Design site has error in the design calculations for RMS >Peak.
In the end the RMS voltage of the transformer should only be slightly more than the LDO Vdc if the optimal parts are selected.
Assuming you want 5V,1A regulated choose LDO regulator with minimum drop otherwise 2.5V typical may require heat-sink , bigger cap. and bigger transformer.
Assuming 50Hz with 100Hz rectifier ripple. produces narrow pulse current in caps (I_rms)
Case 1: LDO drop is 2.5V min. LM78xx, Lm317
Since T=RC the cap discharge time constant is for a voltage drop of (1-1/e) or 63% ripple is excessive but diode charge and load discharge current is equal approximately. If we reduce too much, then diode and secondary current may result in oversized components, thus 5xT is a good tradeoff with <20% sag and charge time is reduced to 1/5 thus peak current x5=5A for normal operation.
Ic=Cdv/dt=1A letting dt=T=10ms, where R is effective load of 5v/1A=5Ω.
Solve the ripple voltage using 20%Vp where Vp=1.414*Vrms-Vf , for diode Vf@5A and estimate Vrms. If centre tap secondary use 1 diode drop per cycle, otherwise full bridge has two Vf drop.
Assume full bridge with no centre tap
Vrms *1.414 - 2Vf =Vp= 5.0V +2.5Vdrop+ 20%Vp
Thus Vp=7.5/80%=9.4 and if Vf=1V,
Vrms=(9.4+2*1)/1.414=8.04Vrms
Full bridge using 2 diodes and centre tap
Vrms=(9.4+1)/1.414=7.36Vrms
If good Schottky diodes, If=0.5V@5A
Vrms=7.0Vrms
If a better LDO with 0.3Vdrop is used, with Vf=0.5V
The half bridge Vrms=(5V+0.3(Vdrop) + 0.5(Vf)/80%/1.414=5.13Vrms
The full bridge Vrms=5.4Vrms
Finally the capacitor size
Recall Ic=Cdv/dt=1A & Vp=1.414*Vrms-Vf
where Vf is doubled for a full bridge, and lower for Schottky vs Silicon
and ripple of rectified DC voltage dv=20%Vp, dt=10ms, I=1A
Thus C= I * dt/dv and my formula for Vp=5V best case parts,
C=1A*10ms/1V=10mF
Vrms =(Vo (DC)+ LDO drop + 20%Vp + Vdiodes)/1.414
Then compute C
in all cases Cap ripple current rating must exceed design ripple current, Idc and voltage ripple %V, where Iripple= 0.7 x Idc/%V
In the end the RMS voltage of the transformer should only be slightly more than the LDO Vdc if the optimal parts are selected.
Assuming you want 5V,1A regulated choose LDO regulator with minimum drop otherwise 2.5V typical may require heat-sink , bigger cap. and bigger transformer.
Assuming 50Hz with 100Hz rectifier ripple. produces narrow pulse current in caps (I_rms)
Case 1: LDO drop is 2.5V min. LM78xx, Lm317
Since T=RC the cap discharge time constant is for a voltage drop of (1-1/e) or 63% ripple is excessive but diode charge and load discharge current is equal approximately. If we reduce too much, then diode and secondary current may result in oversized components, thus 5xT is a good tradeoff with <20% sag and charge time is reduced to 1/5 thus peak current x5=5A for normal operation.
Ic=Cdv/dt=1A letting dt=T=10ms, where R is effective load of 5v/1A=5Ω.
Solve the ripple voltage using 20%Vp where Vp=1.414*Vrms-Vf , for diode Vf@5A and estimate Vrms. If centre tap secondary use 1 diode drop per cycle, otherwise full bridge has two Vf drop.
Assume full bridge with no centre tap
Vrms *1.414 - 2Vf =Vp= 5.0V +2.5Vdrop+ 20%Vp
Thus Vp=7.5/80%=9.4 and if Vf=1V,
Vrms=(9.4+2*1)/1.414=8.04Vrms
Full bridge using 2 diodes and centre tap
Vrms=(9.4+1)/1.414=7.36Vrms
If good Schottky diodes, If=0.5V@5A
Vrms=7.0Vrms
If a better LDO with 0.3Vdrop is used, with Vf=0.5V
The half bridge Vrms=(5V+0.3(Vdrop) + 0.5(Vf)/80%/1.414=5.13Vrms
The full bridge Vrms=5.4Vrms
Finally the capacitor size
Recall Ic=Cdv/dt=1A & Vp=1.414*Vrms-Vf
where Vf is doubled for a full bridge, and lower for Schottky vs Silicon
and ripple of rectified DC voltage dv=20%Vp, dt=10ms, I=1A
Thus C= I * dt/dv and my formula for Vp=5V best case parts,
C=1A*10ms/1V=10mF
Vrms =(Vo (DC)+ LDO drop + 20%Vp + Vdiodes)/1.414
Then compute C
in all cases Cap ripple current rating must exceed design ripple current, Idc and voltage ripple %V, where Iripple= 0.7 x Idc/%V
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