question about chapter 3 voltage multiplier

Discussion in 'Power Electronics' started by Darrell75, Oct 13, 2016.

  1. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    In the second example, captioned Full-wave voltage doubler consists of two half-wave rectifiers operating on alternating polarities, there is a ground at the lower right corner of the circuit. This is probably a really dumb question, but why doesn't this ground negate the voltage across C1? Wouldn't the charge be lost into the ground? If someone could explain to me why that's not the case I'd appreciate it. Thanks!
     
  2. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    Never mind... I see that the ground stays at 0V (at least I think it does). But I do have another question. At the start of the second cycle, why don't the bottom plate of C2 and the top plate of C1 drop down to 0V, bringing the voltage at the output back down from 10V to 5V? There is no resistor between the power source and these plates to slow the drainage, so wouldn't they drain pretty quickly? Thanks.
     
  3. MrSoftware

    Member

    Oct 29, 2013
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    Can you post a schematic for the circuit you're asking about?
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Chapter three of w*h*a*t?
     
  5. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    Sorry about that.... Chapter 3 of Volume III - Semiconductors, from the All About Circuits textbooks.
    I've pasted here the diagram with the chart of voltages at 2, 5, and 3. The text describes it this way:
    "The corresponding netlist is in Figure below. The bottom rectifier charges C1 on the negative half cycle of input. The top rectifier charges C2 on the positive halfcycle. Each capacitor takes on a charge of 5 V (4.3 V considering diode drop). The output at node 5 is the series total of C1 + C2 or 10 V (8.6 V with diode drops)."
    But I'm still a little confused about how the two capacitors work together here along with the ground and D1...
    Thanks!


    circuitdiagram.PNG diagram.PNG
     
  6. crutschow

    Expert

    Mar 14, 2008
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    When the AC source node 2 is positive, the charging path is through D2 and node 3 back to the AC source, charging C2.

    When the AC source node 2 is negative, the charging path is through D1 and node 3 from to the AC source, charging C1.

    Remember that ground is just a reference node for voltage measurements.
    Otherwise it's the same as any other node.
    It has no magical current sink properties.
    Current can only go into the ground node if it also has a path out of he ground node for that same current.
    In your circuit there is only one connection to ground, so no current can go into or come out of that node.
    As far as you circuit is concerned it would work the same no matter where you connected the single ground point. (Of course, the voltages measured to ground would be different, but the relative voltages between nodes would be the same).
     
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  7. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    Thank you! I'm going to have to look at it again with your explanation in front of my, but at first blush, it makes a lot of sense to me. Thanks again!
     
  8. AnalogKid

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    Aug 1, 2013
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    As Wally said, in this circuit the Ground placement is completely arbitrary. As shown, the output is +8.6 V with respect to GND. If you proclaim that node 3 is GND, then the outputs are +/-4.3 V. That's the nice thing about a fully isolated transformer secondary, it gives you lots of options.

    ak
     
  9. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    Again thank you for the explanation, crutschow. I do have another question. Why doesn't the voltage at node 3 (between the capacitors) drop back down to 0 when the generator completes its first cycle? I've tried to depict what I mean in the illustration below. I welcome corrections to whatever mistakes are in my drawings. circuitquestion.jpg
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Why do you think it will drop to 0V?
    Remember the AC source is floating.
    Node 3 voltage will be C1's voltage which after 1 cycle will be equal to the peak AC voltage minus one diode drop.

    You need to get used to viewing voltages as relative, not absolute. :)
     
  11. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    Sorry about that. I think i meant after the first half (where node 2 is positive) of the second cycle. Thanks for your patience!
     
  12. crutschow

    Expert

    Mar 14, 2008
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    But note that the AC voltage at node 2 is relative to node 3, not to ground (between nodes 2 and 3).
     
  13. Darrell75

    Thread Starter New Member

    Apr 19, 2016
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    Thank you for your help. I think I've got it. I have more questions, but I think I'll start another thread for that.
     
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