How does a centre tap transformer enable full wave rectification?

[PainterGuy]

hello everyone,

1:- in the below video i think the transformer used is called centre tap transformer. someone was telling me that the input voltage of 12V is divided between two voltages of 6V across the secondary coils. But at one time only one secondary coil is really conduction because the other one would one wouldn't be able to conduct because the diode is reverse biased.

hence i think at one time entire 12V appear across a single secondary coil instead of 6V.

https://www.youtube.com/watch?v=http://www.youtube.com/watch?v=dzVWhboZ38k

if the video embedding does not work above then please use this link:


2:- actually in this diagram i am trying to understand functioning of centre tap transformer. i think when the light bulb 2 is not conducting, all of 12 volts are appearing across light bulb 1. but when light bulb is also conducting (i mean switch is closed) then only 6 volts will appear across both light bulbs (assuming both bulbs have same resistance). is this what i say correct?

[URL]http://img268.imageshack.us/img268/6240/centretaptransformer.png[/URL]

many thanks for this help. much grateful, have been learning a lot from nice people here.

cheers

 

[I_am_learning]

Sorry, but no.
At any time the Voltage across S1 and S2 are 6 V each, irrespective of whether you connect the bulb or not.
The voltage induced in S1 and S2 are determined by no. of turns in them and in p1 and p2.

 

[Antiphon]

Also, your figure depicts two primaries. This is ok if they are wound on a single core but you did not draw the core. As it is, the figure depicts two separate transformers connected in series. It's not the same thing as a center-tapped transformer (unless you draw the core spanning all the coils.)

 

[PainterGuy]

Sorry, but no.
At any time the Voltage across S1 and S2 are 6 V each, irrespective of whether you connect the bulb or not.
The voltage induced in S1 and S2 are determined by no. of turns in them and in p1 and p2.

Also, your figure depicts two primaries. This is ok if they are wound on a single core but you did not draw the core. As it is, the figure depicts two separate transformers connected in series. It's not the same thing as a center-tapped transformer (unless you draw the core spanning all the coils.)

hello everyone,

thank you for helping me with this. i am a beginner, trying to understand things from practical and basic points.

actually the "1:-" in my post had a video. i think video embedding did not work there. please use this link:

now come to part "2:-" where i draw a diagram myself. when the light bulb 2 is not conduction (i mean the switch is open) then i do not think any voltage will be induced in S2 because the circuit of S2 is not closed. therefore all of supplied AC voltage is appearing across P1 which in turn induces current S1. the voltage can be induced in S1 because the circuit is closed and inducing require investment of energy which will be compensated by drop of volts across P1.

now please tell me your thoughts but please don't complicate things for me. many thanks.

cheers

 

[Jiggy-Ninja]

hello everyone,

thank you for helping me with this. i am a beginner, trying to understand things from practical and basic points.

actually the "1:-" in my post had a video. i think video embedding did not work there. please use this link:

now come to part "2:-" where i draw a diagram myself. when the light bulb 2 is not conduction (i mean the switch is open) then i do not think any voltage will be induced in S2 because the circuit of S2 is not closed. therefore all of supplied AC voltage is appearing across P1 which in turn induces current S1. the voltage can be induced in S1 because the circuit is closed and inducing require investment of energy which will be compensated by drop of volts across P1.

now please tell me your thoughts but please don't complicate things for me. many thanks.

cheers

Your diagram is a little inaccurate, physically speaking. This is the proper symbol for a CT transformer:

There is one primary winding, and one secondary winding, and a wire attached in the exact center of the secondary, which is why it's called center tapped

There are 6 volts across the bottom half of the secondary, whether or not the switch is closed. There will ALWAYS be 6 volts, no matter how much current is drawn from it or the other half of the secondary (assuming ideal transformer and no change in primary voltage). Likewise, there will always be 6 voltage across the top half, regardless of how much current is being drawn from either half.

Just because there is no current doesn't mean that there isn't a voltage.

 

[I_am_learning]

You need to start from basic transformer theory. What you are thinking (that the circuit needs to be closed for voltage to induce and the theory that inducing require investment of energy which is provided by... etc etc) is WRONG. Re-read the theories again. You need to understand simple transformer first before you understand center taps. Center-taps are not such a different thing, knowing the simple transformers automaticallly answers the question.
try these:
http://en.wikipedia.org/wiki/Transformer#Basic_principles
http://sound.westhost.com/xfmr.htm

 

[PainterGuy]

hello nice people,

i have drawn a new diagram:
http://img848.imageshack.us/img848/7963/transformer.png

first let me tell you what little bit of understanding i have about a transformer.

the ration is 3:1, so in the secondary voltage will be 200V but the current will be bigger as compared to the one in the primary.

i don't have any problem (at least till now!) when the switch is closed. so i will only focus on the situation when switch is open. in some previous i said there will be no voltage in the secondary when switch is open.

now i think this.:shy: the primary has frequency of 2hz - two cycles per seconds. which also means polarity changes four times every second. correct? when the electrons are pushed toward "A", "A" becomes negative and as electrons have been pulled away from "B" therefore it becomes positive. when in the next half cycle polarity changes, "B" will become negative and "A" positive. please remember i am only focusing on situation when switch is open. in this case there will be no current (at least no much which can be detected although electrons keep running left and right!). as switch is open, the charge will be pressed against "A" and "B". is this correct? there is voltage but no current, therefore bulb will not light up.

above in some post i also said inducing voltage in the secondary requires energy. i still think so. suppose you have taken "S" or secondary section of transformer circuit away. then all the voltage which is 600V will appear across resistor R1 because the primary coil is doing no work. if you say even when primary is not inducing any voltage anywhere, voltage is dropped in the primary coil, then tell me why is this drop.

cheers

 

[davenn]

i don't have any problem (at least till now!) when the switch is closed. so i will only focus on the situation when switch is open. in some previous i said there will be no voltage in the secondary when switch is open.

and has been pointed out earlier, that is incorrect, there will be a voltage across the secondary regardless of if the switch is open or closed.

think of a battery it has 1.5V, 6V whatever its rated at across its terminals regardless of if its in a circuit or not :)

if you say even when primary is not inducing any voltage anywhere, voltage is dropped in the primary coil, then tell me why is this drop.

in your example the voltage is being dropped across the 2 resistances present,
1) the resistance of the primary coil and
2) across the resistor you have in circuit.


cheers
Dave

 

[PainterGuy]

hello nice people,

i have drawn a new diagram:
http://img848.imageshack.us/img848/7963/transformer.png

first let me tell you what little bit of understanding i have about a transformer.

the ration is 3:1, so in the secondary voltage will be 200V but the current will be bigger as compared to the one in the primary.

i don't have any problem (at least till now!) when the switch is closed. so i will only focus on the situation when switch is open. in some previous i said there will be no voltage in the secondary when switch is open.

now i think this.:shy: the primary has frequency of 2hz - two cycles per seconds. which also means polarity changes four times every second. correct? when the electrons are pushed toward "A", "A" becomes negative and as electrons have been pulled away from "B" therefore it becomes positive. when in the next half cycle polarity changes, "B" will become negative and "A" positive. please remember i am only focusing on situation when switch is open. in this case there will be no current (at least no much which can be detected although electrons keep running left and right!). as switch is open, the charge will be pressed against "A" and "B". is this correct? there is voltage but no current, therefore bulb will not light up.

above in some post i also said inducing voltage in the secondary requires energy. i still think so. suppose if you have taken "S" or secondary section of transformer circuit away. then all the voltage which is 600V will appear across resistor R1 because the primary coil is doing no work on secondary. if you say even when primary is not inducing any voltage anywhere, voltage is dropped in the primary coil, then tell me why is this drop.

cheers

in your example the voltage is being dropped across the 2 resistances present,
1) the resistance of the primary coil and
2) across the resistor you have in circuit.


cheers
Dave

many thanks Dave.

even when the secondary is not taken away and is close to the primary and switch is open, even then the primary is doing no work on the secondary, though voltage is there in the secondary. P=VI, I is zero, so no power. that's why i think all voltage of the circuit of primary will appear across R1. is this true? tell me please.

cheers

 

[sophiecentaur]

That odd looking primary in the first picture was probably drawn that way because of the drawing / simulation package. Death to all simulations - they keep tripping people up.

 

[PainterGuy]

That odd looking primary in the first picture was probably drawn that way because of the drawing / simulation package. Death to all simulations - they keep tripping people up.

hello sophiecentaur,

which picture you talking about. most diagrams were drawn by me in the paint program. and yes one 'good looking' diagram was drawn by Jiggy-Ninja. I'm sure you are not making fun of my talent with the diagrams in paint software!

cheers

 

[sophiecentaur]

Would I ever?
The first picture (couldn't actually see both of them) was drawn with two primaries and no 'core'. It looked as if it had been copied (?) from another source and, indeed, one of the links has a transformer drawn in the same wrong way. That Utube movie doesn't have a centre tapped transformer but two separate transformers. It seems that it was brewed on some limited graphical electronic software that can't handle centre taps. I am having a problem with posts which seem to be treating the outcome of simulators as real events so I put two and two together to make π.
But the hand drawn diagram is charming.

I will, however, have rant about the use, in some of these posts, of 'electrons' whilst describing how a transformer circuit functions. Electrons have their place, of course, when dealing with some solid state device behaviour but they just serve to confuse in a simple straightforward EE discussion. Current, Volts, Resistance etc. are the appropriate terms. Avoid 'pushing' and 'pulling' electrons around and one is less likely to make gaffs about what will happen. Our everyday lives are lived in 'layers' and it is normal to discuss a problem on one level until it is actually necessary to go down a layer. Do we really want to discuss how the magnetic field is brought about by the action of moving electrons every time we do an inductance calculation?

 

[PainterGuy]

But the hand drawn diagram is charming.

some people just lie so beautifully!

and yes i think i 'coped' the idea from that video.

many thanks.

cheers

 

[PainterGuy]

many thanks Dave.

even when the secondary is not taken away and is close to the primary and switch is open, even then the primary is doing no work on the secondary, though voltage is there in the secondary. P=VI, I is zero, so no power. that's why i think all voltage of the circuit of primary will appear across R1. is this true? tell me please.

cheers

will someone please comment on the above text? please.

cheers

 

[sophiecentaur]

I can comment, in general, that you seem to want to learn about electricity with a series of questions and answers. This is a very demanding way of doing it. Demanding on the people who you are asking. And particularly demanding because you don't seem to be taking on board what you are being told. Your responses, to well informed answers, are of the form "Yes, but I look at it this way - how is this wrong?" and someone tells you again what they have already told you.
There are a number of misconceptions which you seem to insist on hanging onto and no amount of comments seem to change your 'beliefs'.

Why not try making an effort and read some ready made source of information like Wikki and try to make sense of it yourself? Text books at all levels are available, too, and they are probably the best way of increasing your understanding.
You need to sort out and learn what is actually meant by Potential Difference and Resistance and all the other quantities involved. We can't do that learning for you although there are many people on this forum (me, for one) who are willing to help with the occasional problem that you might have.

 

[PainterGuy]

I can comment, in general, that you seem to want to learn about electricity with a series of questions and answers. This is a very demanding way of doing it. Demanding on the people who you are asking. And particularly demanding because you don't seem to be taking on board what you are being told. Your responses, to well informed answers, are of the form "Yes, but I look at it this way - how is this wrong?" and someone tells you again what they have already told you.
There are a number of misconceptions which you seem to insist on hanging onto and no amount of comments seem to change your 'beliefs'.

Why not try making an effort and read some ready made source of information like Wikki and try to make sense of it yourself? Text books at all levels are available, too, and they are probably the best way of increasing your understanding.
You need to sort out and learn what is actually meant by Potential Difference and Resistance and all the other quantities involved. We can't do that learning for you although there are many people on this forum (me, for one) who are willing to help with the occasional problem that you might have.

hi sophiecentaur,

first let me offer my apologies if you think that i am asking unnecessary questions. i respect everyone's effort and time. i understand it takes much effort and energy to help someone, especially someone like me (as you say!). but seriously even if you can't notice i have been learning a lot from you guys and your efforts and help is not wasted even it may seem so. for example, in this thread i learned many things and now i have only one confusion left to clear out. you and many other people here are very advanced in your expertise so my questions from your point of you could sound foolish (and perhaps they are) but still i learn from what you say and tell me. some time i have to tell you people why i am thinking otherwise and why i am having difficulty grasping some new concept. i hope you understand what i am trying to say. and thanks for always helping me out in many of my "demanding" questions!

the confusion is:
even when the secondary is not taken away and is close to the primary and switch is open, even then the primary is doing no work on the secondary, though voltage is there in the secondary. P=VI, I is zero, so no power. that's why i think all voltage of the circuit of primary will appear across R1. i am only saying there is voltage in the secondary but no power is being 'taken out' from the secondary, therefore primary is inducing voltage in the secondary but as this induced voltage is doing no work therefore (if i may say so) primary's all power remains with in the primary.


cheers

 

[sophiecentaur]

You are doing exactly what I have described all over again.
Go to Wikipedia and read how a transformer works. Don't try you own home-brewed approach which is yielding nonsense. The conventional electrical theory is straightforward and correct. It predicts very accurately how things will work. Stop messing about and learn some basics and you will realize that all the questions you have been asking are not even valid questions.
There is absolutely no alternative to the framework which is known and loved. If you can't be bothered to learn it then you will never understand this stuff.
Read a blasted BOOK.

 

[Proton Soup]

i think you need to back up a step. if you really want to understand how transformers work (and I'm not sure i entirely do anymore), then you're going to have to learn all about FLUX.

http://en.wikipedia.org/wiki/Magnetic_field#The_magnetic_field_and_electric_currents

this will take a lot of navel gazing about the Right-Hand Rule.

http://en.wikipedia.org/wiki/Right_hand_grip_rule

magnetic flux produced by current is spooky physics. it makes no intuitive sense at all. it simply is. but spend some time pretending that wire runs in the direction of your thumb, and that magnetic flux surrounds the wire moving in the direction of your curled fingers.

now, run your fist around each turn of that primary coil. notice that with each turn, you add more and more flux to the transformer core. that core is like a conductor for flux. flux runs around that core like electrons run through the wire. and the more turns you add to the primary, the more flux you add to the core.

on the secondary side, it is the reverse. every time you wrap a turn on the secondary, you link with more flux. every turn adds more electromotive force (voltage). that force, or voltage, is not current. it is only the potential to produce current. (notice that the words voltage and potential are often used interchangeably). if that force is impeded by infinite (well, almost) impedance (open circuit), then it's just like you pushing as hard as you can against a brick wall (nothing moves, and that means electrons).

but yeah, what [STRIKE]s[/STRIKE]he said. try a physics book. take a class. spend a lot of time thinking about the basic principles and solving textbook problems, or it will always be a big confusing mess. and you may even want to back up a step and learn F=m*a physics, because the analogy of voltage as force makes more sense when you understand that work only occurs when a force actually move that mass through some distance. if your electromotive force isn't moving electrons through a distance, then it's not doing work, either.

 

[Integral]

...

the confusion is:
even when the secondary is not taken away and is close to the primary and switch is open, even then the primary is doing no work on the secondary, though voltage is there in the secondary. P=VI, I is zero, so no power. that's why i think all voltage of the circuit of primary will appear across R1. i am only saying there is voltage in the secondary but no power is being 'taken out' from the secondary, therefore primary is inducing voltage in the secondary but as this induced voltage is doing no work therefore (if i may say so) primary's all power remains with in the primary.


cheers

Much of what you say is correct. No current means no power. However, in the real world, there will be some hysteresis losses in the core.

You are wrong however in your guess at where the voltage will appear. You know the resistance of R1 and the current through it, so use Ohm's law to find the voltage drop. I'll let you do the algebra.

Now, where DOES the voltage appear? It certainly appears across the open switch, source voltage will always be found across an open switch. So what is the source of the voltage?

You will measure voltage between any two points along a current path which includes a secondary coil. The voltage is induced in the secondary by the changes in the primary current.
Edit.
I just looked a bit closer, R1 is in the primary! ? I thought we were talking about the secondary circiut. Everything I said applies to the secondary.

 

[sophiecentaur]

Changes in the primary current.


And "she" 's a bloke.

 

[Proton Soup]

right, di/dt.

sorry, the name threw me.

 

[sophiecentaur]

Too late to change. Bad choice in some ways but it contains the name of my boat. :-)

 

[davenn]

hey gang...

just musing over the comments of the last couple of pages, including my own ;)

correct me if I am wrong.
...an alternating AC (of course) voltage in the primary is going to induce an alternating voltage into the secondary, even when the secondary has no load.

isnt that rising and falling magnetic field coing to induce very brief pulses of current flow in the secondary windings as well ?

yes the current can't flow the full path as there is no load ( complete cct), but would a sensitive AC ammeter in either leg of the winding away from the transformer secondary measure brief bursts of current flow as the (for want of a better description) electrons cycle back and forward ?

I will now sneak back to my Richard Feynman lectures on photons etc ;)

Dave

 

[sophiecentaur]

It won't induce "brief pulses" buthere may be a low value of reactive current flowing in the secondary, due to the self capacitance of the winding which will appear in parallel with the inductance. The waveform will still be sinusoidal if the primary voltage is sinusoidal, though.

You don't need Feynman for this - what did he know?

 

[davenn]

You don't need Feynman for this - what did he know?

hey mate hahaha
no you dont, I was in another thread in the general physics section reading someones queries on the "bending" of light passing through a medium.
Cragar has suggested to him/me to go watch some 1979 lectures he did at the university of Auckland, NZ about photons etc.

cheers
Dave

 

[PainterGuy]

hello everyone,

i think you need to back up a step. if you really want to understand how transformers work (and I'm not sure i entirely do anymore), then you're going to have to learn all about FLUX.

http://en.wikipedia.org/wiki/Magnetic_field#The_magnetic_field_and_electric_currents

this will take a lot of navel gazing about the Right-Hand Rule.

http://en.wikipedia.org/wiki/Right_hand_grip_rule

magnetic flux produced by current is spooky physics. it makes no intuitive sense at all. it simply is. but spend some time pretending that wire runs in the direction of your thumb, and that magnetic flux surrounds the wire moving in the direction of your curled fingers.

now, run your fist around each turn of that primary coil. notice that with each turn, you add more and more flux to the transformer core. that core is like a conductor for flux. flux runs around that core like electrons run through the wire. and the more turns you add to the primary, the more flux you add to the core.

on the secondary side, it is the reverse. every time you wrap a turn on the secondary, you link with more flux. every turn adds more electromotive force (voltage). that force, or voltage, is not current. it is only the potential to produce current. (notice that the words voltage and potential are often used interchangeably). if that force is impeded by infinite (well, almost) impedance (open circuit), then it's just like you pushing as hard as you can against a brick wall (nothing moves, and that means electrons).

but yeah, what [STRIKE]s[/STRIKE]he said. try a physics book. take a class. spend a lot of time thinking about the basic principles and solving textbook problems, or it will always be a big confusing mess. and you may even want to back up a step and learn F=m*a physics, because the analogy of voltage as force makes more sense when you understand that work only occurs when a force actually move that mass through some distance. if your electromotive force isn't moving electrons through a distance, then it's not doing work, either.

many thanks Proton Soup. okay i promise to learn the basic. the problem with the wiki and many other websites is they are difficult to understand for people like me who first want to have some conceptual understanding and then mathematical. english is not my first language. i am learning it with a lot of struggle and giving a lot of attention. so i hope you see the problem. i know things are very simple and basic from your view but they aren't for me and many other people like me around the world. if you were smart when you were a child then that's good. but perhaps i am not smart and need guidance. smart people can do things with very little guidance. there are millions of people like me around the world who need perhaps more guidance than other people. i still try to learn. actually i was bit hurt after reading sophiecentaur's post before yours. well it's okay i can understand sophiecentaur's concern too. his/her advice was that i try to be self sufficient along the way and don;t need to ask questions. okay i understand it.

okay, my thanks for this advice.

Much of what you say is correct. No current means no power. However, in the real world, there will be some hysteresis losses in the core.

You are wrong however in your guess at where the voltage will appear. You know the resistance of R1 and the current through it, so use Ohm's law to find the voltage drop. I'll let you do the algebra.

Now, where DOES the voltage appear? It certainly appears across the open switch, source voltage will always be found across an open switch. So what is the source of the voltage?

You will measure voltage between any two points along a current path which includes a secondary coil. The voltage is induced in the secondary by the changes in the primary current.
Edit.
I just looked a bit closer, R1 is in the primary! ? I thought we were talking about the secondary circiut. Everything I said applies to the secondary.

thank you Integral. it was really encouraging for me where at the start you said most of whatever i said was correct. thanks

hey gang...

just musing over the comments of the last couple of pages, including my own ;)

correct me if I am wrong.
...an alternating AC (of course) voltage in the primary is going to induce an alternating voltage into the secondary, even when the secondary has no load.

isnt that rising and falling magnetic field coing to induce very brief pulses of current flow in the secondary windings as well ?

yes the current can't flow the full path as there is no load ( complete cct), but would a sensitive AC ammeter in either leg of the winding away from the transformer secondary measure brief bursts of current flow as the (for want of a better description) electrons cycle back and forward ?

I will now sneak back to my Richard Feynman lectures on photons etc ;)

Dave

thanks for the input. seriously i also had that thought somewhere in my mind but didn't dare to ask.

cheers

 

[PainterGuy]

This is the proper symbol for a CT transformer:

Hello once again everyone,

Just a small question I hope you would not mind its asking.

I will use Jiggy-Ninja's diagram for my question. As the CTT splits the voltage into 120V around the two secondaries, so I was thinking if it is possible to get some 240V some way. I think this is not possible. By combining the voltage one can get 'overall' increased current as is the case with combining two batteries in parallel.

Have a see here on this link please:-
http://img97.imageshack.us/img97/9539/ctt1.jpg

"B" in the diagram is a bulb and both batteries of 120V connected in parallel.

Is my understanding correct? Tell me please.

Cheers

 

[sophiecentaur]

Connect line 1 to ground / Earth and you have 240V AC. No problem. 120 +120 = 240

 

[PainterGuy]

Many thanks sophiecentaur.

Suppose both primary and secondary carry currents in phase. In the diagram (find link below) R_L is any load such as bulb, C is current and arrows show direction of currents. Do you think is it somewhat correct? And now I am not using electrons!

Diagram:
http://img194.imageshack.us/img194/6163/ctt2.jpg

Cheers

 

[sophiecentaur]

Why should you expect the primary and secondary currents to be in phase? That's a huge assumption.

 

[PainterGuy]

Seriously I don't know.

Jiggy-Ninja told me that if you have windings wounded in the same directions then the currents in the primary and secondary are in phase. That's what I know.

So to keep things simple and understandable for me I assumed they are in phase.

Is that diagram correct? Tell me please.

Cheers

 

[sophiecentaur]

I think he meant that the phases of the two Secondaries are equal - so the voltages add up.
That diagram has an error in it. There is a connection across one of the secondary windings. That is a short circuit. You need to erase the line coming from the centre tap. The secondary is now a single winding of twice the number of turns.
I don't understand why you don't just look at a link about simple transformer theory. Did you ever google it?

 

[I_am_learning]

Hello once again everyone,

Just a small question I hope you would not mind its asking.

I will use Jiggy-Ninja's diagram for my question. As the CTT splits the voltage into 120V around the two secondaries, so I was thinking if it is possible to get some 240V some way. I think this is not possible. By combining the voltage one can get 'overall' increased current as is the case with combining two batteries in parallel.

Have a see here on this link please:-
http://img97.imageshack.us/img97/9539/ctt1.jpg

"B" in the diagram is a bulb and both batteries of 120V connected in parallel.

Is my understanding correct? Tell me please.

Cheers

With that drawing you would explode the transformer.
Actually, the drawing (both Ninja's and Yours) is quite misleading for beginers. The top side voltage should be written as +120, Central as 0 and Bottom side as -120. That should make it clear, that you shouln't join the top side and bottom side, it would be like placing two batteris End to end (i.e. in series) and then short circuiting their the end terminals.

(Yeah, you can get + and - voltage in AC also. That occurs when two voltages are perfectly 180 out of phase.)

 

[Evil Bunny]

Why should you expect the primary and secondary currents to be in phase? That's a huge assumption.

Because they are. He is exactly correct. Those arrows on the secondary will be pointing in the same direction (Now, they may or may not be pointing the same way as the primary) but they will absolutely be pointing in the same direction. In phase!

EDIT: I just read it again. I think you are saying the same as me. The primary and secondary currents might not be flowing in the same direction... my point was that If the arrow is pointing up from L2 to neutral it will also be pointing up from neutral to L1 at the same time. I think you're saying the same thing. My apologies.

 

[sophiecentaur]

I was just going to get shirty with you. Then I read the second half. Ha ha.