"the charger will be looking for progress; ie. the "battery" (dummy voltage source) should be rising in voltage and current sunk tapers off as the charge cycle progresses."
Yes, this is what I need. I will study your diagram and see if it makes sense to me. I thought about using a power supply to simulate the battery voltage but I'm not sure how to do that so the battery charger and the power supply don't fight each other.
Thanks!
The power supply will need to be capable of outputting as much current as your highest DC load test current (same as your DUT?), otherwise, the PS voltage will collapse (constant/over current protection kicks in or fuse blows or self destructs). And if the PS voltage drops to zero, the DUT (battery charger) will stop. Therefore, a switching supply capable of 52V @ say, 20A should suffice?
A fully charged lead acid battery bank (6, 8V cells) is 8.43V*6 = 50.6V.
If the bank is depleted then its voltage is 7.72V*6 = 46.3V.
https://batteryuniversity.com/article/bu-903-how-to-measure-state-of-chargeSo, start with the DUT (charger) off, your PS on and set to 47.3V (we add a volt for the diode drop), and your DC load on and set to 20A. Then, turn on your DUT. Since the DUT now provides the higher voltage causing the diode to be reverse biased, the 20A consumed by the DC load now comes solely from the DUT. Slowly raise the PS voltage and lower the DC load current. PS voltage must not exceed 51.6V (full bank+a diode drop). I'm not sure how long "slowly" should be.