If I am not wrong, the biasing voltage to turn on an NPN transistor is emitter voltage + 0.7 volts.
Similarly, the PNP biasing voltage is emitter voltage - 0.7 volts.
Can someone explain how the two PNP transistors at the XLR input are biased? A voltage divider is not used nor is there a base resistor to ground.
The base resistors don't have to go to ground. They just have to go to a node that's at a sufficiently lower voltage than the base to turn it on. In this case, it's the collectors.
This circuit is easier to understand if you draw the phantom power circuit that's inside the receiving preamp and lay it out in a more conventional manner. The two PNP transistors are in emitter-follower (common collector) configuration.
simulate this circuit – Schematic created using CircuitLab
I have omitted a lot. The nodes labeled In+ and In- connect to the phase inverter (Q5).
The important thing is that the zener diode holds one end of the 22k resistor at 9.1V, which is still well below the supply rail. The big capacitor across the 22k resistor bypasses AC so that the collectors are at ground for AC (thus, common collector).
(Q6 works to supply a regulated ~8.4V supply to the rest of the circuit.)
