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Copper Trace Resistance and Temperature Coefficient

Estimate resistance and voltage drop for a uniform rectangular copper trace, using editable resistivity and correctly converted temperature coefficients.

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Calculator inputs

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Inputs and results stay in this browser tab. Bookify does not upload or store the values you enter.

Formula and method

At25°C, a10cm trace1mm wide and0.035mm thick, with resistivity1.7e-8ohm meters, has resistance0.0485714ohms;2A gives0.0971429V drop.

R = rho L/(width thickness) × (1 + alpha (T − 25 °C)); voltage drop = current × resistance.

Worked example

Enter these known values and leave the other values blank.

Trace width (W)
1 mm
Length (L)
10 cm
Trace thickness (T)
0.035 mm
Ambient temperature
25 °C
Current
2 A
Resistance
0.04857 Ω
Voltage drop
0.09714 V

Assumptions and limitations

  • Assume a uniform rectangular conductor. Resistivity is specified at25°C; the captured copper default is1.7e-8ohm meters and coefficient0.0039 per°C. Both approximate parameters are editable.
  • The coefficient is per temperature interval. A coefficient0.0039 per°C equals0.0021666666667 per°F and0.0039 perK. Absolute temperature retains its usual Celsius/Fahrenheit/kelvin offsets. The captured affine coefficient conversion is repaired locally with a complete pinned selector replacement.
  • The model excludes vias, contacts, skin effect, frequency dependence and self heating. Its linear temperature approximation applies only over a suitable material range.
  • Copper weight in oz/ft² is a captured conventional equivalent-thickness preset, not a length measurement. Choose ordinary length units for trace path length and width.

Common questions

Why does a per°F coefficient differ from a per°C coefficient?

A Fahrenheit degree interval is5/9 of a Celsius degree interval. The coefficient per°F is therefore5/9 of the coefficient per°C; the32-degree absolute offset is excluded.

Can I calculate voltage drop?

Enter current as well as the trace geometry. Drop is current multiplied by trace resistance, without connector or supply impedance.

References

Replace an external trace illustration with the original accessible description of length, width, thickness and current path. The uniform trace model requires positive cross-section and resistivity and a positive linear temperature factor. A temperature coefficient is per temperature interval: °F intervals scale5/9 from °C; kelvin intervals equal °C. Absolute offsets must never enter inverse-temperature conversion. Replace only the complete captured alpha selector; preserve the ambient-temperature affine selector byte for byte.

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