A transmitter sits 200 m from the panel. The loop runs at 24 V DC. The panel builder wants to know whether 2.5 mm² holds the voltage at the far end.
The question
Extra low voltage leaves little room. A drop that would pass unnoticed at 400 V can stop a transmitter at 24 V. The question is the percentage drop over the route, and the length at which that percentage reaches the limit.
The inputs
- Mode: single phase, used here as a DC pair
- Voltage: 24 V
- Current: 0.5 A
- Route length: 200 m
- Power factor: 1
- Conductor: 2.5 mm² copper, PVC insulated, one cable
- Limit: 10 %
The tool models an AC single-phase pair. A DC pair carries the same current out and back, so the resistive part of the calculation matches. Power factor 1 removes the reactance from the result. Treat the answer as an approximation for DC, not an exact figure.
The clause and the tables
Cl 7.5.7 of AS/NZS 3000:2018 covers extra low voltage circuits. The 10 % figure in the example comes from that clause. It is not the 5 % default the tool applies to mains voltage circuits. Cl 4.5 of AS/NZS 3008.1.2:2017 sets the method that combines resistance and reactance into one millivolt figure. Tables 34 and 35 of the same standard hold the conductor data. For d.c. ratings and d.c. mV/A.m, see the d.c. cable size guide.
The steps
The tool reads the conductor data first.
- Resistance: 9.01 Ω/km
- Reactance: 0.102 Ω/km
It combines those at a power factor of 1 and gets 18.02 mV/A·m.
The rest is arithmetic. 18.02 mV multiplied by 0.5 A and by 200 m gives 1.8 V. Against 24 V that is 7.51 %.
The result
7.51 % passes the 10 % limit. The tool reports a pass and no warnings.
The more useful figure is the maximum length: 266.37 m. The loop can run that far on 2.5 mm² at 0.5 A before the drop reaches 10 %. The installed 200 m sits well inside that.
Two points decide whether that margin is real.
The current is the loop current, not the signal current. A 4-20 mA transmitter draws 20 mA at full scale. The 0.5 A here covers a loop that also carries a powered device, a solenoid or a small actuator. Run the calculation with the current the pair actually carries. A 20 mA loop on the same cable drops far less.
The limit belongs to the equipment. 10 % is the clause figure. A transmitter with a 20 V minimum supply and a barrier in the loop may need a tighter budget. Set the limit to the share the field device leaves you, then read the maximum length against that.
Resistance dominates at this size. The reactance of 0.102 Ω/km contributes almost nothing beside 9.01 Ω/km. Doubling the conductor area roughly halves the millivolt figure, and the maximum length roughly doubles with it. Adding a parallel pair does the same thing.
Temperature moves the answer as well. The tabulated resistance applies at the conductor operating temperature. A loop in a hot plant room runs at a higher resistance than a loop in a cool cable tray.
This page is a design aid. Verify every value against the current edition of the standard.