12 V LED Power Supply With Excessive Voltage Drop and a Oversized Primary Breaker
A landscape contractor has installed a 300 W, 120 V input/12 V output LED power supply for a commercial patio. The supply is 10 ft from a 120 V panel and has a nameplate input current of 2.5 A; the nameplate says maximum branch-circuit protection 5 A. The 12 V secondary run is 10 AWG copper, 180 ft to the first fixture, then 12 AWG copper, 50 ft to the last fixture. The connected LED load is 250 W. At the last fixture, the measured voltage is 9.8 V instead of 12 V, and the 12 AWG section is warm. The installer wants to replace the 5 A primary fuse with a 15 A breaker because it keeps tripping when all fixtures start. You must evaluate conductor sizing, overcurrent protection, and voltage drop.
1. Calculate the primary and secondary currents and evaluate the proposed 15 A breaker.
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Primary current at full load is I = P/V = 300 W / 120 V = 2.5 A. The nameplate permits a maximum branch-circuit protective device of 5 A. Replacing the 5 A fuse with a 15 A breaker does not protect the power supply and exceeds the nameplate rating. The branch-circuit conductors may have enough ampacity for 15 A, but the equipment listing controls the maximum protection. A 15 A breaker is unacceptable.
On the secondary side, 250 W at 12 V is 20.8 A. At the measured 9.8 V, a constant-power load would draw even more current, approximately 25.5 A. The 12 AWG section is not adequate for that load and is warm for a reason. The tripping should be treated as a symptom of overload, high inrush, a secondary fault, or voltage-drop-induced current increase, not as a reason to enlarge the primary breaker. Verify the supply's inrush rating and use the listed protection. If nuisance tripping continues, check the secondary wiring and fixture connections; do not defeat the nameplate limit.
2. Is the 12 AWG secondary section adequate for the 250 W load? Show the calculation.
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The connected load current is 250 W / 12 V = 20.8 A. For a continuous load, size conductors at not less than 125%: 20.8 A x 1.25 = 26.0 A. A 12 AWG copper conductor has an ampacity of about 20 A at 60 degrees C and 25 A at 75 degrees C, depending on insulation and termination ratings. It does not have 26 A ampacity in the usual 60 C termination basis, so the 12 AWG section is undersized for the load even before voltage drop is considered.
At the measured 9.8 V, the current for a 250 W constant-power load would rise to about 25.5 A, making the 12 AWG section worse. The 10 AWG home run has more margin, but the 50 ft 12 AWG tail is the weak point. The correct fix is to reduce the load on that tail, enlarge the tail to 10 AWG or larger, or split the fixtures into additional circuits so no 12 AWG section carries the full 250 W.
3. What is the best fix for the voltage drop, and what voltage drop limit applies?
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The NEC does not generally mandate a 3% branch-circuit voltage drop as a mandatory rule, but the informational notes to 210.19 and 215.2 recommend no more than 3% branch drop and 5% total drop for reasonable efficiency. On a 12 V system, 3% is only 0.36 V, so 180 ft at 20 A is difficult with any practical conductor. The measured 9.8 V is an 18% drop, far beyond a good design.
A rough check: 10 AWG copper is about 1.24 ohms per 1,000 ft. For 180 ft, the round-trip length is 360 ft, so R is about 0.446 ohms. At 20.8 A, the drop is about 9.3 V, which matches the measurement. Upsizing to 8 AWG reduces the drop to roughly 5.8 V, still almost 50%, so conductor upsizing alone is not a good fix. Relocate the 12 V power supply closer to the load, use multiple home runs, split the fixtures, or use a higher-output-voltage system with a listed converter at the load. Fix the voltage drop and the undersized 12 AWG section together; do not just replace the primary fuse.