PCB Trace Width Calculator (IPC-2221)
Find the copper trace width you need for a given current and temperature rise. Add the trace length to get its resistance and voltage drop too.
Formula and cautions
The IPC-2221 relation I = k·ΔT0.44·A0.725 is solved for the cross-section A in mil², then divided by the copper thickness to get the width. k is 0.048 for outer layers and 0.024 for inner layers. Inner layers shed heat poorly, so they need more than twice the width for the same current. The formula is conservative, which makes it a safe design starting point.
Worked example: 3 A, outer layer, 1 oz, 10 °C rise
- Outer layer, so k = 0.048. A0.725 = (3)/(0.048 × 100.44) = 22.7, so the cross-section is A = 74.2 mil²
- 1 oz copper is 1.378 mil thick → width = (74.2)/(1.378) = 53.8 mil
- In millimetres that is 1.37 mm. Designers usually draw it a little wider for margin.
PCB trace width by current
| Current | Outer 1 oz | Inner 1 oz | Outer 2 oz |
|---|---|---|---|
| Current0.5 A | Outer 1 oz0.115 mm (4.55 mil) | Inner 1 oz0.3 mm | Outer 2 oz0.0577 mm |
| Current1 A | Outer 1 oz0.3 mm (11.8 mil) | Inner 1 oz0.781 mm | Outer 2 oz0.15 mm |
| Current2 A | Outer 1 oz0.781 mm (30.8 mil) | Inner 1 oz2.03 mm | Outer 2 oz0.391 mm |
| Current3 A | Outer 1 oz1.37 mm (53.8 mil) | Inner 1 oz3.56 mm | Outer 2 oz0.683 mm |
| Current5 A | Outer 1 oz2.77 mm (109 mil) | Inner 1 oz7.19 mm | Outer 2 oz1.38 mm |
| Current10 A | Outer 1 oz7.19 mm (283 mil) | Inner 1 oz18.7 mm | Outer 2 oz3.6 mm |
IPC-2221 with a 10 °C temperature rise.
Using it in practice
Step by step
- Enter the current the trace must carry continuously and the temperature rise you accept (10 °C is a common choice).
- Choose the copper weight (1 oz is about 35 µm thick) and whether the trace is on an outer or inner layer.
- Read the minimum width, then round up to something your board house can make comfortably.
Common mistakes
- Using the outer-layer width on an inner layer. Inner layers cannot shed heat to the air, so IPC-2221 asks for roughly 2.6 times the cross-section for the same rise.
- Designing for the average when there are peaks. Heating follows the RMS current. A motor driver with 3 A peaks and a low average still needs to be checked against its RMS current, and short spikes need sound fuse and layout planning, not just width.
- Forgetting vias and necks. A wide trace that narrows to reach a pad, or passes the current through one small via, is limited by the narrowest point. Use several vias for high currents.
Practical tips
IPC-2221's formula comes from old charts and is conservative for most boards; the newer IPC-2152 data allows somewhat narrower traces when there are copper planes nearby. When space allows, use a polygon pour instead of a trace for power. Also check the voltage drop along the trace: a long, thin trace can be thermally fine yet lose a noticeable fraction of a volt at a few amps.
Worked example: 1 A on an outer and an inner layer
For 1 A with a 10 °C rise in 1 oz copper, the outer layer (k = 0.048) needs a cross-section of 16.3 mil², which is 11.8 mil or 0.30 mm wide. On an inner layer (k = 0.024) the same current needs 42.4 mil², or 30.8 mil (0.78 mm). Halving k does not double the width but multiplies it by about 2.6, because the area enters the formula with the power 0.725.
If the board has room, drawing the outer trace at 0.5 mm instead of 0.3 mm costs nothing and lowers both the temperature rise and the voltage drop.
Resistance along the trace
Width sets heating, but length sets the voltage drop. A 0.30 mm trace in 35 µm copper that runs 50 mm has R = (ρL)/(wt) = 1.72 × 10⁻⁸ Ω·m × 0.05 m ÷ (0.30 mm × 35 µm) = 0.082 Ω. At 1 A that is 82 mV, small for a 12 V rail but 2.5% of a 3.3 V one, and the return path adds the same again unless it is a solid ground plane. Enter the length to get this figure for your own trace; for a sensor ground or a regulator's feedback pin, route a separate thin trace so the load current's drop is not part of the measurement.
Frequently asked questions
How wide should a trace be for 1 A?
With 1 oz copper on an outer layer and a 10 °C rise, IPC-2221 gives about 0.3 mm (12 mil). An inner layer needs about 0.8 mm for the same current.
Does thicker copper help?
Yes. 2 oz copper is twice as thick, so the same current needs roughly half the width. It costs more and makes fine-pitch routing harder.