Power trace width: set by both temperature rise and voltage drop

Width from temperature rise (IPC-2221)

Current through a trace dissipates I²R, and the temperature rises until that heat balances what escapes. IPC-2221 gives an empirical formula for outer layers, I = 0.048 × ΔT^0.44 × A^0.725 (I in A, ΔT in °C, A in mil²). For inner layers the coefficient is half, 0.024.

Solved for 1oz copper (35µm) and a 10°C rise, it gives roughly 0.3mm at 1A, 0.8mm at 2A, 1.4mm at 3A, and 2.8mm at 5A. A 20°C rise cuts the width to about 2/3; 2oz halves it.

With half the coefficient, an inner layer needs 2.6x the width for the same current. But the formula rests on 1950s measurements. IPC-2152 (2009) revised it: an inner layer sheds heat to adjacent planes and the base material, and can carry as much current as an outer layer or more (Figure 1).

Voltage drop scales with length

Length does not appear in the temperature-rise formula, but voltage drop scales with it. 1oz copper has a sheet resistance of about 0.5mΩ/□ (20°C), so a 0.3mm-wide, 100mm-long trace is 333 squares, or 167mΩ: 1A drops 167mV. On a 3.3V rail with a ±3% (100mV) budget, that alone blows the budget.

Keeping 1A over 100mm within 25mV needs R = 25mΩ, 50 squares, and w = 2mm, seven times the 0.3mm from the temperature calculation (Figure 2).

Surface plating brings 1oz copper on an outer layer to 50-60µm finished, so the real sheet resistance is 0.3-0.35mΩ/□ and the voltage drop 30-40% lower.

Short runs by temperature, long runs by voltage drop

In practice, a short run (a few mm to around 10mm) is sized by temperature rise, a long run by voltage drop (Figure 3). The temperature formula is enough from an IC power pin to its decoupling capacitor; a rail crossing the board from a connector needs the voltage-drop calculation.

Copper resistance rises 40% with temperature

Copper's temperature coefficient of resistivity is +0.39%/°C. Resistance calculated at 20°C is 23% higher at 80°C and 39% higher at 120°C.

Correct for the operating temperature (ambient plus self-heating plus nearby heat sources). The trace resistance calculator takes a temperature input.

Via current capacity

A via's conductor is a plated cylinder on the hole wall, 20-25µm thick. At a 0.3mm hole and 25µm plating, the cross-section is π × 0.3 × 0.025 ≈ 0.0236mm² (36mil²), the same as a 0.68mm-wide trace in 1oz copper (Figure 4). The IPC-2221 formula gives about 1.8A for a 10°C rise.

At 1.6mm long, resistance is small (about 1.2mΩ), but current concentrates where the plating is thinnest, so calculate with the minimum thickness (say, 20µm). For 3A use two or three vias; for 5A, four to six, in a row across the trace width.

Copper pours and thermal relief

Above 1A, the standard practice is a pour rather than a thin trace. It removes the width constraint and lowers both voltage drop and heating.

A through-hole pad tied to a pour gets thermal relief (thin cross-shaped spokes). Those spokes (say, four of 0.3mm) become the current path, so for high-current pads widen them, or connect solidly and handle soldering with preheat. (Figure 5).