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### Author Topic: Method for converting HV (static) into usable low-voltage power.  (Read 45265 times)

#### sm0ky2

• Hero Member
• Posts: 3947
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #30 on: October 12, 2009, 03:26:28 AM »
i drew the wires out to the side for visibility, but in actuality they'll have to be along the shaft to keep from interfering with the input charging brushes while they spin.

#### sm0ky2

• Hero Member
• Posts: 3947
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #31 on: October 12, 2009, 07:53:30 AM »
A lot of people have a hard time with evaluating the current in electrostatic systems.
So i felt the need to discuss the curent of an electrostatic discharge.

Now, when we measure electrostatic voltage potential - what we are actually measuring is a charge relative to some theoretical "zero value" being that of the charge held by the object recieving the discharge. We assume that an earth-ground holds a 'true zero' value.

Unlike current from a constant source, such as a generator or battery, the current from an electrostatic discharge is not constant. It takes the form of a symmetrical triangle-wave: increasing from 0 to peak during the first half of the discharge, then decreasing from peak to 0 during the second half. This makes accurately taking physical measurements of the current rather difficult.
Assymetry in this triangle-wave is a function of ionization of the dieletric and will not be included here.

What we will discuss is the method of calculating the peak current rating of a discharge, which occurs at 1/2 of the discharge-time (T)

To do this, we need to know two values:

1) Total Charge ( in Volts)
and
2) Discharge Time ( in seconds)

From this we can determine the rate of discharge, or change in Volts.
Essentially:  Total Charge / Time of discharge
for example:
a charge of 13,020 V and a discharge time of 21.7 microseconds
gives us a rate of discharge of 600v per microsecond

The Peak Current of the discharge is:
A(peak) = 1/2 change-in-V * K
where K is the dielectric constant
in the above example:  1/2 (600V/microsecond) * 1 (the dielectric constant of air)
gives us  300 microamps      below i have drawn how this triangle wave would look on a graph.

The 'mean-current' or average current of the discharge is a bit more complex, one would separate this graph into individual measurements, at say 1us intervals, add them together, then divide by the number of intervals to get the average current throughout the discharge.
which im not going to go through right now, but in this example would be something close to ..... 53 microamps give or take?

#### Steven Dufresne

• Sr. Member
• Posts: 349
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #32 on: October 12, 2009, 04:51:38 PM »
great work there!  may i ask where you obtained the "mesh" you used to make your cylinders??  i've tried to re-created those things using perforated aluminum cans, but i found that cutting and reshaping their cylindrical shape proved to be more of a problem than i could deal with....

Some of my "mesh" is homemade and some is bought. For the homemade ones I've used a hole punch, either pounding the punch with a hammer, or if the metal is thin enough, pushing it in by attaching the punch to a drill press in place of the drill bit and lowering it into the metal. These techniques at the bottoms of these two pages:
http://rimstar.org/sdenergy/testa/potsmk2.htm
http://rimstar.org/sdenergy/testa/potsmk4.htm
I then roll the end of a screwdriver handle (since it's sort of rounded and hard) over the holes to smooth the edges out. Then I roll the resulting sheet over a cylinder to get the right cylindrical shape.

I've also made homemade ones by drilling the holes if the metal is too thick for the above approaches.

The metals I've used are aluminium flashing from hardware stores, thinner metal sheets from craft stores (and ordering them if they didn't have what I wanted):
http://www.ksmetals.com/craftprducts.html
and empty rectangular paint cans (because they're ferromagnetic, and rectangular because at least I'm starting with something flat, also these are ones that I buy empty so that they've never been used, and they're uncoated on the inside.)

For the store bought ones, I've just hunted around stores like Home Depot and craft stores looking at everything. The mesh in:
http://rimstar.org/sdenergy/testa/276disksg2dcpsu3/05_276disk_grids2_DCPSU3_070719.JPG
was a layer from a filter found in Rona, a local hardware store like Home Depot. You usually find them in the air conditioner, stove range hood sections.

I've also have some that I ordered from:
http://www.dickblick.com/products/amaco-wireform-mesh/#reviews
Photos of them can be found here:
http://www.amaco.com/shop/product-368-wireform-metal-mesh.html

Getting the right preforated metal/mesh has always been a difficult thing.

on plauson::      i started out reading about it in some of his litterature, but in his #1540998 patent i believe its mostly discussed near the top, with bits and details down through the center, before he gets into the more complex machinery.

i started thinking along the lines of using the rotating motion of my wimshurst to operate the spinning portion of a plauson-hybrid device. this way i dont need a secondary drive mechanism.
basically what i came up with is an inverted plauson-machine.

That's exactly what I did too. The other difficulty I have is in attaching disks to shafts to get nice perpendicular motion. So I simply bought a Wimshurst machine off of ebay and took it apart:
(Geez... I guess one advantage of publicly documenting most everything is I always have a link  .)

the induced plates and transformer are the stator, and the charged plates rotate straight off the drive-shaft of the wimshurst.
i drew a basic picture of what i have in mind.

Thanks for the drawing. One thing I've learned recently is the issue of leakage between the rotor and stator where high voltage is involved and where your goal is to simply provide a high voltage for purposes of interacting with the electric field without transfering charge. If your stator or rotor has sharp edges and the voltage is high enough then charge will leak from one to the other. The solution is to either avoid sharp edges or use a lower voltage or both. I think that's why the testatika uses wires for it's sectors/rotor. I also think that's one reason my Hyde generator failed. I use wire sectors now (see attached.)
-Steve
http://rimstar.org   http://wsminfo.org

#### Steven Dufresne

• Sr. Member
• Posts: 349
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #33 on: November 07, 2009, 03:26:28 PM »
I just successfully converted the high voltage from my Van de Graaff machine to low voltage DC. See:
http://rimstar.org/sdenergy/testa/testatika_magnets_hv_to_dc.htm
for scope shots, photos, ...

I was producing DC spikes using a spark gap and decided I wanted to turn them into relatively flat low voltage DC so I could measure current. Remembering when I made my 24V power supply, I first used a full wave bridge rectifier to take AC from the wall socket and turn it into pulsed DC. So that's the same sort of thing as the high voltage DC spikes. The next step in the power supply was to smooth out the pulsed DC by putting a capacitor in parallel at that point. But the capacitor had to discharge fully in the time between spikes. There's a simple way of figuring out what size capacitor to use in that case and has to do with something called an RC time constant (again, I was learning as I went along, no EE training here.) Basically you take the time between spikes and divide it by 5. For example, given 200ms between spikes, I want my RC time constant to be 200ms/5 = 40ms. Next, the value you get by multiplying the resistance (R) of the load by the capacitance (C) of the capacitor must be less than or equal to this RC time constant, 40ms. My load was my oscilloscope which had an impedance of 1 Mohm. So if,
RC = 40ms (see above calculation),
C = 40ms / R,
and R is 1 Mohm, so
C = 0.04 / 1,000,000 = 0.00000004 = 0.04 uF (microfarads)
I had some .22 uF, 100V capacitors sitting around, a little high, but I used one of those. The voltage of the spikes was more than 300V but since it was such a brief spike I figured the capacitor would be able to handle it.

And what do you know, I got around 5V DC out. When I used a meter to measure current, there wasn't much though, a few microamps. But that's to be expected of a Van de Graaff machine.
-Steve
http://rimstar.org   http://wsminfo.org

• Hero Member
• Posts: 1257
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #34 on: November 07, 2009, 11:25:20 PM »
Thanks for the drawing. One thing I've learned recently is the issue of leakage between the rotor and stator where high voltage is involved and where your goal is to simply provide a high voltage for purposes of interacting with the electric field without transfering charge. If your stator or rotor has sharp edges and the voltage is high enough then charge will leak from one to the other. The solution is to either avoid sharp edges or use a lower voltage or both. I think that's why the testatika uses wires for it's sectors/rotor. I also think that's one reason my Hyde generator failed. I use wire sectors now (see attached.)
-Steve
http://rimstar.org   http://wsminfo.org

Interesting setup, could you post any more photos? ..I have some similar ideas;  ...I found some 'stained glass foiling tape' which is copper foil with an adhesive backing, I think that this may serve your purposes quite well also.  I found mine at hobby lobby and it requires less work for making sectors and comes in a variety of widths.  this website has some of those materials: www.diamondtechcrafts.com.

http://www.diamondtechcrafts.com/default.aspx?page=itemView&itemsysid=186031

I also figure that the spark gap of the whimhurst machine dictates the voltage difference prior to discharge, and that this distance is directly proportional to the radius / diameter of the rotor ( in some way ).

Additionally, you could create your own adjustable spark gap using a plastic or vinyl pipe and two screws whose inner thread diameter is equal to the inner diameter of the vinyl pipe.  ( just screw one screw into one end of the pipe and once into the other end, attach electrodes and use one of these screws to adjust the distance between the gap.  It may be best to use brass or copper screws for this, as the zinc plating on some screws will eventually blast off the ends and create a mess in the chamber.

#### Steven Dufresne

• Sr. Member
• Posts: 349
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #35 on: November 08, 2009, 12:00:56 AM »
Interesting setup, could you post any more photos?

I've attached a photo from an experiment with all the woven wires but before I'd put the resistors in place. I think I was testing with a diode in the photo. I do all kinds of tests.

..I have some similar ideas;  ...I found some 'stained glass foiling tape' which is copper foil with an adhesive backing, I think that this may serve your purposes quite well also.  I found mine at hobby lobby and it requires less work for making sectors and comes in a variety of widths.  this website has some of those materials: www.diamondtechcrafts.com.

http://www.diamondtechcrafts.com/default.aspx?page=itemView&itemsysid=186031

The edges would still be too sharp to avoid leakage, if that's what you're also after. The wire I use is the same thickness as the wire used for old clothes hangers. And I make sure the ends terminate inside the disk with epoxy insulation all around.

If leakage is not an issue then I often use plumber's aluminium tape, available in most hardware stores.
-Steve
http://rimstar.org    http://wsminfo.org

#### Magnethos

• Hero Member
• Posts: 521
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #36 on: November 21, 2009, 11:58:44 AM »
Another way could be transforming the bidirectional energy (2wires, common) to an unidirectional impulse (1 wire) energy. Then, you can use an ignition coil to receive unidirectional high voltage and transform it to bidirectional low-voltage high current Pulsed DC.

Another way, as showed before, is to use the capacitive transformer. But in both cases you need to use unidirectional impulses as the input energy.
Perreault explains very well the circuitry to do that. Maybe some fractionation techniques must be envolved to accomplish that.

#### Tito L. Oracion

• Hero Member
• Posts: 2203
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #37 on: November 21, 2009, 12:34:36 PM »
hi everyone here is my simple idea to convert HV into L-Volts

note : You must know first the highest voltage AND CURRENT  that you can get ok.

Buy many transformer then connect the primaries in series then connect all secondaries in parallel ok thats how simple it is, cause it will act as a single transformer ok

then everything is self explanatory right ?

TESTED BY ALL OF US OK            LOL

DOES IT NEED PROOF STILL?      JOKE!

ANOTHER SIMPLE IDEA?
don't you notice the chargers of cellphone today?, they are not using transformer but  resistors.

get a lot of resistor of atleast 1M ohms ok then connect them in series of what volts you desire ok
its like series of resistors connected parallel to the source ok.

sample? : how will you connect a LED in a 240 V ? don't you notice that there is resistor there? ok bye .
« Last Edit: November 21, 2009, 01:01:35 PM by Tito L. Oracion »

#### the_big_m_in_ok

• Hero Member
• Posts: 2087
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #38 on: December 09, 2009, 04:23:50 PM »
thinking i could drop 45VK down to 2300, then again down to 115V ... little did i realize, these things are designed for 2-3KV, when i pump 45KV through it, it sparks all over the place and doesnt do its job....
Putting 10-12 of the MOT's in series on the 45 KV side and then lowering the 115 VAC again might have worked.  You didn't have enough of the MOT's to begin with.

--Lee

#### the_big_m_in_ok

• Hero Member
• Posts: 2087
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #39 on: December 10, 2009, 04:39:08 AM »
Putting 10-12 of the MOT's in series on the 45 KV side and then lowering the 115 VAC again might have worked.  You didn't have enough of the MOT's to begin with.
--Lee
Sorry, I did the math wrong in my head:  It's 20-25 MOT's in series with more than one string in parallel to handle the potentially high current if necessary.  A large industrial transformer can lower the 115 VAC to 12VAC.

--Lee

#### wings

• Hero Member
• Posts: 750
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #40 on: December 10, 2009, 03:09:32 PM »

#### Steven Dufresne

• Sr. Member
• Posts: 349
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #41 on: December 11, 2009, 01:17:58 AM »
@wings,
Thanks. I'm quite familiar with that diagram from Moray King's "Quest for Zero Point Energy". He derives it from Hyde's US patent 4897592 (see attached.) Hyde doesn't show any inductors but for the circuit to serve the function that Moray thinks it serves he adds the inductors.
-Steve
http://rimstar.org   http://wsminfo.org
PS. I highly recommend that book for anyone who's pursuing tapping ZPE and needs some ideas or stimulation.

#### the_big_m_in_ok

• Hero Member
• Posts: 2087
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #42 on: December 13, 2009, 01:17:43 AM »
@wings,
Thanks. I'm quite familiar with that diagram from Moray King's "Quest for Zero Point Energy". He derives it from Hyde's US patent 4897592 (see attached.) Hyde doesn't show any inductors but for the circuit to serve the function that Moray thinks it serves he adds the inductors.
-Steve
http://rimstar.org   http://wsminfo.org
PS. I highly recommend that book for anyone who's pursuing tapping ZPE and needs some ideas or stimulation.
Here's the patent:
#4,897,592

Sub-patent references to the main patent above:
#4,622,510
#4,595,852
#4,151,409
#4,127,804
#3,013,201
#2,522,106

--Lee

#### Tito L. Oracion

• Hero Member
• Posts: 2203
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #43 on: January 08, 2010, 05:58:37 AM »
« Last Edit: January 08, 2010, 06:33:03 AM by Tito L. Oracion »

#### Khwartz

• Hero Member
• Posts: 601
##### Re: Method for converting HV (static) into usable low-voltage power.
« Reply #44 on: October 20, 2015, 01:46:41 PM »
A lot of people have a hard time with evaluating the current in electrostatic systems.
So i felt the need to discuss the curent of an electrostatic discharge.

Now, when we measure electrostatic voltage potential - what we are actually measuring is a charge relative to some theoretical "zero value" being that of the charge held by the object recieving the discharge. We assume that an earth-ground holds a 'true zero' value.

Unlike current from a constant source, such as a generator or battery, the current from an electrostatic discharge is not constant. It takes the form of a symmetrical triangle-wave: increasing from 0 to peak during the first half of the discharge, then decreasing from peak to 0 during the second half. This makes accurately taking physical measurements of the current rather difficult.
Assymetry in this triangle-wave is a function of ionization of the dieletric and will not be included here.

What we will discuss is the method of calculating the peak current rating of a discharge, which occurs at 1/2 of the discharge-time (T)

To do this, we need to know two values:

1) Total Charge ( in Volts)
and
2) Discharge Time ( in seconds)

From this we can determine the rate of discharge, or change in Volts.
Essentially:  Total Charge / Time of discharge
for example:
a charge of 13,020 V and a discharge time of 21.7 microseconds
gives us a rate of discharge of 600v per microsecond

The Peak Current of the discharge is:
A(peak) = 1/2 change-in-V * K
where K is the dielectric constant
in the above example:  1/2 (600V/microsecond) * 1 (the dielectric constant of air)
gives us  300 microamps      below i have drawn how this triangle wave would look on a graph.

The 'mean-current' or average current of the discharge is a bit more complex, one would separate this graph into individual measurements, at say 1us intervals, add them together, then divide by the number of intervals to get the average current throughout the discharge.
which im not going to go through right now, but in this example would be something close to ..... 53 microamps give or take?
Hi sm0ky2 ! Just to tell you that you've made a little confusion: you talk about "charge", "discharge" where you should talk about "voltage", "drop of voltage".

Indeed, if you talk about "charge", it means "coulombs" and the ratio of "discharge" versus time is in coulombs/second, thus amps but pretty sure you knew it

Note:

VOLTAGE

is

HOW MUCH A UNIT OF "CHARGE"* IS "ENERGETIC"

in joules/coulomb.

(*a certain number of electrons)

You may not need this note but as I see often basic confusions around, I try to clear up them each time I can

But Very Thanks for sharing these formula, I didn't know

Regards,
Didier