The following power supply circuit uses LD1084#12 voltage regulator, which can provide up to 5Amps with only 1.3v dropout. Input voltage must be around 15v. It is highly recommended that you follow this value exactly, if the input voltage is more, the regulator temperature will increase more.
Calculation for the filtering capacitor:

5i
C = ——–
Vp. f

C: Capacitor value.
Vp: Peak voltage. (”Bridge output max voltage”)
f: Frequency of the AC supply.
i: Load current.

Note that the above equation for 10% ripple voltage.

To calculate the voltage required and the transformer secondary winding we first determine the input voltage for the regulator, which is 15v, plus a 10% of this value for ripple. For a regular transformer we have to consider a bridge rectifier, as a result; we will add 1.4v. So the secondary winding should be 15+1.5+1.4=18.9 lets say 18v @ 5 Amps. Now we will calculate the capacity of the filtering capacitor. By using equation number 1 and assuming that f=60Hz, we will get C=5 x 5 / ((18-1.4) x f) =25,100µF
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Source: http://powersupplycircuit.blogspot.com/2009/04/power-supply-ac.html