Filter Capcitor

This is not a practical one. Your voltage AND amperage point to a BIG value capacitor. The value depends on the actual current you will consume. And the ripple you want to permit to last.

WHAT YOU WANT TO USE IT FOR? You should better think of a bridge rectifier, because your capacitor here will be: difficult, if not impossible to find - say maybe 20.000 microfarad 400 Volts or higher and very expensive. (IF LOW RIPPLE IS REQUESTED) The problem is not only the capacitor, but also the rectifier that needs to deliver the starting current to load the capacitor. A bridge cuts your capacitor value in half. More info please.

Capacitance value is more complex.

Assuming as example: 50Hz, a single-wave rectifier charges the first filter capacitor once every 0.02sec, THEN a double rectifier does it every 0.01sec.

Now since the voltage over a one F (FARAD) capacitor drops 1V per second for each Ampere, YOU can now figure out the size of the capacitor once YOU know the current load. Assuming a charge/discharge cycle of 50%, 10V (pp) ripple and 100mA load, we get For a single-wave rectifier: C=1/10*0.1*0.02*0.5=100MF For a full-wave rectifier: C=1/10*0.1*0.01*0.5=50MF (yep, what might all this be?) This is a start for the example. Do some work now and fill out ALL your numbers, THE VALUES GIVEN ARE NOT EVEN CLOSE TO WHAT YOU WILL NEED... Capacitor size is proportional to the load current.

Basically, this is independent of voltage, but the ripple tolerance will be lower at a lower voltage, so probably a 115V rectifier will need about double the size capacitors, compared to a 250V.

I wasn't going to give more details, because this forum doesn't do your homework. We should also appreciate if you return us your findings, once it is calculated.

A riddle for a riddle.