Diagnosing a 12 V solar study-room lighting fault
A rural school runs a 12 V off-grid lighting system in its after-school study room: three 12 V LED luminaires rated 8 W each, a 150 W panel, a 10 A PWM charge controller mounted beside the battery, and a 12 V 100 Ah lead-acid battery. The panel is roof-mounted 5 m from the controller and is connected with 1.5 mm² copper that has two twisted-and-taped splices in the roof space; the run from controller to luminaires is 18 m one-way in 2.5 mm² copper. Peak sun hours average 4.5 and the system derating factor is 0.7. The controller disconnects the load at 11.5 V. Teachers report that the lamps go out around 03:00 even after a cloudless day, and that during charging the display often climbs to 14.4 V and drops to float within an hour. Measured at 03:00 with a multimeter: 12.1 V at the battery terminals, 11.4 V at the controller's battery terminals.
1. Calculate the expected voltage drop in the battery-to-controller cable at load current, and show what the measured 0.7 V implies.
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Cable resistance. R = ρL ÷ A, with ρ = 0.0175 Ω·mm²/m for copper. The circuit length is 2 × 5 = 10 m, so R = 0.0175 × 10 ÷ 1.5 = 0.117 Ω.
Load current. 3 × 8 W = 24 W; at 12 V that is 2 A.
Expected drop. 2 A × 0.117 Ω = 0.23 V, about 1.9% of 12 V - acceptable on its own.
What the measurement shows. The measured drop is 12.1 − 11.4 = 0.7 V. The cable explains only 0.23 V. The remaining 0.47 V at 2 A corresponds to 0.47 ÷ 2 = 0.235 Ω of extra resistance, which is not in the copper - it is in the two twisted-and-taped splices, roughly 0.12 Ω each. Twisted joints in a warm, damp roof space oxidise and become resistors.
Implication. The controller is not measuring the battery. It is measuring the battery plus a variable resistor whose value depends on current direction and magnitude, so every decision the controller makes - when to stop charging, when to disconnect the load - is based on a false reading.
2. Use the resistance figure to explain both symptoms, and recommend a remedy with a supporting calculation.
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Charging side. A 150 W panel into a 12 V battery charges at roughly 8–9 A. At 9 A the 0.35 Ω total path drops 9 × 0.35 = 3.15 V. The controller sees the battery at 14.4 V when it is really around 11.3 V, concludes the battery is full, and drops to float after an hour. The battery is never fully recharged and gradually sulphates, losing real capacity.
Discharge side. At 2 A the same path drops 0.7 V. The controller sees 11.5 V and disconnects while the battery is still at 12.1 V - roughly 50% state of charge, so the early switch-off is a measurement error, not a flat battery.
Check generation is not the fault. 150 W × 4.5 h × 0.7 = 472 Wh per day against a demand of 120 Wh, so the array is nearly four times the load and cannot be the cause.
Remedy. Replace the run with one continuous 6 mm² cable, crimped and heat-shrunk at both ends, or better, move the controller to within 1 m of the battery and leave the panel on the long run. New resistance: 0.0175 × 10 ÷ 6 = 0.029 Ω, giving 0.26 V at 9 A and 0.06 V at 2 A. Re-check the set points afterwards; the low-voltage disconnect can then be left at 11.5 V.
3. Check the battery against three consecutive cloudy days and state the real risk to the system.
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Demand. 3 lamps × 8 W × 5 h = 120 Wh per day. Three cloudy days = 360 Wh.
Usable storage. 100 Ah × 12 V = 1,200 Wh nominal. At a 50% depth of discharge that is 600 Wh usable - 360 Wh fits with about 40% margin, so on paper the battery is adequately sized.
The real risk. That margin only exists if the battery actually reaches full charge. Because of the sense-lead fault the controller terminates charging early, so the battery lives in a partial state of charge, sulphates and steadily loses capacity. The nameplate 100 Ah is then fiction: within a season the usable capacity can fall below the three-day requirement, and the early 03:00 cut-off will look like a battery that has 'worn out' when the cause is 0.24 Ω of taped joint in the roof.
What I would do. Fix the wiring first, then verify with a hydrometer or a rested open-circuit voltage after a full day of sun, and only then decide whether capacity is short. I would also check whether the room is used at weekends and whether the 50% depth-of-discharge rule is realistic for a lead-acid battery that is rarely equalised - if not, add a second battery rather than a larger panel.