Charging of the capacitor:
Considering the below diagram, with slight approximation the ripple voltage can be supposed as triangular. To the cut-out point from the cut-in point, whatever charge the capacitor achieves is equivalent to the charge the capacitor has lost during the period of non-conduction, that is, from cut-out point to the next cut-in point.
The charge it has acquired
The charge it has lost
Vr p-p * C = Idc *T2
It can be supposed that the time T2 is equal to half the periodic time of the waveform if the capacitor's value is fairly large, or the value of the load resistance is extremely large.
T2=T/2 = 1/2f
Vr p-p =Idc / 2fC
From the above statements, the ripple waveform will be triangular and its rms value is described by
Vr rms =Vr p-p/ 2 √3
Vr rms = Id.c. = 4√3 f C
= V d.c. /4√3 f C RL
Id.c. = Vd.c. / RL
Ripple, ϒ = Vr rms / Vd.c. = 1/4√3 f C RL
The ripple might be decreased via increasing C or RL (both) with a resultant increase in the dc Output voltage.