This whole discussion is a living example of the old saw:Â* Better is the
enemy of good enough.Â* I flew for years with two batteries, two fuses,
and two switches.Â* When battery 1 gets low, flip on battery 2 then flip
off battery 1.Â* Never had an issue.

On 4/15/2020 10:12 AM, NM wrote:
On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
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.
I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).

Tom
Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.

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Dan, 5J