AC fed capacitor circuit vs AC fed capacitor visa sparkgap?

Discussion in 'General Electronics Chat' started by Kojii, Jul 19, 2013.

  1. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    Hi every one, there has been something confusing and making me a great headache lately. I hope my question is as clear as I can get it:
    Note:I know how LC and tesla coils work(or I think I do at least)

    In condensed shortness: What is the difference between:

    1- HV AC source connected in parallel to a capacitor.
    2- HV AC source connected in parallel to capacitor and parallel to a sparkgap.

    Say the HV AC source can put for example 1000v and has sufficient amperage to supply the caps at the desired frequency to reach the voltage.

    What I have learned is that the cap in case 1 charges and discharges to the source´s 1000v each cycle. Good.


    The sparkgap in case 2 is set to fire say at 1000v or 1400v, or any.. air breakdown so in case 2 the capacitor charges every time to 1000v or other voltage decided by the air breakdown and fires to discharge.

    Now what is the difference between the two cases really? and why do we put a spark gap?, I have put so much thought too deep into this that this may actually have a really simple answer?
    It would be such a relief if I got this!

    Thank you
     
  2. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    Just a guess, the cap would allow the spark to jump a larger gap.

    With the cap, it would store a charge until it got enough energy to ionize the air in the larger gap. And then when it finally broke down the gap it would also spark for a longer time.

    Without the cap, it would need a smaller gap, due to a smaller energy.
     
  3. wayneh

    Expert

    Sep 9, 2010
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    Why indeed. You're the one that brought it up. I don't need a 1000V spark gap. Why do you?

    I speculate the capacitor is just a local power supply and helps deliver more juice in a brief instant, intensifying the spark.
     
  4. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    oh no need to be 1000v at all, it could be any voltage at all, I just gave an example.. But wait you may have the answer what you said :

    "I speculate the capacitor is just a local power supply and helps deliver more juice in a brief instant, intensifying the spark" So it is in the discharge that there is difference? Let me see: In both cases, the capacitor charge at the same rate to the voltage in half cycle. In case 1, the cap discharges slowly at the set frequency but in case 2 the cap discharges instantly and does not follow the quarter cycle period of the set frequency of the supply ha?
     
  5. wayneh

    Expert

    Sep 9, 2010
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    Hard to say. The power supply impedance is very low but maybe not as low as the short path from capacitor to spark gaps. So the voltage of the supply would otherwise drop when the current of the spark begins flowing and that's when the current stored in the capacitor can jump in and cross the gap as well.

    That's how it works in my mind. Caveat emptor. ;)
     
  6. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    http://www.richieburnett.co.uk/wavefrms.html

    first part of my answer is there I think? the discharge as seen basically is instant and that is basically is the difference, so instead of a typical AC-source´s slow since discharge, the discharge is immediate with a gap.. hmm there is still more though to grasp
     
  7. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    I have attached picture:

    [​IMG]

    In nr 2, the "red" is simply a wire connecting.

    Would the capacitor charge at all in this nr2 scenario? or would the voltage and current go across the red wire and leave the capacitor alone uncharged?
     
  8. wayneh

    Expert

    Sep 9, 2010
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    It depends. A perfect wire conducts perfectly and there can be no voltage difference across its ends.

    But a perfect power supply maintains a voltage at any load.

    So this is an impossible situation, the outcome of which will blow up the weakest link, ie. melt the wire or blow a fuse in the power supply.

    The capacitor will then side with the winner. If its poles are wired together, they'll be the same voltage. If the power supply still works after the wire has vaporized, the capacitor will be at the PS voltage.
     
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  9. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    So in general there would no voltage across the capacitor?

    I just tried a small 3v circuit and there is only minute voltage across the cap.
    But if we remove the red wire entirely and have only AC source connected to a small capacitance cap. What would happen when the AC source PSU voltage stays the same but the PSU current increase 10 times (or PSU´ impedance decreases to 10 times as can be said in other words as well). Am I correct in saying that the capacitor would then in each half cycle charge and discharge much faster but to the same voltage? So, the voltage across the low capacitance capacitor would look somewhat like this in fig3 ?
    [​IMG]

    Thanks
     
  10. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    The capacitors VOLTAGE cannot arrive at a higher level than the AC source is providing at an EARLIER point of TIME.

    current waveforms are another beastie altogether in this little scenario you have created.
     
  11. wayneh

    Expert

    Sep 9, 2010
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    No, the capacitor would simply track the power supply voltage almost exactly. That would be the case with most power supplies and most capacitors. You would only see an effect with a higher impedance source and a large capacitor. This would cause a shift in phase between current and voltage - voltage being delayed while the capacitor charges - and would lower the peak voltage (and raise the minimum). The voltage waveform would still look like a sine wave, though, as there is nothing to cause the clipping in Fig. 3.
     
  12. #12

    Expert

    Nov 30, 2010
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    Study a relaxation oscillator using a neon lamp. Some impedance delays a voltage build up on a capacitor until the neon arcs. The neon bulb has a definite arc over voltage and stop voltage. The capacitor has that interval to dump its current.

    In your circuit, an oscillator delays a voltage build up on a capacitor until Earth atmosphere arcs. For any given gap, it has an arc-over voltage and a stop voltage. etc. The only difference is that your atmosphere arc is much more difficult to measure with a common voltmeter or oscilloscope.
     
  13. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    Aha, thanks. I got more confused now though, you see.., the way you described it here, that's how I thought it was working until I noticed that rotary sparkgaps can fast charge tank-caps. twice each half cycle to the same NST voltage if you just have enough power in a tesla coil.. ha?
    http://www.richieburnett.co.uk/wavefrms.html
     
  14. Kojii

    Thread Starter New Member

    Jul 19, 2013
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    How can capacitors in a rotary spark gap TC get charged twice each half cycle to the same NST voltage as seen in this output waveform
    [​IMG]

    http://www.richieburnett.co.uk/wavefrms.html


    "The oscilloscope trace opposite shows operation with a 200BPS synchronous rotary spark gap.
    The waveform shows the tank capacitor charging to around 20kV twice in every half cycle of the supply waveform. The 200BPS rotary fires every 5ms. This means that there are now two charging periods and two bangs during each half cycle of the supply waveform.
    I find that this increased break rate gives better spark performance, but it also demands considerably more power. The rotary phase is adjusted to provide two bangs of equal magnitude during each half cycle.
    Timebase = 5ms / div, Vertical = 10kV / div"

    so the power of the AC source was increased which must be the current output I think but I am not sure . If that is the case then it would see that with higher current= faster charging?
     
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