Sumivo · Electrical

PCB Trace Resistance Calculator

Estimate the DC resistance, voltage drop and power loss of a rectangular PCB copper trace.

01 / inputs

mm
mm
oz/ft²
A
°C

02 / result

RESISTANCE

49.557344 mΩ

Resistance
49.557344 mΩ
Voltage drop
0.049557 V
Power loss
0.049557 W
Cross-section
0.03479 mm²

Calculation trace

  1. 1.7241e-8 × (1 + 0.00393 × (20 − 20)) = 1.72410e-8 Ω·m
    1.72410e-8 Ω·m
    Resistivity at temperature
  2. 1 mm × (1 oz × 34.79 µm) = 0.03479 mm²
    0.03479 mm²
    Cross-section
  3. 1.72410e-8 Ω·m × 0.1 m ÷ 3.47900e-8 m² = 49.557344 mΩ
    49.557344 mΩ
    Resistance
  4. 1 A × 0.049557 Ω = 0.049557 V
    0.049557 V
    Voltage drop
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How it works

Estimate resistance, voltage drop and resistive power loss for one rectangular copper trace carrying DC. Cross-section A = width × thickness; R = ρL/A; voltage drop = IR; power loss = I²R. The engine uses ρ(T) = 1.7241 × 10⁻⁸ × [1 + 0.00393 × (T − 20)] Ω·m.

Assumptions & limits

  • Enter length and width in mm, copper weight in oz/ft², current in A and copper temperature in °C. Length, width, copper weight and current must be finite and greater than zero. There is no fixed upper input limit, but calculations outside the finite positive numeric range are rejected.
  • Temperature must be finite and give a positive temperature multiplier: 1 + 0.00393 × (T − 20) > 0 (approximately T > −234.453 °C). This is a mathematical guard, not a material operating-temperature specification. The page starts at 20 °C.
  • This model converts 1 oz/ft² to 34.79 µm of thickness. This fixed conversion is a calculator assumption; enter the nominal copper weight, not a thickness in µm.
  • The calculation uses a uniform cross-section and the temperature you supply. It does not solve self-heating, current capacity, AC impedance, vias, connectors or manufacturing variation.
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Synthetic example

100 mm long, 1 mm wide, 1 oz/ft² copper, 2 A and 20 °C.

  1. Thickness = 1 × 34.79 = 34.79 µm. Area = 1 × 0.03479 = 0.03479 mm² = 3.479 × 10⁻⁸ m².
  2. At 20 °C the temperature multiplier is 1. R = (1.7241 × 10⁻⁸ × 0.1) ÷ (3.479 × 10⁻⁸) = 0.049557344… Ω.
  3. Voltage drop = 2 × R = 0.099114688… V. Power loss = 2² × R = 0.198229376… W.

49.557344 mΩ resistance, 0.099115 V drop and 0.198229 W loss (rounded).

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Boundary case

A width of 0 mm is rejected because the cross-section would be zero. Current of 0 A is also rejected by this version, even though the ideal equations would give zero voltage drop and power loss. A temperature of −250 °C is rejected because the linear model would produce negative resistivity.

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How to read the result

Resistance describes this trace segment only. Voltage drop is the voltage lost along it at the entered current; power loss is the electrical power dissipated in it. These outputs do not establish an allowable temperature rise or a safe current rating. A return trace or connector needs its own contribution.

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Sources and what they support

The copper-weight conversion, input guards and omitted effects above describe this implementation. The synthetic example is arithmetic under these assumptions, not a measured board or a fabrication tolerance.

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FAQ

What temperature should I enter?
Enter the copper temperature for the condition you want to calculate. Ambient temperature is not automatically the trace temperature; this tool does not calculate their difference.
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