How this instrument works
Every copper trace on a printed circuit board has some electrical resistance, and while that resistance is usually negligible for signal traces, it matters a great deal for traces carrying real current — power rails, motor drivers, high-current sense lines. Trace resistance directly determines two practical problems: voltage drop (I x R, subtracting from what actually reaches the far end of the trace) and I2R heating (power dissipated as heat, which can raise trace temperature enough to affect reliability or trip a fuse-like trace failure).
Trace resistance follows the same physics as any conductor: resistance equals resistivity times length, divided by cross-sectional area — R = rho x L / (W x T), where W and T are the trace's width and thickness (their product is the cross-sectional area a rectangular trace presents to current flow). Longer or thinner traces have more resistance; wider or thicker traces have less, exactly as intuition suggests for any wire.
Copper's resistivity is a well-known material constant, about 1.68 x 10^-8 ohm-meters at room temperature (20°C) — this is the default value here, and it's why PCB copper weight (thickness, conventionally specified in ounces per square foot — 1oz copper is about 35 micrometers thick, 2oz is about 70 micrometers) is one of the first specs a designer checks when a board needs to carry meaningful current.
Because trace resistance scales inversely with cross-sectional area, doubling a trace's copper weight (going from 1oz to 2oz copper) or doubling its width both halve its resistance for a given length — either change is a common, practical fix when a calculated trace resistance turns out too high for a power design's voltage-drop or heating budget.
- Enter Trace resistivity ρ (Ω·m) — defaults to copper's standard value, 1.68 x 10^-8 Ω·m at room temperature; change it only if you're working with a different conductor.
- Enter Trace length (m) — the total length the current travels along the trace.
- Enter Trace width (m) — the trace's width as drawn on the PCB layout.
- Enter Trace thickness (m) — the copper thickness, commonly 35 micrometres (0.000035m) for 1oz copper or 70 micrometres for 2oz copper.
- Read Trace resistance (Ω) and multiply by your expected current to estimate voltage drop, or by current squared to estimate power dissipated as heat along the trace.
Worked example — 10cm, 0.5mm-wide, 1oz-copper (35μm) trace
Enter Trace resistivity ρ (Ω·m) = 0.0000000168 (copper's standard value), Trace length (m) = 0.1 (10cm), Trace width (m) = 0.0005 (0.5mm), and Trace thickness (m) = 0.000035 (35 micrometres, standard 1oz copper). Trace resistance (Ω) reads 0.096.
Working through the formula: (1.68e-8 x 0.1) / (0.0005 x 0.000035) = 1.68e-9 / 1.75e-8 = 0.096 ohms. For a trace carrying, say, 1A, that works out to a 96mV voltage drop and about 96mW of heat dissipated along its length — small individually, but exactly the kind of number a power-supply designer checks before committing a layout, since a longer or narrower version of this same trace could easily push those figures into problematic territory.
Questions
How much does doubling copper weight (1oz to 2oz) reduce trace resistance?
Doubling the copper thickness exactly halves the trace's resistance for the same length and width, because resistance is inversely proportional to cross-sectional area (width times thickness) in the R = rho x L / (W x T) formula. Going from standard 1oz copper (about 35 micrometres thick) to 2oz copper (about 70 micrometres) is a common, simple way to cut a power trace's resistance in half without changing the board's footprint or the trace's width.
Why does trace width matter as much as trace thickness for resistance?
Both width and thickness multiply together to form the cross-sectional area the current flows through, and resistance is inversely proportional to that area — so doubling either width or thickness has exactly the same halving effect on resistance. In practice, designers often prefer widening a trace over specifying heavier copper, since trace width is easy to adjust per-net in layout software while copper weight is usually set once for the entire board (or per layer), affecting every trace on that layer.
What is copper's electrical resistivity, and why does it matter here?
Copper's electrical resistivity is approximately 1.68 x 10^-8 ohm-meters at room temperature (20°C) — a fixed material property, not something a PCB designer chooses. It's the default resistivity value in this calculator because copper is by far the most common PCB trace material; resistivity also rises somewhat with temperature, so traces running hot (from their own I2R heating or nearby heat sources) will have slightly higher resistance than this room-temperature calculation predicts.
How do I use trace resistance to check voltage drop on a power trace?
Multiply the calculated trace resistance by the expected current (Ohm's law, V = I x R) to get the voltage drop along that trace. For example, a 0.096-ohm trace carrying 2A drops 0.192V — if that trace feeds a 3.3V rail into a sensitive load, losing nearly 0.2V along the way could push the delivered voltage outside the load's tolerance, which is exactly the kind of check power-supply layouts need before committing to a trace width.
Does trace resistance matter for ordinary low-current signal traces?
Generally no — a typical thin signal trace carrying microamps to a few milliamps has negligible voltage drop and heat dissipation even with resistance in the tenths-of-an-ohm range, so signal integrity concerns for those traces usually center on impedance matching and controlled-impedance routing (a different topic from DC resistance) rather than resistive loss. Trace resistance becomes a real design concern specifically for power rails, high-current sense traces, motor drive traces, and anything else carrying enough current that I x R and I2R actually add up to meaningful numbers.