Charging capacitors at different voltage levels

1. Mar 31, 2004

cala

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. Mar 31, 2004

Averagesupernova

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.

3. Mar 31, 2004

chroot

Staff Emeritus
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

4. Mar 31, 2004

Janitor

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.)

5. Apr 1, 2004

cala

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 )

6. Apr 1, 2004

chroot

Staff Emeritus
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

7. Apr 1, 2004

cala

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
8. Apr 1, 2004

Cliff_J

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

9. Apr 1, 2004

cala

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?

10. Apr 1, 2004

Cliff_J

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

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
11. Apr 1, 2004

cala

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.

12. Apr 1, 2004

Cliff_J

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 $power = \frac {voltage^2} {resistance}$ 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. :)

13. Apr 1, 2004

cala

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" (). 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?

14. Apr 1, 2004

Cliff_J

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

15. Apr 1, 2004

faust9

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
16. Apr 2, 2004

cala

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.

17. Apr 2, 2004

cala

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?

18. Apr 2, 2004

cala

...Again, i'm thrown away to the theory development limb.

19. Apr 2, 2004

Cliff_J

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 $\sqrt{2}$ 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.

Cliff

Last edited: Apr 2, 2004
20. Apr 2, 2004

Staff: Mentor

... 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
21. Apr 3, 2004

cala

I'm now simulating in Spice, and it seems to work!

I post the drawing of the simulated circuit, and the signals on some elements.

Some parameters:

Transformer: L1 = 100 turns, L2 = 2000 turns, K = 1.
Capacitors: C = 3 uF, IC = 5 V.
Switch1: ON = 75 uS, OFF = 10 mS. (Transformer proccess)
Switch2: OFF = 75 uS, ON = 10 mS. (Load branch)

Firstly, i simulated only the AC charging-discharging proccess of the capacitors through the transformer (without the load branch). It seems to work. Adjusting the transformer and the timing of the switch1 we get more or less voltage difference, and more or less time to create this voltage difference on capacitors.

Finally, starting from 5V charged capacitors, i finally get 9V and 4.3V in 75uS of switch1.

Then, I let the voltage difference created on the capacitors discharge through the load. The time of this part depends on the load value chosen. In the example, it takes 10mS to get the initial conditions to start another cycle.

The values on the load are:

V = 4 V peak. Average = 2 V.
I = 4 mA peak. Average = 2 mA.

This is a power of 4 mW.

Of course, this is not much power, it is a proof of the concept. If the loses on the switching and tranformer parts are less than this value, we can consider it overunity or free energy on the load.

I'll continue simulating. Maybe i should improve the switch part and the tricks that we have to do in Spice to simulate. If you have some suggestion or comment, please post it.

I sure you the graphics that seems to work are just here in front of me.

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22. Apr 3, 2004

chroot

Staff Emeritus
How is it overunity? You've lost voltage in the initial capacitor. The system does not return to its initial state.

- Warren

23. Apr 3, 2004

faust9

Do you know how a transformer works? You are fixating on the voltage gain but (and here's the cunundrum) current will go down by the same ratio as voltage goes up. Your circuit doesn't make more power on the secondary side than the primary side is supplying. You are not over unity. You are exploiting the main property of a transformer which is the ability to step AC voltage up or down and likewise step AC current down or up respectively.

From what I see you fiddled around with a circuit using your assumptions and eventually came up with a working circuit. Now my question to you is how do you think that circuit is working? From what I see your charging C3 via the combined power sources and not from C1. Elliminate C1 and see what happens. Another thing is you are using your FETs incorrectly. You don't source a FET through the gate. The gate is used to control the current flow from source to drain. The supply voltage from V2 and V3 is being subdivided through the source and the drain of the FET thus some of the voltage gain on the secondary is a direct result od V2 and not from C1.

I applaude your efforts, but at the same time I question your thought process. Personally, if I were told that a perticular circuit had been developed already and was in use I'd drop my analytical persuits and find documentation associated with the circuit. To that end, I'll tell you step up/step down power supplies were invented a long time ago. Look up complementart transformers on google because that is essentially what you have here. The transformer is doing its job in a physically acceptible manner. You just need to analyze what your sources and loads are actually doing. Draw some current flow paths using some colored penciles and you'll see what I'm talking about.

Now as far as your obsession with overunity is concerned, try as you might your not going to find the magic machine. Transformers have some well know properties (even if you use a virtual lossless XFMR) which apply here. Power on the primary side of a virtual XFMR will equal that on the secondary side which means as V goes up I goes down on the secondary. You can charge a capacitor on the secondary but its not gonna happen the way you think it is I dare say. Do a simple transient analysis with a cap on the primary initially at some voltage and a cap on the secondary at some voltage no other components. In a transient analysis you can specify initial conditions. Watch what happens. Monitor voltage and current through the primary, and secondary of the XFMR. Trust me when I say your not over unity(powerer out !>power in).

24. Apr 4, 2004

cala

I've been playing with coils, capacitors, transformers and loads on Spice.
Then i remembered that i've seen a patent of a electric AC generator based on a simple circuit. Then i tried to simulate it.

I post the simulated circuit. You can see there is no source on the circuit. It is supposed to act like an oscillator, and also powering a ressistive load with AC signal. It's said we need a initial condition on one of the capacitors to make the device work. (a pulse).

Some parameters used on the simulation:

Capacitor C2: 33uF, IC = 1V.
Capacitor C1: 33uF, IC = 0V.
Transformers 1 and 2: K = 1, L1 =8900 turns, L2 = 10 turns.

The values of the transformers sure that the signal will be 50 Hertz AC.

I post also the current and voltage on the ressitive load. As you can see, once the oscillation begins, the load receives 50 Hertz AC signal for 100 seconds!. Is that energy the same than the initial pulse?.

The current is 2mA approx., and the tension of 40 V or so. This is 80 mW.

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25. Apr 4, 2004

faust9

The circuit you described is called a Tank Circuit. Do a google search and you'll find a lot of info. If you notice the circuit oscillates for 100 seconds. Why does it stop? It stops because there are losses in the system associated with intrinsic resistance of the wire and leakage of the capacitor and hysteresis. The actual oscillation is the energy bouncing back and forth between the inductor and the capacitor. To produce a continuous oscillation you have to add enough energy each cycle to make up for the energy lost. So, yes the energy bouncing back and forth PLUS the lost energy is equal to the initial energy put into the system. Think of it like this. You fill cup A with some water. You pour that water into cup B then back to A then back to B..... Each time you pour the water between the cup some splashes out. Eventually(if you do this long enough) the level of water will go down to zero. If you add the water sitting on the counter to what little water is left in the cups you'll have the same ammount of water as you started with; however, you were able to pour the water back and fourth between the cups. You didn't gain water simply because you managed to fill cup B nor did you gain water because you were able to fill each cup multiple times. The same ammount of water was always present just in a different location.

Same thing happens when a tank oscillates. the energy bounces back and forth (for lack of a better term) between the capacitor and the inductor but with each bounce a little electric energy is converted to heat energy.