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Center-tapped rectifier without diodes?

  1. Sep 22, 2012 #1
    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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  3. Sep 22, 2012 #2
  4. Sep 22, 2012 #3
    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).
  5. Sep 23, 2012 #4
    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?
  6. Sep 23, 2012 #5


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    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.

  7. Sep 23, 2012 #6
    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.
  8. Sep 23, 2012 #7
    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).


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  9. Sep 23, 2012 #8
    But I was expecting that you would replace the diodes with resisters (or conducting wires).
    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.
  10. Sep 24, 2012 #9
    You are incorrectly assigning polarity to the transformer output. A.C. input = A.C. output.
  11. Sep 24, 2012 #10


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    yeah I was going to make the same observation :)

  12. Sep 24, 2012 #11


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    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.
  13. Sep 24, 2012 #12
    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.
  14. Sep 24, 2012 #13
    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
  15. Sep 24, 2012 #14


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    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.
  16. Sep 24, 2012 #15
    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.
  17. Sep 24, 2012 #16


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    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.
    Last edited: Sep 24, 2012
  18. Sep 25, 2012 #17
    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)
  19. Sep 25, 2012 #18


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    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.
  20. Sep 25, 2012 #19
    Hi Windadct!

    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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    Last edited: Sep 25, 2012
  21. Sep 25, 2012 #20


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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.
    Last edited: Sep 25, 2012
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