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Cockroft Walton Capacitor charging circuit.

  1. Jun 26, 2010 #1
    Hello. I am trying to find a simple method of charging a bank of 330V, 120uF photoflash capacitors that are in parallel. I would rather not make an extremely complex circuit, as all I want to do is just charge the bank in a reliable and predictable manner.

    I was considering using a three-stage cockroft-walton voltage multiplier and wall DC to charge the bank. The only problem that I have with this method is that the charger will output 360VDC, which is around 8% above the rated voltage of the photoflash caps.
    1- These are Rubycon caps, and I cannot find their tolerance anywhere online. Is 8% over acceptable for safe operation?
    2- If this is not safe, is there a way to create a consistent voltage drop? I know that a resistor's voltage drop is time dependent, so I would rather not just put a resistor in series in case I leave the capacitors charging too long.

    In addition to the questions above, do I need to worry about the bank somehow "shorting out" the CW while charging and causing mayhem? Any other suggestions on how to make a decent charging circuit are always welcome. Thanks.
  2. jcsd
  3. Jun 26, 2010 #2


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    There is nothing safe about what you are doing. That is a deadly power supply and you should operate it only in an enclosed box.

    You could use a voltage doubler to get about 336 volts from a 120 volt AC supply.
    To reduce the charging current, the first capacitor in the voltage doubler could be reduced.

    As an example, in a simulation, I set the input C (C2 in the diagram below, the one in series from the supply) to 100 µF with the second C (C1, across the output) set to 1000 µF (3 of your capacitors) and the peak supply current dropped from about 60 amps to about 8 amps.

    The 1000 µF capacitor took about 1.5 seconds to be fully charged.

    It would be possible to use a 120 volt to 12 volt transformer and connect the 12 volt winding in series with the 120 volt input, but out of phase with it, to get an AC source of about 108 volts. In a voltage doubler circuit as shown, this would give an output of about 302 volts which would be more acceptable for your capacitors.

    [PLAIN]http://dl.dropbox.com/u/4222062/voltage%20doubler.PNG [Broken]
    Last edited by a moderator: May 4, 2017
  4. Jul 1, 2010 #3
    You could use a boost converter too. You really don't want anything like this connected directly to ac power, after it is rectified, every point of the circuit has potential relative to ground and is very capable of killing you.
  5. Jul 5, 2010 #4
    Could you explain this a bit more ?
    How do you hook it up out of phase ?
    And how does a 10% drop [120 to 108] result ?
  6. Jul 5, 2010 #5


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    You would use a circuit like this (iron core not shown for clarity):

    [PLAIN]http://dl.dropbox.com/u/4222062/transformer%20buck-boost.PNG [Broken]

    See how the output voltage depends on the way the secondary winding is connected. One way, it subtracts 12 volts from the output and the other way it adds 12 volts to the output. This gives 108 volts or 132 volts respectively.

    Keep in mind the earlier comments about safety. Connect the secondary one way and then measure if the voltage is higher or lower. If it is the wrong way, then you can reverse the leads to the secondary to get the other voltage.

    Connect your meter before you turn the power on and don't go anywhere near it with the power turned on.
    Last edited by a moderator: May 4, 2017
  7. Jul 5, 2010 #6
    That is a really interesting circuit, vk6kro. I've read a bit about transformers, but I never realized you could connect them like that. Under what circumstances would you use that circuit instead of a standard step down transformer with an appropriate windings ratio? Ie, you could, theoretically, get 108V by having a .9 winding ratio in a single step down transformer, but I'm guessing that there is some sort of practicality reason why you would instead just put 12VAC out of phase. Can you elaborate?

    I ended up just undercharging the capacitors at 240VDC. This way I get slightly higher current in the discharge cycle, which could be useful for magnetic experiments.

    Thank you all for your excellent replies.
  8. Jul 5, 2010 #7
    The accepted safe derating for electrolytic caps is 80%. If the caps are rated for 350v, you should keep the voltage below ~300 volts. Going 20% over is asking for an explosion.
  9. Jul 5, 2010 #8


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    Most electronics stores would carry transformers that give various voltages in the 5 to 15 volt range, while a 120 to 108 volt transformer would be a more expensive specialty item.

    So, it is mainly a matter of convenience and availability. A 120 volt to 110 volt transformer would be OK if you can get one. It would probably be cheaper to just get better capacitors and charge them within their ratings.
  10. Jul 5, 2010 #9
    that's a pretty cool trick!
  11. Jul 5, 2010 #10
    Does the output waveform look identical? This is quite clever.
  12. Jul 6, 2010 #11
    Yes, I predict you'll get a good waveform at a reliable supply voltage. This kind of trick isn't used in commercial products but it is completely reliable.
    Last edited: Jul 6, 2010
  13. Jul 6, 2010 #12
    But you should notice that the dc supply in post #5 has a potentially dangerous catch: the DC is referenced to the Neutral line of the AC supply, which is usually tied to Earth.
    So the DC will short to any available Earth, which means that it is very easy to get shocked. The need for great care has been emphasised in this thread, and I make mention of this need again.
  14. Jul 6, 2010 #13
    There is a second potential problem with the circuit that can be easily eliminated. This is the surge current that can result from applying the AC source when the source is near peak.

    And this kind of circuit, in full wave version, is used in at least one set of commercial products that I am aware of.
  15. Jul 6, 2010 #14
    a nasty surprise
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