LifePo4 Voltage at .5C - .6C Discharge

otrfun wrote:
Appreciate everybody's inputs. Always learning something new.

Called LifeBlue (lifebluebattery.com) on Mon and inquired about the voltage drop with their 100ah LifePo4 battery while under heavy load. Surprisingly, he offered to test one and get back with me. Received an email yesterday with the results. With a fully charged battery, battery voltage was 12.8v at 80a, and 12.7v at 100a. Pretty impressive. Based on the graph FWC posted, I'd guess-estimate battery voltage while would drop to approx. 12.0v at 15% SOC with a 100a load.

Using BB's v/SOC table, 14% SOC is 12.5v, and 9% is 12.0 resting. But you want the loaded voltages while the inverter is running that 100a load to figure this out.

Using BB's full at 13.6v (assumes your batt uses the same--been told here the LFPs all do ?) then initial drop was 13.6-12.7 = 0.9v

You have fat wires to inverter, so call total drop 1v for this exercise. that drop happens whatever the SOC is when you start the run.

Next thing is the drop while running due to the voltage going down as AH are removed from the batt. We have the BB v/SOC table for that too.

Say you run the 100a load for 10 minutes. that is 16.7AH down from whatever SOC you started at.

Staying above the knee so the voltage does not crash from say that 14% at 12.5v (loaded will be 11.5v then, so above the 11v inverter alarm), and 14% is 14AH remaining of your 100, then your 10 minute run has to start at 14 + 16.7 = 30.7 AH ( 13.0v resting and 12v loaded)

If you start the 100a 10 minute run below 30% SOC, your voltage will crash before the run is done. And don't let the furnace come on during the run!
--------------

Meanwhile, what do we know about the internal R of that battery now that affects that initial voltage drop? R = V/I

Note: somebody mentioned that the R of an LFP included the wiring etc with the BMS, so it could be higher than you would expect.

We have 0.9v from 100a so R = 9 mOhm a 100AH AGM is rated at 5 mOhms so 9 seems too high.

We have 0.8v for 80 amps so R = 10 mOms. So try it using the diff between them where 0.1v is 20 amps so R = 5 mOhms.

That is better, same as an AGM, so that means the starting voltage of 13.6 was too high for 100%.

If R is 5 mOhms and I is 100a, then V is 0.5v so 12.7 + 0.5 = 13.2v

That 13.2 looks awfully familiar! Meanwhile BB says 13.2v resting is 70% SOC.

Beats me! You LFP guys can figure it out if you need to :)