Simple push-pull BJT biasing

Explain me please, how biasing exactly works here?

From my view, on positive half of signal, we need about 0,7V + signal voltage (let's say signal is 1V) on NPN base, relative to its emitter=relative to Gnd here. That's simple and clear. I can't figure out, what happens on negative half of signal.

I think, here C1 becomes like Vcc for PNP transistor, so we need Vcc-0,7-1V on its base to drive it open. But how it is possible to get at same time with same biasing circuit 1,7V on upper base AND Vcc-1,7V on lower base?

• You're too focused on the "top voltage rail" of each BJT. You you think C1 is to the PNP as Vcc is to the NPN. But look again. Vcc is connected to the NPN collector but C1 is connected to the PNP emitter. Not the same terminal. In this circuit GND is serving the same purpose for the PNP as Vcc does for the NPN. Your initial examination also seems to ignore the fact the diodes are there. Dec 19, 2021 at 19:17
• This circuit is a NPN source follower and PNP source follower mashed together. Maybe it'd be best if you examine how an NPN source follower works. Then how a PNP source follower works (which is identical but upside down). Then examine how the entire circuit works without the diodes, then with the diodes. Dec 19, 2021 at 19:18
• – G36
Dec 19, 2021 at 19:28
• ZZ, The only good purpose of this schematic is educational. It is otherwise a toy. When your upper quadrant needs its maximum base current due to the signal, this moment is also when the upper quadrant resistor has the least voltage difference across it and is least able to supply the upper quadrant with needed base drive current. Same thing with the lower quadrant. It can be educational to understand how this circuit does, at least, work as a toy. Is that all you want to understand? (Because it's not practical. So you'd only learn how the toy works. Not how to do it "for real.")
– jonk
Dec 19, 2021 at 20:59