String Current Mismatch on a Partially Shaded Array
A 6.4 kW residential array uses sixteen 400 W modules wired as two strings of eight into one MPPT. Module data: Isc 10.9 A, Imp 10.2 A, Vmp 34.2 V, Voc 41.0 V, and Voc temperature coefficient −0.29%/°C. On a clear afternoon, back-of-module temperature is 55°C and plane-of-array irradiance is 900 W/m². A clamp meter reads 9.3 A on string A and 7.8 A on string B. String open-circuit voltages are 322 V and 318 V. The homeowner says a new chimney shades one module on string B from about 1 p.m. until 4 p.m. There are no optimizers or microinverters. The 5.0 kW inverter is producing 4.0 kW AC and has no error code. The array faces 180° at a 25° tilt.
1. Calculate the expected string Imp at 900 W/m² and compare it with the two readings. What does the comparison indicate about string B?
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Expected Imp ≈ 10.2 A × (900/1000) = 9.18 A. String A at 9.3 A is essentially on target (about 1% high, within meter and irradiance uncertainty). String B at 7.8 A is 1.38 A low, about 15% below expected and 16% below string A. Because the strings are in parallel on one MPPT, they must operate at the same voltage, so string B is limiting the array current. A full-module bypass would produce a much larger Voc drop, so this is more consistent with partial shading or a high-resistance connection than with a fully bypassed module.
2. Why does the bypass diode not simply restore string B to 9.3 A, and what would you expect the string Voc to be if one bypass diode group were conducting?
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Bypass diodes protect shaded cells by providing an alternate path around a group of cells once reverse bias reaches about 0.5–0.7 V. They do not restore the shaded module's power; they only prevent overheating and allow the rest of the string to operate. If one bypass diode group, typically one-third of a 60- or 72-cell module, were conducting, module voltage would drop by roughly one-third of Voc-about 13–14 V for a 41.0 V module-so string B Voc would be roughly 322 − 14 = 308 V, not 318 V. The measured 4 V difference is too small for a fully conducting group, supporting partial shading rather than a fully bypassed group.
3. Estimate the DC power lost because string B is at 7.8 A instead of 9.3 A. State your assumptions and one next diagnostic step.
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Use the string operating voltage near Vmp. At 55°C, Vmp is lower than at STC; a rough correction with a −0.35%/°C Vmp coefficient gives 34.2 V × 8 × [1 − 0.0035(55−25)] ≈ 245 V per string. The current deficit is 9.3 − 7.8 = 1.5 A, so the approximate lost DC power is 1.5 A × 245 V ≈ 370 W. This is an estimate because the inverter may not be at the array's true MPPT, and the Vmp coefficient is assumed. Next step: measure each module's Voc and use an IV curve tracer, or isolate the string and use a clamp meter, to locate the shaded or high-resistance module or connector.