You have no base resistors on your TIP142 to prevent Load hogging.Your transformer voltage is 30 volts but after the filter capacitor it will be around 40 Volts.You need to calculate The current output of the Parallel TIP 142 at that voltage .Based on the data sheet the maximum current at the emitters at 40 volts will be around 2.5 amps each. And that's pushing them Hard . You have ballast at your emitters resistors to prevent thermal runaway .But they don't prevent load hogging .You stop load hogging by putting a resistor of equal value at the base of each TIP142 .The current across the collector and emitter is determined by the Beta(Hfe) multiplied by the Base current.This means That is if the BC558 is delivering its maximum current of 100mA then the current out of the TIP 142 will be 0.1A X its Hfe of 500 or 50 amps This far exceeds the TIP 142 Safe Operating Area.Since the Ic (collector current)is around the same as the Ie(emitter current) you only need the formula below.
Ic=Hfe X Ib Ic = 500 X 100mA Ic = 50A To much!
So lets put some reasonable over head on the TIP 142s and try to keep them at an 1A to start.You can go as high as 2A if you like but not suggested.To give 1 amp at the emitter The base only needs 0.002 amps.Based on the above formula.
Now lets calculate the resistor required to limit the base to 0.002 amps. The Vbe is generally the same for most BJTs of around 0.7 volts but can be higher for darlingtons.For the TIP142 the Vbe is 3 volts
Base resistor is equal to the maximum volts of 40 minus the Vbe of 3 volts (darlington) divided by the base current or
Rb = (Vin-Vbe)/Ib Rb = (40-3)/0.002) Rb=18500 Ohms at each TIP142
These resistors will keep your power transistors stable.
This is of the assumption that T1 is is indeed providing a maximum of 100mA but not less than 2mA and you are getting 40volts at the collectors.
You can adjust the base resistor according to the values you have or need but keeping within the safe operating are of the TIP142's
Your circuit analysis is nice, however completely out of reality.
The BC558 is about 8 mA current source. It will never supply more than that.
Your hFE calculation is also good, but has nothing to do with the current limiting. In fact, it is good you have excess of current gain. Trying to design for a maximum current limit based on under-sizing the base drive is very wrong approach. At minimum due to the very very large spread of hFE of each transistor. I do agree that some base resistors should be used when the transistors are not paired for Vbe, but their value two orders lower than you're suggesting, see below.
Also, you do not seem to understand much how the circuit works, as the base drive circuitry (in fact the BC558) can only supply up to some few volts (compliance range of only about 2V*) with regards to the emitter of the TIP142. Putting a 18k resistor in the base will render the circuit completely non-functional, as there would not be a way to supply enough current from the few V limited base drive supply.
Please look closely again at the circuit diagram: The TIP142's collector node is connected nowhere else in the circuit. The base is driven from the aux supply of +-5V, referenced to the positive output terminal. This is a standard widely used topology well established and proven to work in many PSUs.
Current limiting should never be done by limiting the base current. It is a wrong approach. In this case, there is a current regulating loop, that if correctly set, will protect the transistors reliably. In fact, will make the output of the PSU behave as constant current source - which is the wanted property here, obviously.
You might add additional high speed current protection, utilizing another NPN to sense the voltage across the current shunt and to shunt the base current off the TIP142s with it. Personally I consider this unnecessary fort supply such as this, however it is only a few parts added to the BOM and would not discourage anyone against doing so.
Much more realistic values of power dissipation is like 30W per TO247 device maximum for practical reasons. So at 40V Uce, it is
like below 1A. A pair of TIP142s will be barely enough for a 2A 30V output power supply. It will likely not survive a long term output short -the built-in temperature protection is indeed good idea.
That is why most linear PSUs have some transformer tap switching implemented, to lower the power dissipation of the power transistors. Having a relay to switch here between 15 or 30V secondary voltage would be a nice feature - the remaining OPamp could be used to control the relay action.
Please also note, that a 30V transformer secondary voltage is not enough for a reliable 30V DC output. 32-36V range would be more like it. You need to account for mains voltage variation (up to +-10%) and account for the voltage ripple on the filter cap. Making the cap deliberately large won't help much either, will only add excessive load on the rectifiers and transformer (lower power factor, higher VA and effective current pulled from the winding).
Always calculate a base resistor for BJTs even if you don't think one is required or even if a simulator tells you you don't need one.
NO transistor of the same model is exactly alike. Data sheets are based on a few samples and averaged and simulators will only tell you so much based on those datasheets.
Trust in the math based on your measurements.
That is a wrong assumption. Datasheets are completely fine when followed correctly (understood correctly). If a base resistor is not required, do not put one. If you are not sure, use math to be sure. Adding random resistors to circuits "just because" is not the recommended practice.
*I would modify the bias node of the BC558 to increase its compliance range. 2V seems not enough to me, if the Vbe of the darlington plus the emitter shunt voltage is considered. (For example R1=68R, R2=1k and R3=3k3)
//EDIT: Fixed many typos.