Twelve-Volt Floodlight Feeder: When a Bigger Power Supply Is Not the Fix
A workshop extension has six 12 V, 60 W LED floodlights fed from one 12 V, 100 W switch-mode power supply. The feed is a single 2.5 mm² copper two-core cable, 18 m from the power supply to the first and only connection point, where all six floodlights are connected in parallel. On commissioning, the lamps at the far end are noticeably dim and the power supply case reaches 65 °C within ten minutes. With all lamps on, the voltage at the power supply terminals is 11.6 V and at the floodlight terminals 9.1 V. The supply is fed from a 230 V, 6 A circuit. The installer suggests replacing the 100 W unit with a 400 W unit and keeping the existing cable. Copper resistivity is 0.0172 Ω·mm²/m. The floodlights are rated for 12 V ±5%. There is no local driver at the luminaires.
1. Calculate the total load current at 12 V and the volt drop in the 18 m, 2.5 mm² feed. Does the 100 W SMPS have adequate rating? Show your working.
Read the worked answer
Load: P = 6 × 60 = 360 W. At 12 V, I = P/V = 360/12 = 30 A. The SMPS is rated 100 W, so its maximum is 100/12 = 8.33 A. It is therefore undersized by a factor of 3.6.
Cable resistance: R = ρ × 2L/A = 0.0172 × 36/2.5 = 0.248 Ω. Volt drop at 30 A = 30 × 0.248 = 7.43 V, which is 62% of 12 V. The permitted drop for a 12 V luminaire is about 5%, i.e. 0.6 V. The cable is also carrying 30 A, well above the typical 21–24 A clipped-direct rating of 2.5 mm², so this is both a volt-drop and an overload problem.
2. The measured voltage at the SMPS is 11.6 V and at the last luminaire 9.1 V. Reconcile this with your calculation and explain why replacing the SMPS with a 400 W unit alone would not be a compliant fix.
Read the worked answer
The measured drop is 11.6 − 9.1 = 2.5 V. If the cable resistance is 0.248 Ω, the current actually flowing is about 2.5/0.248 = 10.1 A, not 30 A. The SMPS is in current limit or foldback; its output has collapsed, so the lamps see 9.1 V and draw much less than their rated power.
A 400 W SMPS would remove the source limit but not the cable limit. At 30 A the cable would drop 7.4 V, dissipate I²R = 30² × 0.248 = 223 W and overheat; the lamps would receive about 4.6 V. The 2.5 mm² cable is not rated for 30 A. The fix must reduce current or raise the distribution voltage, not just enlarge the power supply.
3. Propose a practical remedial scheme and justify the conductor size or voltage choice with at least one calculation.
Read the worked answer
Best practical fix: run a 230 V feed to a local IP-rated enclosure and use either one 12 V 400 W PSU with very short 4 mm² output tails, or individual 12 V drivers at each luminaire. At 230 V the total current is 360/230 = 1.56 A; a 1.5 mm² feed over 18 m has R = 0.0172 × 36/1.5 = 0.413 Ω and drops only 1.56 × 0.413 = 0.64 V (0.28%), protected by a 6 A MCB.
If 24 V luminaires are used instead, 360 W gives 15 A; 10 mm² cable has R = 0.0172 × 36/10 = 0.0619 Ω and drops 0.93 V (3.9%), which is acceptable, while 6 mm² would drop 1.55 V (6.5%), too much. Maintain polarity, glanding, IP rating and a local isolator.