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[SOLVED] Simple 110vAC to 110vDC Converter

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For 110 VAC you can use e.g. 1N4006 diode or similar with >400 V backward voltage rating.
Capacitor value as well as the regulator depend on load current which you did not indicate.
 

As jiripolivka said, both the diode and capacitor will depend on your current. If your current draw is upto 1 amp in 110v DC, any 1n4004 or higher diode can be used. If your current draw is higher than 1 amp, you need to choose a higher current rating diode, such as 10A10.

A filter capacitor is needed to minimize the ripple voltage. It should be a electrolytic one. Use this formula to find a suitable capacitor.

dU= I/2fC

where dU is peak to peak ripple,
I is load current,
f is operating frequency and
C is capacitor volume in Farad
.
In this way you can choose ripple amplitude and from equation estimate
capacitor volume. For example if you have 1A current and capacitor
4700uF ripple will be

dU= 1/(2*50*0.0047)= 2.12V.

As you can see, capacitor value depends upon the current flow, not the operating voltage. But while choosing a cap, use a voltage rating of more than 250v.
 

Using a full diode bridge and a filter capacitor, the DC voltage, at no load, will be approximately:

Vdc = Vrms * SQRT(2) = 110 * 1.414 = 155.5 V

It is actually a bit less ( 2 diode forward voltage and the dc drop of the capacitor ripple).

To get a regulated 110V DC voltage, the design depends on the load current (its maximum and minimum in the least).
 
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    kam1787

    Points: 2
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while choosing a cap, use a voltage rating of more than 250v.

>250V seems a bit excessive. >> 155V would do. To a degree the higher the better, but cost is a constraint as well as practically.
 

>250V seems a bit excessive. >> 155V would do. To a degree the higher the better, but cost is a constraint as well as practically.

I agree that. :) All needed is to follow a derating factor of minimum 20-30%...
 

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