Ok, please help me to understand this.
I am looking at this diagram that you posted earlier:
http://oi47.tinypic.com/28qt1zq.jpg
I am assuming that "Aux Volts VP-X J2-12" is power monitoring on the VP-X, and not a power source.
Yes correct..
In this circuit, I believe there are two power sources, the aux battery, and the feed coming from the main buss through the resistor and diode.
The node is common with the battery and the charge circuit.
Both of these are feeding the backup bus, correct?
Whichever one presents the higher voltage will be the preferred source?
What you have to remember is that the loads will always prefer the main bus (when it is on and operating properly) over the backup bus since the backup bus will always be one diode drop below the main bus.
At some point the battery and the power coming through the diode will be roughly equal?
Only if the main bus voltage fails to stay in the normal range, if it drops below the aux bus voltage plus the diode drop, the load will transfer to the aux battery seamlessly.
For the sake of discussion, if for whatever reason the aux battery became disconnected, what is the highest current that can safely flow through the diode and resistor to feed the aux bus?
If the backup battery is disconnected and the main bus is on, the loads will still use the main bus since it is still going to be the higher voltage source. If the backup battery is disconnected and the main bus fails, no power will be available. As for the highest current, that goes back to ohms law. If for some reason the power does start to get sourced thru the main bus charge feed, the limiter is the power dissipation capability of 25 watts. I = sqrt(P/R) ... a little over 4 amps max
-Dj