Rooftop Inverter Tripping on Over-Voltage at Midday
A 10 kWp three-phase grid-tied rooftop solar plant on a school has its inverter mounted beside the AC distribution board, 60 m of 4 mm² single-core copper cable away from the net-metering point. Since April the inverter has shut down around 13:00 each day with an over-voltage fault and restarted 10–15 minutes later, losing an estimated 3 kWh per day. At the moment of trip the lineman measures 250 V phase-to-neutral at the energy meter and 258 V at the inverter terminals; grid frequency is 49.95 Hz. The inverter's grid-protection settings are factory-set and sealed. The school's electrician proposes three options: (i) open the inverter and raise the over-voltage trip from its present setting to 265 V; (ii) switch the plant off between 12:00 and 15:00; (iii) replace the AC cable with 16 mm² and ask the utility to check the distribution transformer tap.
1. Is the 258 V at the inverter a grid problem or a plant problem? Show the calculation.
Read the worked answer
Mostly a plant problem sitting on top of a grid that is already high.
- Full-power output current: I = 10,000 / (√3 × 415) ≈ 13.9 A per phase.
- 4 mm² copper has a resistance of about 4.6 Ω/km. The 60 m run is 60 m out and 60 m back, so the loop is 120 m: R ≈ 4.6 × 0.12 ≈ 0.55 Ω.
- Voltage rise along the cable = I × R ≈ 13.9 × 0.55 ≈ 7.7 V.
- Predicted inverter terminal voltage = 250 + 7.7 ≈ 257.7 V, which matches the measured 258 V almost exactly.
So the inverter is not faulty and the meter reading is not wrong. To export, the inverter must sit above the grid voltage, and the undersized cable is adding about 8 V on top of a grid voltage already close to the 253 V upper limit. Both ends need attention: the cable is what the school controls, the 250 V at the meter is the utility's side. The trip is correct protective behaviour, not a nuisance fault, and 49.95 Hz confirms the grid is otherwise normal.
2. Evaluate the three proposals.
Read the worked answer
- (i) Open the inverter and raise the over-voltage trip to 265 V. Refuse. The protection is sealed for a reason; changing it defeats a safety function, almost certainly voids the warranty and the net-metering approval, and would push the school's and the neighbour's voltage even higher. It also hides the real defect - the 8 V cable drop and the losses stay.
- (ii) Switch off between 12:00 and 15:00. This throws away the best generating hours - likely 12–15 kWh a day - for the life of the plant, to avoid replacing a cable. Acceptable only as a one-week holding measure while the cable is procured.
- (iii) Replace the AC cable with 16 mm² and ask the utility to check the transformer tap. Correct on both counts. It removes the local rise, and the tap check addresses the 250 V at the meter. Neither alone is sufficient: with 250 V at the meter, even a perfect cable leaves the inverter close to the limit.
Note that the fault only appears at midday because that is when current, and therefore I × R rise, is at maximum.
3. Calculate the result expected with 16 mm² cable, and state what else you would check and record.
Read the worked answer
Calculation. 16 mm² copper is about 1.15 Ω/km, loop 120 m → R ≈ 0.14 Ω. Voltage rise = 13.9 × 0.14 ≈ 1.9 V, so the inverter would sit at about 250 + 1.9 ≈ 251.9 V - under the 253 V limit with a small margin. 10 mm² would give about 2.9 V, i.e. 252.9 V and effectively no margin, which is why 16 mm² is the correct choice rather than 10 mm².
Also check and record:
- Confirm the inverter's grid-protection settings are standard and the seal is intact before and after the job - never leave a site with modified protection.
- Check the AC isolator, MCB rating, surge protection and the earthing of the inverter frame and module array; a loose neutral or a high-resistance earth can distort the measured voltage.
- Ask the utility to check the distribution transformer tap and the feeder voltage at peak generation. If 250 V at the meter is normal, the tap is the utility's corrective action.
- Record before/after: inverter terminal voltage, meter voltage, current per phase, trip log and daily generation for a week, to demonstrate that the ~3 kWh/day loss has been recovered.