Composite core step down transforme

Well, I didn’t think I was violating any laws of physics at the time I did my experiment.
I was asking a simple question about what would happen if high-voltage pulses such as I described fed to a capacitor bank would increase the energy in the caps, especially if they increased in frequency.
Would you have any idea?

The 9v battery I tickled the caps with is 8v at 500ma. That’s 4 watts. It made a pretty nice arc.
Now if we are using at 7000v pulse at 10ma, that’s 70 watts.
If we progressively increase the frequency of pulsing into the caps, are we going to get an increase of energy stored in the caps?
It’s a fairy simple question.
A yes or no answer would suffice.

SunnySkyguy said:

I'm not an expert on plasma, and the relativistic effects inside a black hole or a solar centre, but the continuous partial discharge of hydrocarbons with magnetic and dielectric materials, has led me to discover the root cause of MVA distribution transformers from contaminants in a transformer factory process cause these effects under high field strength in operation. Such that within the 1 year warranty period the DGA dissolved gas analysis of transformer oil, when it exceeds 4% or the LEL lower explosive level is exceeded by partial discharge, (PD) the transformer is taken out of service. ( It was a multi-million $ factory epidemic liability from very tiny defects.

PD is a result of plasma physics like an "avalanche electric switch" with almost zero discharge time of charges shorting out detonating molecules and breaking down HxCy oil in H2 and longer explosive gas molecules with higher activation energy. My DSO was slow so the current rise time was in picoseconds.

You might want to ask some questions from an expert like Briane Greene to validate your assumptions.

--- Updated Sep 25, 2021 ---

The few Bedini 's experiments I have observed , were disappointing on the non-scalable demonstration of tiny scale electrostatic engines.

Click to expand...

secates said:

The 9v battery I tickled the caps with is 8v at 500ma. That’s 4 watts. It made a pretty nice arc.
Now if we are using at 7000v pulse at 10ma, that’s 70 watts.
If we progressively increase the frequency of pulsing into the caps, are we going to get an increase of energy stored in the caps?
It’s a fairy simple question.

Click to expand...

If you were applying 8V @ 500 mA, what was the initial condition Vc, of the Caps and for how long. Obviously the more time, the more energy either pulsed or continuous, but energy is lost in the ESR of both components when connected.

Re: Oil cooled transformer explosion.
I’ve experienced this personally.
Years ago at a placed of my then employment, an oil cooled pole transformer exploded. It was amazing that it didn’t damage our metal building or break windows.
I think the angle was shallow enough to the building that most of the shock wave bounced away.
I’ve witnessed dynamite explosions that weren’t as load as that one. It’s sobering thinking that the transformers have high voltage electricity going through a bath of oil.

SunnySkyguy said:

If you were applying 8V @ 500 mA, what was the initial condition Vc, of the Caps and for how long. Obviously the more time, the more energy either pulsed or continuous, but energy is lost in the ESR of both components when connected.

Click to expand...

The caps are kept discharged.
I’d rather not die.
The contact was just momentary with a couple of wires.
My scenario about asking about possible energy buildup would also have the cycles coming from the resonator through high-voltage diodes to keep current flowing one direction.

Current thru a capacitor or a megacap like a tiny 9v battery is Ic= Cdv/dt+ΔV*ESR{C&Bat)

I also noticed something interesting, and Don Smith also alluded to in his book.
If you view the input and output specs of the particular transformer he was using the Output VA is always more than the input. It’s a little puzzling?
The particular transformer I have is a neon transformer and the designer manufacturer claims it’s 99% efficient. I don’t know if their claims are true, but it uses 12 votes in at 1A, and the output as I said is 7000V at 10 mA. So, that’s 12 watts input and 70 watts output. Most all of the neon transformers I have looked at have similar specs. Actually I must clarify that, so far I have only looked at automotive 12 V transformer‘s.
My electrical engineer friend, Waldo speculated that possibly it’s just rated like that because of the peak of the voltage pulses, but maybe the actual overall output is not 70W. I’ve always understood (maybe incorrectly) with AC current that the voltage readings would be higher than DC for the same amount of wattage. That’s understandable because you obviously have a peak and trough with time in between as opposed to steady DC current.
But, if we are talking about actual VA output that’s a 580% increase in current. Don Smith postulates that many patented everyday devices have similar characteristics. Microwave magnetrons, the old TV cathode ray tubes, even the devices that fire the shutters in many cameras.

Hi,

A pulse will give some charge into a capacitor.
If we ignore the increasing capacitor voltage, then every pulse will push the same amount of energy into the capacitor.
But to keep the voltage constant you need to draw the same amount of energy from the capacitor as you put in ... in the same time.

If you don't draw energy from the capacitor, it's voltage will increase. It starts with an (almost) contant rate, but the rate will decrease with increasing capacitor voltage.

Usually the higher the capacitor voltage, the less energy you can push into the capacitor at the same time....because the current is dropping more fast.

So if you increase the frequency ... there will be a limit.

In all my years I've experienced the laws of physics are true. But new inventions need new ideas.
If you think it's worth the effort ...

Klaus

SunnySkyguy said:

Current thru a capacitor or a megacap like a tiny 9v battery is Ic= Cdv/dt+ΔV*ESR{C&Bat)

Click to expand...

Where can I look up the symbols in your formula, as some of them are unfamiliar to me.

Hi,

A battery usually is "Ah" rated,, not "A" rated.
So the 8V, 500mA have to be confirmed.

****
To answer the question about increasing frequency:
My answer is:
* yes, below a dedicated frequency
* no, above a dedicated frequency

Klaus

is the change in instantaneous voltage due to internal resistance with a step current = I*ESR

Thus assuming Cap ESR is much greater than the battery, it would rise instantly and battery drops 1V from 9V indicating battery ESR = 2 ohms.

KlausST said:

Hi,

A pulse will give some charge into a capacitor.
If we ignore the increasing capacitor voltage, then every pulse will push the same amount of energy into the capacitor.
But to keep the voltage constant you need to draw the same amount of energy from the capacitor as you put in ... in the same time.

If you don't draw energy from the capacitor, it's voltage will increase. It starts with an (almost) contant rate, but the rate will decrease with increasing capacitor voltage.

Usually the higher the capacitor voltage, the less energy you can push into the capacitor at the same time....because the current is dropping more fast.

So if you increase the frequency ... there will be a limit.

In all my years I've experienced the laws of physics are true. But new inventions need new ideas.
If you think it's worth the effort ...

Klaus

Click to expand...

That makes perfect sense.
In measuring the voltage and amperage output of the caps, would I need a load on the caps to get meaningful readings on my test equipment? I had thought about setting up a big Bank of incandescent lights to try to dump as much current as is needed. If it develops any significant amount of current I may need a lot of lights.
I had mentioned in an earlier post that I found an automotive 10,000x probe I’m hoping has sufficient attenuation to be able to use safely with my friend’s oscilloscope. It’s very difficult and dangerous to come up with safe measurements for people and equipment.
The other very obvious alternative would be to attenuate the voltage input on the neon transformer to significantly lower levels at the beginning and run all the tests at much lower voltages.

SunnySkyguy said:

Khan Academy

www.khanacademy.org

ΔV=Vi−Vc​

is the change in instantaneous voltage due to internal resistance with a step current = I*ESR

Click to expand...

That’s a good question.
With my neon transformer I don’t know. With Rich’s much smaller voltage system that he uses, it has a gate, then a very neat little amplifier in which you can dial in the frequency and the waveform type. For his particular device we were using square waves. The square waves definitely had a higher output. I have heard from several other sources that the corners produced by the square waves pull in a tremendous amount of energy. It has the option of sine waves and triangular. With his demonstration the best results from the secondary coils were at 1.27 MHz. you could move the secondary coils about 12 inches away and they were still very active. They have LEDs on them and we could check voltages.
Obviously, the further away you were the lower the voltage was.
It’s interesting that the voltage is always higher on the output coils then it was on the primary coil.
The coils are the exact same sizes, so there’s no transformer type step up.

Inductors , L, caps, C ( and batteries which are supercaps with an initial voltage. (x kF)
store energy but with series (ESR or DCR) current losses and parallel leakage.. Energy transfer from switching stored it in L and a diode transfers it to C or Bat. and accumulates a small amount each cycle. (aka SMPS) (P * t = E )

Hi,

secates said:

In measuring the voltage and amperage output of the caps, would I need a load on the caps to get meaningful readings on my test equipment?

Click to expand...

For measuring voltage you basically don't need a load. But your measurement equippment most uses some "non_infinite" resustance inner circuitry, like a voltage divider. So, although not needed, there will be some tiny input current to your voltmeter.
Current measurement is different. Here you want to measure current, thus there needs to be some current flow. Like current through a load. You need a closed loop for current to flow. In an un-branched loop the current everywhere is the same, independent where you measure.

Current in a wire is identical to water flow in a hose.
* you can't expect more water to come out than you put into the tube
* in a closed loop of a water hose (hydraulics) the flow rate is the same, independent if it's a thin hose, a thick hose or in the pump.
...unless there is a branch or some unwanted leakage.

A capacitor is similar to a box of water.
* where the flow rate is the current
* the water height is the voltage
You may capture a very low rate of water flow, like a drop every second.
If you wait long enough the box will be full, then you may empty it within half a second.
As some water may escape as humidity from the box, every capacitor has some leakage current.
You can't expect more water to come out of the box than you put into the box.

There are so much analogies with water ... or better say hydraulics .. to electronics
* pump = battery
* hydraulic motor = electric motor
* flow rate = current
* pressure = voltage
* one-way valve = diode
* overpressure valve = zener, MOV
* pressure regulator = voltage regulator
* pressure reservoir = capacitor
* pressure controlled valve = FET
* flow controlled valve = BJT (in the meaning as current multiplier Ic/Ib)
* .....

Klaus

Thanks Klaus
That’s extremely helpful, and very interesting