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doubling voltage on a capacitor

Discussion in 'Electronic Basics' started by lerameur, Jun 20, 2007.

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  1. lerameur

    lerameur Guest

    Hello,

    I have a DC source (lets say 50v) connected to a capacitor with a
    switch in between. everytime the switch turns off and on again the
    capacitor voltage gets hit with 50v and adding it everytime. 50 then
    100, then 150 atc.. how is it possible? what is the physics behind
    this ?
    thanks

    ken
     
  2. Rich Grise

    Rich Grise Guest

    Read some of these links:

    http://www.google.com/search?hl=en&q=cockroft-walton

    Have Fun!
    Rich
     
  3. Eeyore

    Eeyore Guest

    It's not *one* capacitor for starters is it ?

    Graham
     
  4. lerameur

    lerameur Guest

    It is one capacitor with no load. And it is DC , so voltage
    multipliers with Ac will not work.. i think

    k
     
  5. If you have accurately described the circuit components,
    their connections and the result, the physics is not
    anything previously seen in this universe.
     
  6. Chuck

    Chuck Guest

    Sounds like maybe the capacitor's plates
    are moving away, decreasing capacitance
    and increasing voltage.

    Tell us more.

    Chuck
     
  7. Jasen

    Jasen Guest

    What you intend is unclear, but it sounds a little like a Cockroft-Walton
    voltage multiplier or maybe a Marx (high voltage) generator

    A circuit with a battery, switch and, (instead of a capacitor) an inductor can
    generate high voltages, automotive ignition systems were based on this idea.
    this circuit is called a flyback circuit.

    Bye.
    Jasen
     
  8. Chuck

    Chuck Guest

    How about this.

    If the DC source is a constant-current
    source, and the switch is closed only
    for short, repeatable durations, then
    the capacitor's voltage will increase by
    the same fixed amount, doubling each
    time the switch is closed, provided its
    self-discharge rate is negligible in the
    time between switch closures.

    Since each "pulse" provides a fixed
    quantity of charge, and since V=Q/C,
    (delta V) = (delta Q)/C.

    Obviously, this can't go on forever with
    a real capacitor and a real
    constant-current source. But it can
    easily go on for many orders of
    magnitude of voltage doubling.

    Not "exactly" the way the problem is
    described, but close, and it is easily
    demonstrable within the laws of physics
    on this planet. ;-)




    Chuck
     
  9. Joe

    Joe Guest


    Unbelievable. "If the DC source is a constant-current source" is not
    magic. It would have to be a source that increases the voltage to the
    being-charged capacitor. So, a "constant currrent source" would be
    capable of charging any capacitor to voltages approaching infinity, at
    least until the capacitor broke down (exploded?) or until reality took
    over the constant current source, and the constant current source yelled
    "No mas!".

    Too many people have given 1/2 vast explanations that have hindered my
    understanding in the past, so excuse my reaction to this latest
    "explanation".

    --- Joe
     
  10. Chuck

    Chuck Guest


    Hello Joe,

    I assumed that
    would have some meaning. Even the OP
    never said "forever".

    I will admit one more time that real
    electronic components are not ideal:
    real capacitors will have some
    self-discharge, etc. The most accurate 1
    MHz oscillator may never actually
    produce a 1 MHz output, etc.

    With that qualification, you can easily
    design and construct a real circuit that
    will double the charge (and voltage) on
    a capacitor by some fixed amount every
    time you press a momentary contact
    switch for some constant time interval.
    For the physics, the magnitudes of the
    time interval and the voltage increase
    are irrelevant. If the OP were
    interested in absolute accuracies and
    limits on how high voltage could be
    increase, I think he would have said so.

    I'm sorry to hear that your experiences
    with invalid explanations have made you
    hypercautious. But a better approach
    might be to raise specific questions
    about an explanation, rather than to
    reject it outright. If there is some
    aspect of what I suggested that is
    troubling you, I'd be glad to attempt a
    better explanation. Or perhaps someone
    else will jump in.

    Good luck.

    Chuck
     
  11. Joe

    Joe Guest

    <snip>

    The OP said

    "I have a DC source (lets say 50v)..."

    Chuck said

    "If the DC source is a constant-current source..."

    Here, Chuck has already confused the issue by completely changing the
    setup given by the OP.

    Chuck then goes on to say:

    "Since each "pulse" provides a fixed
    quantity of charge, and since V=Q/C,
    (delta V) = (delta Q)/C."

    Which further confuses the issue, because, with a constant-current source,
    no pulsing is necessary or relevant to the OP's original question and
    setup.

    Stating that "Since each "pulse" provides a fixed quantity of charge, and
    since V=Q/C, (delta V) = (delta Q)/C." simply obfuscates that Chuck has
    replaced the constant 50 volt supply by one that is (automatically)
    ramping up its voltage, and thus is increasing the voltage on the
    capacitor.

    The capacitor has little to do with the mechanism by which the
    constant-current source increases the supplied voltage, and so has little
    if any relevance to the OP's question.

    Chuck then tells me:

    "...invalid explanations have made you hypercautious..."

    No, Chuck invalid explanations, especially wordy ones, cause a visceral
    reaction, not related to "caution".

    The plain fact is that the OP most likely had some idea of a voltage
    doubler in mind via a switching arrangement of charging capacitor(s).

    Re-casting the whole discussion into a constant-current source distorts it
    and certainly does not simplify anything relevant to the OP's question.

    Chuck then tells me:

    "...raise specific questions about an explanation..."

    No question raised by me. Rather a comment -- the OP asked about apples
    *, you told him about oranges **.


    * Constant voltage

    ** Constant current

    --- Joe
     
  12. Chuck

    Chuck Guest

    Hello again, Joe.

    It just occurred to me that you might
    have thought my earlier post about
    moving the capacitor plates was one of
    the "1/2 vast" explanations. May I explain?

    There are really only two ways to
    increase the voltage on a capacitor:
    increase the charge or decrease the
    capacitance. In the constant-current
    approach you commented on, the charge is
    increased when the button is pushed.

    In my earlier post, which is more
    humorous than wrong, I imagined that the
    capacitor plates (or whatever) are
    mechanically linked, say by a ratchet,
    to the switch handle. Thus, every time
    the switch is activated, the capacitance
    is (first) reduced by a set amount.
    Every time the capacitance is halved,
    the voltage doubles. This is, of course,
    operated with a constant-voltage source.

    The physics here is elementary, but
    technically, this stretches the rules
    set by the OP because after the first
    switch activation to place a charge on
    the capacitor, subsequent activations of
    the switch lever require no further
    connections to the constant voltage DC
    source. This could be handled
    mechanically, of course, in the switch
    design. The capacitor voltage would not
    change if it were connected to the
    source each time the button is pushed.

    Does that help any?

    Chuck
     
  13. Chuck

    Chuck Guest

    Wow.

    Okay, you're not hypercautious, Joe. But
    I do detect some basic confusion about
    electronics.

    There is no way one can divine from "DC
    source" that the OP's intent is a
    constant voltage source. DC source only
    means not AC.

    A constant current DC source is as much
    a DC source as a constant voltage
    source. Less familiar to some, maybe.

    The fact is that the OP did not specify
    the kind of DC source. So I did.

    But I'll go way out on a limb here and
    say that even with an imperfect constant
    voltage source (i.e., one with a "high
    enough" internal resistance) the circuit
    will function as I described, although
    with an error determined by how high the
    internal resistance is. What I'm getting
    at is that real DC sources constitute a
    continuum from ideal constant voltage to
    ideal constant current.

    But Joe, even if I had blatantly
    violated one of the OP's rules, it
    wouldn't justify such hostility toward
    my post. There was no hidden plan in my
    post to mislead or trick anyone.

    The exercise was most probably posted as
    a riddle intended to foster thought and
    discussion.

    Try to be more calm and enjoy the
    discussion.

    Chuck
     
  14. Joe

    Joe Guest


    Chuck,

    You persist in your word-deluge, and now you have some dishonest editing.

    Here is what the OP said:

    **************
    I have a DC source (lets say 50v) connected to a capacitor with a
    switch in between. everytime the switch turns off and on again the
    capacitor voltage gets hit with 50v and adding it everytime. 50 then
    100, then 150 atc.. how is it possible? what is the physics behind
    this ?
    thanks

    ken
    **************

    Chuck, Chuck, Chuck. You left off the "lets say 50v", so you could say
    that your constant-current setup is consistent with the OP's question.

    Bullshit.

    --- Joe
     
  15. Chuck

    Chuck Guest

    That might be a valid point, Joe.

    I think you're saying that a constant
    current source will not have an
    open-circuit voltage of "say 50v". That
    is true of an ideal constant-current
    source, for which the open-circuit
    voltage is infinite. There is a finite
    open-circuit voltage at the output of
    every real constant-current source. So
    there is NO inconsistency there.

    Also, I didn't think there was any
    significance to the particular voltage
    the OP assumed. The physics ought to be
    the same for any voltage if no solid
    state devices are involved.

    My interpretation of the problem focused
    on doubling the capacitor's voltage each
    time the switch was closed.

    If you believe that the important part
    of the problem was to increase the
    capacitor's voltage above the source
    voltage, then that is fine. I did not do
    that with this suggestion. If you read
    my second two posts in this thread,
    you'll see I never claimed to do that. I
    did do it with my first post, however:
    the one in which the plates moved.

    Hitting the books might be a better use
    of your spare time than continuing to
    vent your hostility on this newsgroup, Joe.

    Thanks for taking the time to let me
    know my postings troubled you and for
    giving me a chance to explain my thinking.

    Good luck.

    Chuck
     
  16. Joe

    Joe Guest


    You are a true bullshit artist, Chuck.

    Please, I'm not "...vent(ing) ... hostility on this newgroup", Chuck.

    Just trying to clear up the bullshit you persist in dropping.

    --- Joe
     
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