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

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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.
 
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null void said:
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.
 
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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?

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Hello NV - When the DIode is fully reverse biased - what are the V on ether side of the diode?
 
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null void said:
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.
 
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 ?
 
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.
 
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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?
 
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.
 
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null void said:
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)
 
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?
 
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.
 
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I am not sure if i am really get your point right now, the -V is the only thing bugging right now.
 
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).
 


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.
 
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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:
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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:

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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.
 
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.
 
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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
 
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
 
NascentOxygen said:
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.
 
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.
 
NascentOxygen said:
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.
 
So the tap play an important role? without it the piv will not be double?
 
null void said:
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 said:
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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.:cry:
 
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You mean this? And the polarity of the the 2 source is same or opposite?
 

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