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Charging capacitors at different voltage levels

  1. Mar 31, 2004 #1
    Imagine two capacitors charged at the same voltage.

    Now, connect one of them to a high frequency switch, and then to the primary of a 1:100 transformer. On the secondary of this transformer, there is a diode bridge, and then, the second capacitor is connected.

    Will the voltage of the second capacitor be 100 times the voltage of the first one with this method?

    If it works, we can "charge" one capacitor to a higher voltage level from a lower voltage capacitor.
     
  2. jcsd
  3. Mar 31, 2004 #2
    You can do this but someone once described an operation similar to what you you are describing as 'all ropes and pulleys'. It does charge the second capacitor up to a higher voltage, but it takes several tries to get it all the way there. There is no real power gain here or anything like that.
     
  4. Mar 31, 2004 #3

    chroot

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    What you're essentially describing is the basis for a switched-capacitor boost converter. These little devices are used everywhere in electronics -- some of them are the size of the ball at the tip of a ball-point pen, and you can buy them for pennies.

    - Warren
     
  5. Mar 31, 2004 #4

    Janitor

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    A couple of comments that may not amount to much.

    Are you imagining the capacitance of the first capacitor to be much bigger than the capacitance of the second, so that the first capacitor doesn't drop much from its initial voltage?

    Is the output of your switch a square wave? If so, is the output of the transformer, in an idealized limit, a sequence of spikes that alternately look like positive and negative Dirac delta functions? If this is the case, why would the 1:100 winding of the transformer limit you to only 100 times the voltage of the first capacitor? (Real world problem lurking: breakdown of dielectric when voltage on second capacitor gets big.)
     
  6. Apr 1, 2004 #5
    Well, it seems that there is no problem with this method, ok.

    I was wondering that a transformer is like a free "water pump" from the point of view of the capacitors.

    I mean: At first, the two capacitors (same capacitance) have the same energy stored. This energy depends on geometry and charge stored. No transfer can happen when the two capacitors have the same voltage, because they have same voltage and energy levels.

    Now, when using the transformer, the energy of one capacitor is wasted on the transformer primary (when the high frequency switch acts), then the secondary creates the same energy at different V-I rates to the second capacitor. So the transformer pass almost all the energy to the second capacitor. I mean: the first capacitor is discharging through the primary, and the second one is charging through the secondary. I think this will stop when the V levels on capacitors match the transformer ratio (1:100).

    Lately ¿What avoid us to connect the capacitors through a load , and restore the initial conditions?

    I mean: finally we have the second capacitor fully charged, and the first one almost discharged. If we connect a load, we will have energy transfer. The load will only affect the time of transference, but finally, the charge on capacitors will become equal, and the transference will stop. But finally, as the energy stored on capacitors depends on charge stored and geometry, we will arrive to initial conditions, and can do another cycle of "free" charging the second capacitor, then powering a load.

    The "free" work becomes from the charging of the second capacitor and discharging of the first one almost for free by transformer operation (99% efficient, and we only need to input power for the switch operation).

    Any mistake?... (Sure :biggrin: )
     
  7. Apr 1, 2004 #6

    chroot

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    I was really hoping that you were not going to turn this into another of your "free energy" crackpot discussions.

    Remember that transformers operate with AC, not DC. The discharge of a capacitor through a 1:100 transformer primary does not create a 100x larger identical waveform on the secondary.

    We're discussing capacitors and transformers, items readily available from Radio Shack. Build one of these circuits and see what it does. Download a trial version of OrCAD's pSpice and simulate the cirucit. See what it does in intimate detail.

    - Warren
     
  8. Apr 1, 2004 #7
    Chroot, my idea was not to discharge the first capacitor all at once or same time. It is discharged on high frequency steps (i mean something like pseudo-AC). Also, the second capacitor goes charging in the same way (little steps of charging).

    Also, i'll try to simulate the idea, but playing with some programs i saw that somethings like induction proccess are not taken into acount, and also, we must use some tricks in some cases to simulate.
     
    Last edited: Apr 1, 2004
  9. Apr 1, 2004 #8

    Cliff_J

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    Nothing is free. A better (although still lackluster) is you have your two buckets of water, and you raised one 100m off the ground on a rope with a pulley. You added the potential energy to the bucket.

    The capacitors could not be the same size -OR- the one attached to the secondary size would increase very little.

    Toridial transformers are 80% efficient, your mileage may vary.

    Here, try this, very simple. You take two capacitors of the same size. One charged to a voltage, the other discharged completely. You connect the two in parallel. Both capacitors (ignoring losses) now have exactly one-half the intial voltage, what is the energy stored?

    Cliff
     
  10. Apr 1, 2004 #9
    Ok Cliff, now following your example:

    Imagine now you took your equally charged capacitors (having V/2 of initial voltage), and then connect them to the transformer proccess, as i explain. Then the first one will finally rise a lower V value than the second one.

    Now connect the two "voltage de-compensed" capacitors again through a load. Then, the charges will redistribute on capacitors, and finally, they will rise the V/2 value. Of course, this V/2 value will not be the same as on initial conditions, because the transformer and switch proccess is not 100% efficient.

    What i wanted to say is that the final voltage level on capacitors seem to be independent of the load connected. If the transference between capacitors on the transformer step were ideal, the capacitors will charge from V and v to (V+v)/2 again.

    Is that incorrect?
     
  11. Apr 1, 2004 #10

    Cliff_J

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    Capacitor energy is determined by:

    Joules = .5 * capacitance * voltage^2

    So we have two 100mF capacitors, one at 0V and the other at 10V. How many Joules do we have:

    (.5 * .1 * 10^2 ) + (.5 * .1 * 0^2) = 5J

    Now we connect in parallel, and if the voltages were half we would end up with:

    (.5 * .1 * 5^2) + (.5 * .1 * 5^2) = 2.5J

    This obviously does not work, we've destroyed energy. So this was a bit of a 'structured' question. The voltage will likely decrease instead by 1/SQRT(2) instead, so we end up with:

    (.5 * .1 * 7.07^2) + (.5 * .1 * 7.07^2) = 5J

    So to answer your original question, there are two answers:

    If the second capacitor is 10,000X smaller than the first one and no losses, the voltage could be 100X greater under complete energy conservation.

    If the capacitors were the same size and again no losses, the best you would end up with would be a 70.7% increase in voltage on the second capacitor.

    Cliff

    P.S. And if my memory is foggy, I'm sure someone will correct my math too. :)
     
    Last edited: Apr 1, 2004
  12. Apr 1, 2004 #11
    Ok Cliff, maybe i'm confused on one point:

    Your example of 0 and 10 V going to 7.07 V is the key.

    My only doubt is:

    If we put a 10 Ohm resistor between them, wil the final voltage be the same 7.07 V?

    If we put 1000 Ohm between, wil the voltage be also that 7.07 or not?

    I know the time of the transfer will change, but what happens with the voltage and energy levels at the end of the proccess?

    Charge and geometry remains the same, so to me, the 7.07 V level and energy asociated should remain the same, no matter the load connected.
     
  13. Apr 1, 2004 #12

    Cliff_J

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    Well, if you have a resistance you will have a voltage drop and there will be power dissipated across the resistor. You'd need to integrate it over time to get an exact value, but here's a quick and dirty analysis. Since [itex]power = \frac {voltage^2} {resistance}[/itex] we can see that increasing the resistance will decrease the power dissipated by the resistor and this makes sense as the current is dimished. Since the capacitor charge/discharge curves are so steep in the begining, my guess is that the 1000 ohm resistor would come closer to the ideal 7.07V than the 10 ohm resistor would. Well, unless you have lossless resistors in the ideal world, then yes they'd be the same. :)
     
  14. Apr 1, 2004 #13
    Yes Cliff, i followed Chroot suggestion and simulated with Spice the charging-discharging proccess of two identical capacitors charged at different voltage levels, and yes, the only change is in time. The final voltage is always the same.

    So then, we can go further.

    Cliff, you said: "Nothing is free. A better (although still lackluster) is you have your two buckets of water, and you raised one 100m off the ground on a rope with a pulley. You added the potential energy to the bucket".

    Ok, on this example, you need to do work to change the potential energy of the bucket. But what is the work the transformer needs to do to change the capacitors potentials? I think transformers doesn't waste (almost) energy, they do not need work to "work" (:biggrin:). I mean, they transform intrinsic parameters of energy (V and I) but doesn't transform energy into another form. The less V on one side, the more on the other, (and the same on the current) so finally, with the transformer, the energy remains (almost) the same, but re-distributed.

    To me, the transformer seems like a "free potential changer", isn't it?
     
  15. Apr 1, 2004 #14

    Cliff_J

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    The transformer is two inductors so its basically doing a electrical --> magnetic + heat --> electrical + heat transformation.

    Its no more a free changer electrical energy than a gearbox is a free changer of mechanical energy. We still end up with losses as heat energy.

    So the transformer can work without any moving parts. So can a transistor. They both get warm when they are 'working' hard.

    Cliff
     
  16. Apr 1, 2004 #15

    ARGH!!!!! Com'on say after me "In physics ther is no such thing as a free lunch." Say that 100 times or write it 100 times or whatever it takes. I'm not trying to attack you, but on the other hand you seem to have this idea that if you look hard enough you'll find that magical device which will be completly lossless. Not gonna happen. Instead of doing these thought experiments alone, you should sit down and draw a circuit and actually do some calculations using real world assumprions. Assume there are heat losses. Assume that transformers are 80% effecient not 99% effecient. Assume the resistors consume power (that is how they work). Assume that wire has an intrinsice resistance that consumes power. Assume that the capacitors have leakage. There are a lot of assumption you can make and a lot of calculation you can do but in the end you find that this particular idea isn't as perfect a solution as you think. These devices do exist. There are DC step up circuits that use cascaded capacitors with transistors between them controlled by switching circuitry, but these circuits are not free lunches. There are losses associated with these circiuts so to overcome the losses, a power source is used. Its not as simple as saying "A fully charged capacitor will charge another capitor to such and such"

    Other things to consider are the RC time constants, LR time constants, LC tank frequencies, etc. If you used a 12V battery in seriec with a 10pF cap in series with a transistor hooked with the base hooked to a function generator (squate wave) in series with the primary of a transformer and the secondary of the xfmr hooked in series to 10pF in series with some load. I bet the output would be a rough sinewave at some frequency less than the control frequency on the primary side. Moreover, the output P-P amplitude would be LESS than the input P-P amplitude. Try it. model it. Download SPICE and play with it. Do the calculations.

    I hate to say this but you have a misunderstanding of how capacitors and inductors interact. You have a misunderstanding of how timeconstanst affect a LRC circuit.

    "There is no such thing as a free lunch in physics."

    PS. the circuit you described is used to turn a square wave into a sine wave.
     
    Last edited: Apr 1, 2004
  17. Apr 2, 2004 #16
    Ok, ok, ok

    I know you all say to me always the same, it's true.

    Just asume that everytime i ask something here is for proposing an overunity or free energy system, so answer my questions or not, but let the "There is no free energy in physics" statement out of my discussions.

    Well, you all are talking to me to take into account the losses and the heat and all the effects that will cause the wasting of the energy.

    That is the question, I mean that in the system I talk about, that losses and heat (on the switch and the transformer) is the ONLY input we need.

    On normal devices, there is the work to do, and also, the losses.

    On this device, it seems to be that the only input needed is to restore the losses.

    As Cliff said, the transformer is a potential changer as a gear is a mechanical changer.

    Do the gear need energy input to do its job? The answer is no (once there is something moving on one side of the gear, the gear will transform RPMs and torque on the other side without work), we only need to go against friction losses on the gear, the "transformation" of mechanical energy parameters is for free.

    The transformer does something similar with electric potentials. It only changes energy parameters almost without work. In the mechanical device I can't see a way to take advantage of the mechanical parameters transformation, but on the electric example, it seems that we can take advantage of the parameter transformation, because the transformer seems to change voltage values for free, and voltage depletion is electric circuits working principle.

    well, finally, you are right: I must simulate all this things to see my errors, because all "no free energy" sentences are not the graphical real thing.
     
  18. Apr 2, 2004 #17
    my last question (and it was the first one also) is:

    Imagine two equal capacitors charged at 5V.

    Could we pass almost all the energy stored on one of them to the other (say the capacitors finally gets charged at 0.1V and 9.9V) using a high K transformer (1:1000 or something) and an AC discharging-charging proccess at high frequency?
     
  19. Apr 2, 2004 #18
    ...Again, i'm thrown away to the theory development limb. :mad:
     
  20. Apr 2, 2004 #19

    Cliff_J

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    For two identical capacitors, without any losses considered, the best you could achieve would be a 41% increase in voltage as the energy follows the formulas above. So if you started with two caps at 5V and transferred all the energy to one cap you end up with 0V and 7.07V. Why?

    Because the energy is the sqaure of the voltage (like power) you end up with results that increase by 41% or decrease by 29.3% because your factor of 2 at teh energy level needs to be changed to a factor of [itex]\sqrt{2}[/itex] at the voltage level. Its NOT a linear relationship.

    Not including losses does make this theory, not reality. I don't assume that any code I write is executed with zero CPU or memory requirements, nor any circuit or belt drive can work without accounting for penalties of operating in the real world. You know what they say about what happens when you ass-u-me things. And over-unity and free energy really fall into something best left to the realm of sci-fi.

    Oh, and a gearbox can easily require more than just frictional losses. Some gear teeth intermeshes cause continous changes in angular momemtum - the effective radius constantly changes as the gears mesh. So I would'nt call it any more free than the roadways I drive on after paying $.30 a gallon gas tax. If you still call it free, a career in marketing or as a political spin doctor might be a more appropriate use of those skills. :cool:

    Cliff
     
    Last edited: Apr 2, 2004
  21. Apr 2, 2004 #20

    russ_watters

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    ... But you knew, didn't you?

    cala, serious question: at what point would you decide perpetual motion is a futile exercise? Or is there even such a point?

    It seriously saddens me that people waste their intellect on such futility. You may actually be able to do some good with it if focused correctly.
    That's true for a lot of systems and its absolutely trivial. Add the losses back in and you may get your system back to zero. That's it. I don't know why you are convinced that -x+x can ever be higher than zero.
    You seem to be quite aware that what you suggest is a violation of the laws of physics - so why would it surprise you to end up in TD?
     
    Last edited: Apr 2, 2004
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