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Parametric amplification of energy on capacitors

  1. Apr 28, 2003 #1
    Parametric "amplification" of energy on capacitors


    My last post was about how geometry affected potential and electric energy transfers. Some people send me interesting information about that, and i've been thinking about parametric power conversion.

    We usually relate E to a variation of Q on a fixed geometry (parameters, like R, C or L), but it's also true that we can obtain E from fixed Q and variation of the parameters (thus geometry).

    If you've got a charged capacitor, the electric energy stored on it is different if you move one plate, or change the surface, or change the dielectric, etc... But this changes takes the same energy as the energy change they cause.

    But imagine two charged capacitors. Connecting them in series is the same as having one capacitor half of each one, and connecting them in parallel is the same as having one double of each one. So we can obtain a factor of 4 on the amount of energy stored only changing connections (with interruptors or transistors), needing almost no energy to do so.

    Now take a look at this page, where i talk about this case:


    The experiment was made with an electronics simulator based on SPICE.

    All of you helps me a lot to understand what concepts sound interesting and what others sounds suspicious, so thanks for your answers about this system and my thoughts.
    Last edited: Apr 28, 2003
  2. jcsd
  3. Apr 28, 2003 #2


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    It sounds like you're working toward an understanding of a capactive boost converter, which is a circuit which uses rapidly switched capacitor topologies to build large voltages from small ones. In effect, it's a pump.

    - Warren
  4. Apr 28, 2003 #3
    Re: Parametric "amplification" of energy on capacitors

    Hi Cala,
    A capacitor is like a low internal resistance battery.

    Take for instance the electric control in my Honda Insight whose motor/generator coupled in two ways (by a very smart chip) to a 144 volt ni-cad battery that discharged when its 12 12-volt batteries were coupled in series and when the motor became a generator the 12 separate batteries were charged while coupled in parallel.

    Also, at a time in early 1960s I had a flash x-ray installation at LANL that powered a 600-kilovolt field emission cathode flash x-ray source. The generator consisted of 12 coax cables. Each of these was several meters long and properly terminated at the far end and with the business end braid grounded and connected to an aluminum 3” diameter ball; the 1/4" dia. central conductor of each coax was connected to a 100 kilovolt supply & ultimately affixed to a similar aluminum 3 inch diameter ball. These 24 balls were separated by 6 or so inches each and arranged alternately in a vertical column. Midway in one of the gaps a small trigger ball that was charged to a mid voltage and was, at the target time, shorted to ground and that avalanched the whole stack with a series coupling that delivered to the cathode load the desired 600 kilovolts. Cheers, Jim
  5. Apr 29, 2003 #4
    I'm astonished with this method!. I've made some calculus, and it looks incredible!.

    Making an analogy with water levels, imagine you have a bottle of water. Then, you throw the water, and the water falls to ground. Now you've got the water scattered into the ground. Now imagine you could "freeze" this water, then change horizontal position for vertical one without spending energy. Now you're again as you started! aren't you?.

    Another water analogy:

    Imagine a big tank of water at 12m high. Imagine two smaller tanks of 12m high, with water at 6m level. Now, open the big tank. The water will fall from the big tank to the smaller ones until the levels rise 12m. The level of the big tank is still about 12m (the change of level is not appreciable). Now, imagine you cost nothing put one of the smaller tanks on top of the other. Now, you've got the same water at a 24m level, so you can open the smaller tank on top, and then the water will flow to the big tank again, just until rise the 12m level as on initial conditions. Now, the smaller tank on top is empty. Now you put the smaller tanks at the ground again, and connect it. Then the 12m of water will divide into 6m and 6m. Now you're on initial conditions... BUT YOU DID WATER TRANSFERS, DON'T YOU?? You could use this transfers to roll a turbine and thus to obtain work!. Of course, put the tank on top costs energy, but this is not neccesary in the electric example of capacitors!, we only have to change electric connections!

    so, here is an electrical example:

    Imagine a 12V and 5W car lamp.

    From this data we can obtain R=28.8 Ohm and I=0.4A.

    Now, imagine you put a 36V battery on one side, and two parallel capacitors of 33uF charged at 23V on the other connection of the lamp.

    We will follow this sequence:


    Initial condition: 36 V battery - 23 V parallel capacitors (13V)
    Final condition: 36 V battery - 24 V parallel capacitors (12V)


    Initial condition: 48 V series capacitors - 36 V battery (12V)
    Final condition: 46 V series capacitors - 36 V battery (10V)

    As you can see, I'm trying to keep the voltage almost constant to the lamp.

    Now, we need to see what time we need in the charging-discharging proccesses to obtain these voltages.

    Making the simulation of the system with a battery of 36 V, a resistance of 28.8 Ohms and a pre-charged capacitors at 23 V, i obtain that we must keep the parallel proccess about 165.6uS, and the series proccess about 73.3uS.

    With this control conditions (with a 4.3478 KHz clock) we can make this lamp work without expending the batteries!!!, only by charging parallel capacitors and discharging the same series capacitors!!

    The system timing must be changed for every different load. The more the resistance, the lower the frequency of the control system.
  6. Apr 29, 2003 #5


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    I hate to burst your bubble <ADMIN EDIT>, but capacitors do not store voltage, they store charge -- such that q = CV. You CAN build a voltage pump out of capacitors -- such switching pumps are called boost converters. A subset of the boost converters manufactured by my company can be found here: http://www.national.com/parametric/0,1850,1767,00.html [Broken]

    In no way have you developed a "free energy" source -- you simply don't know how to do math.

    - Warren
    Last edited by a moderator: May 1, 2017
  7. Apr 29, 2003 #6
    OK, you're right, the capacitors store charge, but this charge stored doesn't create a voltage difference?

    Every different capacitor will create different V values with the same amount of charges Q, but if you take always the same capacitor, the more Q stored, the more potential.

    As you said Q=CV, so V=Q/C

    If C is constant, V is directly related to Q.

    If you connect two equal capacitors in parallel to a V potential, they charge to get the same V.

    Now, if you disconnect the potential, and connect these two equal capacitors in series, Q remains constant, but you've got C/2, so:

    V= 2Q/C = Double the potential than before. It's quite simple.
  8. Apr 29, 2003 #7


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    Actually it's four times. Total charge is conserved. In going from parallel to series topology, you're dividing the total capacitance by four -- from 2 C to 1/2 C. The voltage goes up by four as well so that qfinal = qinitial.

    This is the basic mechanism for a charge pump or boost converter.

    What I don't fathom, though, is how you think you've invented a perpetual motion machine.

    - Warren
  9. Apr 29, 2003 #8
    Suppose two capacitors of C value initially charged at Q charges each one,(that is at V potential).

    When capacitors are in parallel, the battery of 2*V waste Q charges to fill one capacitor, and other Q charges to fill the other at the same time. The capacitors rise to the same voltage as the battery (2*V).

    Then you've got the capacitors charged at 2*V volts.

    Now you disconnect the parallel.

    Now you've got two C capacitors, each one charged at 2*Q. This means that each one of the disconnected capacitors have the same 2*V volts as the battery.

    Now connect them in series. Now you've got 2*V volts of one 2*Q/C capacitor plus 2*V volts of the other 2*Q/C capacitor (series connection). That is 4*V.

    They say that two series C capacitor are equal to C/2 because (in terms of voltage):

    4*V = 2*Q/C + 2*Q/C = 4*Q/C = 2*Q/(C/2)

    Now, it's clear that two C series capacitors at 4*V are the same as a unique C/2 capacitor at 4*V (but charged at half the charge!).

    Why free energy?

    Now, connect the series capacitors to the battery. The 4*V volts will be "dumped" into the battery. Then, the charges stored in the C capacitors will go to the battery, until the potential level of the series capacitors rise 2*V volts of the battery again.

    Now, if you've got 2*V in the series connection, you must have V on each capacitor when you disconnect them. That means that each one of the capacitors is charged at V so the relation of charges is Q/C that is initial conditions.

  10. Apr 29, 2003 #9


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    I do not do real good with math. I understand, can work with it, but try my hardest to avoid it.

    I've played with free energy ideas, with various teachers and on here a while ago.

    The best thing you will ever be able to do is come up with something that cost nothing (little) to operate itself.

    To get any gains, you need something external to add a boost.

    Such as the classic water wheel perpetual motion idea. If a person stood there turning a handle to help the motion a little bit, it would work to some extent.

    But, avoiding the mathematics, you can't get something for nothing.
  11. Apr 29, 2003 #10


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    You said the initial condition was 2V in the first paragraph. Which is it?
    Okay -- so you're shuffling charge around between a battery and some caps. How is this free energy?

    - Warren
  12. Apr 29, 2003 #11

    Tom Mattson

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    Warren, meet Cala. He is our resident disbeliever in thermodynamics, and has been posting these ideas of his here for about a year now. He is not going to be satisfied with your explanation until you pinpoint the exact mechanism of the device that fails, and even then he might not be convinced.

    The standard response that Integral and I give is: Cala, just build it! Then you'll see for yourself that it doesn't work.
  13. Apr 29, 2003 #12
    Sorry dude.... If you have two equal-valued capacitors connected in parallel, they will both have the same amount of negative charge on anode and the same amount of positive charge on the cathode. If you disconnect them and put them in series, the negative charge on the anode of the top capacitor will cancel the positive charge on the cathode of the bottom capacitor. The net charge will be equal to the charge on a single capacitor and the capacitance will have been cut in half... so...

    V = (Q/2) / (C/2) = Q / C.

  14. Apr 30, 2003 #13
    Once again...

    Take V=Q/C

    Now take two equal C capacitors charged at Q (then each one is charged at V)

    Now connect them in series:

    What potential has this series capacitor?:


    Q*(2/C) is the same as Q /(C/2), so that's why two series C capacitors are considered equal to one C/2 capacitor.

    Any comments?
  15. Apr 30, 2003 #14
    Cala, there is complete balance of energy in both processes (charging 2C capacitor and discharging C/2 capacitor back to the battery).

    Indeed, in the first process (charging two parallel connected C+C capacitors initially having V/2 voltage each by battery with voltage V) capacitors gain energy from initial CV^2/4 (total on both)to CV^2, so the net gain is 3CV^2/4. This energy gain comes from the battery which loses the amount int(VI)dt = CV^2 out of which the amount int(RI^2)dt=CV^2/4 is lost into Joule heating of the resistor R, so remaining 3CV^2/4 went into capacitors. Full balance here.

    During rearrangement of capacitors no loss/gain of energy takes place, so capacitors in series have the same energy 2(CV^2/2)= CV^2 as capacitors in parallel (C/2)(2V)^2/2 = CV^2.

    Now discharge of double voltage serial capacitor pack back to the (single voltage) battery. During this process again we have complete balance of all losses/gains. Indeed, capacitors discharge from initial CV^2 to final CV^2/4, i.e. they lost 3CV^2/4. Battery gained back int(VI)dt = CV^2/2, resistor dissipated into heat int(RI^2)dt = CV^2/4, netting 3CV^2/4 - same as capacitors give up amount.

    So, no net gain/loss of any energy is detected.
  16. Apr 30, 2003 #15
    I think there is more than enough evidence here to conclude this topic.
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