PCB Trace Width Calculator

Find the minimum copper trace width for a target current using the IPC-2221 formula, plus the resulting voltage drop and power loss along the trace.

Quick Facts

Formula
IPC-2221: I = k × ΔT^0.44 × A^0.725
k = 0.048 for external (outer) traces, 0.024 for internal traces; A is the required copper cross-section in mil².
Copper thickness
1 oz/ft² ≈ 1.378 mils (0.035 mm)
Trace width = required area ÷ copper thickness for the chosen weight.

Your Results

Calculated
Minimum trace width
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Required copper width, in mils
Minimum trace width
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Required copper width, in millimeters
Voltage drop
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I×R drop over the trace length
Power dissipation
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Resistive loss (I²R) in the trace

Ready

Enter current, temperature rise, copper weight, and layer, then press Calculate.

About the PCB Trace Width Calculator

This calculator finds the minimum copper trace width needed to carry a given current on a printed circuit board without exceeding a chosen temperature rise, using the industry-standard IPC-2221 (formerly MIL-STD-275) empirical formula. It also estimates the voltage drop and resistive power loss along the trace for a given length, so you can check whether a thin trace also causes an unacceptable voltage sag.

The IPC-2221 formula

IPC-2221 relates current-carrying capacity to the trace's cross-sectional area and the allowed temperature rise above ambient:

I = k × ΔT0.44 × A0.725

  • I — current through the trace, in amps.
  • ΔT — allowed temperature rise of the trace above ambient, in degrees C.
  • A — cross-sectional area of copper, in square mils (a mil is one-thousandth of an inch).
  • k — a constant that depends on where the trace sits: 0.048 for external (outer layer) traces exposed to open air, or 0.024 for internal traces buried between layers, which cannot shed heat as easily.

Solving for the required area gives A = (I ÷ (k × ΔT0.44))1/0.725. Dividing that area by the copper thickness for the selected ounce weight (1 oz/ft² ≈ 1.378 mils) converts the area into a minimum trace width in mils.

Voltage drop and power loss

A narrow trace that satisfies the temperature rule can still drop unwanted voltage over a long run. This calculator computes the trace's resistance from its width, thickness, and length using copper resistivity (about 1.7×10⁻⁸ Ω·m at room temperature), then reports the voltage drop (I × R) and power dissipated (I² × R) for the length you enter. For sensitive analog or long power runs, both numbers matter as much as the temperature-based minimum width.

How to get the best results

  • Use the worst-case (maximum) current the trace will actually carry, not the average.
  • A larger allowed temperature rise (ΔT) lets the trace run narrower; 10°C is a common conservative default, while some designs tolerate 20–40°C.
  • Check the manufacturer's minimum trace width and spacing — the IPC-2221 result is a physics minimum, and fabrication limits can push you wider.
  • For high-current power traces, consider 2 oz or heavier copper, which roughly halves the required width compared with 1 oz copper.

Frequently Asked Questions

What formula does this trace width calculator use?
It uses the IPC-2221 (formerly MIL-STD-275) empirical formula I = k × ΔT^0.44 × A^0.725, where I is current in amps, ΔT is the allowed temperature rise in degrees C, A is the required cross-sectional area of copper in square mils, and k is 0.048 for external traces or 0.024 for internal traces. Solving for A gives the minimum copper area needed; dividing by the copper thickness for the chosen ounce weight gives the minimum trace width.
Why do internal traces need to be wider than external ones?
Internal layers are sandwiched inside the board and cannot shed heat directly to open air, so they run hotter for the same current. The IPC-2221 constant for internal layers (k = 0.024) is half that of external layers (k = 0.048); because area scales as (1/k) raised to the 1/0.725 power, that halved constant multiplies the required cross-sectional area by about 2.6 — and the trace is noticeably wider — than an equivalent external trace of the same copper weight.
What copper weight should I use?
Copper weight is measured in ounces per square foot and sets the trace thickness: 1 oz/ft² of copper is about 1.378 mils (0.035 mm) thick. Consumer boards commonly use 1 oz copper; power boards often use 2 oz or heavier to keep traces narrower for the same current. Doubling the copper weight roughly halves the required trace width for a given current and temperature rise.
Does a wider-than-required trace cause problems?
No — the IPC-2221 result is a minimum, not a target. Wider traces run cooler, have lower resistance, and lower voltage drop for the same current, so oversizing a trace is safe. The only downside is board space: wider traces are harder to route densely, and very wide single traces are sometimes better replaced with a copper pour or plane.

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