String Voltage and DC Cable Check for a 5.4 kWp Rooftop Array
A 5.4 kWp rooftop grid-tied system uses ten 540 Wp modules in one string on MPPT input 1. Module data: Voc 49.6 V, Vmp 41.8 V, Isc 13.9 A, Impp 12.95 A, Voc coefficient -0.275%/°C and Vmp coefficient -0.35%/°C. The inverter is rated 5 kW with a 1000 V maximum DC input, a 150-850 V MPPT window and a 15 A maximum input current per MPPT. The site record low ambient is -5 °C and the maximum expected cell temperature is 65 °C. The array-to-inverter cable is 35 m one way in 4 mm² copper, routed in a hot ceiling void; copper resistivity is taken as 0.0175 Ω·mm²/m. The installer proposes using 2.5 mm² cable instead to save cost, arguing that the string current is only about 13 A and that the inverter already contains a DC isolator, so no array-side isolator is needed. The customer wants a written check before the cable is pulled.
1. Check the string voltage at the record low ambient and at the maximum cell temperature. State whether the string stays inside the inverter limits and show your working.
Read the worked answer
Cold-condition Voc
Temperature difference from STC: -5 - 25 = -30 K.
Voc at -5 °C per module = 49.6 × [1 + (-0.00275 × -30)] = 49.6 × 1.0825 = 53.7 V.
String Voc = 10 × 53.7 = 537 V.
This is below the 1000 V maximum DC input, with 463 V margin. The module maximum system voltage must also be at least 1000 V; the calculated 537 V is acceptable.
Hot-condition Vmp
Temperature difference: 65 - 25 = 40 K.
Vmp at 65 °C per module = 41.8 × [1 + (-0.0035 × 40)] = 41.8 × 0.86 = 35.95 V.
String Vmp = 10 × 35.95 = 359.5 V.
This is inside the 150-850 V MPPT window. At high temperature the voltage falls, but 359.5 V still has about 210 V margin above the MPPT minimum. The string current is 12.95 A at Imp and 13.9 A at Isc, both below the 15 A per-MPPT limit. The design passes the voltage and current check.
2. Calculate the DC voltage drop and power loss in the 4 mm² cable at Impp. Decide whether the cable is acceptable, and compare it with the proposed 2.5 mm² cable.
Read the worked answer
4 mm² cable
Loop length = 2 × 35 = 70 m.
R = ρL/A = 0.0175 × 70 / 4 = 0.306 Ω.
Voltage drop = Impp × R = 12.95 × 0.306 = 3.96 V.
Percentage drop = 3.96 / 418 × 100 = 0.95%.
Power loss = I²R = 12.95² × 0.306 = 51 W, also about 0.95% of the 5.4 kWp array rating.
This is acceptable for a typical 1.5% DC voltage-drop limit, but the cable is in a hot ceiling void, so check the conductor operating temperature and derated ampacity. A 4 mm² Cu cable has ample ampacity for 13.9 A, but the thermal rating of the insulation and grouping must still be verified.
2.5 mm² cable
R = 0.0175 × 70 / 2.5 = 0.49 Ω.
ΔV = 12.95 × 0.49 = 6.35 V, or 1.52% of 418 V.
Power loss = 12.95² × 0.49 = 82 W.
The installer’s argument is based only on current-carrying capacity. At 1.52% the 2.5 mm² cable just exceeds a 1.5% design limit, and the hot-route derating makes it a poorer choice. Use 4 mm² as the minimum, or 6 mm² if the route is longer, bundled, or in a very hot void.
3. The installer says the inverter’s built-in DC isolator removes the need for an array-side isolator. Evaluate this and state the practical requirement.
Read the worked answer
The inverter isolator does not replace an array-side DC isolator. The inverter isolator is at the inverter end of the DC cable; opening it leaves the cable and array side live. Anyone working on the cable, combiner, or array cannot safely isolate the DC source unless there is a lockable DC switch near the array.
A practical installation should provide a lockable DC isolator at or near the array, rated for the string Voc (at least 537 V cold, with the actual switch rating above that), the string current, and DC utilisation category. It must be labelled as a DC switch and be accessible to service staff and firefighters. The inverter isolator is still useful for isolating the inverter itself.
Also verify that the DC cables are double-insulated, UV-rated or in conduit, supported, and clearly labelled. The isolator does not replace DC SPDs, insulation monitoring, or correct polarity checks. The array-side isolator is a maintenance and fire-safety requirement, not a convenience.