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24 V DC voltage drop — choosing wire size for remote sensors and devices

A remote 24 V sensor or RS-485 device resets or drops off? Compute the drop from loop resistance, pick a wire size, and avoid inrush and shared-common traps.

Tools for this guideModbus Master Windows program · freeSerial Windows program · freeVoltage Drop & Wire Size Calculator web · no install4–20 mA Loop Load Calculator web · no install

At 24 V DC the same power means more current, and the same voltage drop is a bigger share: 1 V is 0.4 % of 230 V but about 4.2 % of 24 V. Voltage drop is the first thing to check when sensors, remote I/O or RS-485 devices sit far from the panel.

Formula

ΔV = 2 × I × ρ × L ÷ A
  I: current (A), L: one-way length (m), A: cross-section (mm²)
  ρ: copper at 20 °C 1/58 ≈ 0.01724 Ω·mm²/m

The 2 counts both conductors, out (+) and back (0 V). Enter the one-way length (supply to device) and multiply by 2. One kilometre of 1 mm² copper is about 17.24 Ω at 20 °C.

Example: 2 A over 50 m one way

SizeLoop resistanceDropPercentVoltage at device
0.75 mm² (≈ 18 AWG)2.30 Ω4.60 V19.2 %19.40 V
1.5 mm² (≈ 16 AWG)1.15 Ω2.30 V9.6 %21.70 V
2.5 mm² (≈ 14 AWG)0.69 Ω1.38 V5.7 %22.62 V
4 mm² (≈ 12 AWG)0.43 Ω0.86 V3.6 %23.14 V

How much drop is acceptable is set by the device. Take the lower limit from its supply specification (e.g. «24 V DC, allowed range … V») and check the voltage at the device stays comfortably above it. Some supplies let you raise the output slightly, but the drop changes with current, so a thicker wire or a closer supply is the real fix.

What gets overlooked

What it looks like on a communication bus

A device that restarts for lack of voltage does not reply while it boots, so it shows up as «timeouts that come in bursts». Log the times of the timeouts and compare them with when other loads on the same supply (valves, motors, heaters) switch on to tell a power problem from a communication one.

4–20 mA is checked differently

A 4–20 mA signal is a current, so wire resistance does not change the reading. A 2-wire transmitter, however, needs «supply − voltage lost in the loop resistance» to stay above its minimum operating voltage, so calculate the loop load rather than a percentage drop.

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Published 2026-10-23