80 m DC Run on 4 mm² Cable and a 96 °C MC4 Connector
A 10 kWp rooftop system uses two strings of nine 550 Wp modules (Voc 49.6 V, Vmp 41.8 V, Isc 14.0 A, Imp 13.2 A) feeding an 8 kW single-phase string inverter with two MPPTs. The array is on the roof; the inverter and AC distribution board are in a ground-floor room, and the one-way cable route is 80 m. The installer pulled 4 mm² copper DC cable (resistivity 0.0172 Ω·mm²/m) for the combined array-to-inverter run and 4 mm² copper AC cable from the inverter to the service entrance. Six months later the customer complains that his yield is about 12% below a neighbour's identical array. Under steady full sun the clamp meter reads 13.0 A on string 1 and 11.4 A on string 2, the inverter reports a DC input of 296 V, and a thermal camera shows one MC4 connector at the end of string 2 at 96 °C while all the others sit near 42 °C. No error codes are logged.
1. Check the DC cable sizing by calculation and say whether the inverter's 296 V reading supports your conclusion.
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DC drop on 4 mm²
- Loop length = 2 × 80 = 160 m; combined current = 13.0 + 11.4 ≈ 26.4 A.
- R = ρL/A = 0.0172 × 160 / 4 = 0.688 Ω.
- Drop = 26.4 × 0.688 = 18.2 V.
- Array Vmp at operating cell temperature: 9 × 41.8 = 376 V at 25 °C; Vmp falls roughly 0.4 %/°C, so at ~60 °C it is about 376 × 0.86 ≈ 323 V.
- 18.2 / 323 = 5.6 % - well past the usual 2 % DC design target, and about 480 W of heat lost in the cable at full sun.
Required size for 2 %: A = ρLI / (0.02 × V) = (0.0172 × 160 × 26.4) / (0.02 × 323) ≈ 11.2 mm², so specify 16 mm² (10 mm² gives 2.3 %, acceptable only if the route is shorter than stated).
Against the 296 V reading: expected terminal voltage ≈ 323 − 18 = 305 V; the extra ~9 V on top of the cable loss points at additional resistance in the hot connector and joints. So the 296 V is consistent with - and slightly worse than - the calculation, which strengthens the case rather than weakening it. Also check whether 4 mm² is even adequate for 26.4 A in a hot conduit; flexible multi-strand cable in a closed conduit at 45 °C ambient derates sharply.
2. Check the AC cable and the choice of a single-phase 8 kW inverter for this site, and recommend the corrective design.
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AC side on 4 mm²
- 8 kW at 230 V = 34.8 A. 4 mm² in conduit is typically rated around 30–32 A, so it is already thermally marginal.
- Loop 160 m: R = 0.688 Ω → drop at full output = 34.8 × 0.688 = 23.9 V ≈ 10.4 %. That is not a cable, it is a heater, and it will also push the terminal voltage toward the inverter's over-voltage trip.
- 10 mm²: 0.275 Ω → 9.6 V (4.2 %). 16 mm²: 0.172 Ω → 6.0 V (2.6 %). 25 mm²: 0.110 Ω → 3.8 V (1.7 %).
Design verdict: an 8 kW single-phase export over 80 m is the wrong architecture. Convert to a three-phase inverter: 8 kW / 3 ≈ 11.6 A per phase, so a 4-core 6 mm² cable gives roughly 11.6 × (0.0172 × 160/6) = 5.3 V per phase (~2.3 %) - a smaller, cheaper cable carrying a fraction of the current per core. Many DISCOMs also refuse single-phase export above roughly 7 kW, so confirm the connection type before ordering.
If the existing inverter must stay, minimum remedy is 16 mm² AC cable, re-terminated with correct lugs and torque, plus a check that the inverter's grid-voltage window is set correctly rather than widened to hide the drop.
3. Diagnose the 96 °C connector and describe the repair, including the safety steps before you touch it.
Read the worked answer
Diagnosis. A 54 °C rise over its neighbours at 11.4 A means high contact resistance, not normal heating. Typical causes, in order of likelihood: a bad crimp (wrong tool or wrong die), mismatched male/female connectors from different brands (the contact systems are not interchangeable even though they mate), water ingress past a wrongly seated gasket, no strain relief so the cable hangs on the contact, or an undersized connector for 4 mm².
Why string 2 reads 11.4 A against 13.0 A: the resistance is in series with the string, and the MPPT pulls the operating point down; the string is also the one carrying the hottest joint, so part of the loss is real power dissipated in the connector. The 12 % yield deficit is the DC cable loss plus this joint plus the AC loss, not a module fault.
Repair and safety
- Shut down the inverter (AC first, then the DC isolator), lock and tag, and verify dead with a meter rated for the DC voltage present. A PV string is live whenever there is light - there is no 'off' at the module end.
- Cover or otherwise isolate the string if you must work on the array side; never unmate an MC4 under load, since sustained DC arcing does not self-extinguish.
- Cut back the damaged connector, re-crimp a matching branded pair with the correct die, check the gasket and sealing ring, and provide strain relief.
- Re-energise, run at full sun and re-scan with the thermal camera; a good joint sits within a few degrees of its neighbours. Log before/after readings and photographs. Also confirm the combiner fuses and DC isolator are rated for the string Isc × 1.25.