Center-tapped rectifier without diodes?

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.

Attached Thumbnails

 

Uriah_Heep

Hi,

You should look this on-line simulation tool:

http://www.dee.feis.unesp.br/docente...hp?id1=2&id2=0

Select the full-wave rectifier to simulate your circuit. Click "anime" to see operational stages.

Anyway, you should look for basic diode rectifier circuits on the web, this should give you a better understanding on full-wave rectifiers.

Uriah_Heep

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).

ViolentCorpse

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?

davenn

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

I_am_learning

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.

ViolentCorpse

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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I_am_learning

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.
I would be glad to see your corrected diagrams.

pantaz

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

davenn

yeah I was going to make the same observation :)


Dave

AlephZero

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.

Windadct

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.

ViolentCorpse

Oops. That totally slipped out of my mind. Sorry.

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.

Ermm.. wouldn't the current through the middle coil then be 0A due to 0V? And

sophiecentaur

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.

ViolentCorpse

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.

sophiecentaur

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 lets 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.

Windadct

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)