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Overunity Machines Forum



The So-Called Don Smith Generator

Started by Skysabre, April 29, 2014, 03:20:11 AM

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0 Members and 4 Guests are viewing this topic.

Neo-X

Quote from: gyulasun on May 11, 2014, 07:53:45 AM
Hi Neo-X,

Strictly technically speaking,  the spark gap you ask as shown can fire whenever the (positive) AC peak output voltage from the Neon tube driver exceeds the breakdown voltage of the spark gap. In these events the output current available from the Neon tube driver is limited by its own inner impedance and the parallel L1C tank is pulsed by every second half wave via the spark gap.  (I mention the positive peak vs the ground because the output diodes polarity defines that.)

IF you look at the pdf file in Page 27 (what Skysabre uploaded in his first post),  there is a second circuit diagram below the one you refer to where a Neon sign transformer, NST drives a 0.2uF 4000V  capacitor. The AC output from the NST continuously charges up this capacitor alternately to a positive and to a negative peak value and whenever the voltage (either positive or negative, does not matter) in this capacitor reaches the breakdown of the spark gap, the capacitor discharges to the primary coil labeled as L1. In this case the output current is limited mainly by the ESR (equivalent series resistance) of the 0.2uF capacitor which shunts the output impedance of the NST.  Of course an ESR can be much less than the output impedance of the NST so any effect and any output expected by the breakdown of the spark gap can be higher than in the case of the schematic you asked.
Tesla also used charged up capacitors to discharge them via a spark gap because as he explained the amount of the energy taken out from the capacitor could be controlled: the quicker the discharge time is, the more work can be done from the same amount of stored energy. Tesla did not show circuits where he used the output of a HV generator to fire a spark gap,  he charged up a capacitor by a HV generator first and then he discharged the capacitor via a spark gap to utilize its stored energy.

One could speculate that kicking an LC tank (like in the schematic you asked) by half wave peak current is not neccessarily a drawback because the kicked LC tank may 'supply' the other half wave, this can only be answered by tests how efficient this can be vs the second circuit where the spark gap fires twice due to the full wave peak current available from the capacitor for exciting the primary coil. (To be compared energy-wise: kicking an LC tank once within a full AC wave or kicking a primary (i.e. a coupling) coil of an LC tank twice within a full AC wave, the conditions for kicking are not identical mainly due to the differing inner impedances of the HV sources in the two cases, impedance matching is lacking in the first case) I think the second method i.e. discharging the HV capacitor can give higher overall efficiency than the method shown with your question.
Of course I do not know what Don Smith or the 'assemblers' of the PDF file thought originally on the different methods of positioning the spark gap...  ::)

Gyula

Thank you for the explanation now i understand why it has spark gap there.  ;D
I think the circuit in the picture i post or don smith circuit is more efficient than the classical tesla coil. There is a loss of power when current flows in a spark gap. In classical tesla coil the current oscillate most of the time in spark gap but in don smith circuit the current oscillate in closed circuit and the supply current flow in spark gap is in very short period of time only. Its like a class C amplifier with spark gap in series with LC tank.

Farmhand

Neo-X and Gyula, the circuit  Neo posted in reply #17 was in fact tried by at least one competent experimenter  participating in the thread at EF, he found as I did that the arrangement with the spark gap between the tank and the NST did not work very well, the power in the tank and the output was very poor, much less than if the capacitor is charged and the spark gap is placed so as to discharge the capacitor through the L1 coil. Worth anther try though.

Anyway I found a very interesting section of video from the display by Don at the 1996 Tesla Symposium, if you go to 4:30 minutes in the video you can see a guy in a hat asks some questions and ends up getting a thinly veiled threat from Don. The things Don says do not really make much sense, he unplugs the light board saying that the lights burn out quick  ::) in the preceding video's he says and does things that make me suspicious. https://www.youtube.com/watch?v=OHsEoFszeEw

Don also said clearly that NST's were Over Unity "off the shelf", but of course they are not, and his coil setup in the device drawing in reply #17 closely resembles an NST but air cored and tuned.

..

dezeinstein

Quote from: gyulasun on April 29, 2014, 03:32:38 PM
Hi Skysabre,

I went through the pdf file you altruistically uploaded.  I am strongly "amazed" by seeing the wiring diagram shown in pdf Page 27: they use voltage divider resistors to match the alleged 8kV output voltage (stored in the 2uF capacitor) to the input of an off-the-shelf 12V or 24V DC to 110V AC power inverter. Using resistors for dividing voltages makes sense when reasonably small currents are involved because of the unavoidable heat loss in the resistors.

Let's do some simple calculations backwards,  let's choose a power inverter which is able to output just 2kW at 110V AC, ok? Then let's have its own efficiency, say, 90%, so the input DC power to it should be 1.1 x 2kW=2.2kW i.e. 10% higher.

Now if this inverter receives 24V DC input (let's say), then the input DC current demand is about I=2.2kW/24V=91.6A.

Now lets calculate the resistor values for dividing the 8kV DC to 24V. Here is a link I used, it also calculates the power loss in the two resistors:  http://www.bowdenshobbycircuits.info/r2.htm 

Using the example described in the text for finding the series resistance needed for a LED as a sample, I entered for battery voltage 8000, for current 91.6 and for Vout 24, I got 87 Ohm for R1 (and R2 would be represented by the 24V input of the inverter) but with this value the dissipated power in R1 would amount to 730.6kW....  the idea to use resistors to drop the 8000V to 24V is ABSURD, INSANE to say the least. And there is no any other suggestion to reduce the 8kV (or whatever kV coming from the LC tank via the diodes to the 2uF storage capacitor) to 24V or 12V.

The other thing is that while it may be possible to store a few kW power in a resonant LC tank circuit (it would need a coil and a capacitor with extremely low loss i.e. very high Q) but when directly loading such high Q resonant circuit, the circulating power in it would get reduced just because the external load reduces the high Q (the resonant AC impedance of such a high Q LC tank may range from several ten to several hundred kOhm and shunting it say with the 87 Ohm resistor via the inverter input, the high AC impedance gets killed, so the kV voltage reduces accordingly, there is no 'juice' left in the tank to feed the inverter. AND there is no any high wattage resistor shown in the pictures, which would imply they really used them to reduce the kV output...

Now no wonder why this pdf file contains NO any measurement results done by those 'compiling' it. It is a "pathwork", a "botch", taken text and pictures from some sources, and sorry to say I consider it as a hoax.
At other forums and also here, several members attempted to replicate Don Smith setups but nobody succeeded so far (or did not report, that is) and even Don Smith never showed any working device, only talked about the kW output and showed the assembled boards with those components like in the pictures in the pdf file.

Regardless of Don Smith or from this pdf file, the problem to be solved would be to find a circuit topology which does not reduce an LC tank resonant impedance too much and dissipates only a reasonable amount of power while dividing the kVs to a lower more workable value.  This would be challange circuit to design and build for sure, still strongly assuming that the input power to the LC tank could be kept at a lower value than that of the output power received from THE circuit.

Gyula

the resistors work nothing like you calculate. The voltage presented in this system is cold electricity and if you know anything about cold electricity, the resistance is needed to convert the energy from cold energy  to conventional electricity.. The cold electricity, which is basically invisible and non measurable becomes measurable after this conversion,. and, it will do nothing unless it is put through the proper sized resistors. Cold electricity works oppositely with resistances, as Karl palsness has pointed out long ago, the energy out of a high resistance is more power, and it releases no heat as conventional "amperage driven electricity". Since there is no amperage until after this conversion, there is no wattage and no way to burn the resistors. If you use some logic, you will notice that this device is a replica of Tesla's phase conjugate mirror generator, which pumps energy into the ground and returns it as many mirrors of the energy put in, through the ground as a one wire system. If you punched a ground rod into the ground and soldered on an avramenko plug after the energy was continuously pumped into the ground, the ground rod with the two diodes on the other end would read a voltage across each diode to the output side of the resistor. If you connected this before the resistor, youd either get nothing, or smoke your meter. This energy can not be computed or read digitally, it needs analog instrumentation calibrated to it to function and read the energy correctly. Remember, this is the electricity that science did not want use  to know about. It has been suppressed for years, and, no matter how many times we rediscover and try to show it to science, it gets hushed and goes away, along with the folks who bring this to the public table. Get a grip and try to learn about it before comparing it to conventional energy. It is much more powerful, exists in much larger amounts, and, is easily able to be converted. If we use  a spark coil, it is cold electricity. This consists of a already built one wire system capable of powering up and releasing plasma. If we shot the spark of a coil into a copper plate, it would disturb the ambient and allow us attraction of aether energy to a capacitive coupling. In my example, we could put a sheet of mica between a copper and aluminum plate, then, strike it with the spark coil's output. Then, the other side of the plate, we could place an aluminum sheet, smaller than the mica sheet so it can't arc across. Then, twist two hv diodes together at opposite ends and screw them to the aluminum in a spot that doesn't hinder the distance between the plates (keeping them equally tight together)  The other sides of the diodes will now give off energy. One of them can go to ground, and the other can go to a graphene oxide coated aerial copper plate. another plate can be placed near the copper plate, but far enough away so it doesn't arc. connect another avramenko plaug to this plate and test across the diodes installed on all of the plates until you find the energy out. This electricity WILL do work. It can be more powerful than any electrical generator. and, it is cheap enough to buy and make!

Bob Smith

Quote from: dezeinstein on March 15, 2016, 09:54:07 PM
the resistors work nothing like you calculate. The voltage presented in this system is cold electricity and if you know anything about cold electricity, the resistance is needed to convert the energy from cold energy  to conventional electricity.. The cold electricity, which is basically invisible and non measurable becomes measurable after this conversion,. and, it will do nothing unless it is put through the proper sized resistors. Cold electricity works oppositely with resistances, as Karl palsness has pointed out long ago, the energy out of a high resistance is more power, and it releases no heat as conventional "amperage driven electricity". Since there is no amperage until after this conversion, there is no wattage and no way to burn the resistors. If you use some logic, you will notice that this device is a replica of Tesla's phase conjugate mirror generator, which pumps energy into the ground and returns it as many mirrors of the energy put in, through the ground as a one wire system. If you punched a ground rod into the ground and soldered on an avramenko plug after the energy was continuously pumped into the ground, the ground rod with the two diodes on the other end would read a voltage across each diode to the output side of the resistor. If you connected this before the resistor, youd either get nothing, or smoke your meter. This energy can not be computed or read digitally, it needs analog instrumentation calibrated to it to function and read the energy correctly. Remember, this is the electricity that science did not want use  to know about. It has been suppressed for years, and, no matter how many times we rediscover and try to show it to science, it gets hushed and goes away, along with the folks who bring this to the public table. Get a grip and try to learn about it before comparing it to conventional energy. It is much more powerful, exists in much larger amounts, and, is easily able to be converted. If we use  a spark coil, it is cold electricity. This consists of a already built one wire system capable of powering up and releasing plasma. If we shot the spark of a coil into a copper plate, it would disturb the ambient and allow us attraction of aether energy to a capacitive coupling. In my example, we could put a sheet of mica between a copper and aluminum plate, then, strike it with the spark coil's output. Then, the other side of the plate, we could place an aluminum sheet, smaller than the mica sheet so it can't arc across. Then, twist two hv diodes together at opposite ends and screw them to the aluminum in a spot that doesn't hinder the distance between the plates (keeping them equally tight together)  The other sides of the diodes will now give off energy. One of them can go to ground, and the other can go to a graphene oxide coated aerial copper plate. another plate can be placed near the copper plate, but far enough away so it doesn't arc. connect another avramenko plaug to this plate and test across the diodes installed on all of the plates until you find the energy out. This electricity WILL do work. It can be more powerful than any electrical generator. and, it is cheap enough to buy and make!
Nicely put. This is a counterintuitive set of phenomena that makes this possible. Resistance strategically placed in the system becomes a kind of gate which allows cold electricity to enter. Again counter-intuitively, generally the higher the resistance the better.
Bob

gyulasun

Hi Dezeinstein,

It is interesting you returned to this topic which has been dormant for nearly 2 years. I do hope you have achieved something wonderful in practice because most members on the forums do nothing else but talk and talk about the creation of unconventional energy and that is where it all ends and will fall into oblivion.

Looking forward to seeing anything you may have to share, maybe you have a really bulletproof vest...   ::)

Gyula