A classic output short-circuit protection circuit

Power on: The voltage at both ends of C2 cannot change suddenly. The base voltage of Q2 starts to decrease from VCC until it reaches the point where Q2 can be turned on (BE junction voltage drop is 0.7V). This time is about 0.12mS. But at the same time, Q1 is also preventing Q2 from turning on. The time for Q1 to turn on is approximately: 5.87mS, which means that Q2 will not turn on until 5.87mS, but at the same time, C3 is preventing Q3 from turning on. The blocking time is 0.17mS.

Q3 turns on after 0.17MS of power-on, the load is powered, and the C-pole voltage of Q3 reaches about 13.3, forcing Q2 to cut off. This shows that Q1 can be removed.

When short-circuited, the C pole of Q3 is pulled low, Q2 is turned on, forming a self-locking force, forcing Q3 to cut off, and there is no voltage to the load behind Q3. At this time, it does not matter whether there is a load or not, so there will be no output even if the load is removed. If you want to restore the output after removing the load, you can connect a resistor to the CE junction of Q3, about 1K.