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Capacitor connecting in parallel to bridge rectifier

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ash2378

New Member
Hi There,

This is Ash I am new to this forum and electronics. I have a questions about connecting a capacitor in parallel to a bridge rectifier
I require 18VDC constant supply for my circuit and the current rating would be 4Amps. My transformed is supply is 18VAC coming from 400VAC primary transformer 90VA output
Is this capacitor connected at DC side filtering capacitor? Do I need a resistor to be connected in parallel to the capacitor.

Can anyone help with value of filtering capacitor, a resistor and to keep the DC Voltage at 18VDC with load and No load

Thanks for your help in advance

Ash
 

KMoffett

Well-Known Member
The transformer/capacitor/resistor power supply will never provide a constant output voltage under both loaded and no load conditions. The output voltage will rise to ~18V X 1.414 with no load, and may drop to less that 18V with a 4A load. For 18V regulated output you will need a voltage regulator circuit with an input of ~20V minimum input.

Ken
 

electroyas

New Member
since you need 18V constant and current 4A better to use a shunt regulator so at no load condition it will shunt excess voltage via load resistor. Do not just connect a resistor parallel to the capacitor.
 

ronv

Well-Known Member
Most Helpful Member
Do you need exactly 18 volts output voltage? Or could it be 16 volts or 20 volts as long as it is constant?
 

electroyas

New Member
Check this out under shunt regulator **broken link removed** , From a shunt you don't need to find the resistor value it will adjust automatically according to the given voltage.
 

ronv

Well-Known Member
Most Helpful Member
Buy one of these little guys:

**broken link removed**

Then make your supply like the attachment.

Be careful with your transformer. It is a little underrated for 4 amps. Make sure it doesn't get too hot.
 

Attachments

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    18v.png
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electroyas

New Member
How do I connect this resistor? In series or parallel.
To drop the voltage you will need to connect it in series but there should be current through it to lower the voltage. When the load current is varying so does the voltage so not reliable. If you want to drop the voltage from 20 to 18 then 2v should be accros the resistor when 4A is passing through it.R is equal to 2/4 = 0.5ohms .. power dessipated in the resistor is 2*4 = 8W ,
 

MikeMl

Well-Known Member
Most Helpful Member

electroyas

New Member
I was replying electroyas reply

According to the series connection method and constant 4A load you will need more than 8W power resistor in order to be safe from destroying the resistor or changing its value when its heating , Since this is a large value I don't think this is a good method if you concern about the efficiency of the circuit. Total power input is 20V * 4A = 80W ..Then 10% power is wasted as heat.
 

panic mode

Well-Known Member
if it was me, i'd just use old laptop power supply. always have few of them kicking around...
 
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