Good morning all I have a question about low-voltage cut off. I am building a emergency backup system for my home. I have a grid tied solar system already installed in my home. I have two sony boy 5000 watt inverters with a handy power outlet Available. I can pull 1500 W off each inverter in case of a grid power failure. This system works great on sunny days however I wanted to have a nighttime storage solution. Not knowing much I Purchased 4 100 amp hour 12 volt AGM batteries and a 4000 W Pure sine wave inverter. I have wired it into my homes transfer switch which I already had for my generator and it powers some lights and small appliances perfectly. I am charging the system with a 55 amp 3 stage charger Which I keep plugged into grid power. The batteries stay topped off and ready to use. In the event of a power failure I just simply plug the charger into these auxiliary AC power outlet off one of my inverters which remember are powered by my Pv on the roof. I was hoping to be able to power my refrigerator overnight but it seems I could only get 3 to 4 hours before my battery voltage drops too low. My refrigerator pulls 4 amps and I have some lights as well so I’m probably pulling about 600 W or so during my test. I did not let it drop below 12 V But I know upon recharging I was below 50% of my charge capacity. So this is my question how do I install a low voltage cut off? I purchased a 220 amp Victron battery protect however from what I read it seems I should not install that between the battery and the inverter. The inverter will shut off at 9.5 V which seems very low. Also how much capacity do I actually need to run 600 W for 6 to 8 hours