Inverter Clipping and Service Panel Interconnection
A 9.6 kW DC residential array uses twenty-four 400 W modules and a 7.6 kW AC inverter at 240 V single-phase. The main service panel has a 200 A busbar and a 200 A main breaker. The installer proposes a 40 A PV breaker at the bottom of the busbar, opposite the main. The utility limits aggregate inverter output to 10 kW AC and requires feeder voltage rise no more than 2%. The service feeder is 180 ft of 1/0 aluminum with a resistance of 0.13 Ω per 1,000 ft. Annual specific yield is estimated at 1,500 kWh per kW DC before clipping, and the designer predicts 0.6% annual clipping loss. The customer wants the most energy possible and asks whether adding six more modules would help.
1. Calculate the DC/AC ratio and explain what clipping means for this system. Is a 0.6% annual clipping loss plausible?
Read the worked answer
DC/AC = 9,600 W / 7,600 W = 1.26. The inverter caps AC output at 7.6 kW, so on cool, clear days when DC array output exceeds 7.6 kW, the excess DC energy is lost, or clipped. A 1.26 ratio is a moderate oversize; annual clipping depends on climate and orientation. A 0.6% loss is plausible in a location with modest irradiance or if the array is not perfectly south-facing, but in a very sunny, cool climate the loss could be higher. The trade-off is more energy in low-light and cloudy conditions.
2. The main service panel has a 200 A busbar and a 200 A main breaker. May the 40 A PV breaker be installed under the 120% rule? Show the calculation and state the placement requirement.
Read the worked answer
The inverter continuous output is 7,600 W / 240 V = 31.7 A. The PV OCPD must be at least 31.7 A × 1.25 = 39.6 A, so a 40 A breaker is correct. Under the 2020 NEC load-side 120% rule, the sum of the main breaker and the PV breaker may not exceed 120% of the busbar rating: 200 A + 40 A = 240 A, and 1.20 × 200 A = 240 A. The value is exactly at the limit, so it is permitted. The PV breaker must be at the opposite end of the busbar from the main breaker, and the panel must be marked with the PV source. If the busbar were smaller or the main breaker larger, a supply-side connection or a different panel would be needed. Verify the locally adopted NEC edition, because later editions restructured these interconnection rules.
3. Calculate the approximate voltage rise on the 180 ft service feeder and compare it with the utility's 2% limit. Use R = 0.13 Ω per 1,000 ft for 1/0 aluminum and assume the inverter current is 31.7 A at 240 V.
Read the worked answer
One-way resistance = 0.13 Ω/1,000 ft × 180 ft = 0.0234 Ω. For a 240 V single-phase circuit, the round-trip voltage drop is approximately 2 × I × R = 2 × 31.7 A × 0.0234 Ω = 1.48 V. Percent rise = 1.48 V / 240 V × 100 = 0.62%. This is below the 2% limit. The estimate ignores power factor, conductor operating temperature, and service transformer impedance; if the actual feeder is longer or smaller, the result should be recalculated.