Whats the rule for a Secondary of a transformer VS Rectification metho

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Main Question or Discussion Point

I seem to remember my lecturer talking about Transformers and Rectification methods and there ratings for different modes of operation (25 Years ago).

I bought a book called Understanding DC Power Supplies By Barry Davies and I cant find any reference to my question in it.

The question is if you use a Full-wave rectifier using a center tap transformer and 2 diodes, can each of the transformers secondary rating be equal 1/2 the output current needed because each winding is only supping the power for 1/2 the time?
 

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  • #2
vk6kro
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In a full wave rectified transformer, each half of the transformer secondary winding conducts for only half the time.

So, for example, if the winding was rated for 5 amps then it could deliver 10 amps if it was only delivering it for half the time.
The winding can cool down between bursts of conduction.
 
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  • #3
Averagesupernova
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In a full wave rectified transformer, each half of the transformer secondary winding conducts for only half the time.

So, for example, if the winding was rated for 5 amps then it could deliver 10 amps if it was only delivering it for half the time.
The winding can cool down between bursts of conduction.
This is only true if it is a center tap common secondary. A full wave bridge that utilizes the whole secondary at once will be limited to its rating.
 
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jim hardy
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  • #5
try here

there's some additional correction for the non-sinewave current's different RMS value

http://www.hammondmfg.com/pdf/5c007.pdf
Could I assume if I had a Full Wave Center Tap Bridge and loaded the windings 180 ° out of phase I may be able to draw I D.C = 1.27 x Sec. I A.C ?
 
  • #6
The Electrician
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Could I assume if I had a Full Wave Center Tap Bridge and loaded the windings 180 ° out of phase I may be able to draw I D.C = 1.27 x Sec. I A.C ?
Are you really going to have a purely resistive load? Are you not going to filter the rectifier output with a capacitor to obtain nearly pure DC?
 
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jim hardy
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Could I assume if I had a Full Wave Center Tap Bridge and loaded the windings 180 ° out of phase I may be able to draw I D.C = 1.27 x Sec. I A.C ?
i'm not quite sure what you mean by " and loaded the windings 180 ° out of phase "

but a centertap with full wave bridge gives you two supplies, one positive and one negative
provided you use the centertap as circuit common. I think that's what you are proposing ???

I think that'd be equivalent to the diagram labeled " Full Wave Bridge Resistive Load" just above and right. Reason is each half of the transformer winding conducts all the time
Compare that one to your proposed circuit - in your scheme, if positive and negative currents are equal then there's none through centertap so circuits are equivalent and your current available is 0.9 not 1.27, mostly because transformer doesn't get that rest between half cycles.
 
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