Voltage doubler not working

Discussion in 'General Electronics Chat' started by m1ch43l, Sep 19, 2012.

  1. m1ch43l

    Thread Starter Member

    Aug 16, 2012
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    1
    Hello chaps,
    I have a project here. I need to build a small voltage doubler circuit from which I will then expand to suit my needs.
    The problem I'm having is that the basic one outputs 50Vdc whereas I have 240Vrms in. In this case I expect a minimum of 500Vdc. (cant post my drawing for now but I'm using this replica: http://en.wikipedia.org/wiki/File:Greinacher_circuit.svg but without the tx; using mains at 244vrms)
    I am using scavenged capacitors(disc ceramic capacitors -- the small blue ones) and diodes(FR107) for a single stage half wave multiplier.
    I have tried changing the diodes and using brand new 1N4007 whereas I have been through all 30 capacitors I have at my disposal. Non of them appear broken and testing a few has shown nothing contrary to my prior findings. All diodes are in superb condition.
    No component is getting hot and neither are there shorts or funny smells.
    Any ideas where I'm going wrong?

    I need to finish it ASAP like today!
    Thanks in advance.
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    What is the value of your capacitors?
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    This was neither feedback nor a suggestion, so I have moved it to a more appropriate forum.
     
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  4. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    I'm using pairs: each pair is 1000pF, 10000pF and 2200pF.
    Both sets of diodes have a forward voltage of 1000 with diode vdrops at 0.7 and 1.1
    for N series and FR diode respectively.
     
    Last edited: Sep 19, 2012
  5. MrChips

    Moderator

    Oct 2, 2009
    12,447
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    Why the different values?
    You will need capacitors of at least 0.1μF rated at 1kV.
     
  6. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    The pairs here I used were to determine whether the caps were bad...the multiplier is a single stage. I am switching at ~50Hz.
    Ok, how does the 0.1uF 1kV cap influence the output, given that those I used are also 1kV with exception of the 222 caps which are 2kV?
     
    Last edited: Sep 19, 2012
  7. MrChips

    Moderator

    Oct 2, 2009
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    The size of the capacitor will determine how much current you can draw from the output of the voltage doubler. What is your current load?
     
  8. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    The current output is not of concern now. I have turned to the multiplier to explicitly scrub it off for the application I have in mind. If I'm to guess, 0.1A at most. Otherwise the voltage will do. I need bags of voltage irrespective of how much current will be available. Think of it this way: I need a electrostatic field for my app.
    My intention is to switch this tension to my load via a Enhancement N-FET. This way, I further limit my load's current consumption.
     
    Last edited: Sep 19, 2012
  9. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    Someone help please... where am I wrong in this?
    I've been wondering: what effect does a charged non-polar capacitor have on a circuit if its polarity is suddenly reversed/charged with a voltage whose polarity is opposite to the capacitor's?
     
  10. MrChips

    Moderator

    Oct 2, 2009
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    0.1A = 100mA is a lot of current to take from a voltage doubler with such low value capacitors. What are you using to measure the voltage?
     
  11. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    Normal multimeter.
    This current is an estimate. I haven't attached the load yet.
    I can't get the multiplier to work well in the first place. This is where I'm stuck.
    I can't move on.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    1,000pF = 1nF = 0.001uF
    10,000pF = 10nF = 0.01uF
    2,200pF = 2.2nF = 0.0022uF

    You really need much larger caps. They do need to be non-polarized. As already mentioned, voltage doublers are not good for much current; above a few mA, you'll really need something more like a transformer or a switching power supply.
     
    Last edited: Sep 19, 2012
  13. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    I am in need of high voltage and not current for this project.
    What's the difference between connecting the device directly to mains and having the above quoted? I mean is not the output the same, a sinewave?
     
  14. MrChips

    Moderator

    Oct 2, 2009
    12,447
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    Hi Sgt,

    How have you been keeping? Good to see you still keeping in touch.
    btw, your last line is off by a factor of 10.
     
  15. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    Me or sgt?
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    OK. However, your capacitors are so small that the charge is lost due to the turn-off time of the diodes. Get some larger non-polarized capacitors. If you have a hard time finding non-polarized caps, you can use a pair of same-value electrolytic caps back-to-back (both negative terminals together or both positive terminals together) that are rated for a higher voltage than required; if the capacitance of both are the same, then the resulting capacitance will be 1/2 of the rating.

    Do not connect a project directly to mains power; a transformer must be between the mains and your project. A transformer provides galvanic isolation from mains power, and effectively limits the amount of power that can be transferred to your project.

    We do not support discussion of transformer-less power supplies; they are inherently unsafe, and there is nothing one can do to make them safe. Transformers are a remarkable bargain compared to the cost of a funeral.

    MrChips,
    Thanks, nice to "see" you, too. :)
    Yep, thanks - I fixed it. Too many things taking my attention this morning...
    Me. ;)
     
  17. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    :D

    @vk
    I've removed the theory I had in mind.
     
    Last edited: Sep 20, 2012
  18. vk6zgo

    Active Member

    Jul 21, 2012
    677
    85
    Please dont!!
     
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  19. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    Ok.
    Was just a thought in the distro chain relative to the grid input.
     
    Last edited: Sep 20, 2012
  20. m1ch43l

    Thread Starter Member

    Aug 16, 2012
    62
    1
    @sgt
    I tried a simulation with exact values of my caps at 2200pF and the output is what I expect.
    I'm trying with polar caps and 12vrms input. I'll see what's what.
     
    Last edited: Sep 20, 2012
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