Wire Gauge & Voltage Drop Calculator: AC/DC, Single & Three Phase

There are two ways to choose a wire. The first is based on the "this one looks thick enough" principle. The second is to open a cable size calculator, enter the current, and get a number. Both methods can yield the wrong answer if one thing is not considered: the cross-section is selected based not on one criterion, but on two simultaneously, and different situations determine which one rules.

Our cable cross-section calculator calculates both criteria at once. Let's break down how to calculate correctly, the differences between AWG, SWG, and square millimeters, and why a LED assembly running on 12 or 24 volts requires a thicker wire than the same load on a domestic mains network.

Two Criteria for Wire Selection

Calculating cable cross-section by power or current is only half the battle. There are two independent requirements, and the wire must satisfy both. Furthermore, a wire should always be chosen with a safety margin. Primarily, this is for your own safety; a wire that is too thin can lead to a real fire.

Criterion One: Heating (Thermal Constraints)

Current passing through a conductor generates heat because every conductor, except for a superconductor, has resistance. The thinner the core, the greater the resistance and the intense the heating. If the limit is exceeded, the insulation will begin to degrade—first losing elasticity, then cracking, leading to a short circuit and fire.

Hence the concept of allowable continuous current: the maximum at which a wire will not overheat under given conditions. For copper with 70 °C insulation in open air installation at an ambient temperature of 30 °C, this is approximately 19 amperes for 1.5 mm², 26 for 2.5 mm², and 35 for 4 mm². These are maximum values provided without a safety margin; naturally, you should not operate at these limits.

Criterion Two: Voltage Drop

A wire acts as a resistor, and part of the voltage is lost across it. Less voltage reaches the load than left the source. The formula is simple:

ΔU = I × ρ × 2L / S

where I is current in amperes, ρ is the resistivity of the material (for copper, 0.0172 Ω·mm²/m at 20 °C), L is the line length in one direction, and S is the cross-section in mm². The multiplier '2' appears because the current travels to the load and back—the actual length of the conductor is twice the distance.

Voltage drop is assessed as a percentage of the nominal voltage. For LED assemblies, it is customary to stay within 3%, while domestic networks allow up to 5%.

Why Low-Voltage Circuits Are Different

This is where the catch lies! Standard calculators often only consider heating and provide the correct answer for 230-volt networks. However, for 12 or 24 volts, this answer can be dangerously optimistic.

The issue is that voltage drop is measured in volts, but its allowability is measured in percentages. A loss of 0.7 volts in a 230-volt system is 0.3%, which goes unnoticed. The same loss in a 24-volt system is already 3%, the limit. In a 12-volt system, it is 6%, twice the acceptable norm.

Let's look at the numbers. Load is 5 amperes, length is 10 meters, copper material, temperature is 40 °C, installed in a conduit:

• Based on heating, 0.75 mm² is sufficient—this wire will handle 5 amperes with a 25% margin even in a hot conduit.
• However, based on voltage drop at 24 volts, 4 mm² will be required, which is more than five times the cross-section.

What happens if you install 0.75 mm², as suggested by the heating calculation alone? The voltage drop will be about 2.5 volts—that is 10% instead of the allowable 3%. The LEDs (load) will receive 21.5 volts instead of 24, and 12.4 watts will be wasted as heat through the wire, essentially acting like underfloor heating. With the correct 4 mm² cross-section, the drop falls to 0.5 volts (less than 2%), and losses drop to 2.3 watts.

The rule is simple: the lower the voltage and the longer the line, the more likely that voltage drop, not heating, will be the deciding factor. For 230-volt networks on short routes, the opposite is usually true.

Calculating Cable Cross-Section by Power

Often, input data is given in watts rather than amperes: the power of a lamp, heater, or entire assembly is known. Converting one to the other is simple.

For DC and single-phase AC: I = P / U. A 300 W lamp at 24 volts consumes 12.5 amperes. The same lamp powered by 230 volts consumes only 1.3 amperes.

This, incidentally, shows why low-voltage assemblies require thick wires: at the same power, current is inversely proportional to voltage, and wire heating grows as the square of the current. Switching from 24 to 48 volts halves the current and quarters the losses in the wire. This is one of the reasons why industrial solar panels operate at 48 volts.

For three-phase networks, the formula is different: I = P / (√3 × U × cos φ), where U is line voltage and cos φ is the load power factor. For resistive loads, it is close to unity; for motors and switching power supplies, it ranges from 0.7 to 0.95, and this must be taken into account.

A specific warning regarding LED assemblies: power at the driver input and power at the output are different things. A driver with 90% efficiency delivering 100 W output power consumes about 111 W from the mains. You must calculate the input line current based on the consumed power, not the nominal rating of the lamp.

Other Factors Affecting Allowable Current

Table values are given for specific conditions, and real-world conditions almost always differ.

• Ambient Temperature. Tables are compiled for 30 °C. In a grow box, closed panel, or near drivers, temperatures can reach 45–50 °C, and allowable current drops noticeably. According to IEC 60364-5-52, the correction factor is 0.87 at 40 °C, 0.71 at 50 °C, and 0.61 at 55 °C. Thus, a wire rated for 26 amperes will withstand only 16 in a hot panel.
• Installation Method. A wire in free air is cooled by convection. The same wire in a conduit, cable trunking, or bundled with others fares much worse. A correction factor of around 0.72 applies, meaning a quarter less allowable current.
• Material. Aluminum has resistance approximately 1.6 times higher than copper; therefore, at the same cross-section, voltage drop is higher, and allowable current is lower by about 22%. Aluminum is practically never used for low-voltage assemblies.
• Safety Margin. A wire should not operate at the limit of its allowable current—usually, a 20–25% margin is added. This is not over-caution: loads can increase, conditions can worsen, and insulation ages faster the hotter it operates.

AWG, SWG, and mm²: Converting Between Systems

Three designation systems coexist, serving as a constant source of confusion. Wire on AliExpress is sold in AWG, in Europe in square millimeters, and old British documentation uses SWG.

• AWG (American Wire Gauge) is calculated by formula: diameter in millimeters equals 0.127 × 92^((36−n)/39), where n is the gauge number. The logic is counter-intuitive: the larger the number, the thinner the wire. AWG 10 is thicker than AWG 20.
• SWG (British Standard Wire Gauge) is a historical table compiled empirically. Conversion is only possible via a reference guide; it has no calculation formula.
• Square Millimeters represent the cross-sectional area of the core, the most straightforward value. Diameter is derived from it as the square root of 4S/π.

Correspondence Table: AWG to mm² and SWG

Allowable current is stated for copper with 70 °C insulation at an ambient temperature of 30 °C. Power is calculated as the product of current and voltage for quick estimation during cross-section calculation by power.

Please note: correspondence is approximate. The closest gauge to 1.5 mm² is AWG 15, but its actual cross-section is 1.65 mm². The common AWG 16 yields 1.31 mm², which is less than one and a half squares. When replacing European wire with American, this should be considered: a 10–15% difference in cross-section directly translates into the same difference in voltage drop.

Neutral Wire in a Three-Phase Network

This is a separate topic where intuition fails most often.

The classical rule states: with a balanced load, the currents of the three phases are shifted by 120 degrees, vectorially adding up to nearly zero, allowing the neutral to be thinner than the phase wires. This has been the practice for decades.

The problem is that the rule only works for linear loads—heaters, incandescent lamps, motors. Switching power supplies, which include all LED drivers, behave differently. They consume current not as a sine wave but in short pulses, generating harmonics that are multiples of three: third, ninth, fifteenth.

Key point: these harmonics in all three phases are in phase. They are not canceled out in the neutral but are added arithmetically. Consequently, the current in the neutral wire can reach 1.73 times the phase current—meaning the neutral ends up more heavily loaded than any of the phases.

According to IEC 60364-5-52, if such harmonic content exceeds 33%, the neutral is calculated based on its own current, and a derating factor is applied to the cable. For installations with LED drivers, this means: the neutral wire is chosen to be no thinner than the phase wire, and with dense installation, one size thicker.

Typical Mistakes

• Calculating based only on heating. The most frequent mistake. Works for short runs in domestic networks but fails everywhere else.
• Forgetting that length doubles. Voltage drop calculation involves the complete current path—to the load and back. If you only consider the distance, the error will be exactly twofold.
• Taking table current without corrections. Tables assume 30 °C and free air. In a hot panel with wires bundled together, the real allowable current can be half as much.
• Sizing the circuit breaker for the load, not the cable. A circuit breaker protects the wire from overload, not the appliance. If you install 1.5 mm² wire and a 25-ampere breaker, the wire will burn out before the breaker trips. This is a complex topic involving many parameters beyond just current, worthy of a separate article.
• Assuming stranded wire is equivalent to solid wire of the same diameter. A flexible wire core consists of many thin filaments with air in between. The copper cross-section is smaller than what might be deduced from the outer diameter.

How to Use the Calculator

The widget above calculates both criteria simultaneously and shows the result based on the larger requirement.

Set the load current, line length in one direction, and voltage—preset buttons are available for typical values ranging from 5 to 230 volts. Select current type: DC, single-phase AC, or three-phase AC. For DC and single-phase AC, the calculation is identical; a difference appears only with three-phase, where the square root of three replaces the multiplier '2'.

Next, specify conditions: core material, installation method, and ambient temperature. The last two sliders set the current safety margin and allowable voltage drop—default values correspond to standard engineering practice.

The output provides cross-section in three systems, voltage drop in volts and percentages, voltage at the load, power losses, heat emission per meter, and calculated core temperature. The diagram on the left changes with the parameters: wire thickness grows with cross-section, and the color turns red under overload.

Selecting three phases activates a separate block for neutral wire calculation—with a load type switch, because the result is fundamentally different for linear and pulsed loads.

At the bottom, a separate block contains a converter: enter a value in any of the three fields, and the other two will recalculate automatically. It also shows core diameter, allowable current in both installation modes, and copper resistance per meter.

Conclusion

• Cross-section is selected based on the larger of two criteria—heating and voltage drop. Neither can be ignored.
• For low-voltage circuits, voltage drop is almost always the deciding factor, not heating. The lower the voltage and the longer the line, the greater the discrepancy between criteria.
• Tabular allowable current requires corrections for ambient temperature and installation method. In a hot panel, it can drop by half.
• When converting AWG to mm², correspondence is approximate—the closest gauge may differ from the metric nominal by 10–15%.
• In a three-phase network with LED drivers, the neutral wire is sized no thinner than the phase wire due to triplen harmonics.
• Most importantly: any calculation is for reference. Local codes apply to permanent wiring, and when in doubt, it is better to choose a cross-section one size larger—the price difference is incomparable to the consequences of an error.