I am having trouble visualizing/understanding how the current flows in a rectifier of this type... I would seem almost that current on the center tap would be moving in 2 different directions at once.... I have been trying to figure this out on my own but I am at that "head is splitting, having bad dreams about rectifiers" point... Could anyone walk me through the attached schematic from volume 3 chapter 3 "Diodes and Rectifiers" Please!?!

Thanks for you help!

 

when load is symmetrical then current will flow like this:

 

The image you posted:

Notice that the black test probes for both meters are connected to the center tap of the transfomer's secondary winding. This means that the center tap is being used as the 0v reference point, or ground; so all voltage measurements on the secondary side are relative to that 0v point.

If the upper end of the secondary winding is positive, then the lower end of the secondary winding is negative - if the loads are the same value, it will be equal in amplitude but opposite in polarity.

Let's just talk about the upper half of the secondary winding and the center tap for the moment.
Since the upper end of the secondary winding is positive, conventional current flow will go from the top of the winding through the anode of the upper black diode, out through the cathode, through the upper load resistor, and back to the transformer via the center tap.

On the other half of the AC cycle, the upper end of the secondary winding will be negative in respect to the center tap. In this case, the upper black diode will be turned off. Conventional current flow will come from the center tap, down through the lower load resistor, then up at the junction to flow through the anode of the upper blue diode, and into the upper end of the secondary winding.

So, the upper blue and black diodes steer the current flow for the upper half of the secondary winding during the positive and negative portions of the cycle.

The lower half of the secondary winding's current is steered by the lower blue and black diodes, in the same way the upper half is handled.

Does that make more sense to you now?

If you are still uncertain, print out a couple copies of the diagram, mark the polarities of the secondary winding, and use different colored marking pens, crayons, or whatever you have available to trace out the current flow through the transfomer secondary, diodes, and loads.

 

Ok this is helping... So is Jony123's drawing inline with SgtWookie's explanation? What I mean is are you guys in agreement?

So that I can understand what happens simultaneously with both the upper half and lower half of the windingins let me ask a few things?

If the upper half of the winding is positive and the convential current flow is from the upper half of the winding through the upper black diode then back to the winding through the center tap, what does the current flow look like on the bottom half of the windings?

To me it would seem that the current from the upper half would flow through the upper diode through BOTH loads to the lower half of the windings, but if thats the case then how would crrent be flowing for the lower?

It seems like when one half is sending current to the windings via the center tap the other half is pulling it away from the windings via the center tap at the same time, but that can be right?

Maybe it is that I am mistaken to try to think that things are happening on both halves independentally of each other... I seems that I might be trying to smash the phases together or something...

 

First of all if the loads is equal then there is no current in a center tap.
To simplify the analysis replace the transform winding by two voltage sources. And then analysis the circuit once more.

 

Compare the difference between these two circuits, with respective voltages, and see if it helps.

In the first diagram, the same current flows through both loads, and the voltage is divided equally if the loads are the same. In the second diagram, the same thing happens, and no current is flowing to the centre tap, if the loads remain equal. If the loads are not equal, then the difference flows through the centre tap.

This is the basis that the 3 wire 'Edison' system is built upon.

 

Basically the first diagram in post #1 is drawn as a bridge as shown by DeviceInfo, it is just a matter of convention.

 

This may be an odd time to ask this, but so what is the center tap for then? I thought I knew...

I see in the diagrams from GetDeviceInfo that Vcb is negative... but why and when does this occur on the source voltages phase?

In the original picture I posted, each load is attached to a Meter and one shows a positive ripple and one shows a negative ripple... Are these being produced by the rectifier at the same time?

Bare with me guys... I will get it...

 

notice in the drawings I've attached, that Vcb is the inverse of Vbc. Why? because your metering it the opposite way.

The centre tap does a couple of things. One thing is that it carries the unbalanced current, but how is that?

If the loads are unbalance, the voltage division across the loads are not equal.
The resulting voltage at B, is then not equal to the centre tap voltage, which does have equal voltage division respective to the source. Connected together, current then flows to or from point B and the tap accordingly.

The result of this current flow results in the second characteristic of the centre tap, and that is it will cause the voltages across the loads to become balanced.

So what are some practical applications of such a circuit. Well, if you had two balanced loads, you could get away with only two wires. If you had many loads, and split them up so that they where close to balanced, you could get away with two lines sized for the total current, and a much smaller line that only carries the unbalance current.

If you use the centre tap referenced to your common, you then have a negative voltage source, useful in biasing circuits.

As I indicated previously, this circuit configuration is similar to the 'Edison' 3 wire system that commonly feeds our residential AC supply. The 'centre tap' is now referred to a neutral. When we distribute the loads in our homes, we try to balance them between the two hot phases, similar to the second diagram. The neutral carries any unbalanced current, and maintains a balanced voltage. If for some reason the neautral was opened, we'd see an imbalance of the voltage. The higher resistive loads would see higher voltages, and could be damaged as a result.

 

Diagrams show by GetDeviceInfo are not fully equivalently.
In the diagram without centre tap we don't get symmetric output voltage when loads are different.
And when it comes to negative voltage look the diagrams in this post http://forum.allaboutcircuits.com/showpost.php?p=150242&postcount=22

In the original picture I posted, each load is attached to a Meter and one shows a positive ripple and one shows a negative ripple... Are these being produced by the rectifier at the same time?

Yes, because only one current is flow in this circuit.
Look diagram in my first post.

 

I was editing my post on the fly, but you are correct.

I was hoping that the proggression would answer the OP's last question before it was asked.

 

You guys are awesome..

So let me see if I have this right so far...

There is only one flow of current in the circuit at any given time. The electrons will flow through both the loads in the same direction always. The only time current is actually flowing through the center tap is when the loads are unbalanced.

In the picture from my original post you could flips the leads on the meter that is reading negative voltage and it will show positive... this is becasue as i mentioned above the electrons are actually moving through both loads in the same direction...

Dual Polarity FWCT rectifier is equivalent to a Full Wave Bridge rectifier.

How does all this sound?

Now what happens if the loads arent even? (dare i ask)
I mean I know GetDeviceInfo, you said that it is balanced by the center tap... but is there someway I can see some examples? Maybe with numbers? I gues I am not sure what happening exactly at that center tap in terms of voltage and current... but I am getting it cleared up.

If the secondary voltage is 24 volts then are the loads getting 24 volts or 12 volts?

I really appreciate all you guys and your patience... cant afford school and I got no one else to talk to about this stuff.
Thanks again.

 

Referring to the attached;

First 3 diagrams show a balanced load. On the left you have your 24v over a total of 20k for a 1.2mA current flow. The middle diagram breaks your source into to additive 12 volt supplies, for a total of 24v, and still a 1.2mA current. Calculating your voltage drops you can see that each load drops 12 volts, in either case. In the third diagram the centre tap was added. But because there is equal voltage at the tap and at the midppoint of this balanced load, there is no voltage differential, hence no flow. The circuit acts identical to the two previous.

Second attachment indicates an unbalanced load. On your left you have a total resistance of 20k for the same 1.2ma current. However, now the voltage drops are not the same. The first resistor drops 18 volts while the second drops 6 volts. Again, in the second image the source is divided in two. Now what happens when the centre tap is connected.

The upper 12v source will drive a current of 800uA over the 15k resistor, heading into the midpoint between loads. The lower source will want to drive a 2.4mA current over the lower resistor, however, as 800uA is already entering the node, the centre tap will only source 1.6mA. The big difference here is that the centre tap keeps the voltage across the loads balanced.

If anyone sees an error in this, please correct.

 

So let me see if I have this right so far...
There is only one flow of current in the circuit at any given time. The electrons will flow through both the loads in the same direction always. The only time current is actually flowing through the center tap is when the loads are unbalanced.
In the picture from my original post you could flips the leads on the meter that is reading negative voltage and it will show positive... this is becasue as i mentioned above the electrons are actually moving through both loads in the same direction...

looks ok, to me

Dual Polarity FWCT rectifier is equivalent to a Full Wave Bridge rectifier.

But this circuit is a Full Wave Bridge rectifier. Only the differently from of a draw.

 

Awesome, I think I got it now... I would like to spend a little more time after my day job studying what you guys have said and make sure its clear to me... I might have another question or 2...

You guys have been very helpful!!!!