On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.

Andy

KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.

Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.

- Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.