I’m interested in what I can do to improve the likely lifespan of my two Pylontech US3000 batteries. Note that this doesn’t mean I want to necessarily apply the manufacturer’s specs. My aim is to get the longest possible life out of my batteries together with a practical amount of use. As per a previous post I made, I am concerned that the economics of a large bank doesn’t make sense (in my case, with limited back up requirements). Therefore I’d like to keep the two that I have now in service as long as possible, and therefore enabling my PV and backup system to function as intended.
Picking up a few things here and there (of which not all are necessarily correct), this is a list I put together, but would really appreciate some input on:
- Do not use the batteries when they are too hot (how hot is too hot? Above 30C?) or at freezing point (not an issue where I stay).
- Avoiding discharging of the batteries at more than 1C (75A per battery in my case, so 150A in total)
- Charge them up to 100% regularly (at least once a week) and keep them at that level for an hour or two to do some balancing.
- Do not discharge the batteries to less than 20% SoC (80% DoD).
At the moment I am leaving 1 and 2 to the batteries’ BMS to sort out with my Venus GX, so I assume that those are taken care of. For 3, I have put a scheduled charging slot, every day (stop on 20% SoC) between 16:30 and 18:30. My panels are west facing so I still get some sun during that time and the demand in the house isn’t too much, so typically my batteries gets full at around 15:30 and stay there until 18:30, trickling up every now and then with the excess PV, should the batteries demand it.
Another setting I made was to set my minimum SoC (unless the grid fails) at 50%. My reasoning being as follows: 50% DoD gives me 3.5kWh from the batteries. This is at full nominal capacity. When this becomes too little due to capacity degradation, I can move it down to 45% SoC and eventually all the way to 20% SoC. 80% DoD, with 60% nominal capacity remaining, I’d still get (1-0.2)*(0.6)*7 = 3.36kWh from the bank (what I was used to).
I’d really appreciate some input on the above.