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  #31  
Old 08-07-2017, 03:35 AM
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Matthew Jones Matthew Jones is offline
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All of you are contemplating a circuit that does not include switched timing. As soon as a capacitor receives a charge that charge should be moved from the capacitor plate back to the entrance of the circuit and reused. If the capacitor is empty it will not have any resistance in fact it can emulate a ground up to 10% capacity as far as current flow is concerned.
Charging a Capacitor

Stop thinking about a conventional LCR circuit and create the one you want. Any amount of power can be timed and converted.

If you have a bit of energy and you use it to generate more while retaining the original energy minus loss, what do you have left? Stop thinking and do.

Thats it..
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Last edited by Matthew Jones; 08-07-2017 at 03:44 AM.
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  #32  
Old 08-07-2017, 04:24 AM
p75213 p75213 is online now
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Quote:
Originally Posted by Matthew Jones View Post
All of you are contemplating a circuit that does not include switched timing. As soon as a capacitor receives a charge that charge should be moved from the capacitor plate back to the entrance of the circuit and reused. If the capacitor is empty it will not have any resistance in fact it can emulate a ground up to 10% capacity as far as current flow is concerned.
Charging a Capacitor

Stop thinking about a conventional LCR circuit and create the one you want. Any amount of power can be timed and converted.

If you have a bit of energy and you use it to generate more while retaining the original energy minus loss, what do you have left? Stop thinking and do.

Thats it..
^^^^^
This. I haven't got around to the timing yet. Or any of the fine details for that matter. I new it was needed but lack the skills to know exactly what is required. It's on my to do reading list. Everybody seems to be fixated on a tank circuit. It was only my intention to use that circuit to detail my thoughts on the viability of the idea. It was never going to be the final solution.
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  #33  
Old 08-07-2017, 10:12 AM
lotec lotec is offline
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Quote:
Originally Posted by Matthew Jones View Post
If you have a bit of energy and you use it to generate more while retaining the original energy minus loss, what do you have left?
Nailed it right there, but for some of us, some thinking still required.
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  #34  
Old 08-07-2017, 11:07 AM
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Bertoa Bertoa is offline
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Back to the coil/capacitor/switch/lamp theme, this diode-plug circuit is coming to my mind. The coil is here also the generator.
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  #35  
Old 08-07-2017, 03:33 PM
Cadman Cadman is offline
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Quote:
Originally Posted by Matthew Jones View Post
... As soon as a capacitor receives a charge that charge should be moved from the capacitor plate back to the entrance of the circuit and reused. If the capacitor is empty it will not have any resistance in fact it can emulate a ground up to 10% capacity as far as current flow is concerned...
Exactly. Stop charging the cap to it's full capacity.

About 270 years ago Ben Franklin observed this about capacitors or Leyden jars:
“But suspend two or more phials (Leyden jars) on the prime conductor, one hanging to the tail of the other ; and a wire from the last to the floor, an equal number of turns of the wheel (of an electrostatic friction machine) shall charge them all equally, and every one as much as one alone would have been. What is driven out of the tail of the first, serving to charge the second ; what is driven out of the second charging the third ; and so on.” The words in parenthesis are mine.

He also goes on to note that each 'phial' has a reluctance or resistance to charging.
If the first cap is charged to 10% with little resistance, as Matt points out, and the second is charged by the first to 9%, and the third to 8% then what do you have?

One last thing. I firmly believe that the amount of charge in a closed system is finite and cannot be increased unless additional charge is brought in from an external source; such as a battery, generator, from the environment or from the earth.
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Last edited by Cadman; 08-07-2017 at 04:18 PM.
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  #36  
Old 08-08-2017, 10:28 AM
p75213 p75213 is online now
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We have 4 capacitors of equal capacitance. 2 are connected in series to a voltage source and 2 in parallel to another voltage source.
V = Volts
C = Capacitance
Energy of a capacitor = 1/2*CV^2
Parallel capacitance total for 2 capacitors -> 2C
Series capacitance total for 2 capacitors of equal capacitance -> C/2

Energy of the capacitors in parallel: 1/2(2C)V^2 -> CV^2
Energy of the capacitors in series: 1/2(C/2)V^2 -> 1/4CV^2

Therefore the capacitors in parallel have 4 times the energy as those in series. Also for every capacitor in parallel is a duplication of energy. So for n capacitors in parallel we have (n/2)CV^2 joules of energy. For example for 10 equal capacitors we have 5CV^2 joules of energy. Thats 10 times the energy of just one capacitor. Or to put it another way. To increase the energy of a single capacitor just increase the capacitance. Although the charging time would increase accordingly.

An interesting observation. Switch the parallel caps to series for a quicker discharge. Same amount of energy delivered at a faster rate equals increased power.
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Last edited by p75213; 08-08-2017 at 12:16 PM.
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  #37  
Old 08-08-2017, 11:11 AM
lotec lotec is offline
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Heres a circuit that might be relevant. Its not quite finished yet so it is untested.

Its pretty much a pulsed flyback transformer with a small change. The primary pulse helps to charge an ultra capacitor C Ultra. When the top switch is opened, it breaks the primary current, the bottom switch closes, connecting the secondary flyback to C Ultra in series, and helps to discharge C Ultra into the load.




Edit ....... Had a chance to try out the circuit, and it didn't make any gains, so I don't recommend trying it.
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Last edited by lotec; 08-19-2017 at 02:24 AM.
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  #38  
Old 08-08-2017, 11:37 AM
lotec lotec is offline
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Ooops sorry there p75213, it looks a bit disjointed me posting that circuit there after your last post. I wasnt trying to take things in a different direction. You must have got yours in while I was doing mine. Let me know if you want me to delete it.
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  #39  
Old 08-08-2017, 11:52 AM
p75213 p75213 is online now
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Quote:
Originally Posted by lotec View Post
Ooops sorry there p75213, it looks a bit disjointed me posting that circuit there after your last post. I wasnt trying to take things in a different direction. You must have got yours in while I was doing mine. Let me know if you want me to delete it.
Of course not. It all adds to the mix.
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Last edited by p75213; 08-08-2017 at 12:10 PM.
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  #40  
Old 08-08-2017, 01:16 PM
Cadman Cadman is offline
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Quote:
Originally Posted by p75213 View Post
...
Switch the parallel caps to series for a quicker discharge. Same amount of energy delivered at a faster rate equals increased power.
Would it not be better to charge the 3 caps in series and then discharge sequentially?
Initially move 10% of charge from the source, then discharge 10% from the first, then 9% from the second, then 8% from the third.
Total discharge 27% minus losses.
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  #41  
Old 08-08-2017, 01:21 PM
bistander bistander is offline
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Capacitor energy

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Originally Posted by p75213 View Post
We have 4 capacitors of equal capacitance. 2 are connected in series to a voltage source and 2 in parallel to another voltage source.
V = Volts
C = Capacitance
Energy of a capacitor = 1/2*CV^2
Parallel capacitance total for 2 capacitors -> 2C
Series capacitance total for 2 capacitors of equal capacitance -> C/2

Energy of the capacitors in parallel: 1/2(2C)V^2 -> CV^2
Energy of the capacitors in series: 1/2(C/2)V^2 -> 1/4CV^2

Therefore the capacitors in parallel have 4 times the energy as those in series. ...
Hi p75213,

You forget that for the series connection the voltage doubles so the total energy is the same for 2 caps in parallel as 2 caps in series.

bi

{edit}


Energy of the capacitors in parallel: 1/2(2C)V^2 -> CV^2
Energy of the capacitors in series: 1/2(C/2)(2V)^2 -> CV^2
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Last edited by bistander; 08-08-2017 at 01:26 PM. Reason: Added equations
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  #42  
Old 08-08-2017, 09:31 PM
p75213 p75213 is online now
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Quote:
Originally Posted by bistander View Post
Hi p75213,

You forget that for the series connection the voltage doubles so the total energy is the same for 2 caps in parallel as 2 caps in series.

bi

{edit}


Energy of the capacitors in parallel: 1/2(2C)V^2 -> CV^2
Energy of the capacitors in series: 1/2(C/2)(2V)^2 -> CV^2
Each cap in parallel has voltage V and the total voltage is also V.
Each cap in series has voltage V /2. Therefore the total voltage is V.

The series connection would have to have double the source voltage to achieve the same energy output. Or if their were n similar parallel caps it would need n times the voltage source.
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Last edited by p75213; 08-08-2017 at 09:43 PM.
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  #43  
Old 08-08-2017, 09:49 PM
p75213 p75213 is online now
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Quote:
Originally Posted by Cadman View Post
Would it not be better to charge the 3 caps in series and then discharge sequentially?
Initially move 10% of charge from the source, then discharge 10% from the first, then 9% from the second, then 8% from the third.
Total discharge 27% minus losses.
Maybe somebody else would like to comment on this because I don't know. You could be right though.
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  #44  
Old 08-09-2017, 01:32 AM
bistander bistander is offline
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Voltage

Quote:
Originally Posted by p75213 View Post
Each cap in parallel has voltage V and the total voltage is also V.
Each cap in series has voltage V /2. Therefore the total voltage is V.

The series connection would have to have double the source voltage to achieve the same energy output. Or if their were n similar parallel caps it would need n times the voltage source.
You didn't specify V was total voltage and constant. In fact your interesting observation at the end of that post implies a higher total voltage for the series connection, doesn't it?

Quote:
Originally Posted by p75213 View Post
...
An interesting observation. Switch the parallel caps to series for a quicker discharge. Same amount of energy delivered at a faster rate equals increased power.
Why would anyone use series connected capacitors unless it was to get a higher total voltage?

bi
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  #45  
Old 08-09-2017, 01:42 AM
ricards ricards is online now
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Quote:
Originally Posted by p75213 View Post
Each cap in parallel has voltage V and the total voltage is also V.
Each cap in series has voltage V /2. Therefore the total voltage is V.

The series connection would have to have double the source voltage to achieve the same energy output. Or if their were n similar parallel caps it would need n times the voltage source.
bistander is correct in his statement, but you are also correct, because of your given scenario (same source voltage to charge both capacitor arrangement).

I have a similar project where in I have a clock to switch 2 transistor alternately, one to charge the capacitor in parallel and one to discharge them all in series (your idea), the difference is I have Inductors in between capacitor in series, and a bunch of diodes to avoid short circuit and to redirect the "Energy" flow where I want it., But my intention was not so that I will have increase power output (because its the same amount of energy). but to make a High voltage DC pulser without using traditional transformer diode configuration, because that way I can control the frequency better.

If you want I can share the schematics, but I haven't built the actual device yet, so no promise on efficiency.
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  #46  
Old 08-09-2017, 02:47 AM
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Capacitors.

To see what is possible, you need to use a charged capacitor with a "known" charge as the source. Then quickly connect it to several capacitors in series and measure what you have in each capacitor as well as what is still in the original capacitor. Then put ALL capacitors in parallel to send everything back to a second set of capacitors in series. Keep going back and forth with the two sets of capacitors and measure what you get. Are the voltages going up or down?
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  #47  
Old 08-09-2017, 02:51 AM
p75213 p75213 is online now
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Quote:
Originally Posted by bistander View Post
You didn't specify V was total voltage and constant. In fact your interesting observation at the end of that post implies a higher total voltage for the series connection, doesn't it?
bi
If we switch the caps from parallel to series each cap has V volts. Therefore a total of 2V volts for the series circuit. So yes it is a higher voltage for the series circuit. I can see your point. I was just making the observation that it takes a lower voltage to charge parallel caps to the same energy level as the same number of caps in a series circuit.

Why will they discharge quicker in a series circuit as opposed to a parallel circuit. The charge/discharge time constant is RC where R = resistance and C = capacitance. In the series circuit the total capacitance is now C/2 for 2 equivalent caps. Or C/n for n similar caps. Whereas the total capacitance in the parallel circuit was 2C or nC for n similar caps.

Why do we get more power. Because power is defined as the rate of using energy. In the series circuit the energy is being used at a faster rate.
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Last edited by p75213; 08-09-2017 at 02:56 AM.
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  #48  
Old 08-09-2017, 05:31 AM
bistander bistander is offline
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RC

Quote:
Originally Posted by p75213 View Post
...
Why will they discharge quicker in a series circuit as opposed to a parallel circuit. The charge/discharge time constant is RC where R = resistance and C = capacitance. In the series circuit the total capacitance is now C/2 for 2 equivalent caps. Or C/n for n similar caps. Whereas the total capacitance in the parallel circuit was 2C or nC for n similar caps.
...
Hi p75213,

For any number of identical capacitors in any series or parallel combination, the equivalent time constant for the group will be RC, the time constant of the individual capacitor.

2 in parallel -> (R/2) * (2C) = RC

2 in series -> (2R) * (C/2) = RC

bi
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  #49  
Old 08-09-2017, 05:46 AM
p75213 p75213 is online now
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Quote:
Originally Posted by bistander View Post
Hi p75213,

For any number of identical capacitors in any series or parallel combination, the equivalent time constant for the group will be RC, the time constant of the individual capacitor.

2 in parallel -> (R/2) * (2C) = RC

2 in series -> (2R) * (C/2) = RC

bi
Your right. I forgot about R. It was looking pretty good their for awhile.
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  #50  
Old 08-09-2017, 06:59 AM
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Quote:
Originally Posted by Turion View Post
To see what is possible, you need to use a charged capacitor with a "known" charge as the source. Then quickly connect it to several capacitors in series and measure what you have in each capacitor as well as what is still in the original capacitor. Then put ALL capacitors in parallel to send everything back to a second set of capacitors in series. Keep going back and forth with the two sets of capacitors and measure what you get. Are the voltages going up or down?
I do not understand

Charge cap 1 with 20vdc supply say then charge several caps in series then
send that to a second bank of 1 cap and several series caps but send only
using a parallel connection path?

If I don't understand your very important post, there are lots who have
no clue. I want to stress how important I consider your entries but have
found out from past exchanges that only a picture can solve my inability.

I think I understand some of it but not sure and I do not want to let this
go by without asking to find out for sure.

I picture 2 banks of caps to send energy back and forth in certain
connection paths using a single source to start out with? Again I am
unsure. Or is it that each time a test is made I am to use the same
known source?

But then I could not send that charge over to a second bank from the
first bank and make any sense out of my reasoning so I conclude charging
the first set of 1cap and several (say3 or 4 or 5) in series is done with a
known source?
Which one is right?

@everyone
I want everyone to know I consider Turion a great researcher so when
he talks, I try to understand.
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Last edited by BroMikey; 08-09-2017 at 07:03 AM.
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  #51  
Old 08-09-2017, 08:56 AM
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Quote:
Originally Posted by Turion View Post
To see what is possible, you need to use a charged capacitor with a "known" charge as the source. Then quickly connect it to several capacitors in series and measure what you have in each capacitor as well as what is still in the original capacitor. Then put ALL capacitors in parallel to send everything back to a second set of capacitors in series. Keep going back and forth with the two sets of capacitors and measure what you get. Are the voltages going up or down?
Does this look right so far?

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  #52  
Old 08-09-2017, 01:20 PM
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Ufopolitics Ufopolitics is offline
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Quote:
Originally Posted by Cadman View Post

One last thing. I firmly believe that the amount of charge in a closed system is finite and cannot be increased unless additional charge is brought in from an external source; such as a battery, generator, from the environment or from the earth.
Hello Cadman,

You are correct, however:

You could increase the amount of charge in ANY Closed System, without absolutely any direct connections (wiring) to any external sources like a battery, generator, etc, just by adding a "receiver" core-coil to the closed system....then INDUCING a charge (EMF) through a Magnetic Field from a second exciting (transmitter) coil circuit.

Closed system will remain "physically" closed, and energy will enter through Spatial Electromagnetic Induction.

But still, I understand energy would be coming from an "external source"...


Regards


Ufopolitics
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