Guide
How wide should a PCB trace be? IPC-2221 current capacity, explained
Undersized traces overheat; oversized traces waste space. A practical explanation of the IPC-2221 trace-width rule — copper weight, temperature rise, and the numbers that actually matter.
A PCB trace is a thin ribbon of copper, and like any conductor it has resistance. Push too much current through a trace that is too narrow and it heats up; push enough and it acts like a fuse. The industry rule of thumb for how wide a trace needs to be is IPC-2221, and it is worth understanding rather than just plugging into a calculator.
The three inputs that matter
- Current — how many amps the trace carries continuously.
- Allowable temperature rise — how much warmer than ambient you'll let the trace get, typically 10 °C.
- Copper weight — how thick the copper is, quoted in ounces per square foot. 1 oz copper is about 1.37 mils (35 µm) thick; 2 oz is about 2.75 mils.
The IPC-2221 formula
IPC-2221 gives the current a trace can carry for a chosen temperature rise as a function of its cross-sectional area:
I = k × ΔT^0.44 × A^0.725
I = current (amps)
ΔT = temperature rise (°C)
A = cross-sectional area (mils²) = width × thickness
k = 0.048 for external (outer-layer) traces
0.024 for internal (inner-layer) tracesTwo things fall straight out of this. First, outer-layer traces carry roughly twice the current of inner-layer ones for the same width, because they shed heat to the air (the k constant is double). Second, to find the width you need, you solve for area at your target current and temperature rise, then divide by the copper thickness.
Tip: Because area is width × thickness, doubling your copper weight from 1 oz to 2 oz lets a trace of the same width carry noticeably more current — often cheaper than widening every power trace.
A worked feel for the numbers
On standard 1 oz outer-layer copper with a 10 °C rise, a trace around 0.25 mm (about 10 mil) wide comfortably handles roughly an amp, and you scale up from there. Signal traces carrying milliamps can be far thinner — routing density, not current, sets their width. It is the power and ground traces where this math matters.
Don't forget voltage drop
Ampacity keeps the trace from overheating, but a long thin trace can also drop meaningful voltage. Copper's resistance is about 1.7 × 10⁻⁸ Ω·m; a long 5 V or 3.3 V rail through a skinny trace can sag enough to matter for sensitive analog or RF. For power distribution, check both the temperature rise and the end-to-end voltage drop.
The trace width calculator does all of this — enter current, copper weight, and temperature rise and it returns the minimum width plus resistance and voltage drop. When Banana Board routes a board, its power traces are sized against these same rules automatically.
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Frequently asked
What is the IPC-2221 trace width formula?
I = k × ΔT^0.44 × A^0.725, where I is current in amps, ΔT is the temperature rise in °C, A is the trace cross-sectional area in mils², and k is 0.048 for external traces or 0.024 for internal ones. You solve for area at your target current, then divide by copper thickness to get width.
How thick is 1 oz copper?
About 1.37 mils, or 35 µm. That figure — copper 'weight' in ounces per square foot — is the thickness input to the trace-width calculation, so 2 oz copper (about 2.75 mils) carries more current at the same width.
Why do outer-layer traces carry more current?
Because they can shed heat to the surrounding air, while inner-layer traces are buried in the board. IPC-2221 reflects this with a k constant of 0.048 for external traces versus 0.024 for internal ones — roughly double the ampacity for the same width.
Do signal traces need to follow this rule?
Rarely. Signal traces carry milliamps, so their width is set by routing density and manufacturing minimums, not current. The IPC-2221 math matters most for power and ground traces.
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