# Center-tapped rectifier without diodes?

• ViolentCorpse
In summary: Vcc, then -...-....-....-....-....-....-....-....-....-....Vdd and so on.In your second diagram you have labelled the coil voltages vi wrongly.
ViolentCorpse
Hi everyone. I'm back again with another silly question. Please bear with me.

Below is a figure of a basic center-tapped full-wave rectifier circuit and my question is; why can't we use anything other than diodes to get a unidirectional output current? For example, what difference would it make to the output if I use two resistors instead of the diodes D1 and D2?

If we imagine that there are resistors in the place of diodes, the way I see it, during the positive portion of the AC input cycle (with the polarities as they're shown in the figure below), the direction of conventional current through the load resistor R is from right to left. And to me it looks like that should still be the case during the negative portion of the AC input cycle. The polarities would constantly be changing in the top and bottom resistors, but my object of interest is the resistor in the middle, which doesn't seem to be (to me) changing polarities any time during the entire cycle.

So what do semiconductor diodes add to this circuit that other components (or simple conducting wires, for that matter) can not?

Much thanks and apologies for my ignorance.

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Consider the voltage over the upper coil as v1 and over the other coil as v2.

The center tap of the transformer can be considered a reference point. So, if you simply "ground" it you will understand that that point is always with 0 volts (as it is the reference). Then, when the voltage supply is at its positive half-cycle, v1 is positive and D1 enters conduction. On the other side, D2 is blocked because it is reversed polarized. The voltage over the resistor is equal to v1 (v1 on the right side of the resistor and ground on the left side of the resistor).
During the negative half-cycle, D1 gets blocked and, as v2 polarizes D2, it conducts. On the right side of the resistor there is v2. On the left side there is the ground.

As v1 = v2 (that is a center tap, right ? So both voltages are equal), the voltage across the resistor is always positive, never gets negative (you can see that using the simulation tool I mentioned).

Thanks for your response. But I think you didn't understand my question.

My question is, why are diodes necessary for achieving full-wave rectification in this circuit and why can't we simply use conductors. Wouldn't the voltage across resistor still always be positive without semiconductor diodes, since the current path for the resistor is only from right to left?

ViolentCorpse said:
Thanks for your response. But I think you didn't understand my question.

My question is, why are diodes necessary for achieving full-wave rectification in this circuit and why can't we simply use conductors. Wouldn't the voltage across resistor still always be positive without semiconductor diodes, since the current path for the resistor is only from right to left?

because diodes ONLY conduct in one direction. Resistors conduct in both directions and would not provide rectification. No the current is not only from left to right. its from both directions. its direction changes with each half cycle of the AC

Each diode only conducts on one half of each cycle those 2 conductions are then combined to give an overall positive raw DC output which is then smoothed with capacitors.

Dave

ViolentCorpse said:
Since the current path for the resistor is only from right to left?
How so?
I would be glad if you could Draw two diagrams:
1. Make the Top of coil +ve and bottom -ve. Draw current directions in all branches and Voltage magnitudes at all nodes.
2. Same as 1, but with Polarity of the Coil reversed.

I_am_learning said:
How so?
I would be glad if you could Draw two diagrams:
1. Make the Top of coil +ve and bottom -ve. Draw current directions in all branches and Voltage magnitudes at all nodes.
2. Same as 1, but with Polarity of the Coil reversed.

I edited the diagram I attached on my original post and drew current directions quite lazily. I don't know the voltage magnitudes, so I left that part out (I hope that's not a big problem).

Thanks!

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ViolentCorpse said:
I edited the diagram I attached on my original post and drew current directions quite lazily. I don't know the voltage magnitudes, so I left that part out (I hope that's not a big problem).

Thanks!

But I was expecting that you would replace the diodes with resisters (or conducting wires).
Anyways,
In your second diagram you have labelled the coil voltages vi wrongly.
Starting from top it would be - + - +.
Even better do this:
Assuming a total coil voltage of 20Volts;
In the first diagram, starting from top node of the coil, the voltages would be
Top Of coil = 10V
Middle of coil = 0V
Bottom of Coil = -10V
In the second diagram
Top of Coil = -10V
Middle of coil = 0V
Bottom of Coil = 10V

Now, use these voltages and draw the current diagrams; once using diodes and once using conducting wires (or resisters) instead of diodes.
When you use diodes remember that it prevents current from flowing in the reverse direction.

You are incorrectly assigning polarity to the transformer output. A.C. input = A.C. output.

pantaz said:
You are incorrectly assigning polarity to the transformer output. A.C. input = A.C. output.

yeah I was going to make the same observation :)

Dave

ViolentCorpse said:
My question is, why are diodes necessary for achieving full-wave rectification in this circuit and why can't we simply use conductors.

You don't necessarily need diodes, but you do need some components that have a nonlinear response, i.e. the graph of current against voltage is not a straight line through the origin. For "conductors" (I assume you really mean resistors) the graph is a straight line - that's what Ohm's law says.

The reason why diodes are used far more often than any other nonlinear component is because they come close to having the "ideal" nonlinear behaviour that is required.

Hello VC - I think the issue you are facing is the "+" and "-" symbols used in your diagrams. The top coil is not always Positive (+), it only is when the primary coil or Vi is Positive. When the Vi goes negative all of the polarities in the system reverse.

That is why "+" and "-" symbols on transformers is misleading - and preferably just use a black dot on both the Primary and secondary - indicating their relationship to one another. Not that the coil is always positive.

I_am_learning said:
But I was expecting that you would replace the diodes with resisters (or conducting wires).
Oops. That totally slipped out of my mind. Sorry.

Anyways,
In your second diagram you have labelled the coil voltages vi wrongly.
I assume you mean the CT voltage polarities? Yeah, I didn't bother changing it because, well to be honest, I have no idea what a center-tap actually is. I'm just assuming that it acts as a ground to complete the circuit. I don't know what the + and - specified above and below it in the diagram really mean.

Assuming a total coil voltage of 20Volts;
In the first diagram, starting from top node of the coil, the voltages would be
Top Of coil = 10V
Middle of coil = 0V
Bottom of Coil = -10V
In the second diagram
Top of Coil = -10V
Middle of coil = 0V
Bottom of Coil = 10V
Ermm.. wouldn't the current through the middle coil then be 0A due to 0V? And

If one end of a resistor is at 10V and the other is at 0V, won't you expect to get a current flowing? It's PD that is responsible for current flow and not just Potential.
I think, when one has a problem in understanding something that is as tried and tested as a basic rectifier circuit, the temptation to conclude that the circuit is wrong should be resisted. One should ask oneself "what have I misunderstood about this?" and not try to find reasons why one might not be wrong.

Eh, I'm not being presumptuous. I know that the fault is in my understanding and I don't think I've implied otherwise. I'm sorry if you felt that way.

If one end of a resistor is at 10V and the other is at 0V, won't you expect to get a current flowing? It's PD that is responsible for current flow and not just Potential.
I said earlier that I don't know what the center-tap really is and I am assuming it to be a ground a.k.a a potential of 0V. Starting from that assumption, I think it may not be so wrong to say that you're going to get 0A current.

ViolentCorpse said:
Eh, I'm not being presumptuous. I know that the fault is in my understanding and I don't think I've implied otherwise. I'm sorry if you felt that way.

I said earlier that I don't know what the center-tap really is and I am assuming it to be a ground a.k.a a potential of 0V. Starting from that assumption, I think it may not be so wrong to say that you're going to get 0A current.

Sorry but there have been 'explanations' earlier on in the thread and you don't seem to have taken them on board. Anyway-
It can be at any potential you care to take it. You could attach it to the output of a 2kV transformer if you wanted to. But it's the Potential at the other ends of the connected components that determine the current flow. Wherever it was connected, the other 'ends' of the secondary would be at potentials, respectively 10V above and 10V below (or in antiphase, to be more precise). That (potential difference), as I said before, is what determines the current. If two equal resistors are connected in series between the two ends, then the potential (of course) at their junction would be the same as that of the centre tap (the potential divider situation). No net current will flow into the centre tap when connected via a resistor R. Is this what you were saying? It's certainly why you need to have diodes, which have (approximately) zero and infinite resistance (i.e. unequal resistances), depending on the polarity of the connection so they act like switches so that on alternate cycles, it is the forward conducting diode that let's current flow from the end with the positive PD wrt the centre tap and NOT into the other end, against the 'off' diode.

The arrows in those two diagrams early do not show what's happening very well. The arrow 'into' the reverse biased diode should be zero in length (or at least very short) to show the process whereby the current always flows 'into' the CT.

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VC - it still seems that you are looking at this as the "+" of the coil is always positive V relative to the center tap - like a battery (Direct Current =DC) - but it is not, the actual voltage in this case varies from +V to -V; Transformers use Alternating Current (AC) ; as in the Sine Wave on the left of your original diagram. For the first half of the wave the Polarity is positive then for the second half all of the polarities are reversed ( in this simple case)

To original poster...current flows one direction (from source) during the positive sign of the sin wave.
On the negative side of the sine wave current flows the other direction. (from source)

Without one way diodes you get crappola.

The current must be stopped in one direction to produce a sin wave that just "rides" on the top of the x axis. With the diodes, The load (resistor) only gets current in one direction thru it. This is how direct current is made from AC. The current MUST be stopped in one direction twice per cycle (full wave) or you get nada.

VC - it still seems that you are looking at this as the "+" of the coil is always positive V relative to the center tap - like a battery (Direct Current =DC) - but it is not, the actual voltage in this case varies from +V to -V; Transformers use Alternating Current (AC) ; as in the Sine Wave on the left of your original diagram. For the first half of the wave the Polarity is positive then for the second half all of the polarities are reversed ( in this simple case)

Yes I'm aware that the input is AC and the top coil is not always positive (disregard my drawings. That was an embarrassing attempt. I'll stop being lazy and make another one). What I am saying is that, my poor and fallacious understanding (for which I apologize again) tells me that if I would be using three resistors, one resistor on top, one in the middle (that would also be my output), and the the third on the bottom, the current through the MIDDLE resistor should always be - according to my fallacious understanding - in the same direction.

I have no problem understanding that the current through the top and bottom resistors would indeed be an A.C current, but the circuit geometry of the common center-tap rectifier circuit suggests (only to me, of course) that the resistor in the middle should always be getting a current in one direction only, irrespective of the changing polarities of the source.

I don't know if it is possible for current to flow FROM the center tap TO the right-middle node of the resistor during some particular portion of the full cycle, but if it is, then that's problem solved for me. Now, is it?

I really appreciate all the replies I've got (especially I am Learning), but unfortunately, I haven't been able to quite catch it and I admit that is my problem. The moderators can lock this thread if they want to.

Edit: Attaching a little more accurate drawing of what I am trying to say here. The arrows show the current direction.

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You are (still) very selective in your assimilation of information about this!. If you read my earlier post you would see that the mid point of the two resistors (the ones replacing the diodes) is at the same potential as the centre tap (assuming the resistors are equal). No current will flow through R in that case because, at all times, the potential difference is zero across it. That arrow you have drawn should have ZERO length in each case.
This is, in fact, a trivial version of the well know Bridge Circuit where no current flows when the ratio of impedances on each side is the same. Look up Wheatstone Bridge.

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Oh I'm so sorry sophiecentaur, I missed your last post before this one and just read it now that you mentioned it. I think I'm finally getting it now.

I actually did say, after I am Learning gave me node voltages to work with that the current should then be 0A through the middle coil, remember?

And yes I have studied Bridge Circuits and have even solved a number of problems related to it. Thanks for mentioning that, it makes it easier to understand this now.

Thanks a ton sophiecentaur, Windadct, I am Learning and each and everyone of you. And I'm sorry for being a little... slow.

I've just got one more question, what if the resistors weren't equal (feel free to not answer this one, as my main problem has been solved.)

I've just got one more question, what if the resistors weren't equal (feel free to not answer this one, as my main problem has been solved.)

The current would split according to the ratio of the resistors...and the resistor in middle would get half the voltage of the outside resistor. This is essentially the case in your breaker panel in your home if you live in USA. Most loads are getting 120 volts...other loads are getting 240 volts.

Also, if using your 120 or 240 like in a USA house...you would almost never have more than 1 resitor in series. Putting resistors in series splits voltages. Not good...appliances and most things like their full voltage. That is why the original picture you show only shows one resistor!

Also, you would not get DC...you would get full AC thru the three resistors (loads) you are proposing.

YOur home USA electrical distribution to your house is almost identical to your original picture...just no diodes in that case because you actaully want the AC!

Sophiecentar is a busy man...he doesn't have time for this:)

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Well back to the original post - the Resistor attached to the Center Tap - is actually the "load" in which we would be looking for DC power. So for half of the cycle the current flows through the top Diode and back through the Center Tap (load) and for the Other Half cycle the current flows through the bottom diode back through the Load - the current in the load is allways one direction ( DC ) if the actual wave form is a Square Wave ( not a sine) then the DC Voltage will be more consistent.
As your latter posts are asking - If all of the resistors ( actually just the top and bottom) are equal - but no diodes, there will be no current in the Center Tap ( ideally) - but then there is no voltage across the Resistor at the center tap - which per the original post is really supposed to be the load - so this is a completely different case.

Maybe this will help. Using the Analog Circuits applet at falstad.com, I created a simple simulation of the transformer with resistors (no diodes). The sim animates the flow of electrons.

This link will open the Java applet with the circuit running. The three "oscilloscope" displays are for the top, middle, and bottom resistors.
http://tinyurl.com/8hyds3v

Here's their sim of a basic full-wave rectifier:
http://tinyurl.com/9yp7fpj

pantaz said:
Maybe this will help. Using the Analog Circuits applet at falstad.com, I created a simple simulation of the transformer with resistors (no diodes). The sim animates the flow of electrons.

This link will open the Java applet with the circuit running. The three "oscilloscope" displays are for the top, middle, and bottom resistors.
http://tinyurl.com/8hyds3v

Here's their sim of a basic full-wave rectifier:
http://tinyurl.com/9yp7fpj

I can't decide what the 'throbbing colours' are supposed to represent on the first sim. They seem to imply that, even with equal values of 'series' resistors, current will flow in the load. This is clearly not the case because the 'bridge' is balanced and that resistor shouldn't be throbbing.
Also, why not just call it Current, rather than electron flow? You are just adding another level of (irellevant) complexity; all the electrical calculations I have ever come across use V and I (defined in a way that there is no misunderstanding).

psparky said:
Also, you would not get DC...you would get full AC thru the three resistors (loads) you are proposing.

Hmmm. I see.

Thank you everyone! :)

With only linear components, you can't produce a signal with anything other than the original frequency components (including f=0).

ViolentCorpse said:
Also, you would not get DC...you would get full AC thru the three resistors (loads) you are proposing.

Hmmm. I see.

Thank you everyone! :)

Also, no one in there right mind would hook up the three resistors you show in a center tapped transformer. Instead they would feed a main panel with the two hots and one common (center tap) and then they would use three different breakers or feeders from their panel. Two 2-pole breakers at 240 and One 1-pole breaker at 120 volts. However, to make rectified DC from AC, they certainly would use your configuration, but they would use two diodes on the outside and one resistor in the middle:)

Keep asking questions. Everyone on this message board without exception learns something whether they are asking questions or answering questions.

ViolentCorpse said:
I said earlier that I don't know what the center-tap really is ...

The center tap just litterly taps the center of the coil instead of the outside of the coil. Let's just say you are talking about two wires and using the 240/120 example.

If you wire a load from the two outside ends of the coil you get 240 volt. If you wire a load from one end of the coil to the middle...you get 120 volts. It only gets half the voltage. This is exactly what's going on with your receptacles, lights or 120 Volt loads in a USA home.

I said earlier that I don't know what the center-tap really is ...[QUOTE/]

Each turn of the secondary coil provides a small voltage and all the volts are in phase. The turns are in series so the volts add up over the total length of the coil. Half way up, the volts are half (wrt the 'bottom end') than the volts at the top. PDs are all relative so, using the centre tap as reference, the two ends have equal volts and are in antiphase. It can be handy to connect the centre tap to Ground and then you have half supply volts to run your lamps etc. on - more safely. When you need some serious power, you can use the full 'end - to - end' volts without needing an embarrassing amount of current (thick wires); the volts are double so the power can be double for the same current.
I have recently come to realize that this has some advantages over the UK (all at 230V) system because the engineering of plugs and sockets can be much more sloppy / cheap than what's used in the UK.

sophiecentaur said:
I can't decide what the 'throbbing colours' are supposed to represent on the first sim.
The description of the applet says:
"The green color indicates positive voltage. The gray color indicates ground. A red color indicates negative voltage. The moving yellow dots indicate current."

They seem to imply that, even with equal values of 'series' resistors, current will flow in the load. This is clearly not the case because the 'bridge' is balanced and that resistor shouldn't be throbbing.
IMHO, the simulator is not for serious design and analysis, and since that particular circuit really doesn't follow normal design rules, the extraneous "throbbing" is probably just an artifact of the underlying Java code.

If you want an accurate answer, you would have to ask the applet's creator.

Also, why not just call it Current, rather than electron flow? You are just adding another level of (irellevant) complexity; all the electrical calculations I have ever come across use V and I (defined in a way that there is no misunderstanding).
I apologize for any confusion. It was almost 1:00 am, I was very tired, and I chose my words poorly.

The reason that I took exception to the applet, looking again, was that there was an implication that the centre tap voltage changes. I now realize that the secondary circuit actually has no ground, so the voltages are a bit meaningless as they are not referenced to anything. It would have helped if there had been a ground connection on the CT and then the colour of the middle resistor could have been grey all the time (which is surely what the whole thing is about). Any ground connection would have helped. The non-moving charges in the middle leg make a good point.

Sorry for the grumpy bit about electron flow but I have a 'thing' about it. Bringing it into every electronics explanation is guaranteed to confuse at least a few of the readers. It ranks with 'the photon explanation' as something that can do as much harm as good in most circs.

Violent Corpse,
AlephZero made a comment you have not taken on board.

Armed with your newfound knowledge I suggest you revist his post.

He said you do not have to use diodes to achieve rectification.

However resistors, (or capacitors or inductors) will not do.

You could replace the diodes with switches. If you could switch them on and off at appropriate intervals you could achieve rectification.

This is not as silly as it sounds as in the past this was actually done when suitable diodes were unavailable for power reasons.

Studiot said:
Violent Corpse,
AlephZero made a comment you have not taken on board.

Armed with your newfound knowledge I suggest you revist his post.

He said you do not have to use diodes to achieve rectification.

However resistors, (or capacitors or inductors) will not do.

You could replace the diodes with switches. If you could switch them on and off at appropriate intervals you could achieve rectification.

This is not as silly as it sounds as in the past this was actually done when suitable diodes were unavailable for power reasons.
Yes sir, I have taken that comment on board. I understand that if we can make any other non-linear component work as a switch, it should work as a rectifier. However, my main concern was, with the original circuit diagram still in mind, why couldn't we use ANY component for the purpose?

Thankfully, my question has been answered for equal resistors. All the posts have been very informative, and I've been religiously following this thread for any new comments. :D

Studiot said:
You could replace the diodes with switches. If you could switch them on and off at appropriate intervals you could achieve rectification.
And to put a picture with it:

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