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new original voltage doubler circuit

Discussion in 'General Electronics Discussion' started by ratstar, Aug 3, 2020.

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  1. ratstar

    ratstar

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    Aug 20, 2018
  2. WHONOES

    WHONOES

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    ratstar likes this.
  3. ratstar

    ratstar

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    Aug 20, 2018
    Yes indeed voltage doublers have been around since the birth of electronics right? But my method is slightly different, and it runs off a dc source, but it has a spark gap on the high voltage line, so its like an oscillator anyway. And I think (but I havent tested it yet.) I can run mine without the diodes, but I have to test it yet. for a diodeless voltage multiplier. =)
     
  4. Alec_t

    Alec_t

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    I cant't tell from your drawing. Are those supposed to be switches in the bottom rail, or is it just a wavy line?
     
  5. ratstar

    ratstar

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    Its just a wire, excuse the dodgy drawing. :)
     
  6. AnalogKid

    AnalogKid

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    Very low efficiency - way too many resistors.

    AND - it won't work. Add reference designators to *each* component and values to each resistor, and I'll explain why.

    ak
     
  7. ratstar

    ratstar

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    The caps need to be rated at the voltage post multiplication, but they can be any farad - the lower the farads the less amps in the plasma and the more frequent it is.
    The resistors can be very low ohmage, and only the resistors on the blue wire limit the current on the output - the charge resistors (on the white wire) doesnt matter so much what the ohmage is, just slows the charge down a little.

    The resistors are to divert the current to make the capacitors charge one at a time, and the diodes are to stop the capacitors from discharging on the charge lines. The resistors also double to spread the high volts over time to make a continuous spark.

    btw, its already tested. I got the multiplier on the discharge line with a voltmeter.
     
    Last edited: Aug 3, 2020
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    As @AnalogKid stated: It won't work. Not with a static DC voltage at the input as indicated by your schematic.
    The idea of a voltage multiplier in ladder form is not new. But it requires an AC input signal (or at least a pulsed signal).
    "one at a time" is indicative of a pulsed input voltage, otherwise which "time" would you be referring to?

    I simulated with 100 nF and 100 Ω without being able to create a multiplied output voltage. Give us concrete values so we can follow your line of thought in more detail.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    My first thought was that this was a Marx generator, but you need spark gaps for that (and no diodes).
     
  10. ratstar

    ratstar

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    It uses varying voltage division to charge the capacitors one at a time.
    The first empty capacitor in the chain gets the majority of the current until it fills, then the next capacitor takes over. Then the capacitors are charged to the battery volts, and discharged in series for times x volts.

    See how the led's go on one at a time - not at the same time? Thats what happens with this circuit.
     
  11. AnalogKid

    AnalogKid

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    I think I know. The first cap charges almost instantly, as there is no series resistance: V+ > diode > cap > diode > V-. The next cap charges through a (top secret value) resistor, so the upper diode's anode voltage increases much more slowly. The third cap doesn't start charging until the 2nd cap's voltage is high enough for the third cap's diode to start to conduct. So there is a ripple quality to the timing of when the caps begin charging. However, it is not the case, except for the first cap, that one cap charges all the way up, then the next one starts. Depending on the (top secret) source voltage, many of the caps can be in some portion of their charging ramp at the same time.

    ak
     
  12. ratstar

    ratstar

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    The resistors ohmage can be anything, 10ohms all the way to 10,000 ohms probably works, as long as your consistent. It will all produce the ripple.
    The first cap has the least resistance, but the rest are all at least "bare ended" on the negative side, so they follow suit just like the first cap, except with a little voltage loss through voltage division. (its like a lot of voltage dividers in a chain)

    I think it only charges them up to 50%, so theres 50% loss on the system. The source volts could be anything as long as it doesnt overdrive the capacitors?
     
    Last edited: Aug 4, 2020
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