# Dropping solar panels' input voltage to a solar inverter by approximately 10 V

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.

• " the DC Input limit of the solar inverter is 450V DC." - What make and model is your inverter? Sep 20, 2021 at 4:34
• 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? Sep 20, 2021 at 10:31
• A really beefy 10 V Zener diode in series? Preferably with an array of 12 V light bulbs in parallel. Sep 20, 2021 at 11:05
• Or remove one panel. Keep it as a spare or use it (via suitable charge controller) for charging car/RV/boat batteries.
– user16324
Sep 20, 2021 at 12:39

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

This will reduce the voltage to about 230V.

• 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. Sep 20, 2021 at 13:42

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.