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

[null void]

In the full wave centre tap rectifier, piv of diode require is twice of the input current of the transformer. I can't understand why is it twice when the source is only 1V-peak.

 

[NascentOxygen]

In the full wave centre tap rectifier, piv of diode require is twice of the input current of the transformer. I can't understand why is it twice when the source is only 1V-peak.

Hi null void.

Have you drawn the schematic? If so, mark voltages on the secondary terminals, showing CT as ground, upper end +1V, and the lower end as -1V.

Then proceed to examine the diode inverse voltages from there.

 

[null void]

https://www.filesanywhere.com/fs/v.aspx?v=8b6a648c596070b5a3ac
That link contain a centre tap schematic circuit, not sure if there is anything i missed out.


Thanks NascentOxygen, but I don't really get it, what is CT?

 

[null void]

This is my just my assumption:

I consider the voltage drop at the resistance is same as the voltage drop at the reversed biased diode...am i right?

 

[Windadct]

Hello NV - When the DIode is fully reverse biased - what are the V on ether side of the diode?

 

[NascentOxygen]

This is my just my assumption:

I consider the voltage drop at the resistance is same as the voltage drop at the reversed biased diode...am i right?

CT is the centre tap, and that is grounded here. When one winding on the transformer has a voltage V volts, the other winding has a terminal voltage of -V volts. The conducting diode has no voltage loss across it. Mark these on your diagram, and you'll see what the voltage is on the reverse biased rectifier.

The load voltage is determined by the conducting diode.

 

[null void]

Opps accidentally replace the image of previous post. So i have to consider the forward biased diode as a jumper, and point a has voltage of V, and point B has voltage of 0V ?

 

[NascentOxygen]

Between point B and ground is a winding of the transformer! At the instant that A has a voltage of V volts (wrt ground), the voltage at point B has a voltage of -V volts (wrt ground) courtesy of its half of the transformer winding.

 

[null void]

Is that correct to view the circuit independently like in the diagram? How does the ground wire which direct connect to the transformer affect the circuit?

 

[sophiecentaur]

How are you defining the volts from the transformer? Is the "V" you have put on the diagram the volts referenced to Earth (i.e. is the transformer a V-0-V secondary)? It isn't clear from your sketch and it's vital for a proper answer.
But you have, basically a loop of wire from one end of the transformer to the other (you can ignore the resistor as it just appears in parallel with one of the windings) with a gap in it (the reverse biased diode). The voltage across that gap will be the same as the volts across the ends of transformer winding (=2V for a V-0-V transformer). Simples.

 

[null void]

sorry my V is the peak voltage of the transformer

 

[sophiecentaur]

sorry my V is the peak voltage of the transformer

Which "Peak Voltage"? There are three connections on it. (See my edit on my previous post)

 

[null void]

in a case of a series circuit with a transformer and a diode only, the voltage across the diode when it is reversebiased is also 2V_peak of the ac supplier source? or the transformer needto be grounded at the middle to create such condition?

 

[sophiecentaur]

If the total PD across the transformer are +V - (-V) =2V then it would have to be 2v.
This is why I was being picky about your use of the 'unqualified' value V. Do you take my point?
There is also the issue of Peak and RMS - which is why I slipped in the root 2 factor in my earlier post.
Grounding the centre tap is of no consequence, aamof, all that counts is the (vector) subtraction of the voltages at each end.
Note. If you only use one diode, there is a short circuit on the reverse cycle and what you say has no meaning. It's a totally different circuit that's required. You'd need to draw it out and it would be clearer.

 

[null void]

I am not sure if i am really get your point right now, the -V is the only thing bugging right now.

 

[NascentOxygen]

There are 2 identical secondary windings, and at any moment in time they produce identical voltages. So if you say there are V volts across one, then there must also be V volts across the other (regardless of whether there is current being drawn or not).

So for the situation depicted, with V volts across the lower winding, this places B at -V volts with respect to the other end of that winding (which being connected to earth, we call 0 volts).

 

[vk6kro]

As you can see, the center tap of the transformer is always at zero volts because it is grounded.

So the opposite ends of the transformer winding swing positive and negative relative to this ground connection and produce a peak voltage of 1.414 (√2) times the RMS voltage.

If you take the moment when the top winding produces a maximum positive voltage and the bottom winding produces a maximum negative voltage, there is a combined maximum reverse voltage across the bottom diode.
This is when it is most likely to fail and it has to be rated for at least this reverse voltage.

 

[null void]

There are 2 identical secondary windings, and at any moment in time they produce identical voltages.

this sound like what my lecturer said, he split the ac source into 2:

And if i am not wrong, each of the splited source can produce 0.5 of the V_peak. And is it because the circuit is somehow not completed, the voltage differnce across the the gap is same as the v_peak, like in this diagram:

 

[NascentOxygen]

EDIT: oops, I see you have it right after all.

Because the - of the upper source is earthed, then the upper winding delivers +V volts. Because the + of the lower is earthed (it shares the same centre tap), then the lower winding terminal delivers -V volts.

 

[NascentOxygen]

You have to get it clear in your mind and be consistent, either view it as one long winding producing 2V volts with a centre tap, or view it as two windings, each producing V volts.

If it's a 36V winding centre-tapped, then each half winding delivers 18V, and the output is 18V full wave rectified. This is the same as viewing it as a pair of 18V windings, with one end of each joined together and earthed (and phasing correct). Your second sketch with one winding showing no voltage is quite wrong.

 

[null void]

I didn't notice this at the beginning of the post, but after reading the posts you guys have posted, i slowly catch the idea. Thanks everyone for spending times on my problems :D

 

[null void]

I didn't notice this at the beginning of the post, but after reading the posts you guys have posted, i slowly catch the idea. Thanks everyone for spending times on my problems :D

 

[sophiecentaur]

There are 2 identical secondary windings, and at any moment in time they produce identical voltages. So if you say there are V volts across one, then there must also be V volts across the other (regardless of whether there is current being drawn or not).

So for the situation depicted, with V volts across the lower winding, this places B at -V volts with respect to the other end of that winding (which being connected to earth, we call 0 volts).

I'm sure you cannot mean that. The two windings of a centre tapped transformer are in antiphase - the terminal voltages are always of opposite polarity. (Otherwise there would be no point in having twice as many turns on the former). If one terminal is at +Peak volts, the other terminal is at -Peak volts.

 

[NascentOxygen]

I did mean that, because I was picturing a meter across each to measure the amplitude. I didn't want to complicate with phasing there, but mentioned it in parentheses later just in case OP was wondering.

 

[sophiecentaur]

I did mean that, because I was picturing a meter across each to measure the amplitude. I didn't want to complicate with phasing there, but mentioned it in parentheses later just in case OP was wondering.

But surely it's the phasing that makes the whole thing work. It is very misleading to imply that V is the same as -V, which is what you were implying. If you don't make it clear which end of the 'meter' is connected to what, it would naturally be assumed that the voltages would be referred to Ground. The two terminals have opposite signs of PD (relative to Ground). This is an idea that I have found missing throughout this thread and I think it's the reason that the OP has taken so long to understand things. It seems that he is now starting to get it - so it's been worth while I think.

There is, of course, the extra point that needs to be emphasised and that is the fact that the Earth and Resistive load connections are not, in fact, relevant to the the voltage drop situation 'around the loop' of transformer windings and diodes when calculating the PIV on each diode.

 

[null void]

So the tap play an important role? without it the piv will not be double?

 

[sophiecentaur]

So the tap play an important role? without it the piv will not be double?

Do you understand about transformer taps? A tap which is half way along the winding will have no effect on the total voltage across the winding (as long as there is not excessive current drawn via some path). As it happens, a centre-tapped transformer is normally specified as, say 100V-0-100V, meaning there is 200V across the whole winding and two 100V , antiphase outputs - relative to the centre tap (often, earthed). This is the configuration used in many power supplies with 'full wave' rectification (like yours).
I keep saying this but you don't seem to be taking it on board. It's the total voltage across the whole length of the winding that appears across the two series diodes. Chop off the Earth and resistor connections and that is obvious (?). I think the resistor and Earth connections must be confusing you but that diagram, earlier, with the two batteries, will also show my point. Disconnect the resistor and Earth and you have two batteries' worth of PIV on the open circuit diode. Re-connect them and you still have the same situation. On one end of the diode you have -V and the other end, connected to the resistor, has +V - making 2V across it. So it happens with or without the tap.

 

[null void]

Same as this? Voltage inversed = 24v ?

 

[sophiecentaur]

Same as this? Voltage inversed = 24v ?

Where does the 24V come from? You start with only one 12V battery so how can you get more? Have you ever come across Kirchoff's Laws?
Your transformer has TWO! voltage sources and not just one.

 

[null void]

You mean this? And the polarity of the the 2 source is same or opposite?

 

[NascentOxygen]

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.

 

[sophiecentaur]

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".

 

[NascentOxygen]

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?

 

[sophiecentaur]

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.

 

[NascentOxygen]

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.)

 

[sophiecentaur]

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?

 

[null void]

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.

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):

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.

 

[sophiecentaur]

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.

 

[null void]

Do u mean this:

blue arrow is the 'direction' of + volt

red is the 'direction of - volt

 

[sophiecentaur]

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.

 

[null void]

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...

 

[sophiecentaur]

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!

 

[null void]

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

 

[sophiecentaur]

"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?

 

[davenn]

it definitely doesn't help when he's talking about transformers and having them look like batteries and visa versa

Dave

 

[null void]

Ok, I will go back to the basic.

 

[sophiecentaur]

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?

 

[davenn]

I agree and don't call me "shirley" haha (cant remember which Naked Gun movie the shirley thing was from)

D

 

[Windadct]

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.

 

[null void]

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.