Design half wave rectifier circuit as fallows specification

Arrive at a design specification of the diode and filter capacitor of half wave rectifier rated 2A forward current.
I did not get how to design this Half wave circuit only using current rating.... and I dont know what is exactly current rating meant for? can any one please solve and provide answer for it.

Poorly worded question.

It asks for a design specification then says "rated 2A forward current". The current is through the diode. There is a big clue to the minimum current rating there. All you need to add is PIV of the diode which is minimum of twice the peak AC.

Nobody can tell you the capacitor specification without knowing the frequency, DC load current and permissible ripple.

Brian.

Hi,

I assumd this is only a part of an exercise. And I assume the missing informations are given in the earlier part(s) of the exercise.

Klaus

Thank you so much Brain sir....!
But Is it solvable only on given data in the question ? if not which all specification still we require to arrive the design of 2A rated current?

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No it was given as Interview question for me ! But i failed to arrive design. Thank you KlausST

A common rule of thumb is for every Ampere, add 1000 uF capacitor value. It is not a calculated answer but it may be sufficient to show that you know a few design shortcuts.

To make a calculation, you want to find a capacitor value which is high enough so it drives the load during idle gaps.

The RC time constant is the time required for voltage to fall 63 percent. Say your spec is for 5% ripple (voltage drops 5 percent during idle gaps). If mains is 50 Hz, then 1/50 second is idle time in a half-wave power supply. Derive the time constant. Approximately we have voltage falling 63 percent in 12/50 sec. Approximate is good enough.

Divide 12/50 by load resistance and the result is your capacitor value.

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Namely the capacitor value that yields 5% ripple with that load resistance, in a half-wave power supply at 50Hz.