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

1. May 23, 2013

### 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.

2. May 23, 2013

### Staff: Mentor

Hi null void. [Broken]

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.

Last edited by a moderator: May 6, 2017
3. May 23, 2013

### null void

4. May 23, 2013

### 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?

5. May 23, 2013

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

6. May 23, 2013

### Staff: Mentor

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.

7. May 23, 2013

### 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 ?

8. May 23, 2013

### Staff: Mentor

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.

9. May 23, 2013

### 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?

10. May 23, 2013

### 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.

Last edited: May 23, 2013
11. May 23, 2013

### null void

sorry my V is the peak voltage of the transformer

12. May 23, 2013

### sophiecentaur

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

13. May 23, 2013

### 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?

14. May 23, 2013

### 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.

15. May 23, 2013

### null void

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

16. May 23, 2013

### Staff: Mentor

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).

17. May 23, 2013

### vk6kro

[Broken]

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.

Last edited by a moderator: May 6, 2017
18. May 23, 2013

### null void

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:

19. May 23, 2013

### Staff: Mentor

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.

20. May 23, 2013

### Staff: Mentor

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.

21. May 23, 2013

### null void

I didn't notice this at the begining 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

22. May 23, 2013

### null void

I didn't notice this at the begining 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

23. May 24, 2013

### sophiecentaur

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.

24. May 24, 2013

### Staff: Mentor

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.

25. May 24, 2013

### sophiecentaur

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.