What formula calculates PCB trace width?

IPC-2221: first the area A = (I / (k × ΔT^0.44))^(1/0.725), then Width = A / (copper_oz × 1.378). k is 0.048 for external layers and 0.024 for internal.

What trace width do I need for 1 A?

For 1 oz copper, external layer and a 10 °C rise, about 0.30 mm is enough. Keeping it a little wider for thermal margin is good practice.

What is the k constant and why differ inside/outside?

k is an empirical constant. External layers cool via air (0.048); internal layers are buried and cool worse, so 0.024 is used (about twice the width needed).

How many mils is 1 oz copper?

1 oz/ft² copper ≈ 1.378 mils ≈ 35 µm. 2 oz is twice that.

Does trace width matter for signal lines?

The current is tiny, so width is not critical; it matters most on power and ground. On high-speed signals, width instead affects impedance.

How to Calculate PCB Trace Width (IPC-2221 + Example)

PCB trace width is the most important factor deciding how much current a copper trace on a printed circuit board can safely carry. A trace that is too narrow overheats, causes a voltage drop and, in the worst case, melts and opens. Trace width is calculated from the current and the allowed temperature rise using the IPC-2221 standard.

Why does trace width matter?

A copper trace is not an ideal conductor; it has a small resistance and heats up as current flows (I²R loss). The narrower and thinner the trace, the higher its resistance and the more it heats. On power and ground rails, an under-sized trace causes voltage drop, efficiency loss and thermal damage. On signal lines the current is tiny, so width is not critical — it matters most on power/ground.

The IPC-2221 formula

First find the required cross-sectional area:

A = (I / (k × ΔT^0.44))^1/0.725

Symbol	Meaning	Unit / value
A	Required cross-sectional area	mils²
I	Current through the trace	Amperes (A)
ΔT	Allowed temperature rise	°C
k	Layer constant	External: 0.048 · Internal: 0.024

From area to width

The area is divided by the copper thickness to get the width. Copper thickness is usually given in ounces (oz): 1 oz/ft² ≈ 1.378 mils (35 µm).

Width (mils) = A (mils²) / (thickness_oz × 1.378)

Conversion: 1 mil = 0.0254 mm.

How to calculate it, step by step

Determine the maximum current the trace will carry (I).
Choose the allowed temperature rise (ΔT, typically 10–20 °C).
Pick k based on external vs internal layer (0.048 / 0.024).
Compute the area: A = (I / (k × ΔT^0.44))^1/0.725.
Choose the copper thickness (e.g. 1 oz).
Convert to width: A / (oz × 1.378) → mils → mm.

Worked example

1 A of current, a 10 °C temperature rise, external layer (k = 0.048), 1 oz copper:

A = (1 / (0.048 × 10^0.44))^1/0.725 ≈ 16.3 mils²
Width = 16.3 / (1 × 1.378) ≈ 11.8 mils ≈ 0.30 mm

Approximate trace width by current

For 1 oz copper, external layer, ΔT = 10 °C, roughly:

Current	Approx. width
0.5 A	~0.15 mm
1 A	~0.30 mm
2 A	~0.70 mm
3 A	~1.15 mm
5 A	~2.2 mm
10 A	~5.6 mm

Internal vs external layers

External (surface) layers touch the air and cool better (k = 0.048). Internal layers are buried in the dielectric and cool worse; for the same current they use k = 0.024 and need roughly twice the width.

Practical tips

Add a thermal margin; draw high-current rails wider than the minimum.
For very high current, use copper pours or parallel layers.
Keep ΔT low on sensitive or hot designs.

Use the calculator

To instantly find the trace width from current, temperature rise, layer and copper thickness, use our PCB Trace Width Calculator.