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

[vk6kro]

AC voltages are usually given as RMS voltages. This is the DC voltage which has the same heating effect as the AC sinewave.

However, the PEAK or maximum voltage is what matters for working out peak inverse voltages of diodes.

This is covered in most books of Electrical Engineering, or Wikipedia, but it is 1.414 times the RMS voltage.

So, this may be why you are having trouble working out the voltages.
A 10 volt RMS sinewave has a peak voltage of 14.14 volts.

If you put a large capacitor across your load resistor, this would charge up to almost the peak value of the voltage on each side of the transformer centre tap, so it is an important value to know.

 

[sophiecentaur]

My main confusion is when the secondary winding is only 24V, and when divided into 2, each of the lower and upper has 12v. My understanding about the piv is only 24v, this is what i think, (lets say the lower diode is reversed bias), and i consider it as a gap. The right 'terminal' of the gap is from the upper source which is 12v, and the left terminal of the gap is from the lower source which is -12v. And the potential difference across the gap is -12v -12v = -24v

But most of the reference said it is double of the secondary voltage of the transformer which is 48v, i am so clueless where is another 24 v come from.

You are mis-understanding the quoted voltage of the transformer. As I pointed out in my first post to you, your initial diagram just shows a single value of voltage "V", floating around near the middle of the transformer symbol. A tapped transformer must be specified in terms of both windings (V1:0:V2) - otherwise there is no way to distinguish a centre tapped transformer from a transformer with extra taps (at one end) for voltage adjustment.
I already made this point strongly, about a hundred years ago and you ignored it. If you ask a question on PF then you should really expect to read the answers and take them seriously (unless they are strongly refuted in other posts - nonsense answers are quite possible and the 'peers' are there to minimise damage)
Afaics, there is no problem with any of this if you read the transformer spec properly.

 

[null void]

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?

 

[vk6kro]

The left diagram has two points labelled as zero volts, but they are not at the same voltage.

Choose one, like the negative output of the bridge, and then you can work out the other voltages relative to that one.

 

[sophiecentaur]

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]

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]

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]

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/devices-circuits/center-tapped-full-wave-rectifier#axzz2UUgY5TQC

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

http://www.electronic-factory.co.uk/full-wave-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]

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]

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/devices-circuits/center-tapped-full-wave-rectifier#axzz2UUgY5TQC

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

http://www.electronic-factory.co.uk/full-wave-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]

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.