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

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SUMMARY

The discussion centers on the peak inverse voltage (PIV) in a full-wave center tap rectifier, specifically why the PIV requirement for the diode is twice the input voltage from the transformer. Participants clarify that in a center-tapped transformer, when one winding produces a voltage of +V, the other produces -V, resulting in a total PIV of 2V across the diode. The grounding of the center tap does not affect the voltage across the diode; rather, it is the vector subtraction of the voltages at each end of the transformer winding that determines the PIV. Understanding the relationship between the transformer windings and the diode's operation is crucial for accurate circuit analysis.

PREREQUISITES
  • Understanding of full-wave rectification
  • Knowledge of transformer operation and center tap configurations
  • Familiarity with diode characteristics and peak inverse voltage (PIV)
  • Basic circuit analysis skills, including voltage referencing
NEXT STEPS
  • Study the operation of full-wave center tap rectifiers in detail
  • Learn about transformer phasing and its impact on voltage outputs
  • Explore the concept of peak inverse voltage (PIV) in various rectifier configurations
  • Examine practical applications of center tap transformers in power supply design
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Electrical engineers, students studying power electronics, and anyone involved in designing or analyzing rectifier circuits will benefit from this discussion.

  • #31
sophiecentaur said:
But surely it's the phasing that makes the whole thing work.
Indeed. But I sensed that OP may not have a good grasp of "phasing". That's why I referred to it as polarity, as you'll note in my 1st and 2nd posts in this thread.

There is, of course, the extra point that needs to be emphasised and that is
Many points may be emphasized to give a more thorough analysis, but I sensed that OP does not have too solid a grounding here. I kept to just the essentials to answer his* question, lest he get mired in indigestible detail.
 
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  • #32
null void said:
You mean this? And the polarity of the the 2 source is same or opposite?

"Polarity" relative to what? There are three correct answers.
The same, relative to the bottom connection.
Opposite, relative to the mid point.
The same relative to the top.

If you live on the ground floor, everyone is 'above you'.
If you live half way up, some are above and some are below.
If you live on the top floor, everyone is below you.

You have to specify a reference and, in a circuit with an Earth connection, it is normal to use that as a reference. The centre tap is connected to Earth in your first diagram but that makes no difference to the voltage across the two ends. You could put a tap on the winding that is 1/4 way along it and connect that to Earth instead. The voltage across the winding would still be the same (2V). The lower terminal would be at (-)V/2 and the upper would be at +3V/2.

There is still something nagging at the back of your mind about this. I wish I could spot what it is. Then I could think of the magic words to sort it all out for you. Perhaps it's the word "Relative".
 
  • #33
null void said:
You mean this? And the polarity of the the 2 source is same or opposite?
The polarity is as you have them drawn. (I'm not going to say whether that should be termed "the same" or "opposite" because it depends on how you view them.)

But the polarity is as you have shown them.

So you can see that the PIV is equal to the sum of both windings?
 
  • #34
NascentOxygen said:
Indeed. But I sensed that OP may not have a good grasp of "phasing". That's why I referred to it as polarity, as you'll note in my 1st and 2nd posts in this thread.


Many points may be emphasized to give a more thorough analysis, but I sensed that OP does not have too solid a grounding here. I kept to just the essentials to answer his* question, lest he get mired in indigestible detail.

Ah well, in that case he should not get involved with centre tapped transformers. Dual DC power supplies might be something to start off with.
 
  • #35
sophiecentaur said:
Perhaps it's the word "Relative".
Indeed. It means something to me, but is unlikely to add clarity to OPs understanding.

(I actually used "relative polarity" in my last reply to poster, but then edited it out for being unlikely to mean much.)
 
  • #36
NascentOxygen said:
Indeed. It means something to me, but is unlikely to add clarity to OPs understanding.

(I actually used "relative polarity" in my last reply to poster, but then edited it out for being unlikely to mean much.)

In that case, I have to put it in a different way - like 'where you put the red and black leads on your voltmeter and the sign you will get on the display'. Without some concept of polarity there is no way that even a diode function can be described.
Are we, perhaps, over-medicating him?
 
  • #37
Yeah i admit i didn't get the mining of phasing and relative. For the 'relative', let's say i have 3 charge, 1st is + and 2nd is - and 3rd is +, so the 1st is relatively same as 3rd? and the 1st and 3rd is relatively opposite with 2nd?

And from this post, i feel like i have some miss-concept even in a simpler circuit...

You could put a tap on the winding that is 1/4 way along it and connect that to Earth instead. The voltage across the winding would still be the same (2V). The lower terminal would be at (-)V/2 and the upper would be at +3V/2.

M.aspx?v=8b7067875a62b57b9f67.jpg


This might my main problem to understand this idea. I thought the the V_A = 4V (by summing both of the voltage source in series) , but how come it is either (1/4)8V or (3/2)8V , this make me feel like it somehow related to the earthing in middle of the sources.

And i still really can't understand where the negative voltage come from, if it is in a simple circuit like this(ignore the 12 voltage at the diode, this is just from my previous drawings):
M.aspx?v=8b706786619db47bae6b.jpg


there is no negative volts at either end of the diode right? I just have to consider the reversed biased diode has infinite resistance and the voltage dropped completely, so piv = V only.
 
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  • #38
Do you know about potential dividers?
Also, do you not realize that if one point A on a circuit is positive referenced to an earthed point B and you then re-connect Earth to point A, then B is negative referred to earth? This is such basic stuff that you could never understand the rectifier circuit without it.
 
  • #39
Do u mean this:

M.aspx?v=8b7067875e8f71a8a56e.jpg

blue arrow is the 'direction' of + volt

red is the 'direction of - volt
 
  • #40
Bordering on the right idea but not properly expressed.
The blue bits (but not the bit through the resistor) are at a Positive Potential (ref Earth)
The red bits (but not the line from the bottom battery to Earth) are at a Negative Potential. How could you have the same voltage on both ends of the lower battery?
The difference (the Potential Difference) is 4V.
Have you ever looked at other circuits and the way the voltages are put on them? Have you ever seen lines drawn to show where the voltages 'go'? No you haven't. Voltages are shown where there is a voltage difference. Why do you want to try and use your own notation? It's just not right. Only a genius can hope to invent a totally new approach to circuit theory and get anywhere with it.

Do you realize just how inefficient this method of trying to learn things is? Why don't you go to any number of web pages with the whole thing explained to you (or even a Text Book?) - starting with the basics. For example, this link.
 
  • #41
M.aspx?v=8b7067885a9aa6a6a99e.jpg


M.aspx?v=8b7067885b9fabaaaea3.jpg


If i want to split the circuit, is this the right way? And I am not very sure if the points of voltage in there is correct...
 
  • #42
null void said:
M.aspx?v=8b7067885a9aa6a6a99e.jpg


M.aspx?v=8b7067885b9fabaaaea3.jpg


If i want to split the circuit, is this the right way? And I am not very sure if the points of voltage in there is correct...

only one battery and the Earth connected to the resistor? Does that look right to you? Try some thinking on your own before sending out random nonsense. And do some reading too!
 
  • #43
Sorry about that, but i think this link will not be wrong :
http://www.falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+1.0+50%0Aw+464+256+464+288+1%0Ag+272+208+320+208+0%0Aw+272+160+272+176+0%0Aw+272+240+272+288+0%0Aw+272+288+320+288+0%0Ad+320+288+432+288+1+0.805904783%0Aw+272+160+272+96+0%0Aw+272+96+336+96+0%0Ad+336+96+416+96+1+0.805904783%0Aw+416+96+464+96+0%0Aw+432+288+464+288+0%0Aw+464+96+464+112+0%0Aw+464+192+512+192+0%0Aw+512+192+512+272+0%0Ar+512+272+512+336+0+100.0%0Ag+512+336+512+368+0%0Aw+464+192+464+112+0%0Aw+464+192+464+256+0%0Av+144+240+144+176+0+1+40.0+5.0+0.0+0.0+0.5%0Av+272+208+272+176+0+0+40.0+5.0+0.0+0.0+0.5%0Av+272+240+272+208+0+0+40.0+5.0+0.0+0.0+0.5%0Ao+5+64+0+35+10.0+9.765625E-5+0+-1%0Ao+1+64+0+35+7.62939453125E-5+0.1+1+-1%0Ao+5+64+0+35+10.0+9.765625E-5+2+-1%0A

It is a link to a circuit calculator coded in java, the reason it is so long is because it contain the information about the circuit to draw. You can point them to see the voltage value

This was I initial understanding about this kind of circuit, but it doesn't seem that the reversed bias diode has the double inverse voltage. There are 2 voltage source, each with 5 v, so the the voltage across the reverse bias diode is 5+5 = 10 , but i just can't understand where is another 10v come from
 
  • #44
"Another"?
You need to sort out some basic circuit theory or just give up on this, I'm afraid. You have not changed since the OP. What is the point of this?
 
  • #45
it definitely doesn't help when he's talking about transformers and having them look like batteries and visa versa

Dave
 
  • #46
Ok, I will go back to the basic.
 
  • #47
davenn said:
it definitely doesn't help when he's talking about transformers and having them look like batteries and visa versa

Dave

It's not that bad. It is just considering what happens at the peak of one cycle of AC.
What is wrong is that, like a lot of these 'serial question askers', he seems to think that understanding will come as a result of a series of random, self generated, questions rather than approaching it in a linear way, which has been developed and used successfully for most, if not all, students of Electricity. As you can see, he more or less repeats his very first question, 43 posts into the thread, which makes my point, I think.

Shirley you must agree with me . . . . . Shirley, are you there?
 
  • #48
I agree and don't call me "shirley" haha (cant remember which Naked Gun movie the shirley thing was from)

D
 
  • #49
A good way to relate the two AC sources is to "take a snapshot" - a single moment in time where, for example the top source is at it's peak voltage. Say 12V - now - the lower source is what? Also at it's peak, but the opposite polarity ( relative to the center tap/ grounded point).

So DC case is the OK model IMO - as then the jump to thinking about AC, can be daunting.
 
  • #50
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.
 
  • #51
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.
 
  • #52
null void said:
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.
 
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  • #53
Sorry for not understanding you. But I still a little bit confuse.
M.aspx?v=8b70678b5960adb4a16c.jpg


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?
 
  • #54
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.
 
  • #55
null void said:
Sorry for not understanding you. But I still a little bit confuse.
M.aspx?v=8b70678b5960adb4a16c.jpg


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.
 
  • #56
null void said:
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.
 
  • #57
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.
 
  • #58
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?
 
  • #59
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).
 
  • #60
null void said:
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
 

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