ive used Schottky Diodes before for dual inputs before but im wondering if the "switch" is immediate? Trying to provide power to a gps module until it gets a gps lock then power up rest of system without the gps module "rebooting"...

here's the circuit:

 

That looks like a good solution.

There will be a brief period of overlap of conduction as the input to one anode drops below that of the other, but that is a good thing. There should be no dropout of power to the GPS module. I have used that method several times, getting into a complicated switch-over circuit is not worth it to save a few hundred millivolts, at least in my applications.

You may want to raise the flight controller's 5V supply or lower the battery backup voltage by a hundred millivolts or two and keep the diodes in thermal contact with one-another to assure that no significant current is drawn from the battery. Alternatively use a Schottky for the flight controller 5V connection and silicon junction diode for the backup supply.

 

That looks like a good solution.

There will be a brief period of overlap of conduction as the input to one anode drops below that of the other, but that is a good thing. There should be no dropout of power to the GPS module. I have used that method several times, getting into a complicated switch-over circuit is not worth it to save a few hundred millivolts, at least in my applications.

You may want to raise the flight controller's 5V supply or lower the battery backup voltage by a hundred millivolts or two and keep the diodes in thermal contact with one-another to assure that no significant current is drawn from the battery. Alternatively use a Schottky for the flight controller 5V connection and silicon junction diode for the backup supply.

I know it's been a couple years since I originally posted this question, but I have just recently thought about finally doing it...

You mentioned that I should have 100mv+ difference between the two supply voltages in order to "keep the diodes in thermal contact with one-another." Can you elaborate/explain more on that? Do you mean that if both of the input voltages are exactly the same one or both of the diodes could heat up? Regardless though I will make sure to double check that the voltages are slightly different, but most likely they will be since I'm sure neither of them are going to be exactly 5.0

Thanks

 

If the ‘switching’ between supplies is not fast enough, increasing the capacitance of C1 should solve this.
As a rule of thumb, the capacitor voltage will decay according to the following equation:-

∆V = (I x t)/C

Where I= current in amps
t = time in seconds
C = capacitance in farads