Switch circuit analysis.

This is a key switch circuit, and the two opposing transistors are interlocked, not differential amplification. The realized function is to press switch S, OUT has output, and press again, OUT does not output.

Assuming that the output is dead:

Q2 is turned on due to B+ and R3, C16, and R7, and the C output of Q2 is low, then Q1 is turned off;

At the same time, there is voltage at both ends of C5 (due to the role of R12, R5, and through Q2);

When S is pressed, C5 and C6 will be discharged, then Q2 will be disconnected, and C of Q2 will output high level, which means that Q1 will be turned on;

Q3 will also be turned on, and the voltage across C16 will be further pulled down to make it discharge and accelerate the turn-off of Q2;

At the same time, C6 starts to charge and keep Q3 on.

In addition, as you said, there is indeed no lock function.

Personal opinion, I don’t know if the analysis is right.

The cleverness of this one-key switching power supply circuit is that it obtains a simple way to manipulate the negative pulse of the bistable circuit state transition.

Two transistors Q2 and Q3 form a bistable circuit. Capacitor C16 brings a slight asymmetry to the circuit.

Before power on, the initial voltages of capacitors C5, C6, and C16 are all zero.

When the power is turned on for the first time, because the initial voltage of the capacitor C16 is zero, Q2 is forced to be turned off through R7, and Q3 is turned on, so that C16 is discharged without charging, and Q2 is kept off. The grid of the power switch tube Q1 is turned on by obtaining a high potential through R4 and R9, connecting the B- terminal of the battery to GND, and the power is turned on.

The conduction of Q3 causes C5 to be charged to a level close to B+ through R12 and R5, with up positive and down negative;

The cut-off of Q2 makes the lower end of C6 close to the high level of B+ and is not charged.

Click the button switch S, the left end of the resistor R12 (the upper end of C5, C6) is momentarily shorted to B-, which means that the lower end of C5 acts on the base of Q3 through D2 with a negative pulse whose amplitude is close to B+, forcing Q3 to cut off, and C16 After charging, after a slight delay, Q2 turns on, Q1 turns off, and the power supply turns off.

The conduction of Q2 causes C6 to be charged to a level close to B+ through R12 and R6, up positive and negative;

The cut-off of Q3 makes the lower end of C5 close to the high level of B+ and is not charged.

Click the button switch S again, the left end of the resistor R12 (the upper end of C5 and C6) is instantaneously short-circuited to B-, which means that the lower end of C6 acts on the base of Q2 through D3 with a negative pulse with an amplitude close to B+, forcing Q2 to cut off, resulting in Q3 is turned on immediately, C16 is discharged quickly, and Q2 is kept off.

Q2 is off, Q1 is on, and the power is turned on again.