Measuring Switch State using PIC

The output high voltage from the emitter of Q1 will never exceed Vb-Vbe. If the maximum base drive voltage is 5V, then the emitter will never be able to rise above approximately 4.3 volts when Q1 is saturated.

If the power supply increases to 5.9 Volts, then the transistor is operating in a linear region and the difference of 5.9-4.3 is dissipated across the CE junction. As soon as the emitter tries to rise up, it clamps off the base and shuts down a little until it reaches equilibrium. If the trace you show is of the actual circuit and not of a simulation, then the ringing is most likely cause by some noise source in the circuit as 1Hz is to slow for it to be a transient response.

The upper end of R1 never floats. It is bound by the Vbe junction of Q1.

If you wish to use a high side driver, then perhaps consider a PNP device. If you want a low side driver, then use an NPN device. this will free R1 from Vbe.

If I got the requirement correctly then this can work.

^{simulate this circuit – Schematic created using CircuitLab}

The mosfet is an enhancement mode P type.

While the main battery is >13v LiPo saver outputs that voltage to the gate keeping Vgs close to 0 and the mosfet off.

When the main battery voltage drops below 13v LiPo saver turns off (output floating) and the gate is pulled down from resistor R2 (Vgs of about -15v), the mosfet turns on and the relay switches to the backup battery.

Note that normaly the max Vgs is about 20v so 15v is a high as you want to go, if you expect Vgs to go higher (if the battery can be more than 15v) then you should use a voltage limiter (like a zener).