The 80Ohm bypass is required to power the MCU and other electronics when the switches are turned off to both batteries (which is the starting condition). Note that when turned on, the switch will carry over 100A. For the 12V supply, I found the TPS7A4001 (up to 80V input). From this 12V supply, I can drive a 3.3V regulator (not shown this way in my schematic).
Note that since there are two such batteries connected in the same way, it needs to operate safely when the batteries are at different voltages. This was the reason for the optoisolator for the control signal in the drain-to-drain configuration of post#7. If the PCB is at 55V due to connection to the other battery not shown, and the battery shown is 60V and not switched in, then there would be a 5V delta across the 80ohm, and the control signal would not be a suitable voltage for the lower NFET.
crutschow, that is an interesting circuit, do you know if it has been tested in real life? I am trying to figure out how the gate charge current makes it's way back to the battery if the NFET is off (KCL). By leakage current? This could mean a slow turn-on time depending on the FET properties?
FvM, your circuit looks like the same idea. I suppose a large-value resistor could be added between the source-source point and the battery ground to provide a better gate charge current path.
I'm going to study this circuit further.