Let's first look at its circuit working principle

Initial state
When the circuit is powered on, due to the voltage division of R97 and R99, the base voltage of Q63 is pulled up and is in the cut-off state. At this time, the gate voltage of MOS tube Q62 is close to the source voltage (+24V). Since the gate-source voltage Vgs is less than the threshold voltage (due to the existence of the voltage regulator diode D28, the gate voltage is clamped at about +12V), the MOS tube is in the cut-off state, and there is no output at the output terminal OUT.

 applied to

the base of Q63 (for example, through an external control circuit), Q63 is turned on and its collector voltage is pulled low. At this time, the gate voltage of the MOS tube Q62 is grounded through R98, the gate-source voltage Vgs is greater than the threshold voltage, the MOS tube is turned on, the power supply is supplied to the load RL through the MOS tube, and the output terminal OUT outputs a +24V voltage.

How do you protect the circuit?

Resistor R95 is a current-sense resistor with a resistance of 0.7 Ω. When the load current exceeds 1A or the output is short-circuited, the voltage drop across the current-sense resistor R95 will exceed 0.7V. At this time, Q63 is in saturation, and the saturation voltage drop between its collector and emitter is about 0.3V. Since the emitter and collector of the triode are connected in parallel with the source and gate of the MOS, the gate-source voltage of the MOS tube Q62 is clamped at about 0.3V at this time, and Vgs no longer meets the turn-on condition. The PMOS tube Q62 is turned off, cutting off the output of the subsequent stage, thereby effectively protecting the load and the subsequent stage circuit.