The 15v zener is fed via the 4K7 resistor from the input. With 0.6v across the base - emitter of the PNP transistor, this establishes a voltage of 14.4 across the 3K3, so there must be a current of nearly 4.4mA through the 3K3. This current is available to feed the emitter of the transistor - provided that the diode D1 doesn't pinch any of it. However with no load on the output, some of this current flows through the PNP to turn on the BFR39 and the 2N6254 which will conduct.
As there is no load, the only place the power transistor can get any current through D1 so it will pinch enough current so that what remains of the 4.4 mA is only just enough to keep it working.
Since there is 14.4v across the 3K3 and about 0.6v across the diode there will be 15v present across the output.
If the output is loaded the voltage drops by a fraction - this reduces the diode current, allowing more current to flow into the output pair to keep the output voltage up.
When the output pair is using all the available 4.4mA there will be no current through the diode so any extra load will cause the output voltage to drop.
If it drops by 1.2 volt (two diode drops) the second diode will start to conduct. This pinches current from the 15z v zener so its voltage reduces, reducing in turn the current through the 3K3 thus reducing the drive to the output so the output collapses - the supply has tripped off. However it is not quite off: even with the output shorted out a tiny current flows in the PNP so that the circuit will reset when the load is removed.
To alter the output voltage - chose a different zener voltage. This will alter the current available to the PNP so you will have to alter the 3K3.
More Detail for the Simple regulated power supply with overcurrent trip Please see
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