Choosing the width of a power trace is one of the first decisions in PCB layout. Too narrow and the trace heats up, drops voltage and in the worst case burns open; too wide and it wastes routing space. The standard starting point is the IPC-2221 relationship between current, cross-section and temperature rise, which this calculator rearranges to give the width you need.
Enter the current the trace must carry, the copper weight and how much hotter than ambient you allow it to get. The required cross-section follows from A = [I ÷ (k ΔT^0.44)]^(1/0.725), and dividing by the copper thickness gives the width. Inner-layer traces need roughly twice the cross-section of outer-layer traces under IPC-2221 because they cannot cool to the air – a conservative assumption that the later IPC-2152 standard relaxes.
With a trace length the calculator also gives the resistance at operating temperature, the voltage drop and the power the trace dissipates, which often matter as much as heating for low-voltage rails. Round the width up to your fabricator’s grid and add margin for etch tolerance, and keep the result within the chart range (up to 35 A, 400 mil, 10–100 °C rise).
How to use the PCB Trace Width Calculator
- Maximum current
- Copper thickness — 1 oz/ft² = 1.378 mil = 35 µm.
- Maximum temperature rise
- Trace location
- Ambient temperature
- Trace length
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the current, copper thickness, allowed temperature rise and layer, plus the ambient temperature and trace length. The calculator solves the IPC-2221 equation for the cross-section, divides by the thickness for the width and computes resistance, voltage drop and power from copper resistivity.
Worked example
For 1 A on an outer layer in 1 oz copper with a 10 °C rise: A = [1 ÷ (0.048 × 10^0.44)]^(1/0.725) = 7.566^1.379 = 16.30 mil², so W = 16.30 ÷ 1.378 = 11.8 mil (0.30 mm). An inner-layer trace needs 30.8 mil. At 35 °C a 5 cm trace of that width has 86.6 mΩ, dropping 86.6 mV and dissipating 86.6 mW at 1 A.
Width for 1 oz copper, 10 °C rise (IPC-2221)
| Current | External | Internal |
|---|---|---|
| 0.5 A | 4.5 mil (0.12 mm) | 11.8 mil (0.30 mm) |
| 1 A | 11.8 mil (0.30 mm) | 30.8 mil (0.78 mm) |
| 2 A | 30.8 mil (0.78 mm) | 80.0 mil (2.03 mm) |
| 5 A | 109 mil (2.77 mm) | 283 mil (7.19 mm) |
Inputs explained
| Field | What to enter |
|---|---|
| Maximum current | Units: A, mA |
| Copper thickness | Units: oz/ft², mil, µm, mm. 1 oz/ft² = 1.378 mil = 35 µm. |
| Maximum temperature rise | Units: °C, °F, K |
| Trace location | Choose one: External (outer layer), Internal (inner layer) |
| Ambient temperature | Units: °C, °F, K |
| Trace length | Units: cm, mm, in, mil |
What each result means
- Required cross-section = ([Maximum current]÷([k constant]×[Maximum temperature rise]^0.44))^(1÷0.725)
- Minimum trace width = [Required cross-section]÷[Copper thickness]
- Trace temperature = [Ambient temperature] + [Maximum temperature rise]
- Resistance = 1.72e-6×[Trace length]÷([Required cross-section]×6.4516e-6)×(1 + 0.00393×([Trace temperature] – 20))
- Voltage drop = [Maximum current]×[Resistance]
- Power dissipation = [Maximum current]^2×[Resistance]
Formula
A = [I ÷ (k × ΔT^0.44)]¹/0.725) (mil²) · W = A ÷ T (mil) = A ÷ (1.378 × copper oz) · k = 0.048 external, 0.024 internal · R = ρL ÷ A × [1 + α(T − 20 °C)]
Worked example
Inputs — Maximum current: 1 A; Copper thickness: 1 oz/ft²; Maximum temperature rise: 10 °C; Trace location: External (outer layer); Ambient temperature: 25 °C; Trace length: 5 cm.
Results — Required cross-section = 16.296 mil²; Minimum trace width = 11.8258 mil; Trace temperature = 35 °C; Resistance = 86.6214 mΩ; Voltage drop = 86.6214 mV; Power dissipation = 86.6214 mW; IPC-2221 chart range = Within the IPC-2221 chart range.
Key terms
- IPC-2221
- Generic standard on printed board design, including trace current curves.
- Temperature rise
- Allowed increase of trace temperature over ambient.
- k constant
- 0.048 for external and 0.024 for internal traces in the IPC-2221 formula.
- mil
- One thousandth of an inch (0.0254 mm).
Frequently asked questions
How wide should a trace be for 1 A?
About 12 mil (0.30 mm) on an outer layer with 1 oz copper and a 10 °C rise; about 31 mil (0.78 mm) on an inner layer.
How wide for 3 A?
Around 54 mil (1.37 mm) on an outer layer with 1 oz copper at a 10 °C rise, or about 27 mil (0.68 mm) with 2 oz copper.
What is IPC-2221?
The IPC generic standard for printed board design, which includes the classic current-capacity curves and formula.
Is IPC-2152 better?
It is based on newer, more detailed tests and gives less conservative internal-layer results, but uses charts rather than one formula.
Should I add margin?
Yes – allow for etching tolerance (±10–20% of width), ageing and nearby heat sources.
Does trace length affect the width?
Not for heating (it is a per-length effect), but long traces have more voltage drop, which may call for a wider trace.