Electrical – Selecting low power mosfet for battery monitor

But let's make it more complicated by assuming I can't guarantee that the 200 volts across the cap is +200 and 0. It might be +100 and -100 (or even +10 and -190), whereas my battery might be +12 and 0 relatively speaking.

You can pick any node in your circuit you want and declare it to be "ground" for that circuit. So you can pick the more negative terminal of your capacitor and call that ground. Then you know the positive terminal is +200 V (relative to the ground that you defined).

If I simply connect the 12 volt battery right into the circuit, will its low potential automatically match up (Float?) to the low potential on the high voltage circuit, thus guaranteeing a 12 volt difference to between the Gate and Source of the mosfet?

Yes. If you connect a net of an isolated circuit into your circuit, you then make the potential of the two connected wires equal. In this case, you make the voltage of the negative terminal of the 12 V battery 0 V by definition.

And for a more advanced system: What if the 12 volts isn't even supplied by a battery? What if it's just another capacitor bank?

What really matters is Kirchoff's Voltage Law. If you add up all the voltages of the branches around a loop, they must add up to zero. The law doesn't care which node is labelled as ground.

And for a really complicated system: ...

Without a schematic of what you connected, nobody can tell you why that situation caused a problem.

First of all, thank you for specifying exactly which MOSFET you are using. It would also be very helpful to have a diagram showing exactly how you connected the MOSFET. Full schematic of the system from end-to-end would be best. I know you think you have described the circuit clearly, but the language you used is ambiguous, and I am forced to re-read several times, and then make assumptions.

I also think you might as well just explain why you would use one battery to charge another of the exact same type. Of course I can think of some possible reasons why you would do this, but since it is such an odd thing to do, I think you should explain it. If the setup you describe above is not your end goal system (that is, if you are systematically working your way up to develop some other type of system, you should explain what that is, too). For example, if you eventually want to replace the first battery and boost converter with a solar panel, then the whole thing is at least somewhat logical. But don't make us guess. Otherwise you will get all kinds of suggestions for more efficient ways to do what you are currently doing (maybe you could get rid of the boost and charge control and just make a simple current limited boost-mode DC-DC converter with no voltage output control or charge control, since it is impossible to over-charge the battery, assuming the discharged battery is at 20 or 30% capacity when you start).

I think there are three likely explanations for why things are not working as you expect. The first possibility is that you have wired the MOSFET incorrectly. A schematic and/or a picture of the test setup might help determine if this is the case.

Another likely explanation is that the MOSFET on resistance is large enough to foil the whole thing. The MOSFET you chose has Rds(on) specified as 250 mOhm typical at Vgs = -4.5V. But you only have, at best, 4.2V available to turn it on (if I understand correctly). So it might even be a bit higher. That may be a bit too high for the boost converter to work correctly.

Another possible explanation is that the battery being charged is nearly full. So the low current you see is just a result of low demand from the charger.

I can't think of any other explanations at the moment. The basic plan of putting a power MOSFET between the first battery and the boost converter is perfectly sound. The input bypass capacitor for the regulator should be between the MOSFET and the regulator.

Good luck! If you figure it out, be sure to come back and post an update.