Electrical Engineers help me?

Any capacitor will reduce the ripple.  The larger the capacitor, the more it smooths it.  But you can never get it all the way to zero.

The size of the capacitor depends on the expected load current, and the desired amount of ripple.

Use the following formula:

C = I / 2 x f x Vpp

where I = load current

f = input frequency of AC

Vpp = the permissible ripple voltage.

Depending on DC current load, here is the simple way....

for under 1A load, use 1000uF and at least 400Vdc capacitor, under 2A use 2000uF and so on.

Remember: output peak voltage becomes as high as 322Vdc at no load condition.

To find the capacitance for a smoothing capacitor you calculate the amount of charge it has to supply during the half cycle and relate that to how much the voltage of the  capacitor would drop by. Then by setting the voltage drop, you effectively set the capacitance required.

Q = Iload t . . . . . . . [where t is half the period]Q = C ΔVC = Q/ΔVC = IIoad t/ΔVC = Iload / 2f ΔV