Choosing capacitance for recifier
Using Q = C * V, charge = capacitance times voltage, we take the derivatve with respect to time, to produce

dQ/dT = C * dV/dT + V * dC/dT

Then assume C is constant, making the 2nd term become zero, and assume dQ/dT =  I, and we have

Icurrent = dQ/dT = C * dV/dT







simulate this circuit &ndash; Schematic created using CircuitLab

Now you need 3 variables: I (one amp), T for 50Hertz (50 cycle per second, 0.02 seconds) and V the ripple voltage looking like a sawtooth slow fall and abrupt recharge time of 1 volt

And rearrange the I = C * dV/dT to become

C = I * dT/dV

C = 1 amp * 0.02seconds / 1volt = 0.02 Farad or 20,000 microFarads

If you use full wave rectifier at 60Hz, thus 120Hertz recharge events per second, the math is

C = 1 amp *( 1/120seconds ) / 1volt = 0.00833 Farads, or 8,333 microFarads

If 0.1 amp load, and your regulator accepts 2 volts ripple, then

C = 0.1 amp * (1/120seconds) / 2volt =  416 uF (470 is a standard value)