The inefficiency of all linear ac-dc 50Hz must be reduced otherwise the series drop V * I represents a huge bypass loss for lower V out.
I dont know what transformer you plan or have or budget limitations, but let's shoot for 30V 3A supply with a 24V 100VA transformer
Thus if your requirement is 90W for 30V @3A which represents a 10 Ohm for a linear load, the diode peak current will be 5x or 15A with 20% ripple.
Using 4 of these SMT Schottky
diodes
MBR745 = 45Vpiv 840mV @ 15A but can withstand 100A for startup cycles to charge up Cap.
Note from VI curve these diodes have an ESR loss from the slope above from 10 to 20A = 0.1V/10A = 0.01 Ohm which is far superior to what you selected.
Diodes drop in voltage when they get hot but a good design limits this to 85'C not 125'C even though they withstand this.
Simulate your Transformer with a series resistance.
If P=V^/R = 100 VA and efficiency is 95% then 5W loss represents 1.67V at 3A.
Then simulate this unregulated design first with a variable load to see Vavg, Vmin vs I or R from 10 to 1K Ohms. Vmin will be the absolute highest Vreg you can achieve with 2N3055's or even better a <=100mOhm P-type MOSFET mounted on a CPU cooler heatsink.
Then you can rethink how to regulate it simpler. using single supply Op Amps and NPN driver to P-MOSFET, YOu can add CC current limiter adjustment and V regulator adjustment without having to use a simple LDO which drops 2.5V Instead the only drop is past the low RdsON resistance of the MOSFET , and of course the diode bridge and minimum ripple peak Vmin.
From your schematic I see 10mF so a 10 Ohm load will produce an RC= 10*10= 100ms time constant. which for the half cycling in 50Hz rectified is 10ms giving only 10% ripple. . This would suggest, you can use a bigger power transformer or smaller Cap if you can tolerate 20% ripple and get 30Vmin out of the bridge with 10 OHm load.
Try that first or adjust parameters to suit your budget and goals.