24V DC Control Cable & Voltage Drop Sizer
Calculate two-wire DC loop resistance, conductor voltage drop, and terminal operating voltage for industrial sensors, PLCs, safety scanners, and solenoid valves. Verifies that field device voltage stays strictly above the mandatory 20.4 Vdc (-15%) operating threshold.
Results
Worked example
24.0 V supply, 35 m one-way run (70 m loop), 2.0 A sensor load, 0.75 mm² cable at 40°C:
Resistivity at 40°C = 0.0175 × [1 + 0.00393 × (40 - 20)] = 0.01888 Ω·mm²/m
Loop resistance = (70 m × 0.01888) / 0.75 mm² = 1.76 Ω
Voltage drop = 2.0 A × 1.76 Ω = 3.52 V (14.7%)
Terminal voltage = 24.0 V - 3.52 V = 20.48 V (Passes 20.4 V limit with 0.08 V margin)
⚠️ Educational design aid. For safety-critical control circuits (E-Stops, light curtains), always allow at least 1.0 V operating headroom above the manufacturer's low-voltage cut-off to accommodate supply ripple and power supply aging.
Regulation reference: BS EN 61131-2 specifies DC power supply tolerance zones for industrial automation. BS EN 60204-1 §6.4 and §9.4 govern control circuit voltages and safe disconnection under earth fault conditions.
Frequently Asked Questions: 24V DC Control Loops
Why is voltage drop critical on 24V DC industrial control circuits?
Unlike 230V mains where a 5% drop is 11.5V, a 5% drop on 24V DC is only 1.2V. Relays, contactor coils, and PLC digital inputs commonly drop out below 20.4V DC (-15%), causing intermittent control failures and machine downtime on long cable runs.
How do you calculate voltage drop on 24V DC circuits?
Use loop length (double the one-way route distance): V_drop = (2 * Length * rho * Current) / CSA, where rho is copper resistivity (0.0175 Ω·mm²/m at 20°C, temperature-corrected for operating conditions).