Electronic – Circuit Analysis with opposing batteries

Actually, it will draw 1/10 of the current at 1/10 voltage, producing 1/100 (1%) of the original power and heat!

This is because of Ohm's law: voltage = current * resistance (E=IR). Using algebra, I=E/R. Because the voltage, E is 1/10 while resistance, R is constant, I is 1/10.

Power = voltage * current. Because both voltage and current are 1/10 of the original value, power is 1/100 of the original value.

Your heater will produce 1% of the heat at 12V. Sorry.

There are chips that will do this. Linear Technology make at least one. Anyway, the problems with your circuit: -

• The diodes will drop maybe 0.4 volts to 1 volt (diode dependant and load dependant) so you are making the problem even worse using this technique.
• Even if you could find zero volt drop diodes (hint: as rare as unobtanium) the MOSFETs aren't really bringing anything to the party because the diodes would "sort out" which power supply is the "winner" based on highest voltage.
• If the mosfets were a good idea (which they don't appear to be) the gate voltage has to rise above a the source voltage by a few volts to adequately turn it on and, given the source is at 4.1 volts, it's unlikely but that's what the hot swap controller chips do - they have an internal oscillator and voltage shifter that produces an internal supply that is significantly higher than the incoming battery voltage.

So, my advice is look for hot-swap controller chips. Here's the sort of thing I mean: -

And here's another: -

There will be logic pins on these controllers that allow the use of an external voltage comparator so that when one power supply drops to a certain limit, the other power supply can take over.