PCB trace resistance: the formula, sheet resistance and squares, temperature, and a worked example
Copper is a good conductor, but a trace is thin: 35 µm on a 1 oz board. A 0.3 mm trace 100 mm long is 165 mΩ at room temperature, which drops 165 mV at 1 A, and a long sense line or a 3.3 V rail can lose more than its tolerance that way without ever getting warm. This calculator gives a trace's resistance from its width, length, copper weight and temperature, shows it the way layout designers count it (sheet resistance times squares), and with a current gives the voltage drop, the power lost, the rise IPC-2221 predicts and the width that keeps the drop under a limit.
How to use the trace resistance calculator
- Type the width and the length of the trace in millimetres and pick the copper weight in the menu row.
- Set the copper temperature: 20 °C for the data-sheet figure, or the board's ambient plus the trace's own rise.
- Add the current for the drop, the loss and the heating, and a drop limit for the width that meets it. The Widths tab compares the standard widths.
The formula
R = ρ L / (w t) ρ = 1.72×10−8 Ω·m × (1 + 0.00393 (T − 20))
Resistance is resistivity times length over cross-section, and the cross-section is the width times the copper thickness: 35 µm per ounce. Copper's resistivity is 1.72×10−8 Ω·m at 20 °C and rises 0.39 % per degree. Plating adds a few microns to outer layers and etching removes a little width, so measured values sit within about 10 % of this.
Sheet resistance and squares
Divide the formula by the width and it becomes (ρ / t) × (L / w): a sheet resistance that depends only on the copper, times the number of squares along the trace. A square of 1 oz copper is 0.49 mΩ at 20 °C whether it is 0.1 mm or 10 mm on a side, because a bigger square is both wider and longer. So a layout designer estimates a trace by counting squares: a 0.25 mm trace 25 mm long is 100 squares, about 50 mΩ; a corner counts as roughly 0.56 of a square because current crowds the inside of the bend. Half-ounce copper is about 1 mΩ per square, 2 oz a quarter of a milliohm.
Temperature
Use the copper's own temperature, not the room's. A trace carrying its rated current in a 50 °C enclosure with a 20 °C rise is at 70 °C and has 20 % more resistance than the 20 °C figure; at 100 °C a third more. The effect runs both ways: the extra resistance makes a little more heat, which is why ratings leave margin.
Drop versus heating
Two different things set a trace's width. Heating is covered by the trace width calculator: a 0.3 mm trace carries 1 A with a 10 °C rise. But at 1 A over 100 mm that trace drops 0.17 V, far too much on a 3.3 V rail or a sense line to an ADC. For low-voltage rails and precision signals the drop sets the width, and it is usually the larger figure; the calculator's drop limit gives it directly: w = ρ L I / (t V).
Your trace, step by step
- Copper at 50 °C: ρ = 19.2 nΩ·m; 1 oz is 35 µm, so 553 µΩ per square.
- Squares: 100.0 mm / 0.30 mm = 333.3; R = 184 mΩ (165 mΩ at 20 °C).
- At 1 A: 184 mV drop, 184 mW, IPC-2221 rise 10 °C; 0.55 mm would keep the drop under 100 mV.
Worked example: 0.3 mm × 100 mm in 1 oz copper at 50 °C
At 50 °C copper's resistivity is 1.72×10−8 × (1 + 0.00393 × 30) = 19.2 nΩ·m. The cross-section is 0.3 mm × 35 µm = 0.0104 mm², so R = 19.2 nΩ·m × 0.1 m / 0.0104 mm² = 184 mΩ. Counted in squares: 100 / 0.3 = 333.3 squares of 553 µΩ each, the same 184 mΩ. At 1 A the trace drops 184 mV and dissipates 184 mW, with a rise of about 10 °C by IPC-2221. To hold the drop under 0.1 V it would have to be 0.55 mm wide, far more than the heating alone asks for, which is the usual story on a low-voltage rail.
Questions
Does the resistance change with frequency?
Above a few hundred kilohertz the current crowds into the surface (skin effect) and the effective resistance rises; at 1 MHz the skin depth in copper is 65 µm, so a 35 µm trace is barely affected, but at 100 MHz it is several times the DC value. This calculator is for DC and low frequencies.
How accurate is the copper thickness?
Base foil is specified to about ±10 %, and outer layers gain 20–30 µm of plating on a plated-through board, which lowers their resistance. For a precise figure use the fab's finished thickness.
Can I use a trace as a current-sense resistor?
Only roughly: the tolerance and the 0.39 %/°C drift are poor. It works for a crude overcurrent trip; use a real shunt for measurement.
What about a copper pour or plane?
A plane is a very wide trace: a 12 V plane 50 mm wide over 100 mm is two squares, a milliohm. Its resistance is set by the narrowest neck and by the vias in and out, not by the plane itself.