# Flux Capacitor Circuit Diagram

Discussion in 'Architecture & Engineering' started by Layman, Nov 25, 2014.

1. ### leopoldValued Senior Member

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17,455
no, i don't have any experience with time machines.
my comment was in regards to specifically high voltages and generally to electronics.
what's disturbing is that you might not know what you are dealing with
high voltages and currents exist almost everywhere.
those big metal latticed towers you see can easily carry upwards of 400,000 volts.
a voltage tripler can be built to get over a million volts.
dangerous?
yes, if you don't take the right precautions.
free energy?
no such thing.
all energy comes with a price.
you are trying to make money off of this?
do you realize the number of scientists that will be breathing down your neck???
not to mention world governments and military leaders.
your attempts at deception will not succeed.
i can't imagine anyone here helping you be deceitful
you can call it your aunt fannys girdle as far as i care.

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3. ### LaymanTotally Internally ReflectedValued Senior Member

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1,001
It was a joke.
Now I might have a better idea now that I ran it through the simulation. Suffice to say, you guys don't seem to be much help trying to understand it further.
How long does it take to get that high of voltage from a low power source? Does it act in a few milliseconds, or does it take a long time for that much power to go in from the source?
It does require an input voltage, but I think theoretically it could have a feedback loop that can be activated once power is put in it; then the power source could be removed. It would kind of be like money, and it takes money to make money. Then it would take electrons to make electrons. That is why I think random particle pair production would be required for the universe to exist. It would create the pair of free particles (even though having one positive particle and one negative particle would be considered to be zero energy to most) in order for them to be intensified. Then the expansion of space itself could have acted like the voltage regulator in the circuit diagram.
Large sums of money are always nice, but I don't see how if the only thing you cannot patent by law is a free energy device. If such a thing is possible, it seems very un-American if you ask me. Without a patent, any company already established could mass produce it and use it in any of their devices that I would never be able to compete with. The government wouldn't be able to legally stop foreign powers from producing it.
I figured they wouldn't be able to help themselves once I claimed it was free energy.
That does have a nice ring to it, but I wouldn't want to be known as the only goof ball that invented an electronic circuit.

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5. ### billvonValued Senior Member

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In simulation? You can easily do it in milliseconds.
No, electromagnetics doesn't work like that.
However, it is not possible - which is why they won't issue patents for one. They have learned it is a waste of their time.

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7. ### Aqueous Idflat Earth skepticValued Senior Member

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Since energy=power × time (like kW-h) then power= energy ÷ time. Rewrite this power = energy × (1/time) and pick a short time, like a millisecond; power = energy × (1/0.001) or power = energy × 1000.

As you see, power can be multiplied simply by using the available energy in a short time. Explosives do this. They burn a large number of molecules at about the same time.

Conversely, to multiply energy, the time span over which the power is harnessed must be extended (multiplied).

Since electrical power (DC) = voltage × current, then voltage = power ÷ current. If the current remains the same, then as power multiplies due to releasing the available energy in a short time, the appropriate circuit can produce a high voltage simply by releasing the available energy in a short time.
Start with 21.1 watts. Store it for 1 second in a battery. Discharge that energy in one nanosecond and you develop the 21.1 GW needed to power the DeLorean. And it will run for one nanosecond.

Thoughts: the law of conservation of energy cannot be repealed.

Suggestions: Advance yourself to the understanding that energy = power × time. From that point forward, remember that energy is conserved, even when high power is developed.

8. ### LaymanTotally Internally ReflectedValued Senior Member

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I checked back to the circuit, and the current being produced on the output is zero amp ... It stays zero amps during the whole 80ms it takes to get to 108MV

9. ### billvonValued Senior Member

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15,114
So the net power produced is zero.

10. ### LaymanTotally Internally ReflectedValued Senior Member

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That just means that it is not being used to do anything. That would be the case in most basic circuit block diagrams. I was just pointing out that it was not the case that Aqueous ID had suggested. The output voltage is not a result of rapid current flow, since it is zero. Then there would be current flow if the output was used as the input of another circuit. That would completely depend on what circuit was then using that voltage.

It actually acts like a "voltage stair climber". I just tweaked it so it isn't as apparent. Then after the negative voltage died down to zero, it would climb to a higher voltage yet again. It would just take a really long time to do that, and it is hard to see it with the simulator lagging out after that point. I tried adding a relay system to touch the negative voltage to a positive source, but since it is drawing so much negative voltage it makes the relays activate at the wrong times. Then I couldn't figure out how to make a relay system work as intended under those conditions to make it operate faster at extreme voltages. If given long enough, the voltage would keep climbing like a staircase.

11. ### billvonValued Senior Member

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15,114
Right. And once a significant load was being drawn, that high voltage would collapse very rapidly.

There are a lot of ways to do that. Some of the simpler ways are the boost converter and the charge pump, They are often used to generate the high voltages needed for flash tubes and the higher voltages needed for RS-232 ports.

12. ### Aqueous Idflat Earth skepticValued Senior Member

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Think about that for a moment. To get a larger than normal voltage, charges (or holes) need to flow onto the surface where the voltage is going to be measured (with respect to ground). That flow is called "current". Electrons can't move in zero time, but they can do it PDQ. The rate of flow is governed by the physics of the circuit involved. My point still applies: you begin with a fixed amount of available energy, and to the extent you can harness it quickly, you can develop a large short pulse of power. But energy is conserved always, and the circuit itself consumes some of that energy (i.e. converts it to heat/EMR).

In the case where a high voltage develops as a consequence of breaking a circuit across a gap (spark gap, etc.) you will notice that the arc occurs as flash. Why is that? Because the current flowing during the arc quickly depletes the energy available at that node. No matter how you cut this, you can't get around this limitation. And that is because energy is always conserved.

Also don't forget that the fundamental chargeable devices (capacitors/batteries) need time to charge. Capacitors are governed by an exponential law, which says they begin to charge quickly and then the rate declines as they approach max. voltage. Same with inductors used in developing a spark. They need time to develop the current which is flowing at the moment the switch (e.g. distributor/points in an engine) opens. The same exponential law applies for them with regard to current. So you're stuck with conservation of energy no matter how you slice or dice this.

That's a separate issue. But when you look into that, the available energy will be less than what you started with, since the large voltage can never support more than a small current over a long time, or a large current over a short time -- minus your circuit losses. Again, energy is conserved. Energy = power * time (for purposes of this discussion) and you are limited by that. No way around it. Not until you begin to understand that energy is the area under the curve (power vs time) and you work with your design to achieve some desired result. And then you are still limited by the area under the curve. Power * time is the area under a rectangle -- the simplest case -- in which power is assumed to be constant. You should learn this first and worry about the calculus (area under the curve) later.

You won't be able to produce the power to actuate a conventional relay, nor will you be able to do more than to send a pulse too short to actuate the relay, unless you start with enough energy, that is, the energy needed to hold the relay at a given amount of power, over a given amount of time. You have to start with enough to account for all losses as well. Be sure to add a margin to account for other scenarios (for example, fluctuation in a line voltage due to transients, etc.)

Again, energy is conserved. Be sure you understand that this is a law of Nature which can not be repealed.

That was the wh0le point to my post. Be careful that any errors, granularity, assumptions, or idealizations in the simulator you are playing with don't convince you otherwise. That would be counterproductive.

13. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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5,051
The basic principle is pretty straightforward: find yourself a tall building and wrap a bare wire around the highest lightning rod on the roof during an electrical storm.

14. ### LaymanTotally Internally ReflectedValued Senior Member

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What short pulse of power? I already said the output power is zero. I know what current is, and I don't see how any of this validates your point.
Did you even look at the circuit? It doesn't have a spark gap in it. The reason why there is a spark is because air has a high resistance. Then once the voltage reaches a certain threshold, it acts if there is little resistance. This doesn't have anything to do with conservation.
I actually thought about capacitor charge times many times while making the circuit. I kept their values very low, so the circuit would charge up faster, even though capacitors with that low of values wouldn't be able to handle that much voltage. Again, I have no clue as to why you think this applies to conservation.

The available energy could be any value you wanted it to be, just depending on the resistance of the load to ground. No calculus would even be needed. E=IR; I = V/R. Then there would be absolutely no way whatsoever that you could get the same current from +10V and -10V that would equal the same current as 108MV!

That has absolutely nothing to do with the cause of the problem I mentioned. I wanted them to activate so when I got a couple of negative thousand volts it would turn on, but they would activate at positive voltages as well, and when I blocked postive voltage to them with a diode they would still develop a positive and negative voltage across their inductor.
So I take it that you believe that it is perfectly fine that the circuit can develop millions of volts even though it only has about a 20v difference in potential to begin with, and that complies with conservation of energy with no real problem?

I don't think that conservation would be violated, because the simulation produces such high voltages. I thought conservation would be violated by creating such a circuit, and that is the whole reason why I made it that way to begin with. Then it just so happens that the physics running the simulation agrees with me, and it creates more energy than is supplied into it.

Current and voltage are directly proportional to each other. If given the same resistance, more voltage would mean that you have more current or power! It would be impossible for the voltage to be different and the current to be the same under the same resistance. It just isn't going to happen mathematically.

15. ### billvonValued Senior Member

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15,114
Sure. 20 volts at 1 amp is 20 watts. Ten million volts at 1 microamp is 10 watts. If you built such a converter, then it would be only 50% efficient.
Conservation is not violated in either case.
No.
No, they are inversely proportional if you hold power constant.
Correct. That means power is proportional to voltage, and power is proportional to current. They are not proportional to each other.

16. ### LaymanTotally Internally ReflectedValued Senior Member

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1,001
What is this? Pseudo-conservation? You think that the same voltages from different sources would only be able to provide different amounts of power? How ridiculous...
I haven't seen anyone make up their own physics to show otherwise.
Do you even know what directly proportional and inversely proportional even mean?
E=IR
Voltage is represented by E and is in the numerator. Current is represented by I and is in the numerator. Therefore, since both "E" and "I" are in the numerator of each side of the equation, they are directly proportional to each other.

Say I rearrange the equation to be, I = E/R. Then current and resistance are inversely proportional to each other. "I" is in the numerator of one side of the equation, and "R" is in the denominator of the other side of the equation. Then as your resistance increases the current decreases, and if the resistance decreases, then the current increases. It doesn't obey any pseudo-conservation laws that are biased to the voltage source.
Power is current. They are not proportional to each other because they are the same thing. If you have a certain voltage with "x" resistance it will produce the same current as another voltage with "x" resistance no matter how that voltage was acquired.

17. ### billvonValued Senior Member

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15,114
?? No. But if you start with a certain amount of power, the circuit will only be able to output that amount or less. Never more. (All of this, of course, is averaged over time.)
Yes, you have written Ohm's Law, which we weren't talking about. Now try the formula for power given current and voltage.
Incorrect. Power is current times voltage, and in that case current is inversely proportional to voltage at a given power.
No, they're not. Try the following:
Calculate the power given by 1 amp at 1 volt. Now calculate the power given by 1 amp at 100 volts. See if the power is the same.

18. ### TrippyALEA IACTA ESTStaff Member

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10,890
Where are you going to get an Op-amp capable of handling 1 gigavolt?
Where are you going to get a 1 gigavolt power supply?

19. ### TrippyALEA IACTA ESTStaff Member

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10,890
Because you're measuring the current drawn by an open circuit...

20. ### Aqueous Idflat Earth skepticValued Senior Member

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6,152
Refer back to the 1 ns of power for the DeLorean.

I thought this was supposed to be your idea of a power supply for a magical DeLorean. What good is a power supply that produces zero watts?

As soon as you tell me current=dq/dt I will guardedly believe you. But at that point you will have to retract the statement that no current flowed as the device went from zero to Vmax.
Correct.
No. You made it clear that that would be a waste of time, as soon as you suggested that your strategy is to repeal the laws of nature.

Not until you pass billvon's electrical safety review, and Trippy's device feasibilty review. How far away are the high and low conductors? BTW is this immersed in oil? Suppose it's OK in dry air. At what relative humidity would you feel safe using it to power a device that vacuums gasoline at the site of a large spill?

And the charge dumps quickly, dropping the voltage below breakdown voltage, and the large current ceases after a short time. That's the only way to multiply power, as I said before.
It's central to this discussion. Until you understand what we mean by "area under the curve", (here referring to energy being the area under the power curve) then you simply cannot move forward with your design. You will never command nature to violate conservation of energy. Therefore you are limited to the area (energy) you started with and when you try to eke a few more kV out of this, you are only robbing Peter (current) to pay Paul ( high voltage) unless you plan to deliver high power (the 1 GW for the DeLorean) in which case you have scrunched the rectangle (area under the curve) into a short pulse (1 ns per Joule of available energy, per second). I'm pretty sure you don't understand, but that's why this is so relevant. This is your main stumbling block.
Not really. But your whole circuit becomes a pair of electrodes at some point. You can look this up in the electrical safety codes.

Lower capacitance reduces the useful power of this circuit too, but I guess that's OK since you're just giving the DeLorean the equivalent of one spark in one cylinder of a Pinto or whatever.

The reverse is true. You have no clue why conservation of energy limits you.
No, the available energy is the number of Joules your energy provider furnishes to crank the DeLorean, which is one billion times the number of seconds the starter is cranked.

That's the kind of thinking that creates the above mentioned stumbling block.
You mean "V=IR" or "I=E/R" Now if you can just learn the other laws that apply you should see the light.
Ignoring the error at the end of that statement (voltage is current) it would appear that you are on the verge of learning how conservation of energy limits you.
I've been ignoring your random approach to design, thinking the aforementioned breakthrough will remove your stumbling block.

It's your beliefs that are at odds with Nature.
Energy = qV, not V. Learn that, and you will understand why this is moot/meaningless.
Conservation of energy cannot be violated. You simply need to beware that the simulator you are using has limitations. You need to be able to recognize what they are, and to not to make invalid inferences from invalid assumptions about somebody else's algorithm. This is why critical thinking skills are a big piece of science. That, and access to sufficient information.
If you don't have a power source at the input, and a load at output, then you really didn't accomplish that.
No that's incorrect. You said it develops no power. Therefore the net energy is negative.
You said the current is zero. So now what?
No way. Power is limited to the capacity of the power supply, averaged over time, as billvon said. If you want more power per second than your supply furnishes, you will have to settle for short pulses. This is a due to COE, as explained above.

What's impossible is to create energy from nothing. A step-up transformer uses, not creates, energy.

You need the relevant math.

Last edited: Dec 7, 2014
21. ### leopoldValued Senior Member

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17,455
one of the very first things you need to do is determine the difference between power, voltage, and current.
they are not the same and you can't interchange them.
in the first part of the above quote you talk about voltage, then you switch horses and start talking about "that power".
anyway, to answer your question:
depends on what type of multiplier you use.
a transformer will deliver that voltage instantly.
a voltage multiplier (doubler, tripler, etc) will take a few cycles.
yes, correct.
the catch is circuit resistance.
without a power source, your circuit will lose all power within, i don't know, a second or so.
the only exception i know of is superconducting materials at cryogenic temperatures.

22. ### LaymanTotally Internally ReflectedValued Senior Member

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Then I see no reason why I should have to read the rest of what you just wrote.

23. ### LaymanTotally Internally ReflectedValued Senior Member

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Looks like you didn't even bother to look at the circuit either. No wonder why I can never prove anything on these forums, when no one bothers to look at the information.