Inputs
System (DC, single-phase AC or three-phase AC), conductor material, nominal voltage, operating current, one-way route length, permitted voltage drop in per cent, and for AC the power factor cos φ.
Electrical engineering · IEC 60364-5-52
Cable cross-section and voltage drop
Calculate the cross-section a circuit needs so its voltage drop stays inside the permitted limit – and what the next standard cross-section actually achieves. The calculator covers single-phase AC, three-phase AC and DC, and reports the voltage drop in volts and per cent, the voltage at the load and the power loss.
Only the voltage-drop criterion is evaluated, in a resistive model. Current-carrying capacity for the installation method, the protective device, the disconnection condition and the protective conductor must be verified separately; this calculator does not replace a qualified electrician.
AΔU ≤ 3 %
02
Cross-section and voltage drop
Set the circuit and calculate.
Inputs and method
Cable cross-section from voltage drop
Determine the required cable cross-section from the permitted voltage drop and check what the selected standard size actually achieves.
Calculation
The voltage drop follows from ΔU = k · L · I · cos φ / (κ · A) with k = 2 for DC and single-phase AC (out and return conductor) and k = √3 for balanced three-phase. Solved for A with the permitted ΔU = U · p/100 this gives the required cross-section; it is rounded up to the next standard size of the IEC 60228 series, from which the actual voltage drop is calculated back.
Example
230 V single-phase, 16 A, 30 m of copper and 3 % permitted give A = 2·30·16/(56·6.9) = 2.48 mm²; 2.5 mm² is selected, at which the voltage drop is 6.86 V or 2.98 %.
Sources and limits: Resistive voltage-drop model as used in German installation practice with κ = 56 m/(Ω·mm²) for copper and 35 for aluminium at 20 °C; cross-section series per IEC 60228; the limits from DIN 18015-1 (3 % in dwellings) and DIN VDE 0100-520 supplement 3 (3 % lighting, 5 % other circuits) are an input, not an assumption. The inductive part of the line impedance and the current-carrying capacity are not covered.
Cable cross-section from voltage drop in detail
Determine the required cable cross-section from the permitted voltage drop and check what the selected standard size actually achieves.
What is voltage drop?
Every cable has an ohmic resistance. When current flows, part of the voltage drops across it, so less arrives at the load than was fed in. This voltage drop grows in proportion to current and route length and falls in inverse proportion to cross-section and conductivity. Installation rules limit it so that equipment works inside its permitted voltage range and losses stay acceptable.
Formula and variables
A = k · L · I · cos φ / (κ · ΔU_permitted)
DC and single-phase AC: k = 2Three-phase AC (balanced): k = √3ΔU_permitted = U · p / 100ΔU = k · L · I · cos φ / (κ · A)
| Symbol / input | Meaning |
|---|---|
| L | One-way route length; the return conductor is contained in the factor k. |
| I, cos φ | Operating current and real-power factor of the load. |
| κ | Specific conductivity of the conductor, 56 for copper and 35 for aluminium at 20 °C. |
| p, ΔU | Permitted voltage drop in per cent and the resulting limit in volts. |
| A | Required conductor cross-section, rounded up to the IEC 60228 standard series. |
Choose the inputs correctly
System (DC, single-phase AC or three-phase AC), conductor material, nominal voltage, operating current, one-way route length, permitted voltage drop in per cent, and for AC the power factor cos φ.
How to use the calculator
Choose the system first – it sets the prefactor of the formula and the matching nominal voltage. Then enter the actual operating current (not the rated current of the fuse), the one-way route length and the permitted voltage drop in per cent. For AC systems the load's cos φ is added. The calculator states the required cross-section, rounds it up to the standard series, and shows the voltage drop that cable really achieves.
Worked example
230 V single-phase, 16 A, 30 m of copper and 3 % permitted give A = 2·30·16/(56·6.9) = 2.48 mm²; 2.5 mm² is selected, at which the voltage drop is 6.86 V or 2.98 %.
How should the result be read?
The selected standard cross-section is a lower bound, not a finished answer: it satisfies exactly one of several requirements on a cable. What is informative is the comparison between the calculated requirement and the selected size – if the requirement sits just below a standard step the margin is small, and a warm conductor may already breach the limit. The power loss additionally shows what the cable costs in energy over time.
Voltage in volts, current in amperes, length in metres, cross-section in mm², conductivity in m/(Ω·mm²). The permitted voltage drop is entered as a percentage of the nominal voltage.
Typical applications
Sizing long supply cables to a garden house, workshop, wallbox or pump, checking existing installations with a noticeably low voltage at the load, and dimensioning 12 V and 24 V DC circuits in vehicles and solar systems, where the low voltage makes the drop almost always the governing criterion.
Assumptions, limits and common mistakes
Resistive voltage-drop model as used in German installation practice with κ = 56 m/(Ω·mm²) for copper and 35 for aluminium at 20 °C; cross-section series per IEC 60228; the limits from DIN 18015-1 (3 % in dwellings) and DIN VDE 0100-520 supplement 3 (3 % lighting, 5 % other circuits) are an input, not an assumption. The inductive part of the line impedance and the current-carrying capacity are not covered.
Common mistake: Enter the route length once, not twice – the return conductor is already in the factor 2. For three-phase use the line-to-line voltage of 400 V, not 230 V. Do not enter the fuse rating as the operating current, or the cable becomes needlessly large. And the key point: a cross-section found here is sufficient only from the voltage-drop standpoint – the current-carrying capacity for the actual installation method may force a larger one.
Frequently asked questions
Why √3 instead of 2 for three-phase?
With DC and single-phase AC the same current flows out and back, so the drop occurs over twice the length. In a balanced three-phase circuit the neutral carries no current; the governing voltage is the line-to-line voltage, and the phasor geometry gives the factor √3 ≈ 1.73.
Is the calculated cross-section enough for the installation?
No, it is only one of several conditions. The current-carrying capacity for the actual installation method and ambient temperature, coordination with the protective device, the disconnection condition under fault and the protective conductor size must all be satisfied as well. The governing size is the largest one from all conditions.
Which conductivity should I use?
The stored 56 (copper) and 35 (aluminium) m/(Ω·mm²) apply at 20 °C. A warm conductor conducts worse – roughly 0.4 % per kelvin – so copper at 70 °C is closer to 48. To model the less favourable operating temperature, enter that value as a custom conductivity; the required cross-section rises accordingly.
Why does a smaller cos φ reduce the voltage drop?
Because this simplified model considers only the resistive line resistance, at which only the in-phase component of the current produces a drop in phase with the voltage. From roughly 50 mm² upwards the inductive part of the line becomes relevant and partly reverses this effect; there the simplified formula is no longer on the safe side.
Why is voltage drop so critical at 12 V?
Because the limit is a percentage of the nominal voltage: 3 % is twelve volts at 400 V but only 0.36 V at 12 V. At the same current and length a low-voltage circuit therefore needs many times the cross-section of a 230 V circuit.