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We installed a solar panel array at our house. The total output of the solar panels is around 465 V DC with open circuit. the DC Input limit of the solar inverter is 450 V DC.

I want to lower the incoming voltage from the solar panels to the solar inverter by around 10 V so the inverter can actually work without displaying an error message the whole time.

I need as much wattage as I can, so removing 1 solar panel is not an option. The current that would be passing through is around 19-23 A max.

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  • \$\begingroup\$ " the DC Input limit of the solar inverter is 450V DC." - What make and model is your inverter? \$\endgroup\$ Sep 20, 2021 at 4:34
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    \$\begingroup\$ Sounds like a bad design to begin with. Chose a more suitable inverter. You’ve got to ask yourself if working around a bad design is faster/better/cheaper than just getting the right equipment? \$\endgroup\$
    – Kartman
    Sep 20, 2021 at 10:31
  • \$\begingroup\$ A really beefy 10 V Zener diode in series? Preferably with an array of 12 V light bulbs in parallel. \$\endgroup\$
    – winny
    Sep 20, 2021 at 11:05
  • \$\begingroup\$ Or remove one panel. Keep it as a spare or use it (via suitable charge controller) for charging car/RV/boat batteries. \$\endgroup\$
    – user16324
    Sep 20, 2021 at 12:39

3 Answers 3

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Do two strings in series of half the panels connected to the inverter in parallel.

This will reduce the voltage to about 230V.

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  • \$\begingroup\$ This piqued my curiosity and I spent an hour or so studying whether this approach was acceptable or even safe. I found nothing to fault it, it's apparently a common solution, although it's mostly individual PV cells connected in parallel, not strings of series connected cells in parallel with each other. @SolarMike, please elaborate, or provide some links to support the claim, because this answer deserves to be less hand-wavy. \$\endgroup\$ Sep 20, 2021 at 13:42
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If Voc is 465v then Vmp will be no more than 420V and probably under 400V. What you need is something to load the panels down to say 440 V or about 25V drop. Even if IMAX of 23A was dissipated you'd have 25 X 23 or about 600 Watts dissipation. In practice the current required to load the panels down about 25V from Voc will be usefully less than that.

A simple solution that does not waste energy in normal operation is a resistive load that is switched in when the voltage is too high and switched out when the inverter operates and voltage drops. This could be eg a comparator driving a MOSFET and a resistor of about R = V/I = 450/15 = 30 ohms.

The high voltage and current may make an IGBT better than a FET, and if this only operate when the inverter is off load then a suitably rated contactor may be better.

If this approach is attractive more information can be provided.

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Assuming that the present configuration of solar cells cannot be changed, there are two options to reduce the voltage.

  1. NOT RECOMMENDED. Dropping the voltage using resistors/diodes 10V*20A = 200W. Note that the drop-out will change depending on the total current (dependent on lighting conditions)
  2. Using a buck converter to step-down the voltage, very costly affair but provides better efficiency. If you are brave, you may try building a buck converter yourself with this power rating

As @Solar Mike suggested, changing the array configuration is the most efficient way for your application

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