How does the transistor slow-start circuit evolve from power-on to steady-state operation?

When power is on, C160 is charged, which leads to the Q5 package and CE conduction, VGS "VTH, the MOS tube is not conductive, with the C160 charging package and Q5 exits from the package and state to the loaded state, the G pole of the MOS tube pulls Low to ground level, so that the MOS tube is turned on. And it is a positive feedback process. Once it starts conducting, the potential of the D pole rises, causing the base electrode of Q5 to rise (the capacitor voltage of C160 cannot be abrupt), driving Q5 to cut, the G pole tends to fall, and the MOS tube is turned on.

The design idea is good, but it feels a bit complicated. The process is such that at the initial stage of power-up, the PNP tube is directly turned on under the action of the charging current of D1 and C160, which causes the Q4 gate bias to be 0 and no output.

As the charging current decreases, to stop charging, the PNP current tends to decrease until it finally stabilizes at a CE voltage of about 6V determined by the two base voltage divider resistors, which just meets the reasonable gate voltage of Q4 but not too high.

Here C161 165 constitutes AC negative feedback, which is used to limit the output voltage rise speed after delayed start. In the middle of soft start, C160 is first charged to a suitable voltage, and the PNP tube current begins to decrease, and the output voltage rises. This is a positive feedback. The voltage will increase the entire capacitor voltage due to the output increase. At this time, C160 again Failure to discharge in time will produce a relatively high BE junction back pressure. The role of D1 can effectively isolate the value of C160 that may be higher than the reverse voltage of the be junction through the output voltage superposition. Subsequently, the C160 only discharges the charging voltage through the resistor.