A 3 A Power Track on 1 oz Copper
A 5 V, 3 A supply feeds a payload board through a single 100 mm long, 0.30 mm wide track on the outer layer of a 1 oz (35 µm) FR-4 board. Copper resistivity is 1.72 × 10⁻⁸ Ω·m and the temperature coefficient is 0.0039 per °C; 1 mil = 0.0254 mm. The layout was checked against the IPC-2221 external-conductor relation I = 0.048 · ΔT^0.44 · A^0.725, with I in amperes and A in square mils. The board sits in still air at 55 °C and the payload draws its 3 A continuously. In test the laminate discolours above the track and the payload browns out on load steps. The contractor proposes simply widening the track to 0.5 mm and closing the issue.
1. Calculate the track cross-section, its dc resistance at 20 °C and the voltage drop at 3 A. Is the drop acceptable on a 5 V rail?
Read the worked answer
Cross-section: A = 0.30 mm × 0.035 mm = 0.0105 mm². In mils: 11.81 mil × 1.378 mil = 16.3 mil².
Resistance: R = ρL/A = (1.72 × 10⁻⁸ × 0.100) / (0.30 × 10⁻³ × 35 × 10⁻⁶) = 0.164 Ω
Drop at 3 A: V = 3 × 0.164 = 0.49 V, i.e. 9.8 % of a 5 V rail, dissipating 1.47 W in 100 mm of track.
And the 20 °C value flatters it: once the track is hot (say 110 °C), R rises by a factor 1 + 0.0039 × 90 = 1.35, giving 0.22 Ω and a 0.66 V drop. A payload that expects 4.75 V minimum will already be marginal at 0.49 V and will brown out on transients. So the drop alone rules the design out; the thermal question is a second, independent failure.
2. Use the IPC-2221 relation to estimate the temperature rise of the 0.30 mm track and of the proposed 0.5 mm track. Is widening to 0.5 mm sufficient?
Read the worked answer
0.30 mm track (A = 16.3 mil²):
16.3^0.725 = 7.55 → 3 = 0.048 × ΔT^0.44 × 7.55 → ΔT^0.44 = 8.28 ΔT = 8.28^(1/0.44) = ≈ 122 °C
With 55 °C ambient that is a 177 °C conductor - above the Tg of ordinary FR-4, which is exactly why the laminate discoloured.
0.50 mm track (19.69 mil × 1.378 = 27.1 mil²):
27.1^0.725 = 10.9 → ΔT^0.44 = 3 / (0.048 × 10.9) = 5.71 ΔT = 5.71^2.273 = ≈ 52 °C
Final temperature ≈ 55 + 52 = 107 °C. Better, but still too hot for continuous duty inside an enclosure, and the resistance rise at that temperature pushes the drop back up. The proposal reduces a gross error to a marginal one; it does not close the issue.
3. What width (or copper weight) would give a 10 °C rise, and what else would you consider given the 3 A continuous duty?
Read the worked answer
Target ΔT = 10 °C, so ΔT^0.44 = 10^0.44 = 2.754:
3 = 0.048 × 2.754 × A^0.725 → A^0.725 = 22.7 → A = 22.7^(1/0.725) = 73.7 mil² ≈ 74 mil²
- On 1 oz copper: width = 74 / 1.378 = 53.7 mil = 1.37 mm
- On 2 oz copper (70 µm, 2.756 mil): width = 74 / 2.756 = 26.9 mil = 0.68 mm
So a single outer trace needs roughly 1.4 mm on 1 oz, or 0.7 mm on 2 oz, for a 10 °C rise. Practical recommendations, in order:
- Do not use a trace at all: run the 5 V feed as a copper pour/plane on an outer layer, or split the current across two layers with a via field at each end (check via current rating).
- If a trace must be used, use 2 oz copper at ≥ 1 mm and keep it on the outer layer where convection is best.
- Add local bulk capacitance at the payload to ride load steps, since the drop is a resistive, current-proportional error.
- Re-check the return path - a high-current ground return through a thin trace is the classic companion of this fault.