detailed method when the design is close to the limit.
Worked example
Example: 3 A, 10 °C rise, 1 oz external trace
A board needs to carry 3 A continuously on an external 1 oz copper layer. The design allows a 10 degrees Celsius rise above ambient.
- Use k = 0.048 for an external layer.
- Area = (3 / (0.048 x 10^0.44))^(1 / 0.725).
- 10^0.44 is about 2.754; 0.048 x 2.754 is about 0.1322.
- 3 / 0.1322 is about 22.69; 22.69^(1 / 0.725) is about 74 square mils.
- Width = 74 / (1 x 1.378) = 53.8 mils, or about 1.367 mm.
- The same trace on an internal layer uses k = 0.024 and needs 140 mils.
Reference table
| Current | External width | External area | Internal width | Internal area |
|---|---|---|---|---|
| 1 A | 11.8 mils | 16 mil² | 30.8 mils | 42 mil² |
| 2 A | 30.8 mils | 42 mil² | 80.0 mils | 110 mil² |
| 3 A | 53.8 mils | 74 mil² | 140.0 mils | 193 mil² |
| 4 A | 80.0 mils | 110 mil² | 208.2 mils | 287 mil² |
| 5 A | 108.9 mils | 150 mil² | 283.2 mils | 390 mil² |
Limits
- A conservative curve fit, not a thermal simulation of the actual board.
- Does not include copper planes, nearby traces, airflow, enclosure temperature, or connector heat.
- Does not check voltage drop, impedance, high-voltage spacing, or via current capacity.
- Assumes still air and a single trace. IPC-2152 or a thermal study is better for designs near the limit.
Common mistakes
- Using the external constant for an internal trace, which underestimates the width.
- Treating 1 oz copper as 1 mil thick instead of about 1.38 mils.
- Entering the temperature rise as an absolute temperature instead of a rise above ambient.
- Sizing only for the average current and ignoring startup, stall, or fault current.
- Forgetting that a connector, a via, or a narrow neck can be the real hot spot.
- Using the exact calculated width with no margin for etching, plating, or manufacturing tolerance.
Try it yourself
Quick checks. Answers sit tucked under each question.
A 5 A trace runs on an external 2 oz layer with a 10 °C rise. What width is needed?
The area is the same 150 square mils as the 5 A, 1 oz case. Width = 150 / (2 x 1.378) = 54.4 mils.
A 2 A trace runs on an internal 1 oz layer with a 20 °C rise. What width is needed?
Using k = 0.024 and a 20 °C rise gives about 72 square mils. Width = 72 / 1.378 = 52.5 mils.
What is the internal width for 3 A, 10 °C, and 1 oz copper?
About 193 square mils, which gives 140 mils, or 3.556 mm.
Questions
What temperature rise should I use?
Ten degrees Celsius is a common target for ordinary electronics. A tighter rise needs a wider trace; a higher rise saves space but reduces margin and can affect reliability.
Why do internal traces need to be wider?
The laminate around an internal trace traps heat, so the same current produces a larger temperature rise. The IPC-2221 internal constant accounts for that.
Is IPC-2221 still valid?
It is still widely used as a conservative starting point. IPC-2152 is newer and accounts for more board and environment variables. For a design near the thermal limit, use the newer method or a simulation.
How thick is 1 oz copper?
About 1.38 mils, or 34.8 micrometers, after processing. The exact finished thickness depends on plating and the fabrication process.
Does this include the via current?
No. Vias, connectors, and narrow necks need their own current check. A wide trace can still fail at a small via or a poor connector interface.
Can I use a narrower trace if I have a copper plane?
Often yes, because the plane spreads heat, but the IPC-2221 formula does not credit the plane. Keep the calculated width as a baseline and verify the real thermal path.
Sources
- IPC-2221B: Generic Standard on Printed Board Design, table of contents
- Purdue University: New Correlations Between Electrical Current and Temperature Rise in PCB Traces