Question 1

How do I calculate PCB trace width?

Accepted Answer

The IPC-2221 method first finds the cross-sectional area A in mils² from A = ( I / (k · ΔT^0.44) )^(1/0.725), where k = 0.048 for external (outer) layers and 0.024 for internal layers. The copper thickness in mils is the copper weight in oz times 1.378, and the width equals A divided by that thickness.

Question 2

Why are external and internal traces different?

Accepted Answer

Internal traces are buried inside the board and cannot shed heat to the air, so IPC-2221 uses a smaller constant (k = 0.024 instead of 0.048). For the same current and temperature rise an internal trace needs about twice the cross-sectional area of an external one.

Question 3

What does 1 oz copper mean?

Accepted Answer

Copper weight is how many ounces of copper are spread over one square foot. One ounce equals about 1.378 mils (35 µm) of thickness, 2 oz is roughly 2.76 mils, and so on. Thicker copper lets you use a narrower trace for the same current.

Question 4

Is IPC-2221 conservative?

Accepted Answer

IPC-2221 is the older, widely used chart-based formula and tends to give wider, safer traces. The newer IPC-2152 standard accounts for board material and surroundings and usually allows narrower traces. Treat this calculator as a quick first estimate and verify against your board house's design rules.

Question 5

What temperature rise should I use?

Accepted Answer

The temperature rise ΔT is how much hotter the trace is allowed to get above the board's ambient temperature. A common starting point is 10 °C for signal-class traces and 20–30 °C where some heating is acceptable. Larger allowable rises let you use a narrower trace.

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