Every electrical cable has a resistance, across which a voltage drop occurs when current flows. If the voltage drop is too large, loads do not receive the full rated voltage: motors run hot, lights flicker, electronics shut down. DIN VDE 0100-520 limits the voltage drop in residential installations to 3% of the nominal value (= 6.9 V at 230 V). The voltage drop calculator checks whether your cable meets this limit.
Step by Step: How to Use the Voltage Drop Calculator
- Enter the cable length: The one-way distance from the distribution board to the load in metres. The calculator automatically applies the factor of 2 for the live and neutral conductors in single-phase AC systems.
- Enter the current: The operating current of the load in amperes — or the circuit breaker rating for a worst-case calculation.
- Select the cross-section: Enter the existing or planned conductor cross-section in mm². Standard sizes: 1.5 / 2.5 / 4 / 6 / 10 / 16 mm².
- Select the conductor material: Copper (ρ = 0.0178 Ω·mm²/m) is the standard; aluminium (ρ = 0.028 Ω·mm²/m) is used for house service connections and large cross-sections.
- Check the result: Voltage drop in volts and percent. Green: below 3%. Red: increase the cross-section or reduce the cable run by adding a sub-distribution board.
Practical Examples
Example 1 – Garage with long cable run: 28 m copper 1.5 mm², 16 A. ΔU = (2 × 28 × 16 × 0.0178) / 1.5 = 10.6 V = 4.6% → too high! With 2.5 mm²: ΔU = 6.36 V = 2.8% → just OK. With 4 mm²: 3.97 V = 1.7% → comfortable margin.
Example 2 – Wallbox 32 A supply cable: 15 m copper 6 mm², 32 A. ΔU = (2 × 15 × 32 × 0.0178) / 6 = 2.85 V = 1.24% → well within the limit. For a 25 m run, 10 mm² would be required.
Example 3 – Basement lighting: 22 m copper 1.5 mm², 6 A LED lighting. ΔU = (2 × 22 × 6 × 0.0178) / 1.5 = 3.14 V = 1.36% → no problem, even with 1.5 mm² the limit is comfortably met.
Voltage Drop Calculation per DIN VDE 0100-520
Formula (AC): ΔU = (2 × L × I × ρ) / A. Limit: 3% of 230 V = maximum 6.9 V. Example: 20 m cable, 16 A, 1.5 mm² copper: ΔU = (2×20×16×0.0178)/1.5 = 7.6 V → too high! Use 2.5 mm²: ΔU = 4.6 V → OK.
Frequently Asked Questions (FAQ)
Does the 3% limit apply to the entire installation or only to final circuits?
DIN VDE 0100-520 recommends 3% for final circuits (from sub-distribution board to load) and a maximum of 5% for the complete path from the house service connection to the final load. If the incoming service cable is long, less than 3% budget remains for the final circuits.
Why is a factor of 2 used for AC circuits?
Because the current flows through both the live conductor (phase) and the return conductor (neutral) — both contribute to the resistance. For three-phase systems a different formula applies: ΔU = √3 × L × I × ρ / A (for a balanced three-phase load).
What can I do if the voltage drop is too high and I cannot replace the cable?
Possible solutions: 1) Install a sub-distribution board closer to the load; 2) Reduce the protective device rating and replace the load with a more efficient one; 3) For PV systems: relocate the grid connection point. Replacing the cable with a larger cross-section often remains the only clean solution.
