Why does the PIV in centre tap rectifier is twice of the V source?


by null void
Tags: centre, rectifier, source
sophiecentaur
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May27-13, 05:26 AM
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Quote Quote by null void View Post
Sorry for not understanding you. But I still a little bit confuse.


Both the transformer are induced by a same ac source, same number of coils in primary coil, and both transformer are in same phase which producing peak voltage. Is there anything missing in the diagram or there is some voltage labeled wrongly?
When you say both transformers are induced by the same source, what has that to do with the secondary voltages? Do you know how transformers work? Your statement is meaningless.
Both those circuits give the expected DC output volts. As I have said before, you need to do more basic study before leaping into somewhere that is relatively advanced. It's not a free ride.
NascentOxygen
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May27-13, 05:40 AM
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Quote Quote by null void View Post
Both the transformer are induced by a same ac source, same number of coils in primary coil, and both transformer are in same phase which producing peak voltage. Is there anything missing in the diagram or there is some voltage labeled wrongly?
You are comparing two different circuits. Back in the olden days, when transformers were cheap (and saving weight was not a big consideration) and rectifier diodes very expensive, the full-wave rectifier using 2 diodes was popular. Two 10V windings (i.e., 20V centre-tapped) give 10V output (roughly* speaking).

Today, bulky transformers are relatively expensive but diodes are really cheap, so it is more common to use a smaller lighter transformer (having only half the secondary turns) and to use 4 diodes in a bridge rectifier arrangement. Again, the rectification is full-wave, but this time a single 10V winding gives 10V output.*

There are other differences, but not as major as this.
sophiecentaur
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May27-13, 06:43 AM
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Actually, for a given power handling, there won't be much to choose. Imo, the reason against centre tapped is the extra difficulty in winding.
null void
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May27-13, 07:22 AM
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In the "electronic Device - conventional current device, ninth edition" , page 54, it show a several equation to find the piv in center tap rectification.

V_peak(out) = V_peak(secondary) / 2 - 0.7V // i still understand this line; then in the next equation
V_peak(secondary) = 2 * V_peak(out) + 1.4v //why is it necessary to times 2 on every term?
//is it because the secondary coil is divided by 2?
and finally it get:
PIV = 2 * V_peak(out) + 0.7v

http://www.daenotes.com/electronics/...#axzz2UUgY5TQC

that page says: PIV = 2Vp(sec) + 0.7 V

http://www.electronic-factory.co.uk/...ve-rectifiers/
this page give the same equation as in the book.

And if i am not wrongly copied what my teacher wrote, it is :
PIV = 2Vp(sec) + 0.7 V

Which source is correct?
sophiecentaur
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May27-13, 07:38 AM
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Why do you keep asking the same question again and again? Look at the basic theory of transformers. Look at how a centre tapped transformer is specified. Apply Kirchoff's second saw. Do some thinking for yourself instead of asking all the time.
You have to "times two" every time because the full end-to end voltage on the secondary is applied - which is twice the output from half of the secondary.
You will not find any extra answer to your question. Your problem seems to be that you are expecting some other answer that will sort out your misunderstanding. There is not another answer. You need to interpret what has been given to you (many times, already).
NascentOxygen
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May27-13, 09:44 AM
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Quote Quote by null void View Post
In the "electronic Device - conventional current device, ninth edition" , page 54, it show a several equation to find the piv in center tap rectification.

V_peak(out) = V_peak(secondary) / 2 - 0.7V // i still understand this line; then in the next equation
V_peak(secondary) = 2 * V_peak(out) + 1.4v //why is it necessary to times 2 on every term?
//is it because the secondary coil is divided by 2?
and finally it get:
PIV = 2 * V_peak(out) + 0.7v

http://www.daenotes.com/electronics/...#axzz2UUgY5TQC

that page says: PIV = 2Vp(sec) + 0.7 V

http://www.electronic-factory.co.uk/...ve-rectifiers/
this page give the same equation as in the book.

And if i am not wrongly copied what my teacher wrote, it is :
PIV = 2Vp(sec) + 0.7 V

Which source is correct?
They are all correct, except for the last one. The error is tiny & trivial, but it should be
PIV = 2Vp(sec) - 0.7 V
sophiecentaur
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May27-13, 09:54 AM
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Quote Quote by NascentOxygen View Post
They are all correct, except for the last one. The error is tiny & trivial, but it should be
PIV = 2Vp(sec) - 0.7 V
Yes. Kirchoff 2 applies here. How anyone could imagine that a diode actually supplies emf escapes me.

All of this can be resolved by properly specifying the (multiple) output voltages of the centre tapped transformer.


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