PCB Trace Width Calculator

Trace width for current is set by how much heating you will tolerate. The IPC-2221 chart fit is

$I = k\,\Delta T^{0.44} A^{0.725} \qquad\Longrightarrow\qquad A = \left(\frac{I}{k\,\Delta T^{0.44}}\right)^{1/0.725}$

Symbols and the units this empirical fit demands:

• $I$ — current, amperes (A)
• $\Delta T$ — allowable temperature rise above ambient, degrees Celsius (°C)
• $A$ — trace cross-sectional area, square mils (mil², where 1 mil = 0.001 in)
• $k$ — $0.048$ for an external (surface) trace, $0.024$ for an internal trace, which is buried and cannot shed heat as well

Width then comes from the copper thickness:

$w = \frac{A}{t} \qquad t = 1.378\ \text{mil per ounce of copper weight}$

so 1 oz copper is $1.378$ mil ($35\ \mu$m) and 2 oz is $2.756$ mil.

The operating assumption: a bare trace in still air on a board that is not otherwise hot, carrying steady DC.

For a controlled-impedance signal the width is set by geometry, not heating. The IPC-2141 microstrip approximation is

$Z_0 = \frac{87}{\sqrt{\varepsilon_r + 1.41}} \ln\!\left(\frac{5.98h}{0.8w + t}\right)$

with $Z_0$ in ohms, $h$ the dielectric height and $w$, $t$ the trace width and thickness, all in the same length unit. FR-4 has $\varepsilon_r \approx 4.3$ at low frequency. A wider trace or a thinner dielectric gives lower impedance.

State this plainly: IPC-2221 is a conservative curve fit from the 1950s, not a thermal model of your board. It ignores adjacent copper pours, board material, via heating, altitude and airflow. IPC-2152 supersedes it with data that accounts for those effects and often permits narrower traces.

Use these results to understand and to sketch a first pass. The width that actually goes on a manufactured board must be confirmed against IPC-2152, your fabricator's stack-up and impedance data, and any product safety standard that governs spacing and creepage. A calculated number is not a released design.

• Feeding millimetres into the IPC-2221 formula — it is an empirical fit in mils and square mils. Convert at the end: $1$ mil $= 0.0254$ mm.
• Using the external $k$ for an inner layer — an internal trace needs $k = 0.024$ and comes out roughly three times wider.
• Confusing area with width — the formula returns a cross-section in mil²; divide by the copper thickness to get width.
• Treating copper weight as a width — 1 oz describes thickness, $1.378$ mil, and says nothing about how wide the trace is.
• Forgetting plating — outer layers are usually plated up beyond the base foil weight, so the finished thickness differs from the stack-up nominal.
• Ignoring the return path — impedance depends on a continuous reference plane; a split under the trace invalidates the microstrip model.
• Sizing for average current on a pulsed load — heating follows the RMS value.