This is a zombie thread now, but theThe OP's original post has formulae reversed. TheThe ripple for half-wave rectification is larger than that for full wave. Easy
Easy to think about: if you charge a cap with a half a wave of voltage and current, it will discharge less if you have another charging half wave right after the first, and more if you have to discharge the cap another half wave with no input. Ripple
Ripple with full wave rectification is I / (2 f C)\$I / (2 f C)\$, and with half wave is I / (f C)\$I / (f C)\$. Probably
Probably poor practice to use the R\$R\$ from the filter in the equation: what matters is load current which will reflect both filter AND load resistance.