Thank you for helping me out!
np
1. Why do we need s NPN transistor AND a P-ch MOSFET? Can we simplify the schemantics by using a sole N-ch MOSFET? Can then we just connect active-high signal from GPIO pin directly to GATE pin of MOSFET (assuming that it's the logic level input, we don't need a resistor?). SOURCE pin of the MOSFET then would be +5VDC source, and load would be connected to DRAIN pin? Can this schemantics work, or I'm missing something? (sorry, as you can see I lack the basic knowledge of electrical engineering).
No, this will not work. The MOSFET controlled by voltage difference between source and gate, when Gate is above (more positive than) Source by threshold voltage, the transistor starts to turn on. N-Channel MOSFET will work fine in the schematic "A" (replace pins like this base-gate, emitter-source, collector-drain), but this will disconnect ground wire from the device. Not sure if this is acceptable.
P-Channel MOSFET controlled similary, just reverse all voltage polarities.
2. Speaking of MOSFETs, I'd probably like to use a dual channel one to control both modules and save some space. If I have a chance to select from different MOSFETs (i.e. can order any MOSFET made by FairChild Semi), what are the most important characterictics to look for? Obviously, I'd like to minimize current leakage in OFF state and voltage drop under load in ON state. Does it mean that the lower value of "Drain-to-Source ON resistance" is the better?
At first, you should look at drain current and select a device wich allows at least Imax *1.2. Most modern MOSFETs have negligible "Drain-to-Source ON resistance" but it wouldn't hurt to verify volatge drop: Vdrop = Imax * Rdson (0.1V = 0.5A * 0.2Ohm).
Please note, that Rdson depends on gate voltage when it near gate threshold voltage and this depedency always represented graphically in datasheets. As for leakage in the off state - this will be very small value if transistor is closed.
What do you think about using FDS6910? It's a dual N-channel MOSFET in SO-8 package. It has "Drain-to-Source ON resistance" of only 0.017 Ohm which if I understand it correctly would give it 2.5VA * 0.017 Ohm = 0.0425V voltage drop @ 500mA load?
A bit different: Vdrop = 0.5A * 0.017Ohm = 8.5e-3V (by Ohm's law).