24V 2A Variable Power Supply

What do you mean by used?
T2 should not get hot unless you have a considerable load current.

Have you connected the output stage correctly? With 0V on the base of T1, the output should be 0V.

 

You still havn't said what you mean by used. Was it driving an led at 2mA or charging a battery at 14V 5A?
Measure the voltages on T2 and measure output current.

Much more information is required.

 

So you're drawing 100 mA out of the regulator, which is set for an output voltage of 15V.

What's the voltage on the collector of T2?

T2 is being used in its linear mode. You can calculate the amount of power it will dissipate by multiplying the current flowing through it (0.1A) by the voltage across it (that is, the collector voltage minus the emitter voltage). This gives a figure in watts.

How big is the heatsink? Can you upload a photo?

 

You will need a much bigger heat sink if you are going to utilise the full capability of the circuit.

T2 must not have a large negative voltage base/emitter. A higher output voltage than the base is not a good thing. Reducing C5 would allow the output to drop faster but I would not do this, instead I would put a diode in series with the base lead so that only a turn on current is allowed.

Edit. The maximum permitted reverse b/e voltage is 7V.

 

A rough calculation suggests that T2 could dissipate up to 70W. A BIG heat sink would be required for this.

The added diode is there to protect the transistor, you do not want to pop off another.

A 10k load on the output would drop the voltage quicker as you suggest. The time constant will be R * C and the voltage will drop to about a third in this time.

 

Confirmed.

 

Yes you can do that, as well as adding a resistor across the output as previously discussed.

Yes, and the best way to prevent damage is to reverse-connect a diode across the transistor's base-emitter junction, not to connect a diode in series with the base (that will reduce the maximum output voltage unnecessarily). Just add a 1N400x with its anode to T2's emitter and its cathode to T2's base. And I would do the same for T1 to protect it, and add a resistor of around 1k between the wiper of the voltage setting potentiometer and T2's base, to protect the potentiometer!

 

Yes, if a parallel diode is used, then both transistors need to be protected.

 

I always like to have a bleed resistor across the output to ensure the capacitor is discharged. Adding the two diodes is a very good idea but there could be a residual voltage on the output when the diodes stop conducting. It won't be much and maybe of no concern but I thought I would mention it just in case the value of capacitance is changed to a much larger value in the future.
Adam