With above R4=4k7 the TL431's current will be from 10mA to 4mA for 3V/5A to 30V/5A output voltage/current..
Mind it is an simulation only, with values and parts taken as an example..
With 60V input and 3V/5A output the Q1 will dissipate 20W and Q2 260W of heat
We must take into account the case of output voltage set to 25V with maximum load.
In such case, when using a real transformer, the rectified voltage would drop several volts due to the transformer losses and the ripple, so the minimum rectified voltage has to be greater than:
> maximum Isense voltage drop (about 0.7V) +
maximum ballast voltage drop (in case of paralleled power devices i.e. about 0.7V) +
maximum compound Vbe drop (1.5-2V) or ce saturation voltage (whichever is greater) +
peak-to-peak ripple (1-6V depending on the value of the capacitor(s) and the maximum output current)
And then the resistor feeding the base of the power devices should provide at least 1mA through the TL431 plus the base current for the power devices... let's make it a couple of mA in case of just 1A max output current, that makes : 2e-3 * 4.7e3 = 9.4V...
That's a lot... at least 9.4W wasted by the power devices for no real purpose.
Besides that the task of reducing the input ripple would be totally on the shoulders of that poor TL431...
So now we have 3 options:
-1) use a low minimum voltage constant current source (making the TL431 more prone to oscillations)
-2) use a transformer with voltage much higher than required (wasting further 10W per A of output current)
-3) use a charge pump (2 diodes + 2 capacitors voltage doubler) to get a high voltage rail
The 3rd options might be the only viable in case of a TL431 PSU.
So in this case the minimum rectified voltage can be as low as 28V which would allow to get more than 50V (more likely 53-54V) on the high voltage (low current) rail.
If we just need 2mA (like in the above example) we must use a (50-28)/2e-3 = 11k base-feed resistor.
But we could better use 2 resistors, let's make it 3.3k and 8.2k (or 4.7k and 6.8k) and place a 100µF electrolitic cap in the resulting node to ensure both soft start and ripple reduction.
The maximum current through the resistor would be achieved with minimum output voltage and no load. In that case the high voltage rail might get as high as 70V-71V that would make less than 7mA...
still safe for the TL431 with just around 17mW of dissipation while it would be better to use 2 1W resistors (although they would dissipate about 1/4 W).
Of course, in case of higher maximum output current, the resistor values (and power) should be recalculated.