Problems with Battery Tester Circuit

[ChrisP58]

Ron, how about taking your constant current circuit from post #33 and changing V3 to a pot. But drive the top of the pot from a 2nd opamp that is configured as a comparator looking at the battery voltage.

That way the current will go to zero when the battery voltage falls below the setpoint.

 

[ronsimpson]

Vbattery starts out at 14.5V and discharges down to 0 then back up. (test condition)
Current IR1 (green) holds at 570mA until the battery voltage drops to 9V. Current=0. Battery voltage must get up to 10V before current starts again.
The (9V off/10V on) is important because the battery voltage will jump up when the current is removed. You might need more than 1V.

U1 tests the battery voltage. R10 & D1 sets hysteresis.
R9 is a pot and sets the battery turn off voltage.
Point "A" is 12V if battery is charged and is 0V if battery is low.
R6 is a pot to set current. 0-2V This is the voltage across R1. Current since resistor.
R11 needed to be added to keep U2(+) slightly above ground.

Remember R-R input and R-R output opamp. Micro Chip has some.

 

[ADWSystems]

Wow. That's a lot of work, Ron. But this is supposed to connect to a data logging and control system to run the tests. I'm not sure how to swap R6 out for something computer controlled and don't see how to change the cutoff voltage to coincide with the number of cells in use.

Additionally I still lost on why we are exploring all of these other circuits. What is wrong with the circuit built in the OP but change to use a FET with a Vgs of 30V, 35V, or 40V?