PCB Design
PCB Trace Width Calculator
IPC-2221 trace width for a given current and temperature rise, plus the trace resistance, voltage drop and power loss you get once you have chosen a width.
Cross-section, drawn to scale
Conditions
minimum trace width
The IPC-2221 formula
IPC-2221 gives an empirical relationship between current, conductor cross-section and steady-state temperature rise:
A = ( I / (k × ΔT0.44) )1/0.725
where A is the cross-sectional area in square mils, I is current in amperes, ΔT is the temperature rise in °C, and k is 0.048 for external traces and 0.024 for internal ones. Internal traces get half the constant because they are buried in laminate with nowhere to shed heat by convection.
Width follows from the copper thickness: one ounce of copper per square foot is 1.378 mils (35 µm) thick, so width = area / (1.378 × oz).
Where the formula is optimistic
IPC-2221 comes from measurements on isolated traces on a bare board in still air. Real boards are not that. Treat the result as a floor, not a target, and widen when any of the following apply:
- The trace runs next to other hot traces. Adjacent conductors heat each other; the curves assume isolation.
- The board is in a sealed enclosure. Ambient inside a box can be 20 °C above the room, and that adds directly to your trace temperature.
- The trace passes through a thermal relief or a necked-down section. The narrowest point sets the limit, and that is usually at a pad or a via transition.
- Current is pulsed with high peaks. The formula is for steady state; short pulses tolerate far more, but RMS heating is what matters over a cycle.
- Voltage drop matters more than temperature. On a 3.3 V rail, 100 mV of trace drop is 3% of your budget — often the binding constraint long before the copper gets warm.
Copper weight in practice
Standard fabrication is 1 oz outer, sometimes 0.5 oz inner. Going to 2 oz roughly halves the width you need for the same current, but it costs more, limits how fine your minimum trace and space can be (etching a thick layer undercuts more), and increases the board's weight and stiffness. For most designs it is cheaper to route a wider 1 oz trace, or to use a copper pour, than to specify heavy copper.
Vias carry current too
A trace sized for 3 A is pointless if it passes through a single 0.3 mm via. A plated via's current capacity depends on barrel diameter and plating thickness; as a rough guide, a 0.3 mm via with 25 µm plating handles about 1 A for a 10 °C rise. Use multiple vias in parallel for any significant current, and remember they also form the thermal path — via stitching under a hot part does double duty.
Voltage drop and resistance
Copper's resistivity is about 1.72 × 10−8 Ω·m at 20 °C and rises roughly 0.393% per °C. The calculator above accounts for this at the trace's working temperature — ambient plus your allowed rise — which is why the resistance figure it gives is slightly higher than a naive room-temperature calculation. For long runs on low-voltage high-current rails, that drop is usually what forces the trace wider, not the thermal limit.
Quick sanity figures (1 oz copper, external, 10 °C rise)
| Current | Minimum width (mm) | Minimum width (mils) |
|---|---|---|
| 0.5 A | 0.12 | 4.6 |
| 1 A | 0.30 | 11.8 |
| 2 A | 0.78 | 30.8 |
| 3 A | 1.37 | 53.8 |
| 5 A | 2.77 | 109 |
| 10 A | 7.19 | 283 |
Beyond about 5 A a plain trace becomes impractical and you should be thinking about copper pours, multiple layers stitched with vias, or a bus bar. These figures are guidance only — the fabricator's capabilities and your thermal environment are what actually decide.
Frequently asked questions
How wide should a PCB trace be for 1 amp?
On a 1 oz copper outer layer with a 10 °C rise, IPC-2221 gives about 0.30 mm (11.8 mils). For a 20 °C rise it drops to roughly 0.19 mm. Internal traces need about twice the width because they cannot shed heat by convection.
What is the IPC-2221 trace width formula?
A = (I / (k × ΔT^0.44))^(1/0.725), where A is cross-sectional area in square mils, I is current in amps, ΔT is temperature rise in °C, and k is 0.048 for external traces or 0.024 for internal ones. Width is then area divided by copper thickness in mils.
Why do internal traces need to be wider?
An internal trace is buried in laminate with no air contact, so it can only lose heat by conduction through the board. IPC-2221 halves the constant k to account for this, which works out at roughly double the width for the same current and temperature rise.
How thick is 1 oz copper?
One ounce of copper spread over one square foot is 1.378 mils, or about 35 µm. 2 oz is 70 µm, 0.5 oz is 17.5 µm. Copper weight is quoted per layer, and inner layers are often lighter than outers.
Should I design for voltage drop or temperature rise?
Whichever is tighter — and on low-voltage high-current rails it is almost always voltage drop. A 5 V rail can tolerate a lot more heating than it can tolerate 200 mV of trace drop. Calculate both and take the wider result.
How much current can a via carry?
As a rough guide, a 0.3 mm via with 25 µm plating handles about 1 A for a 10 °C rise. Use several vias in parallel for anything more — a wide trace that necks down to a single via defeats the whole exercise.