Hi,
The calculation is relatively simple.
1F = 1 As / V, or C = I x t / V.
It says: you need 1F for a load current of 1A for 1second of time and a voltage (drop) of 1V
Example:
You have a frequency of 50Hz, 0.1A of load current and you want the ripple(peak-peak) to be about 2V.
50Hz gives 20ms for a half wave rectifier (period time = max discharge time)
Then C = I x t / V = 0.1A x 0.02s / 2V = 0.001F = 1mF = 1000uF.
( C = capacitance in F, I = average load current in A, t = discharge time in s, V is voltage ripple (pp) in V)
With this formula you are on the safe side, because discharge_time is less than 20ms.
20ms = mains_frequency_period_time = charge_time + discharge_time
Klaus