Relay Chatter on a 180 ft Gate Control Circuit
A contractor is troubleshooting a 24 VAC control circuit for a sliding gate at a self-storage facility. The transformer is 24 V, 40 VA, located in the office 180 ft from the gate operator. The run is 18 AWG stranded copper, two conductors. The operator relay is rated to pick up at 21 V minimum. Measured voltage at the transformer is 24.1 V; at the operator, it is 19.2 V with the operator idle and 17.8 V when the gate motor starts. The relay chatters and the gate reverses. The maintenance tech replaced the relay and transformer once. The contractor suspects the wire is too small, but the owner says “24 volts is low voltage, so voltage drop doesn’t matter.” The load draws 0.9 A when the relay and photo-eye heater are on. There is no spare conductor.
1. Calculate the expected voltage drop and determine whether the 18 AWG wire fully explains the measured 19.2 V.
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18 AWG copper has about 6.5 Ω per 1,000 ft per conductor. The loop is 360 ft, so R ≈ 6.5 × 0.36 = 2.34 Ω. At 0.9 A, drop ≈ 2.1 V; at a 24.1 V source the operator would see about 22.0 V, above the 21 V pickup. The measured 19.2 V implies a loop resistance of about (24.1 − 19.2) ÷ 0.9 = 5.4 Ω, more than twice the expected wire resistance. So the wire is marginal but not the whole story: check splices, terminals, transformer regulation, and whether the 40 VA transformer is loaded near capacity. The 17.8 V during motor start shows the supply sags further. The owner’s “low voltage” statement is wrong: low voltage does not mean no voltage drop.
2. What Article 725 and power-supply principles apply to this Class 2 control circuit?
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The circuit is a Class 2 control circuit under Article 725. The 40 VA transformer is a listed Class 2 power supply, and the load must not exceed its rating. Voltage drop is a performance issue, not just an ampacity issue: the relay’s 21 V pickup is a minimum, and motor inrush causes a transient dip. Use the manufacturer’s data and measure under worst-case load. Do not solve the problem by installing a non-Class 2 supply or by mixing Class 2 wiring with power conductors; keep the Class 2 separation and listing intact.
3. Recommend a repair sequence and the measurements that should be recorded to prove the fix.
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Best fix: move the transformer to the operator location, or replace the 18 AWG run with 16 AWG (about 4.1 Ω per 1,000 ft; loop 1.48 Ω; drop 1.3 V at 0.9 A) and verify at least 21 V at pickup with the heater on. If 16 AWG is still marginal, use 14 AWG or a local 24 VAC supply. After repair, measure at the operator under idle, relay-pickup, and motor-start conditions, and record the values. Also check that the transformer is not backfed and that the relay coil is the correct voltage. The goal is not merely a higher idle reading; it is reliable pickup during the worst-case load.