Free Energy | searching for free energy and discussing free energy

Solid States Devices => solid state devices => Topic started by: ayeaye on September 11, 2014, 11:50:58 PM

Title: Negative discharge effect
Post by: ayeaye on September 11, 2014, 11:50:58 PM
This experiment, it doesn't work any more, it looks like as if it were some paranormal phenomenon https://archive.org/details/ndischarge (https://archive.org/details/ndischarge)
Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 05:51:51 AM
Hi folks! I succeeded to repeat the experiment. The effect was less than at previous time though, and appeared on another frequency. What it depends on, i don't exactly know. The video about this repeating the experiment is the last on that page, number 4. The circuit is also now simpler:

    -uuuu-
    |        |-
    \        =
    |        |+
    -->1---

https://archive.org/details/ndischarge (https://archive.org/details/ndischarge)
Title: Re: Negative discharge effect
Post by: TinselKoala on September 27, 2014, 08:29:57 AM
I wish I could figure out what you are getting at. You are using a frequency generator and a mosfet to charge a capacitor? Can you please explain, and can you please put a real drawing of your actual circuit somewhere where it can be examined? The text diagram doesn't mean much to me.

The mosfet's own capacitances will pass power quite well from Gate to Source or Drain, even if the Gate isn't leaking, when the right frequencies are used. It's a mistake to assume that a mosfet Gate drive signal is completely isolated from the load side of the circuit.
Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 10:04:26 AM
The circuit diagram is in the video, but i will try to upload the original diagram also to the archive.org page.

Basically yes, there is a capacitor, a coil and a diode, and there is only switching, no external power source, yet voltage appears on the capacitor.

I'm well aware of the mosfet leaking, or maybe what i call leaking, you call capacitances passing power, but this is what i mean by mosfet leaking.

I did everything to measure the mosfet leaking, you see if you see the videos. I even tried to measure the input current of the mosfet, but it is infinitesimal, thus difficult to even detect. I used a circuit which had only one capacitor, no coil, and by measuring the voltage on the capacitor found that any mosfet leaking which there can be, can only cause more than 100 times less power than the power generated as an output. This was also done with the same frequency and duty cycle as the experiment.

But whatever mosfet leaking there is, all will end up in the capacitor. The mosfet's output capacitance empties itself to the circuit in each switching cycle. So the more frequent is the switching, the more the mosfet leaks, there is nothing more special to it in my simple mind. Thus with the ten times lower frequency the mosfet cannot leak more than with the ten times higher frequency.

To be more certain, one certainly may measure the leakage on different frequencies, but i don't think that it gives much more. In the last video i made a leakage test only by measuring the voltage on the capacitor with the frequency on which the effect clearly didn't work. Why i made only that test, is because i don't want to make people to do a test with only a mosfet and capacitor. Because this is dangerous and creates a condition similar to short circuit, which is a quick way to burn the mosfet and maybe even the signal generator.

Mosfet is very good for such applications. Some say mosfet switches faster than a transistor and its input power and leakage is much less than that of a transistor. When meaning by leakage both the capacitance leakage (or passing power) and current leakage.
Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 12:23:03 PM
Tried to draw it up, you can check it.
Personally I think its drawing from your micro controller but just my opinion.
Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 12:59:10 PM
Just my opinion, but I havent watched all your vids.
Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 01:34:49 PM
Yes the circuit is correct, thank you Dave45.

At left is a microcontroller, yes. KL25Z, a 32 bit ARM microcontroller, but can be any microcontroller.

The weird effect is, the source capacitor charges negatively (minus is up) the same as the charged capacitor, and they have almost the same voltage. But at that, back emf can only go to the charged capacitor, because the mosfet is closed during back emf. So it looks like that the coil still takes current from the source capacitor, when there is a voltage spike, but when that capacitor has no positive charge, it will be charged negatively.

This is the circuit diagram in the first videos, in the last video the circuit diagram was that:
Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 01:53:15 PM
I think the pos bemf is passing through the transistor's diode and charging the backside of the cap.

Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 02:03:19 PM
An npn transistor passes neg to the coil and into the neg side of your caps.
The bemf from a neg pulsed coil is pos and charges the pos side of your caps.
Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 03:04:02 PM
irf 630 is avalanche rated, meaning its diode will pass pos current
http://www.vishay.com/docs/91031/sihf630p.pdf
Just my opinion but work it out for yourself
Check everything,
Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 03:55:03 PM
I will, i had no idea that there is any kind of diode inside.
Title: Re: Negative discharge effect
Post by: Kator01 on September 27, 2014, 04:16:35 PM
Hello,

you are pumping gate-charges via the Miller-Capacitance into the Drain-terminal , pumping it via the inductance to capacitor ( LC-tank)
if pulse is shut off, negative puls is created ( collapse-spike ). After a while you will damage your electrolythic capacitor.


 http://electronics.stackexchange.com/questions/83712/gate-capacitance-and-miller-capacitance-on-the-mosfet (http://electronics.stackexchange.com/questions/83712/gate-capacitance-and-miller-capacitance-on-the-mosfet)

also see Vladimir Utkin-paper how to gather negative charge. Here starting on page 58, but be sure only to use
metal-foil-capacitors

http://www.free-energy-info.co.uk/VladimirUtkin.pdf (http://www.free-energy-info.co.uk/VladimirUtkin.pdf)

Regards

Kator01




Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 04:29:58 PM
I see.

But why the effect appeared on different frequency when i repeated the experiment? While everything remained the same, the mosfet, the coil, the diodes, the capacitors. What changed, the mosfet, the coil's core, or the capacitors?
Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 04:34:53 PM
Quote
Here starting on page 58
Yes the buck converter generates neg energy.
A pos pulsed coil has a neg bemf.
Title: Re: Negative discharge effect
Post by: Kator01 on September 27, 2014, 05:03:43 PM
@ayeaye: (http://www.overunity.com/profile/ayeaye.79983/)a capacacitor is frequency-sensitive. Different frequency means different reactance, different reactance means different amount of charge transfered per period via Miller-Capacitance.
See here: http://www.sengpielaudio.com/calculator-XLC.htm (http://www.sengpielaudio.com/calculator-XLC.htm)
Capacitive reactance XC = 1 / (2 · π · f · C)
It´s not worth to persue IMHO, exept what Utkin proposes. I have to say however that I could not replicate his results with the negative charged capacitor. Energy in form of negative voltage was always less than existed before in the cap as positive voltage.



Kator01
Title: Re: Negative discharge effect
Post by: Dave45 on September 27, 2014, 05:11:53 PM
Most circuits out there, if you study them are either boost converter's, buck converter's or a variation thereof,
Look at the topology, how their switched, from which rail, how the pulse is hitting the coil, is the output inverted.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 27, 2014, 07:54:27 PM
@ayeaye: (http://www.overunity.com/profile/ayeaye.79983/)a capacacitor is frequency-sensitive. Different frequency means different reactance, different reactance means different amount of charge transfered per period via Miller-Capacitance.
See here: http://www.sengpielaudio.com/calculator-XLC.htm (http://www.sengpielaudio.com/calculator-XLC.htm)
Capacitive reactance XC = 1 / (2 · π · f · C)
It´s not worth to persue IMHO, exept what Utkin proposes. I have to say however that I could not replicate his results with the negative charged capacitor. Energy in form of negative voltage was always less than existed before in the cap as positive voltage.



Kator01

The voltage on a capacitor in a ringing circuit reverses polarity with every half-cycle of ring.  There is no such thing as "negative" or "positive" voltage as such, because voltage polarity and magnitude are always relative to some reference level. If a ringing cap starts out with a positive peak voltage, the next half cycle will reverse the polarity of the voltage on the cap as it discharges to zero and then _recharges_ to the negative peak voltage, which, sure enough, will be less than the original positive peak because of losses in the circuit. Let it ring long enough without resupplying energy and you get that beautiful exponential decay ringdown, voltage reversing polarity on every half-cycle. So if you arrange your snubbing circuitry properly you can "shut off" the discharging ringing capacitor with either polarity of charge and at whatever level, below the initial charge energy, you like.

Consider two 12volt batteries connected together in series. Measure across the stack. What is the voltage and polarity at the most positive terminal? It is 24 volts (nominally) positive WRT the most negative terminal. Now... what is the voltage and polarity at the "center" terminal where the two batteries are connected together?

SO... I think the voltage on the cap in the device under test will be sensitive to the frequency for at least two reasons: First, the capacitive and inductive reactances depend on the frequency as you note, and also the mosfet is going to be turned on and off at different points in the "ring cycle" by the signal and thus will trap different amounts of energy on the cap, with polarity that could be either way depending on the frequency of drive and its relationship to the resonant frequency of the tank circuit. I think. I haven't done the experiment yet but I may fire up the kit later on today if I have a chance.

Title: Re: Negative discharge effect
Post by: TinselKoala on September 27, 2014, 08:15:45 PM
The 1n5399 diode is a HV rectifier diode, rated 1 kV and 1.5 A. But it is _slow_, having a typical reverse recovery time of 2 microseconds.

UF4007 is similar in voltage, a little lower in current handling but is quite a bit faster at 75 nanoseconds max.



Title: Re: Negative discharge effect
Post by: TinselKoala on September 27, 2014, 08:22:40 PM
I see.

But why the effect appeared on different frequency when i repeated the experiment? While everything remained the same, the mosfet, the coil, the diodes, the capacitors. What changed, the mosfet, the coil's core, or the capacitors?
The mosfet or the capacitor or both could be changing. Mosfets can fail progressively especially if they are subjected to lots of avalanching, and the equivalent series resistance of the electrolytic caps can also change if they are subjected to short HV spikes that could exceed their dielectric rating, even as the cap still works as a capacitor.
I doubt if the core of the coil could be changing under your experimental conditions, but the mosfet and caps can definitely change. But I don't think that component changes are needed to account for your results. I'll know more after my scopes warm up.

I just pulled a handful of IRF630s out of an old monitor chassis, by strange coincidence...
 ;)
Title: Re: Negative discharge effect
Post by: TinselKoala on September 27, 2014, 08:37:39 PM
I will, i had no idea that there is any kind of diode inside.
All ordinary mosfets have a "body diode" that is the result of the manufacturing process. This is often omitted from the mosfet symbol but it is there. It is a reverse-biased Zener diode with anode at Source and cathode at Drain. If your diode-check function on your DMM has enough range you can check this diode in your mosfets yourself.

http://en.wikipedia.org/wiki/Power_MOSFET#Body_diode
http://hephaestusaudio.com/media/2008/11/mosfet-body-diode.pdf

http://www.youtube.com/watch?v=RBJGOOTEwfU
Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 09:12:38 PM
Ahh, a Zener diode, but then the backward current through it is also somewhat restricted. Drain-source breakdown voltage? It's 200 V by the datasheet. When its open then i think it should be on its internal drain-source on resistance, 0.4 ohms. But i still didn't understand, is that internal diode active also when the mosfet is closed, because during the back-emf the mosfet is closed.

Yes i ripped all my components from my old CRT monitor. The core is the deflection yoke core, and the wire i also got from the degaussing coil of the monitor.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 27, 2014, 09:20:52 PM
Ahh, a Zener diode, but then the backward current through it is also somewhat restricted. Drain-source breakdown voltage? It's 200 V by the datasheet. When its open then i think it should be on its internal drain-source on resistance, 0.4 ohms. But i still didn't understand, is that internal diode active also when the mosfet is closed, because during the back-emf the mosfet is closed.

Yes i ripped all my components from my old CRT monitor. The core is the deflection yoke core, and the wire i also got from the degaussing coil of the monitor.

At the end of the video I show the body diode of the mosfet being checked. Select "diode check" function on the DMM and place the Positive probe on the Drain pin of the mosfet. Make sure the mosfet is off by touching the Negative probe from the DMM to the Gate pin. Then put the Negative DMM lead on the Drain pin and the Positive lead on the Source pin. The DMM should read the fwd voltage drop of the body diode, something around 0.45 or 0.5 volt.
Title: Re: Negative discharge effect
Post by: ayeaye on September 27, 2014, 10:39:40 PM
Ok, i understood now that the internal Zener diode is also active when the mosfet is closed. But we are talking about reverse current. So as i understand, this Zener diode opens when the reverse voltage in the circuit is more than the drain-source breakdown voltage of the mosfet, 200 V in the case of IRF630.

I don't know whether the back-emf is more than 200 V. But it's not only that. As i understand, a Zener diode acts reversely almost the same way as a normal diode acts when there is a forward current. That is, it always has the voltage drop equal to the breakdown voltage, which i understand is 200 V. But this does not explain why the voltage on both the source capacitor and the charged capacitor was almost equal, that is, the absolute value of the voltage was slightly greater on the source capacitor. If this were the reason of the negative charge, then likely the source capacitor had to have much less voltage on it, than the charged capacitor.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 28, 2014, 01:07:06 AM
Can you confirm the inductance value of the coil?

Title: Re: Negative discharge effect
Post by: ayeaye on September 28, 2014, 10:22:54 AM
Can you confirm the inductance value of the coil?
The problem is, i have no good tools to measure. I measured it by connecting the coil and a potentiometer in series to a 12 V AC wall adapter, it was really 14.4 V. Then i adjusted the potentiometer so that the voltage on that was half of it, 7.2 V AC. Disconnected the potentiometer, measured its resistance and calculated the inductance by the 2. method there http://www.wikihow.com/Measure-Inductance , when changing the resistance instead of frequency. And with that method i got the inductance 6.6 H.

But then i tried to check that method and measure a known inductance. I had only a small known inductor though, and i don't have so small potentiometer, so i had to use small resistors. I also had to use a 100 ohms resistor to restrict the current of the adapter, to not to burn the adapter. This 100 ohms resistor went so terribly hot that i thought i can use it as a soldering iron. I measured the voltage of the inductor and a combination of resistors in series, and choose resistors so that their voltage was half of that voltage. I got that way 80 mH, but what was written on the inductor was 8,2 mH , that is, 8, what looked like a short vertical line, and 2. I thought that means 8.2 mH but i'm not entirely sure that the thing in between there was a dot, so maybe it was 82 mH.

The core is the core of the deflection yoke of a CRT monitor, and there is 900 turns of a 26 gauge magnet wire on it. I calculated the inductance, and got only 0.16 H. Assuming that the relative permeability is 100, but for such ferrite core it is likely much higher.

So this is what i could get with my primitive tools. As much as i remember,  the potentiometer was 2.2 k when the total voltage was 14.4 V, and there was 7.2 V on the potentiometer. Frequency should be 50 Hz. I measured in the 200 V AC range of my multimeter. So this is how it is, with primitive tools it is not possible to get too good results.
Title: Re: Negative discharge effect
Post by: nelsonrochaa on September 28, 2014, 12:39:46 PM
The problem is, i have no good tools to measure. I measured it by connecting the coil and a potentiometer in series to a 12 V AC wall adapter, it was really 14.4 V. Then i adjusted the potentiometer so that the voltage on that was half of it, 7.2 V AC. Disconnected the potentiometer, measured its resistance and calculated the inductance by the 2. method there http://www.wikihow.com/Measure-Inductance , when changing the resistance instead of frequency. And with that method i got the inductance 6.6 H.

But then i tried to check that method and measure a known inductance. I had only a small known inductor though, and i don't have so small potentiometer, so i had to use small resistors. I also had to use a 100 ohms resistor to restrict the current of the adapter, to not to burn the adapter. This 100 ohms resistor went so terribly hot that i thought i can use it as a soldering iron. I measured the voltage of the inductor and a combination of resistors in series, and choose resistors so that their voltage was half of that voltage. I got that way 80 mH, but what was written on the inductor was 8,2 mH , that is, 8, what looked like a short vertical line, and 2. I thought that means 8.2 mH but i'm not entirely sure that the thing in between there was a dot, so maybe it was 82 mH.

The core is the core of the deflection yoke of a CRT monitor, and there is 900 turns of a 26 gauge magnet wire on it. I calculated the inductance, and got only 0.16 H. Assuming that the relative permeability is 100, but for such ferrite core it is likely much higher.

So this is what i could get with my primitive tools. As much as i remember,  the potentiometer was 2.2 k when the total voltage was 14.4 V, and there was 7.2 V on the potentiometer. Frequency should be 50 Hz. I measured in the 200 V AC range of my multimeter. So this is how it is, with primitive tools it is not possible to get too good results.

Hi ayeaye,
I make a similar tests and all my mosfet and igbt burn after some time working.
I think that BEMF will fry any mosfet in a common configuration like you illustrate because the gate is not isolated.The diode protection of mosfet in the peak of collapse will not work because the static is so much higher
that diode will let flow the current backward and didn't make the job.
I try the commutation with relay and see what happens. https://www.youtube.com/watch?v=pf_qUlwSZl0
https://www.youtube.com/watch?v=DfxEAQNOjp0.

Good tests

 
Title: Re: Negative discharge effect
Post by: ayeaye on September 28, 2014, 03:15:07 PM
Yes mosfet has a body diode in parallel to it, which always conducts positive current from source to drain, even when the mosfet is closed. I tested it, connecting positive probe of the multimeter to the source and negative to the drain when the mosfet was closed, in the diode range, and it conducted. This has no importance though for that circuit, because the diode in the circuit is opposite to that diode, and thus when the mosfet is closed, nothing can go through the circuit for that reason. As i said earlier, it is only important for that circuit that the body diode is a Zener diode.

But the real reason why i thought about negative discharge, was because without it, because of the diode, the circuit is disconnected for the forward current in the coil, and this is the only thing that can be caused by switching. And without switched forward current, there cannot be back-emf, and when there is no back emf, then nothing can go to the charged capacitor. So to say in short what i mean, is that negative discharge is the only way how the circuit can be closed (connected) for the forward current. And for the opposite current the main circuit is only closed when the back-emf goes through the mosfet's Zener diode, as we talked.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 28, 2014, 04:13:07 PM
Circuit elements are not perfect. It seems to me that you are modelling your components in your mind as "perfect". But, for example, the diode you specify is very slow, it has a reverse recovery time of 2 microseconds.
This means it will be ineffective in blocking fast spikes in the reverse biased direction.  Allow me to suggest that you compare the system's performance with different diodes. Try the UF4007 for example and see if your capacitor voltages are the same as when you are using your present diodes. CRT monitors have some fast HV diodes in them, check the internet for the diode data sheets for the part numbers that you find in your scavenging.

The mosfet is not a perfect switch either. For gate voltages near the threshold (usually around 4 volts) the mosfet will be operating in a "linear conductance region" where its on-state resistance is linearly related to the charge on the gate. It is more like a variable resistor than a switch when the gate charge is in this range. So if you have a gate drive that is low voltage and slowly changing, or has only the ability to deliver small currents (filling the gate capacitance slowly) the mosfet is no longer going to be switching cleanly.

Furthermore if there is no way for charge to _leave_ the gate then the mosfet will not turn _off_ cleanly. This is one reason that you sometimes see mosfet gates being driven with AC signals: the reversed (negative) voltage sucks charge out of the gate and turns the mosfet off faster than simply bringing the gate voltage to zero.  Also "pulldown" resistors may be incorporated from gate to source, to allow the charge to leave the gate when the mosfet is supposed to be off.

I speak of course of N-channel mosfets; P-channel are the same but polarities are reversed.

You can make a good inductance meter with an Arduino (or other microprocessor system) and a few other components. You do not need the LCD display, the Arduino can report its data over the serial (usb) line and display on your computer monitor.

http://www.youtube.com/watch?v=S6N8ys8FiA4 (http://www.youtube.com/watch?v=S6N8ys8FiA4)

http://www.youtube.com/watch?v=SCxypoN8-xc
Title: Re: Negative discharge effect
Post by: ayeaye on September 28, 2014, 05:01:18 PM
Talking about microcontroller. My microcontroller board is kl25z, and this is a 32 bit arm microcontroller. This means that its logical 1 voltage is 3.2 volts, different from 5 volts in microcontroller boards such as arduino uno. But i tested it, switching the mosfet on and off, and this 3.2 volts switches the mosfet fully on, so that on the resistor in series there is a voltage almost equal to the source voltage.

And yes i controlled the microcontroller, that is generating pulses, through usb, using the minicom terminal emulator.

About pulses, i think that what matters is the exact length of the pulses, not frequency, frequency is important only because when the frequency is higher, the things are changing faster. But i may be wrong about that particular thing.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 28, 2014, 05:54:32 PM
Talking about microcontroller. My microcontroller board is kl25z, and this is a 32 bit arm microcontroller. This means that its logical 1 voltage is 3.2 volts, different from 5 volts in microcontroller boards such as arduino uno. But i tested it, switching the mosfet on and off, and this 3.2 volts switches the mosfet fully on, so that on the resistor in series there is a voltage almost equal to the source voltage.

And yes i controlled the microcontroller, that is generating pulses, through usb, using the minicom terminal emulator.

About pulses, i think that what matters is the exact length of the pulses, not frequency, frequency is important only because when the frequency is higher, the things are changing faster. But i may be wrong about that particular thing.

No, at 3.2 volts the IRF630  mosfet is not turning fully on, especially if you are sourcing the gate current directly from the microprocessor. The gate _threshold_ voltage of the IRF630 is between 2 and 4 volts but it will not be fully on until the gate voltage is near 8 volts. See the graphs below, taken from the Vishay data sheet.

And no, frequency is always important, since the mosfet has a finite switching time, the reactances vary with frequency, the rise time of the drive pulses probably varies with frequency, etc etc.

You really need an oscilloscope monitoring the mosfet drain voltage to see what you need to see in this experiment.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 29, 2014, 09:39:56 AM
OK, I have built the first circuit (with a few substitutions that shouldn't matter much.) IRF630 mosfet, 1n2071 rectifier diodes, 160 uF 330V fhotoflash caps (Thanks Pirate!).
I'll give a better report a little later. Scope below shows the junction of the two caps and the coil, wrt mosfet source.
Title: Re: Negative discharge effect
Post by: ayeaye on September 29, 2014, 03:07:46 PM
OK, I have built the first circuit
Oh great, thanks :)

So what it shows, a 3 V voltage spikes when the gate is switched on. These are likely higher, the scope may not show the full height. What is weird is the constant coil ringing, maybe due to a slow zener diode of the mosfet. But seemingly no back-emf. As i have seen, the effect occurs only with a certain frequency and duty cycle, which may even be different with the same components. And only then there is a significant back-emf.

Oscilloscope, great thing. I cannot get one, the old oscilloscopes they sell here seem to be only the ones which are so hopelessly failed, that no one can repair them. I even tried to make an ascii oscilloscope with my microcontroller, wrote the code, and even tested it. 500 ksps and 16 bits is the best one can get with that anyhow. Because of high voltage spikes it cannot be made without an op amp. And even with an op amp there are too many interferences, so who knows what else should be made to protect against these. It's like stable, but then it jumps to some who knows what value, then jumps back. So i found it too difficult, and not good enough.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 29, 2014, 07:07:19 PM
No, I could find no trace of a narrow, higher voltage spike. This surprised me because I expected it to be there.  I'm using a yoke that I wound for another purpose, it doesn't have quite 900 turns on it. I tried another coil of around 3 H and it behaved almost the same except with a much lower ring frequency at the TP1. Still no higher voltage, narrower spike showed up. So today I'll wind a proper 900 turn yoke (I only have #27 wire though).

I deliberately used the slowest diodes in my box. I am going to start changing diodes to see if faster diodes make any difference. I have a bunch of diodes from TV and monitor chassis but I don't recall seeing your exact diode in there, I'll have to look again. I'll also change to a faster scope, in case the 60 MHz Tek 2213a is missing something.

I know money is tight these days but you really need an oscilloscope for this kind of work. You should be able to get a decent analog scope in the 60 MHz bandwidth range for 200 dollars US or so on Ebay, and I have recommended the Hantek PC-based DSO several times as a reasonable choice for a "first scope", as it is only about 100 dollars or even less and comes with 2 decent probes.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 30, 2014, 07:41:05 AM
OK, I've wound a proper yoke with 900 turns of #27 magnet wire. I wound it all on half the yoke using my recently-built coil winder. It makes quite a lump on the yoke half!

Then I reassembled the yoke and measured its inductance on the ProsKit MT-5210 RLC meter. It measures 600 milliHenry. Is it possible that you misplaced a decimal point when you calculated the inductance using your measurement method? (That method is valid, I have checked it myself and when it's done properly it produces a good answer.)

If I am understanding your negative discharge effect properly I am able to reproduce it with this setup. We are in the non-linear behaviour region of the mosfet for sure! The oscilloscope reveals so much interesting behaviour that I am going to have to make a video to show it all. I'll be putting that together over the next little while.

I'm using the second circuit, as below.
Title: Re: Negative discharge effect
Post by: TinselKoala on September 30, 2014, 08:19:03 AM
The IRF630 data sheet from Vishay has the characteristics of the body diode.
At the moment I am tending to believe the following: The device might actually be picking up power from the house mains! Here's what I'm seeing. The 3.2V signal to the gate is indeed critical. If I go a little bit above this value the voltage does not accumulate on the capacitor and the signs of line power pickup on the scope decrease. If I go a little bit below this value the mosfet isn't stirred up enough to allow the voltage to accumulate on the cap. Pulse width is also critical but the effect is harder to explain, I'll just have to show it. The pulse width determines the shape of the spike response and if it's too wide the peak spike droops and the rate of voltage accumulation on the cap decreases or stops. I think that the coil acts as the pickup for the line power and the mosfet in its just-barely-turned on state allows the negative discharge effect to happen on the capacitor.
Maybe. I'm not sure about any explanations yet. I am sure that at low Gate drive voltages, right around and below the 3.2 V value, I definitely am picking up power from the mains somehow. I don't know if this is what is being stored on the capacitor though.
Title: Re: Negative discharge effect
Post by: nelsonrochaa on September 30, 2014, 11:08:19 AM
The IRF630 data sheet from Vishay has the characteristics of the body diode.
At the moment I am tending to believe the following: The device might actually be picking up power from the house mains! Here's what I'm seeing. The 3.2V signal to the gate is indeed critical. If I go a little bit above this value the voltage does not accumulate on the capacitor and the signs of line power pickup on the scope decrease. If I go a little bit below this value the mosfet isn't stirred up enough to allow the voltage to accumulate on the cap. Pulse width is also critical but the effect is harder to explain, I'll just have to show it. The pulse width determines the shape of the spike response and if it's too wide the peak spike droops and the rate of voltage accumulation on the cap decreases or stops. I think that the coil acts as the pickup for the line power and the mosfet in its just-barely-turned on state allows the negative discharge effect to happen on the capacitor.
Maybe. I'm not sure about any explanations yet. I am sure that at low Gate drive voltages, right around and below the 3.2 V value, I definitely am picking up power from the mains somehow. I don't know if this is what is being stored on the capacitor though.


Hi ,
if you put a cap of 200nf in parallel with the coil the result will be better .:) and the output in large capacitor have to be a load , without a load the circuit will not work optimal and maybe will fry the mosfet because the collapsed magnetic of the coil.
If you you monitor the charge in the cap when the voltage increase you will observe a defective commutation of the mosfet . 
When the coil collapse will fill rapidly the cap , but if the cap  is already full i observe that high voltage peake produce in the coil will try to find the path to the ground and return by the gate create a mistaken pulse  and the fault of the mosfet.
Another think that i test with good results : if you put a full bridge rectifier after the fast diode and connect the big cap in the bridge you will see that cap will charge more fast and the mosfet will work better.If you put the full bridge without the diode the result will be poor 
So far my best result is with a coil  3 ohm resistance. Can i ask how much resistance have the coil in your tests ?

thanks
Title: Re: Negative discharge effect
Post by: ayeaye on September 30, 2014, 03:01:31 PM
So far my best result is with a coil  3 ohm resistance. Can i ask how much resistance have the coil in your tests ?
Thank you for repeating.

The static resistance of my coil, measured in the 200 ohms range, is 20.5 ohms. But the resistance of the crocodile wires is 1.3 ohms, so the resistance is 19.2 ohms.

I measured 80 mH on an inductor on which it was likely written 8.2 mH. I also calculated that the inductance of my coil is 0.16 H, assuming that the relative permeability of that core is 100. So the inductance of my coil may indeed be 0.66 H, which is 660 mH. I still cannot understand how could i make a mistake one decimal point. Then the relative permeability of the core is approximately 400, which is possible, but quite low for ferrite. The relative permeability of steel is 100 and, as i understand, the permeability of ferrite 3C90, common for toroid cores, is 2300 http://www.ferroxcube.com/FerroxcubeCorporateReception/datasheet/3c90.pdf .
Title: Re: Negative discharge effect
Post by: ayeaye on October 01, 2014, 03:29:41 AM
What concerns measuring the inductance with a resistor in series, i have seen 3 different equations, yes 3, and all claimed to be about sine. But in my simple mind, coil's impedance with a sine waveform is 2 * pi * f * L , so R = 2 * pi * f * L and thus L = R / (2 * pi * f) , simple. So this i think is the right equation of these 3. Measuring inductance with an AC adapter is not so reliable though, because it assumes that the waveform in the main power supply is sine, when it differs from it anyhow, then there would be an additional coefficient. So wrong equation together with not so perfect sine, likely caused in my case a 10 times error.

Nelsonrochaa, if you did an experiment, then you may put the results up somewhere. The best in a video, so that everything would be known about the experiment, some small details may be important.
Title: Re: Negative discharge effect
Post by: TinselKoala on October 01, 2014, 05:12:07 AM
http://www.youtube.com/watch?v=qUMb6e6QQIA (http://www.youtube.com/watch?v=qUMb6e6QQIA)
Title: Re: Negative discharge effect
Post by: ayeaye on October 01, 2014, 04:02:33 PM
Many thanks TinselKoala, and looks beautiful too. I added a reference to this image of your circuit to the description of my video, and said that you are not conclusive yet about where the energy comes. This i found the best way to say what the state of the things are so far.

My video https://archive.org/details/ndischarge (https://archive.org/details/ndischarge)
Title: Re: Negative discharge effect
Post by: ayeaye on October 01, 2014, 09:08:08 PM
TinselKoala, i saw you made a video too https://www.youtube.com/watch?v=qUMb6e6QQIA&list=UUZFlznLV3IyePfbc2TfDetA , thank you, great video, great work :) This video helped to learn a lot of new about that circuit. And the presence of the effect is now confirmed, an effect is often considered valid only after repeated the first time by another person, so it's like a part of finding it. It is all now about how it is interpreted, and which of these interpretations is more correct.

Now maybe i would like to stick to my interpretation, forgive me, as this was the idea why i started it at the first place. So this can be taken as my interpretation, others have different interpretations of their own. But this is what i would like to say.

Notice these things. It picks the frequency out of the mains, right, some kind of induction. But this does not occur when the circuit is not in resonance. Also the voltage on the multimeter almost does not rise when the gate voltage is low, and there are no spikes in the coil. Even in spite that this induced mains voltage sometimes goes negative.

The ringing of the coil, 20 kHz, this is caused by the mosfet's output capacitance, the amplitude of that is very small, and this is likely as much there is leakage from the mosfet, this voltage is not much and not likely to cause the effect.

But the main thing is, there is almost no back-emf on the coil. And thus it can be no way explained how the capacitor can charge with the polarity how it does. The voltage on the capacitor is related to the voltage of the spikes, because these are the only things which can provide high enough voltage. So this is where comes my suggestion of negative discharge. It is about that a positive voltage spike in the coil, causes current to flow from the capacitor, and when there is no positive charge in the capacitor, it charges negatively. It may seem naive but, this is nothing bout the first thing i thought when i first saw the effect.

There are always many interferences, and if we say that when there are any of them present, then this must be the cause of the effect, then we can dismiss whatever effect. I also succeeded to get higher voltages, up to 2.6 volts, which is much higher than the amplitude of that mains induction, but i could not repeat it on the same frequency and duty cycle.

But one more thing which this inducing voltage by the mains may say. It says that there are things which interfere with the circuit. This may be another reason why the effect may occur on different frequency and duty cycle even when all the components remain the same. It may be enough that the frequency of your mains power changes slightly, and the effect is not on the same frequency any more.
Title: Re: Negative discharge effect
Post by: ayeaye on October 02, 2014, 04:36:32 PM
Maybe i should explain more what i mean by negative discharge. For example when charging three capacitors in series, then the middle capacitor charges by negative discharge. Because the plates facing each other of two capacitors have zero charge, but after charging they have instead equal and opposite charges. And they take that charge from each other, like when taking positive charge, as a result the plate where they take that charge, would have a negative charge. So this is not charging, but negatively discharging.

  | +
  =
  | 0
  | 0
  =
  | -

  | +
  =
  | -
  | +
  =
  | - 
Title: Re: Negative discharge effect
Post by: TinselKoala on October 03, 2014, 10:17:56 PM
Well, isn't it obvious that any two plates that are connected together (Plate B of Cap 1 is wired together with Plate A of Cap 2, etc) have the same charge?

At any rate I've been testing a few different diodes. Most interesting is when the "direct" setting -- no diode -- is used.

Title: Re: Negative discharge effect
Post by: ayeaye on October 04, 2014, 07:00:09 PM
It's really great what you do, TinselKoala, and i'm thankful to you, but then with your experiments you create more questions than answers...
Title: Re: Negative discharge effect
Post by: TinselKoala on October 04, 2014, 10:14:29 PM
It's really great what you do, TinselKoala, and i'm thankful to you, but then with your experiments you create more questions than answers...

I think that's the nicest thing anyone has said to me in a long time!

 :)

Title: Re: Negative discharge effect
Post by: ayeaye on October 07, 2014, 09:53:15 PM
Thanks.

TinselKoala, i don't know that much about oscilloscopes but, are you sure that this mains voltage is not caused only by high sensitivity of your oscilloscope? I mean, when the mosfet is almost closed, the current is very low, and then the oscilloscope is open to all kind of interferences. I have not succeeded to measure any mains voltage in my coil, directly without any switching at least.
Title: Re: Negative discharge effect
Post by: ayeaye on November 12, 2014, 07:49:18 PM
TinselKoala, what was the rising time of your function generator? I know it is Interstate F43, but i couldn't find anywhere what its rising time is. It is remarkable that it was capable of duty cycle 5%, the minimal duty cycle of the cheap function generators is some 25%. The cheap function generators have rising time 100 ns and 25 ns for TTL, but TTL is i guess 50% duty cycle, so it is important to know whether they are capable of the necessary rising time.

I cannot use the computer's sound device as an oscilloscope, neither can i use any USB oscilloscope, because of possible coupling problems. I use microcontroller for generating pulses, but it is connected to the computer with USB. So i cannot use anything else connected to the computer, because it is likely in a way or another connected to the computer's ground. But i want the oscilloscope ground to be after the mosfet, to see the voltage on the coil when the mosfet is closed.
Title: Re: Negative discharge effect
Post by: TinselKoala on November 12, 2014, 09:43:09 PM
TinselKoala, what was the rising time of your function generator? I know it is Interstate F43, but i couldn't find anywhere what its rising time is. It is remarkable that it was capable of duty cycle 5%, the minimal duty cycle of the cheap function generators is some 25%. The cheap function generators have rising time 100 ns and 25 ns for TTL, but TTL is i guess 50% duty cycle, so it is important to know whether they are capable of the necessary rising time.

I cannot use the computer's sound device as an oscilloscope, neither can i use any USB oscilloscope, because of possible coupling problems. I use microcontroller for generating pulses, but it is connected to the computer with USB. So i cannot use anything else connected to the computer, because it is likely in a way or another connected to the computer's ground. But i want the oscilloscope ground to be after the mosfet, to see the voltage on the coil when the mosfet is closed.

The F43's rise time is around 15-30 ns for rectangular pulses, depending on frequency. The minimum duty cycle I have been able to set is a bit under 5 percent, perhaps 3.5 percent or so. This is a "high voltage" FG that is capable of 40V p-p, but it only goes up to about 3 MHz.

I also have a DataPulse DP-101 pulse generator that gives a rise time, into a 50 ohm properly terminated load, of 5 ns (spec) and about 7 ns measured on my bench, and it can do very very short duty cycle pulses, much much less than 1 percent at slow frequencies, if required. 20V p-p into 50 ohms with independently adjustable positive and negative peaks, and up to 10 MHz frequency.  Very short fast pulses of course require proper cabling and terminations in order to make it to the device under test.

You really really should get yourself a decent analog oscilloscope. I am sure you can find good serviceable scopes for under 200 dollars US. For example:
http://www.ebay.com/sch/i.html?_sop=12&_nkw=tektronix+2213a&_frs=1 (http://www.ebay.com/sch/i.html?_sop=12&_nkw=tektronix+2213a&_frs=1)
Title: Re: Negative discharge effect
Post by: ayeaye on November 13, 2014, 03:58:29 AM
The F43's rise time is around 15-30 ns for rectangular pulses, depending on frequency.
Thanks a lot.

The maximum output rise time of my microcontroller is 36 ns, by its specifications. I did not choose microcontroller for generating pulses, without a reason. It's easily adjustable, but it also has lower rise time than simple electronic oscillators such as an astable multivibrator or 555 timer.

The cheapest function generators are likely made based on the 555 timer, so their rise time is 100 ns, the same as the 555 timer. A new function generator capable of 35 ns rise time, one can get for less than 200 euros, with shipping. But new function generators with a rise time 25 ns, already cost twice as much.

We need a fast kick there, so just whatever will most likely not do.
Title: Re: Negative discharge effect
Post by: ayeaye on November 17, 2014, 07:17:26 PM
Then some may have that problem, where the energy comes from, when it really can be shown that it doesn't come from any known source. The first law of thermodynamics and everything, though i don't know how exactly were magnetic fields and electric fields considered in thermodynamics. I think more fundamentally the conservation of energy is about rhythm, there are closed loops of change. What is not always known though, is in what way.

How everyone can see overunity, is by Tesla radiant energy receiver. No these are not radio waves, because it is measured that the energy comes in very short pulses, radio waves don't propagate that way. This is a bit difficult though, as one should have a whole roof of a house to light a LED.

In my earlier experiments with magnet motors i found, that there is overunity in magnetic field, but not enough to cause continuous rotation. Because there is a path which a pole of a magnet can go through, without any repulsion. And the same should be true about electric field, both are asymmetric fields, and every asymmetric field can be made to do work. This is the drawing i made during my magnet motor experiments, so think where the energy comes from.
Title: Re: Negative discharge effect
Post by: ayeaye on July 24, 2015, 07:38:43 PM
I only wanted to say, now i think the name negative discharge effect is a kind of misleading, this is just how it seemed to me at first. This should be called an asymmetric current overunity experiment. Because i now think how it works, is that when the mosfet opens, the leakage current of the diode goes from the capacitor to the coil. This induces a voltage spike and current in the opposite direction, which then goes through an open diode to the capacitor. Thus the reason why it works, is that the current is asymmetric, lower current in one direction through a diode leaking, and greater current in the opposite direction. The diode does not have to be fast for that i think, it has enough time to open during the voltage spike. What has to be fast is opening the mosfet, which causes a quick change in current, by the diode leakage current starting to go through the coil.

I still see that evidently the energy doesn't come from any known source. The induction in the coil of radio waves, or mains voltage, which i have not been able to find, or the mosfet leaking through its output capacitance, all look like a hundred times less than the effect. Of course i can do further experiments, like i can put the whole thing in the faraday cage, to make sure that nothing outside induces anything in the coil. I even bought an oscilloscope, but because of difficulties in my personal life i have not been able to do any further experiments, and i don't know whether i will be able to do them in any foreseeable future.

But who wants to do an overunity experiment, t think this is a good experiment to start from. It is simple, the circuit is very simple, and all the parts are cheap (in fact all ripped from an old crt monitor for free), and easily obtainable. I wanted to do an experiment which an absolute beginner can do, who has only a multimeter and some parts. It turned out that this experiment is better for some more advanced people. Getting that deflection yoke core from an old crt monitor or tv, takes a bit of effort, but then good coil is a great thing for many other experiments too. Who has a lot of money, can buy a big toroid ferrite core instead of course, but such thing is too expensive for me. Considering the quite low frequency where the effect appears, i think arduino can be used for generating pulses. Arduino nano should be good for the purpose, and one can buy it from ebay for only $2.50 with shipping. But, it's also good to have an oscilloscope for such experiment, at least 20 mhz two channels i think. And this is for people who are a bit more advanced in electronics. A bit more advanced, considering that i myself am almost a beginner.

I bought an old oscilloscope, not very old, from 1980's. I would say, the biggest problem was to clean it. It appeared to be fully working, except the contacts needed some switching many times, before they started to work well. The biggest problem was that it was held a long time in who knows where, in some dusty and moist storage room. All the dirt was so stuck to the case, that it was difficult to remove it with a strongest cleaning agent. Yet i could do that, so that it now looks decent. This is a big problem, you don't really want to do some advanced things, with some dirty equipment.

If you want to replicate that experiment, good luck, and have some real fun :)
Title: Re: Negative discharge effect
Post by: guest1289 on August 17, 2015, 04:39:57 AM
Quote
Guest1289, talk about the coils there http://overunity.com/14925/negative-discharge-effect/#.VdFQ5tcuviY , not a right place here.

This  thread  goes straight over my head,  but I'll do my best,  in order to see if this thread can be  translated  for other members in this  website.
 
(  I can't even fully understand the 'water'  analogy with electricity,   they say that increasing the speed at which water flows through a pipe,     is the same as increasing the current in a wire,    and yet  electricity always flows at the exact same speed.      I suspect they mean that either the amount of electrons actually travelling in a  cross-section  of the wire at any one time is increased,  or,  that the frequency at which these electrons are  pulsed through the wire is increased .     But,   I will never need to understand it anyway )

(  To  me  induction  is either,    current is travelling in a Coil,  and a wire is placed through the Coil,  so an electrical current in generated in the wire,   OR,  if the wire is first carrying a current,  and it induces a current in the Coil  )

Quote
In brief, the voltage induced, depends on the speed of switching the current, not on the strength of the initial current, e = df / dt, or such, this is what it is about, very basic, and this is what induction is. Now the induced current should remove the magnetic field, and it will, but these processes have a certain inertia. And this is why there is a voltage spike.

I think your'e saying that the  frequency/amperage  of the initial current,  determines the voltage of the induced current.     
But I'm totally lost when you say   that this  induced-current  should remove the  magnetic-field . 
You probably meant to say that the  induced current,  will have a lesser  magnetic-field than the  original current (  that,  that is what you're trying to achieve  )
I did not know it is scientifically possible to have an electrical current that does not produce an electromagnetic field .   [  I SUSPECT THAT YOU HAVE METHODS OF PLAYING WITH THE PROPERTIES OF THE INITIAL CURRENT,  OR ,  OF THE INDUCED CURRENT,  IN ORDER TO REMOVE THE  MAGNETIC-FIELD  ]   
________________

If  you're onto something in this  thread ( if it has some worthwhile validity ),  then,  to get more people to recognize it's  potential,   you really should :

( 1 ) -  Explain,  in the simplest language possible,   'What the problem is that you are trying to solve'

( 2 ) -  And , in the simplest language possible,  explain how your solution  attempts to solve the problem in point ( 1 ),  or if you have solved it,  then how.

( 3 )  And,  in the simplest language possible,  explain the proof you have recorded ( hopefully replicate-able  )  that your solution to the problem is having any effect

___________________

You had  stated,  that you have given up on using  'Permanent-Magnets',  and have replaced them with  Coils/Induction,  and yet you stated 
Quote
Now the induced current should remove the magnetic field
  ,   it seems like a contradiction in aims.

Title: Re: Negative discharge effect
Post by: ayeaye on August 17, 2015, 02:29:22 PM
and yet  electricity always flows at the exact same speed.

I have never heard that. I try to explain it in as simple way as possible.

I used to know that current is the speed of the electrons. It is like you are pushing a wheelbarrow. The speed with which the wheelbarrow moves, is current, and the force you use to push it, is voltage. Voltage is also called electromotive force. When you use the same force to push the wheelbarrow, it moves faster on an even road, and slower in a dirt, or when it has to go over rocks. The conditions that slow down the wheelbarrow, are called resistance. So with the same voltage, there is less current, when the resistance in the circuit is higher. What matters is the resistance in the whole circuit, that is literally in the circuit in which the electrons move. Because electrons are tied to each other by repulsion, the resistance to one electron or to some electrons, also resists all other electrons in the circuit. Something like, many people in a row push wheelbarrows in a circle, when one wheelbarrow slows down because of some obstacle, all wheelbarrows slow down.

There are two kind of inductions in a coil, the induction of the magnetic field in the core of the coil, and the induction of electromotive force or voltage by the change of the magnetic field in the core of the coil. The induction of the magnetic field happens by the Ampere's law, which says that the relation between the current in the coil and the magnetic field, is linear. That is, the greater the current, the greater the magnetic field, and vice versa, the strength of the magnetic field is always the current multiplied by some constant.

The induction of the electromotive force in the coil happens by the Faraday's law. This says that the electromotive force e = df / dt (the letters are not correct), where df is the change of the magnetic field, and dt is the period of time. This in the other words says that the strength of the induced electromotive force depends on the speed by which the magnetic field in the coil increases. And by the Ampere's law, it depends on the speed by which the current in the coil increases. Thus it does not depend on the strength of the current, it only depends on the speed by which the current increases. This induction of the electromotive force is responsible for the rising half of the voltage spike.

Now the electromotive force induced in the coil, is by the Lenz law such that the current that it causes, is opposite to the current that induced that electromotive force. That is, the current caused by the induced electromotive force, induces a magnetic field that is opposite to the initial magnetic field, which means decreasing the magnetic field until it eventually becomes zero. This is responsible for the falling part of the voltage spike. Because of these forces working against each other, any current going through the coil should instantly be reduced to zero. But there is a voltage spike with some duration, because the processes that cause that, have some inertia.

Now why are we talking about these "spikes"? Because what we do, is that we periodically switch the circuit on and off, using a mosfet. You may think about mosfet as an electronic switch. The signal at the gate of the mosfet is generated by a function generator, or arduino. Arduino because, arduino is cheaper, arduino nano (costs $2.50 when bought from ebay with shipping) causes a sharp rise of the signal, and enables to adjust the signal easily. Now all is about that in every cycle, the current is switched on very sharply. Which causes the induction of greater electromotive force, by the Faraday's law.

This periodic switching, you may think about it as an equivalent in solid state devices, of rotation in the mechanical devices. Because both of these are a cyclic change, which is the general case. If we want something continuous, we usually need something cyclic. And the cyclic processes in electronics are so common, that oscilloscope is made to measure only such cyclic processes. Thus oscilloscope is very useful for experiments like this, and enables to see better what happens, but the experiment can also be done without an oscilloscope, using only a multimeter (in ebay costs $4 with shipping).

Now what we switch on every time? Well, we have a circuit, consisting of the coil, the mosfet, a capacitor, and a diode, that's all there is in the circuit. We switch on the leakage current of the diode. What is leakage current, it is the backwards current of the diode. A diode is an electronic component that conducts well in one direction, forward, but less in the other direction. There is a leakage current because of some charge there is in the capacitor.

Now because of that every time we switch the mosfet on, we cause a fast increase of the current through the coil, which induces a voltage spike, that causes a greater current which goes to the capacitor, and in that direction the diode conducts well. And thus every time the voltage on the capacitor increases. The voltage on the capacitor can only go as high as the maximum voltage of the voltage spike.

Now the frequency and the duty cycle with which we switch. Duty cycle is the percentage by which the signal is on, so a relative length of the positive pulse. It should not depend on that at all, because all the induction is only caused by switching on the mosfet. Yet it depends, because mosfet, when switched off, is like a small capacitor and a resistance. And this capacitance (output capacitance) is in series with the capacitor in our circuit. So there is a certain self-oscillation when the mosfet is switched off, caused by the coil and the total capacitance, the coil "rings", so to say. This self-oscillation is not great, because of the diode that greatly decreases the current in one direction, and the mosfet's resistance. Yet it matters in that, when we switch the mosfet on when the current in the circuit is opposite to the backward current of the diode, there would not be much increase of the current. So the frequency and duty cycle of switching has to be in resonance with that self-oscillation, and thus the effect, what i now call an asymmetric current induction effect, is there only with a certain frequencies of switching.

You may ask, does the capacitor have to be somewhat charged in the beginning? No this is not necessary. Short circuit the capacitor before the experiment, to make sure that it is empty. Because the voltage at the gate of the mosfet, induces some charge on the output capacitance of the mosfet. In spite that it is very small, it still causes some self-oscillation as described above. Now when the mosfet is switched on at the right moment, there is some charge in the capacitor with the right polarity, and the fast decrease of the mosfet's resistance causes a fast increase of the current the same as it normally happens. Whenever it starts, the process can go ahead only in one direction, and thus the voltage on the capacitor starts to increase.

Now what concerns the overunity, it is the difference between the output power and the input power. What concerns the input power, there are two possible sources, which both should be measured or estimated, to determine the overunity. First is the induction of voltage in the coil from some outside sources, due to electromagnetic radiation. This should be practically zero, because the radio waves or any other electromagnetic radiation, can likely at best induce power which is a thousand times less than the generated power. To make absolutely sure that no output power is caused by the induction in the coil though, an experiment should be done where all the circuit and the multimeter measuring it, is enclosed in a Faraday cage. Like by wrapping an aluminum foil all around it.

The other is the leakage of the mosfet. It is that the voltage on the gate of the mosfet, somewhat causes the output capacitance of the mosfet to charge, which then discharges to the circuit. The power of switching the mosfet is infinitesimal though, so in the experiments and by calculations it was estimated that this leakage power should be a magnitude less than the output power. To make it more clear, it should be calculated and measured more. This is what all this experiment is about, to measure the output and input power as well as possible.

This experiment is very basic, enables to learn about the very basic things in nature. When doing it, you may think that you are a new Faraday, this is a basic thing next to the Faraday's first induction experiment. Faraday was a very religious man, who wanted to find out how the nature works, to see how great it is. And he did. Faraday also, could almost do no math, he just wanted to find out what happens, by doing experiments.

So i explained it now as well as i can. Feel free to ask questions if anything was unclear, or there is something that you still need to understand.

Title: Re: Negative discharge effect
Post by: guest1289 on August 17, 2015, 06:18:22 PM
Well,   everything you have typed is the exact reason why  I'm only interested  in magnetic-fields  (  or other  similarly simple  subjects which appear in this  website  )  ,    it's the exact reason why I'm not interested in the  properties of electrical currents  .

It's good that you typed it though,  so maybe sometime if I'm able to,  I can go back to that post,  and fully understand  everything  you have  typed. 
(  Normally,  on normal days,  my level of comprehension is definitely not high enough to fully comprehend everything you have typed,    which is why I have no interest in complex things like the  properties of electrical currents    )

(  Yes,  I did know how  permanent-magnets  can induce an electrical-current  in a conductor  )
__________

I just found a  wikipedia  page :   https://en.wikipedia.org/wiki/Speed_of_electricity  ,   and now I realize that even the little I knew about   electrical-currents ,   is totally wrong,   but at least now I know I'm wrong.   
     I also never new that the  individual electrons travel  much-much  slower than the  'Electromagnetic Waves'  in a circuit,  and that the actual energy in the circuit is actually the 'Electromagnetic Waves',   not the electrons.  (  I should not be surprised though,  since  'Electromagnetic Waves',  is all I'm interested in,  although I'm only interested in the  fields of  permanet-magnets  ).     
       It's all pretty obvious though,  it's just that I never had to think about it before.
(  Now I think that a person struck by lightning,  would be damaged by the  'Electromagnetic Waves',  and not by the  actual flow of electrons themselves,  could be both though   )
__________

   Your earlier post on  'November 17, 2014, 07:17:26 PM'  ,  with your diagram of the magnetic-field  of a   'permanent-magnet',   should have been posted in the   Permanent magnet motor    thread.   
      (  That diagram immediately reminded me of the diagrams  I  posted on  'August 15, 2015, 05:30:30 PM',   in that thread   ).
       I may have interpreted  the  2 small  'No-Repulsion-Zones'  in that diagram incorrectly,   but I think  that what you have  identified   gets  right to the core of the problem which was being discussed in  the   Permanent magnet motor    thread ,  of  what is  the  exact  difference  between  the  field  of an  electrical-conductor,  and that of a permanent-magnet ( in order to achieve a  Non-Electric-Permanent-Magnet-Powered-Faraday-Motor  ) .   
But what you have  identified,   will not be a factor in   all  designs,  I  can imagine designs  bypassing  the  'No-Repulsion-Zones'.

       I identified a possible difference  between  the  field  of an  electrical-conductor,  and that of a permanent-magnet ,   I think  that  field  of an  electrical-conductor  may also contain an  'Electric-Field' (  even though the  'Electric-Field'  and  'Magnetic-Field'  were unified in the  special-theory of relativity  ),   and that that   'Electric-Field'  is also  moving in some direction I am not sure of.   
       But,  I assume that even a  permanent-magnet  has an  'Electric-Field',  even though it will be much smaller and very different to that of an  electrical-conductor  .
_________________

     Hopefully, you, and the other members dallying in your field( and closely related fields ) will sometime get an  insight  into exactly what it was  that  'Moray'  discovered  (  'Moray Valve'  ).
     A  recent  new thread (  rightly suggesting that  SHAPES  are an  Ultimate-Key  in collecting energies,  that either have not been collected before,  or  not effectively   ).    The  thread  suggested  that the  Shape  of the  antenna  of the  'Moray Device'  may have been  the actual reason why his device was able to  receive the energy.   
        (   But in historical accounts I have read of  Moray  demonstrating  his device,  I read an account of how the device was  driven to the remote  countryside,  and various components of the device were removed,  and it kept on functioning,  and I am sure that it stated that one of the components   removed ,  was the  antenna  )
      [  CORRECTION :  That thread    http://overunity.com/15969/physical-shapes-that-cause-potential-differences-to-natural-energies/#.VdIdkrKqqko   ,    was actually referring to  "the "antenna" on Tesla's Pierce Arrow"    ]
        The  strange  instinct  that I get about moray's device,  was that there  was a  special  'spacial gap'  in one of the components such as the   'germanium-transistors( or were they diodes or resistors )',   and that  that   special  'spacial gap'  created an  irresistible potential  through which  the energy  he harnessed had to flow
        (  And possibly that a catalyst for that  'special-process'  was the  radio-active  material  the components  also  contained,  a bit like the following -  the radio-active material  emits  some type of  particle,   and then some force suddenly appears to chase that particle.    The problem with this  extra  catalyst  idea is that the  radio-active  material only emits  some type of  particle, on an intermittent basis,  very unlike  the very unusually good current that his device produced (  assumedly very high frequency or something   )
__________________

       I assume that  'Negative discharge effect'  is  not at all related to  'Moray's'  discovery,  they seem to not share any common principles etc.
__________________

       I always worry that people inventing  Solid-State-Overunity-Devices,  may often  mistake   'Persistent-Current' (  https://en.wikipedia.org/wiki/Persistent_current  ,  an actual phenomenon )  or  external types of radiation which travel through all materials,  as being a discovery of a  slight  overunity .
_______________

     It's amazing that two years ago they discovered how to  turn-off   the  magnetic-field  of a permanent-magnet,   just by using  'Polarized light'.   
     Its so difficult to  re-find that discovery on the internet.   
     (  It is a type of  reversal of the  'Faraday Effect' concerning light  )
_______________

       
Title: Re: Negative discharge effect
Post by: ayeaye on August 17, 2015, 06:49:13 PM
       I always worry that people inventing  Solid-State-Overunity-Devices,  may often  mistake   'Persistent-Current' (  https://en.wikipedia.org/wiki/Persistent_current  ,  an actual phenomenon )  or  external types of radiation which travel through all materials,  as being a discovery of a  slight  overunity .

Well, the persistent current is in nanoamperes, this may give power in nanowatts. But the effect i described, can give power in milliwatts. This is a huge difference in magnitude.

The field, well, the electromagnetic field is the movement of photons. Because photons are the gauge bosons of the electromagnetic field. Every field is the movement of gauge bosons, well, likely except gravity. The only interaction is when a gauge boson hits a particle, and its properties cause the particle to change speed or other properties. Thus electrons must emit photons all the time, not only when they move. Thus the main difference between the electrostatic and magnetic field, are the properties of the photons. The photons of the electrostatic field should have no frequency, while the photons of the magnetic field have some frequency, as they are emitted by moving electrons that orbit the nucleus of an atom. Weirdly, knowingly no one has yet discovered photons with no frequency. Likely because they have little effect other than attracting or repulsing charged particles. But by the quantum theory, they should exist.

The electric things are complicated, well yes, somewhat more than permanent magnets. Not though when we go to extreme with permanent magnets, trying to achieve a continuous rotation, which is very difficult to do, if not impossible. This is why i say, i think your effort is better used when you deal with simple electric things, than when you deal with permanent magnets. And what you do would finally be less complicated as well. But it sure is complicated, as there simply is some complication which we just cannot escape. All we can do, is to make it as simple as possible. Thus i think you would not regret when you go into electric things, in spite somewhat more complicated, they are much more awarding. Do it gradually, learn step by step, and you find that you can manage it, and these complicated things will look much simpler for you. Everyone started that way, and i am rather a beginner in electronics, thus not so long way to go.

My aim was to do a *simple* overunity experiment. Now what it appears, it is too complicated. Yet i know nothing more simple.

Title: Re: Negative discharge effect
Post by: ayeaye on August 27, 2015, 04:00:14 PM
I think this and permanent magnet motors, is all about getting energy out of these fire wheels, which are called atoms.

How powerful should a free energy device be, well, a simple answer, 500w. Why, because the smallest electric radiators are 500w, such devices can give power to almost all house appliances. Also electric scooters 500w are useful things. But we can get only milliwatts, so a long way to go there. But it is only about research anyway.
Title: Re: Negative discharge effect
Post by: ayeaye on September 20, 2015, 08:56:01 PM
I hope now i finally figured out, the only way how switching on the mosfet can cause any current in the opposite direction of the diode in my circuit. And indeed the current should be exactly such.

When the mosfet is closed, the capacitances of the diode and the mosfet will be charged so that the total voltage on them is equal but opposite to the voltage on the capacitor, because they are connected in parallel to the capacitor. Say some half of that voltage on one and half on the other, because both are some 50 pf.

This voltage stays on the diode, as the diode  is directed opposite. But this voltage does not stay on the mosfet, because its body diode is in the same direction. Yet the voltage that is equal to the threshold voltage of the body diode, i don't know how much it is, maybe 1 volt, stays.

Then when the mosfet switches on, the current starts to flow from the diode's capacitance to the + plate of the capacitor, and it rapidly grows. This current tries to increase the voltage on the diode's capacitance, to become equal to the voltage on the capacitor. Because opening the mosfet short circuited the voltage on its capacitance, and thus the capacitances in parallel to the capacitor no longer have the voltage equal to the voltage on the capacitor.

The diode's leakage current is extremely small, some 1 microampere, and it doesn't change, so it likely can be disregarded.

I had to consider the mosfet's and diode's capacitances, and even the threshold voltage of the mosfet's body diode, things which people usually don't consider when thinking about circuits.

Maybe i were lucky in using a large rectifying diode, that has a great junction capacitance, as it could not work with a smaller diode.

Now when using a relay instead of the mosfet in that circuit, some 50 pf capacitor may have to be connected in parallel to the relay's output, as weird as that may sound.

Neither may there be an escape from the self-oscillation, which happens with the frequency determined by all the capacitances in series, which is very small, thus the oscillation should have a quite high frequency. This self-oscillation should mostly only change the voltages on the diode's and mosfet's capacitances, which are very small, and almost not at all the voltage on the capacitor. Capacitors in series work that way, find that out if you don't believe. And because of that self-oscillation the effect would occur only on certain frequencies. Because the self-oscillation should start after the end of the voltage spike. Now when the periods of time from that to switching off and then to switching on are such that the voltages on the diode's and mosfet's capacitances are again such as they initially should be, then everything should work as described above.

Why it self-starts, i have no other explanation, than that the voltage on an electrolytic capacitor after short-circuiting it never goes completely to zero.

The input power of that circuit is that necessary for charging the mosfet's input capacitance, some 500 pf, to the voltage on the gate, that is 5 volts when using arduino, once in every switching cycle. I see no way how any of that charge goes to the capacitor in that circuit, all the mosfet's capacitances will be discharged and i think all their energy is just wasted. Yet formally for overunity, the power at the mosfet's gate in this circuit has to be considered the input power. And whether there can be any overunity also considering that, i'm not quite sure. But i don't exclude that either.

Now this all may sound complicated, but it may be the only way to achieve an asymmetric current.

Sorry for using this thread as a kind of blog, but i think this information is very important for understanding how that circuit works.
Title: Re: Negative discharge effect
Post by: TheComet on September 21, 2015, 11:12:23 AM
You guys realise the power is coming from the microcontroller, right?
Title: Re: Negative discharge effect
Post by: ayeaye on September 22, 2015, 09:47:38 PM
You guys realise the power is coming from the microcontroller, right?

Why asking in that tone, only want to find out things, knowingly not your enemy.

Cannot exclude anything. Have you realized that saying that there is overunity, is an offense? The same as saying that anything is true beyond doubt. Like Higgs boson has been found. It is an offense because it does not allow anyone to think differently, does not allow anyone to doubt. This is why it is an offense. All one can say is what evidence there is.

I see no way how any charge from the mosfet can go to the capacitor, the only way how the power can come from the microcontroller in that circuit. Because it can only happen when the voltage on the mosfet's capacitance is greater and opposite to the voltage on the capacitor. By the same direction i mean in the same direction in the closed loop, like in the Kirchoff's law. Like two capacitors in parallel, when can one increase the voltage on the other, think about it. But how can that happen there when the voltage on the gate-source capacitance, which is the greatest voltage there and the only voltage created by the microcontroller's output (the input voltage), is in the *same* direction as the voltage on the capacitor. If you disagree, what you can do, show how can the charge on the mosfet's input capacitance created by the input voltage, go to the capacitor in that circuit. Everyone has a burden of proof, otherwise some have no responsibility.

That but, i will also measure the currents there before the voltage spike and during the voltage spike. Whether the former is less than the latter. This shows whether there is overunity in the coil, when we look at the coil separately. Showing overunity in the coil is important, for theoretical reasons, and this is all what i'm really interested in.

Overunity in the coil though doesn't necessarily mean overunity in the circuit. Because the circuit has to be properly considered everything including the mosfet, including the current at the mosfet's gate. One may also go further and say that the input power is the power consumed by the microcontroller. The power of generating pulses has not been tried to be made minimal in that experiment, for lower power the pulses have to be generated by cmos gates, which is though more difficult to adjust. But anyway, whether it is is or is not considered overunity in that sense, is not really important, what is important is the research.
Title: Re: Negative discharge effect
Post by: TheComet on September 23, 2015, 12:08:37 AM
Quote
Why asking in that tone, only want to find out things, knowingly not your enemy.

I apologise. Let me explain in better detail what's going on here.

The circuit can be simulated and analysed, there is a very logical explanation as to why the capacitor slowly charges. Attached you will find the simulation file which you can run yourself using the free program LT Spice from linear technology: http://www.linear.com/designtools/software/ (http://www.linear.com/designtools/software/)

I modelled the circuit according to attachment 0.png. As I mentioned earlier, the power is coming from the micro controller. Micro controllers have a push/pull output stage which can deliver up to 25mA typically. This means the pin can sink or source 25mA at whatever output voltage (typically 3.3V).

If you look at attachment 1.png you will see the relevant measurements. The micro controller is outputting a square wave at 10kHz with 4% duty cycle. If we measure the voltage over the coil, we see spikes. Why? This is basic boost converter theory and can be further researched here: https://en.wikipedia.org/wiki/Boost_converter (https://en.wikipedia.org/wiki/Boost_converter)

As to where the voltage is coming from in the first place, if you measure the current on all three pins of the MOSFET you will notice that current is being transferred from the the gate to the other pins via the MOSFET's parasitic capacitances, i.e. the current is coming from the micro controller. For a more accurate model of the MOSFET, see this image: http://powerelectronics.com/site-files/powerelectronics.com/files/archive/powerelectronics.com/images/Fig5-1-0515.jpg (http://powerelectronics.com/site-files/powerelectronics.com/files/archive/powerelectronics.com/images/Fig5-1-0515.jpg)

If we look at the current going through the diode, we see that there is an average current of about 220nA. This means the capacitor is being charged with that current, and we are able to make a prediction about how much voltage will exist over the capacitor after a specific amount of time.

Is it overunity? No, it is not. We can prove this by impedance-matching the output of the circuit with the micro controller's power source and measuring the energies. We see that the micro controller has output way more energy than what was consumed by the load. Therefore, this device is not overunity.
Title: Re: Negative discharge effect
Post by: ayeaye on September 23, 2015, 12:24:19 AM
There is no action that generates additional power.
Is there?

The only such action can be induction in the coil.

I just thought about what TheComet said. There is still one way how some of the energy from the mosfet's gate can go to the capacitor. First there is a voltage on the gate, this fills the input capacitance. It is charged in the opposite direction to the capacitor, so it causes a current which discharges the capacitor. This is the same direction as the current which causes the voltage spike. But this voltage is there only for a very short time, as the mosfet opens very quickly, and when it opens, it short circuits the input capacitance. That is, the gate to source and gate to drain capacitancies in series.

Thus it is not about the charge in the input capacitance going directly to the capacitor, but helping to cause the voltage spike, which then causes current in the opposite direction, that charges the capacitor. Nevertheless, some of the input energy such way goes to the capacitor. But it should be very small, because there is such voltage on the mosfet only for a very short time. Measuring the current with the oscilloscope should show how great that current actually is and how long it lasts.

The imput capacitances are between the gate and source, and gate and drain. For the circuit they are in series, and the gate and drain capacitance is not between the input and ground. So as i think the voltage on the gate-source can be greater, and this influences the circuit.

You are bombing me, TheComet. So by now, i just ignore your emulation results. Emulators also don't explain, and may also not correspond to the reality. I prefer to understand what happens, not to trust emulators. Your explanation is vague, you just say "mosfet's parasitic capacitances", while i take every effort to understand what capacitances and how. Mosfet has three electrodes and thus three capacitances, between every pair of electrodes. Gate is insulated from the semiconductor with a dielectric. No current goes through the mosfet, except the current between source and gate, when the mosfet is open.

PS Mosfet is tricky, but the only way is to understand it. Mosfet sounds strange actually, i would prefer to call it a mos transistor. The transistors in the integrated circuits are mostly mos transistors. Metal oxide semiconductor, yes. Semiconductors i also like to call rather n-conductors and p-conductors, this prefix "semi" there gives a very wrong idea, they are special types of conductors.
Title: Re: Negative discharge effect
Post by: ayeaye on September 23, 2015, 06:19:19 AM
Is it overunity? No, it is not.

Didn't i tell you that saying that something is in a certain way beyond doubt, is offensive? Saying that there is overunity is offensive, saying that there is no overunity is offensive. Because it does not allow people a possibility of doubt. Think about it.

In ancient Greece the teachers said, i'm teaching you a certain doctrine, you must learn, but you can doubt in it. Today the teachers say, i'm teaching you how the things are.
Title: Re: Negative discharge effect
Post by: TheComet on September 23, 2015, 09:25:46 AM
Quote
Didn't i tell you that saying that something is in a certain way beyond doubt, is offensive? Saying that there is overunity is offensive, saying that there is no overunity is offensive. Because it does not allow people a possibility of doubt. Think about it.

Sure, but there is no doubt here. The energy is with one hundred percent certainty coming from the micro controller. Unless you can find a flaw in the measurements I provided, there is no way to dispute that fact.

If I put two apples in front of you, would it be an offence to say there are two apples there? Would you doubt something like that? If so, why?
Title: Re: Negative discharge effect
Post by: ayeaye on September 23, 2015, 11:10:24 PM
If I put two apples in front of you, would it be an offence to say there are two apples there? Would you doubt something like that? If so, why?

Yes right. This is a good analogy of saying that something is the way it was said, beyond doubt. Like when we say that there is the Eiffel Tower and describe how it looks like, then indeed one can go and see the Eiffel Tower, it is indeed there and looks like the way it was said. In fact, i have climbed to the highest floor of the Eiffel Tower, and can confirm that it is really there. It is not so clear with other things though, what you cannot go and see by yourself. Like when did they last time show the Higgs boson to the people in the Madison Square Garden?

First you did not simulate the later version of my circuit on the first drawing below, which was the version of the circuit also replicated by TinselKoala. I used a more complex circuit when i found the effect, true, but then i simplified it to remove unnecessary complexities and to make things more clear. So the circuit of that experiment should be the circuit on the first drawing below. Also now i call the effect the asymmetric current induction effect, as the first name of the experiment was only based on my first impressions.

You said that my circuit works like a boost converter. Look at the drawing of my circuit (first below) and the drawing of the boost converter circuit (second below). You may notice that these circuits are different. The current in the boost converter is not asymmetric, when the switch is closed, while in my circuit the current is always asymmetric. Because of that, the boost converter completely wastes the first voltage spike in the coil. The boost converter uses the second negative voltage spike to generate output. My circuit uses only the first positive voltage spike, and not the second negative voltage spike, it cannot use it, because as the positive voltage spike is utilized, there almost is no negative voltage spike. So because of that, the circuit of the boost converter is likely more inefficient.

I currently have no time to do any circuit simulations, and check the accuracy of your simulation, but i concentrated on some of the information you provided about the results of your simulation, to concentrate on the most important. On your figure which shows the voltage over inductor, you wrote "Spikes are caused when the MOSFET switches off". Now the oscilloscope in the TinselKoala's replication of my experiment https://www.youtube.com/watch?v=qUMb6e6QQIA (https://www.youtube.com/watch?v=qUMb6e6QQIA) clearly shows that the voltage spikes start when the mosfet switches **on** , 9:15 in that video. So what did you simulate, a boost converter?

Title: Re: Negative discharge effect
Post by: TheComet on September 24, 2015, 01:04:41 AM
Quote
Yes right. This is a good analogy of saying that something is the way it was said, beyond doubt. Like when we say that there is the Eiffel Tower and describe how it looks like, then indeed one can go and see the Eiffel Tower, it is indeed there and looks like the way it was said. In fact, i have climbed to the highest floor of the Eiffel Tower, and can confirm that it is really there. It is not so clear with other things though, what you cannot go and see by yourself. Like when did they last time show the Higgs boson to the people in the Madison Square Garden?

No, it isn't clear to YOU because you haven't taken the time to study everything required to understand the proof behind the Higgs Boson. Just because YOU don't understand something doesn't mean said something is wrong. Do you really have to know how everything works before you believe it does? If everyone had to know everything we wouldn't progress as a human race. We'd get stuck at a certain point because one human can only learn so much.

Here's my problem, ok. There's no point in explaining this circuit because:
1) It's senseless. You may as well throw random parts together and you'd still somehow be able to make outrageous claims about it because you have a lack of understanding.
2) There's no explanation on how the circuit is supposed to work and what existing physical laws support said working. What's the point of even discussing it when it's obviously garbage?
3) You're not doing real science. If you were actually disproving an existing physical law through experimentation I'd be highly intrigued and would be all ears. But you're not. This circuit is meaningless. It's bits of wire hanging in a non-sense manner in mid air, and it just so happens some of the components are arranged in such a way as to allow reception of some electromagnetic waves.

What claims are you trying to make? I watched the video and he explains perfectly where the voltage is coming from: EMI from the mains in the house and power from the signal generator. Case closed.

It's the equivalent of duct taping a fork to the ceiling and measuring its temperature and getting excited when the fork slowly starts getting warmer because surprise: Hot air rises, making the top of your room slightly warmer. But that's obviously not what's happening, the fork must be getting warmer because of <insert reason with no grounds>, therefore we can't exclude the possibility of overunity even though no existing physical law permits it.

If you want to test this circuit correctly you'll have to:
1) Perform the measurements in a room with no electromagnetic waves (e.g. Faraday cage).
2) Show that the energy is NOT coming from the signal generator and/or micro controller.
3) Show that the energy is NOT coming from the multimeter.
4) Show that the energy is NOT coming from chemical based effects inside the capacitor itself.
5) Show the capacitor slowly charging.

Then we can start talking.
Title: Re: Negative discharge effect
Post by: ayeaye on September 24, 2015, 02:17:40 AM
Do you really have to know how everything works before you believe it does?

No, but i have to see that it works, and i also have to see what works. Don't you, do you just believe when someone says that something works?

Higgs boson, the evidence of its existance is only based on statistics. On the data which is stored only in one place, in the big computers of cern. So one has to trust this only one source of information, cern, to be convinced that Higgs boson exists. Btw i think that it likely exists, but i brought it just as an example how something is said to be beyond doubt, which everyone cannot see by themselves. I'm interested, btw, and i'm going to talk to people who work at cern, to find out what is really going on there. But then, none of them knows everything there, everyone mostly knows only their part of the work.

> This circuit is meaningless. It's bits of wire hanging in a non-sense manner in mid air

Thank you. Thank you for showing your ignorance.

> I watched the video and he explains perfectly where the voltage is coming from: EMI from the mains in the house and power from the signal generator. Case closed.

The EMI from the mains power is something which i have not seen, and have not been able to measure it in the coil. Neither does my emf meter show any emf near the coil. I don't say that there isn't any, but i cannot see any. Also how can mains power induce any significant emf in the coil? An antenna has to be half the wave length, so calculate, for 60 Hz it has to be 2500 km long. Not so bad though, the induction just significantly decreases the shorter is the antenna. Yet you need quite a long antenna to get any significant energy out of mains power. This is why no one has succeeded to get any energy from the power lines by induction, in spite that many have tried.

Much more likely, TinselKoala has some mains power interference on his bench, likely due to some old equipment, which interferes with his oscilloscope. When he says he nears his hand to the coil, in fact he also nears his hand to the oscilloscope's probe close to that coil. I know that when i sit near my computer and near my hand to my emf meter some distance away, it also detects some emf. Which is infinitesimal though, but emf meter is very sensitive, and so is oscilloscope.

> 1) Perform the measurements in a room with no electromagnetic waves (e.g. Faraday cage).

I will do that, i will put all circuit in the Faraday cage. I already thought about doing that, you didn't have to advise me.

> 2) Show that the energy is NOT coming from the signal generator and/or micro controller.

I will try to measure the current at the mosfet's gate, though it is very difficult to measure because it likely comes in pulses that are very short. But even without that, is is quite clear what the input energy is, it is the energy necessary to charge the mosfet's input capacitance, some 500 pF, with the output voltage of the microcontroller (5 V on Arduino), once in every switching cycle. I already said that, but you sure don't read what i wrote previously.

But more importantly, i will try to measure the current in the circuit before the voltage spike. This may indicate whether there is overunity in the coil.

> 3) Show that the energy is NOT coming from the multimeter.

Multimeter, when it measures voltage, is completely passive, and uses no external power. Multimeter uses an analog to digital converter to measure voltage. This usually has a small capacitor, and the voltage on it is compared to various voltages using the comparator. Anyway the coltage from that capacitor is used only as an input, and in the process it can only discharge and not charge.

4) Show that the energy is NOT coming from chemical based effects inside the capacitor itself.

This is ridiculous, it is known how much the chemical processes can increase the voltage of the capacitor, and this is very small. Discharge an electrolytic capacitor and you see that after that the voltage slowly comes back, but it is mostly in microvolts, these chemical processes cannot create more emf.

5) Show the capacitor slowly charging.

I have already shown that. But sure you didn't watch my video, neither did you read anything that i wrote before.

Haha, i see now that your intention was to come here and discredit. Which i foresaw from the very first moment, which you may have noticed. How can one trust a person with such intention, anything one says or provides as evidence, which needs a certain amount of trust to believe it?

In contrast, my intention is just to do research, not to make claims, which you so much want to emphasize. In spite that i never made any claims. But you have to make it look like i did, because this is the necessary element the way you want to discredit me, that i'm ignorant and dishonest. Well and, what enables you to present yourself the way you want yourself to be seen, knowledgeable and competent, but you failed in that.

Title: Re: Negative discharge effect
Post by: TinselKoala on September 24, 2015, 07:10:56 AM
I'm pretty sure that the energy in the capacitor is coming from the function generator and is leaking through the mosfet's capacitance. There may be some effect from any slight voltage offset of the FG's baseline (zero) voltage if such is present. There might be a very slight contribution from ambient EMI, but aye-aye is right that the ripple I showed in the demonstration is coming from an _effectively_ ungrounded probe during the time when the mosfet Gate signal is below the threshold. I don't think it's contributing to the charging of the capacitor much if any. When the Gate signal exceeds the threshold and the mosfet actually turns on, then this spurious mains pickup goes away.  However, the point of that part of the demonstration is that the scope signal can _look_ like there is a real, negative, voltage there during the low parts of the 60 Hz cycle, if the scope is not interpreted properly.
Title: Re: Negative discharge effect
Post by: ayeaye on September 24, 2015, 08:12:27 AM
Thank you Tinsel.

Ok, this has to be found out.

This, talking about the "leaking through the mosfet's capacitance" is pretty vague. It's like stumbling in darkness. We have to know how the leakage happens, because before doing the measurements we have to know how it works, then we can also understand the measurements.

Basically i think there are two capacitances that matter, the gate-source capacitance and the gate-drain capacitance. When the mosfet is switched on, they are constantly charged by the input voltage, say 5 V on Arduino. These two capacitances are connected in series for the rest of the circuit. But, they are connected to the open mosfet with a quite low resistance. So, can these two capacitances constantly leak to the circuit, and if, then how much. This needs to be figured out. As i already said, they may leak more once, at the moment when the mosfet opens, during the time when its resistance is still greater. But because mosfet opens very fast, this leakage may be very small.

As you see, i don't know everything, i need to figure things out. What it is all about, no one researches something which is already well known. And when i find out more, i will write it in this thread.

Title: Re: Negative discharge effect
Post by: ayeaye on September 29, 2015, 01:51:28 AM
Ok, i did the simulation of the circuit with ngspice. Why ngspice, because ngspice is the successor of berkeley spice, that is, ngspice is berkeley spice today, so the most classic. And also open source. It is in batch mode, the same as the berkeley spice is in batch mode. The netlist was the followig.

* Spice netlister for gnetlist
.SUBCKT irf530n 1 2 3
**************************************
*      Model Generated by MODPEX     *
*Copyright(c) Symmetry Design Systems*
*         All Rights Reserved        *
*    UNPUBLISHED LICENSED SOFTWARE   *
*   Contains Proprietary Information *
*      Which is The Property of      *
*     SYMMETRY OR ITS LICENSORS      *
*Commercial Use or Resale Restricted *
*   by Symmetry License Agreement    *
**************************************
* Model generated on Sep 21, 01
* MODEL FORMAT: SPICE3
* Symmetry POWER MOS Model (Version 1.0)
* External Node Designations
* Node 1 -> Drain
* Node 2 -> Gate
* Node 3 -> Source
M1 9 7 8 8 MM L=100u W=100u
.MODEL MM NMOS LEVEL=1 IS=1e-32
+VTO=3.79209 LAMBDA=3.64034 KP=81.097
+CGSO=8.9e-06 CGDO=1e-11
RS 8 3 0.056205
D1 3 1 MD
.MODEL MD D IS=9.06112e-12 RS=0.00982324 N=1.13042 BV=100
+IBV=0.00025 EG=1.2 XTI=3.54513 TT=0.0001
+CJO=7.4e-10 VJ=1.52632 M=0.693198 FC=0.5
RDS 3 1 1e+06
RD 9 1 0.0219755
RG 2 7 4.4648
D2 4 5 MD1
* Default values used in MD1:
*   RS=0 EG=1.11 XTI=3.0 TT=0
*   BV=infinite IBV=1mA
.MODEL MD1 D IS=1e-32 N=50
+CJO=9.44672e-10 VJ=0.5 M=0.9 FC=1e-08
D3 0 5 MD2
* Default values used in MD2:
*   EG=1.11 XTI=3.0 TT=0 CJO=0
*   BV=infinite IBV=1mA
.MODEL MD2 D IS=1e-10 N=0.400249 RS=3e-06
RL 5 10 1
FI2 7 9 VFI2 -1
VFI2 4 0 0
EV16 10 0 9 7 1
CAP 11 10 1.29816e-09
FI1 7 9 VFI1 -1
VFI1 11 6 0
RCAP 6 10 1
D4 0 6 MD3
* Default values used in MD3:
*   EG=1.11 XTI=3.0 TT=0 CJO=0
*   RS=0 BV=infinite IBV=1mA
.MODEL MD3 D IS=1e-10 N=0.400249
.ENDS irf530n
**************************************
*      Model Generated by MODPEX     *
*Copyright(c) Symmetry Design Systems*
*         All Rights Reserved        *
*    UNPUBLISHED LICENSED SOFTWARE   *
*   Contains Proprietary Information *
*      Which is The Property of      *
*     SYMMETRY OR ITS LICENSORS      *
*Commercial Use or Resale Restricted *
*   by Symmetry License Agreement    *
**************************************
* Model generated on May 30, 03
* MODEL FORMAT: SPICE3
.MODEL 1n4007 d
+IS=7.02767e-09 RS=0.0341512 N=1.80803 EG=1.05743
+XTI=5 BV=1000 IBV=5e-08 CJO=1e-11
+VJ=0.7 M=0.5 FC=0.5 TT=1e-07
+KF=0 AF=1
R2 0 n5 10
R3 n4 n3 10M
D1 n5 n4 1n4007
C1 n4 n3 470uF
L1 n2 n3 660mH
V1 n0 0 dc 0 pulse 0 5V 0 25ns 25ns 48us 1ms
R1 n0 n1 10k
X1 n2 n1 0 irf530n
.END

The following transient simulation was done.

tran 0.1ns 0.5ms

The figures below are the circuit diagram, voltage on the coil, voltage on the diode, and currents. The voltage polarities are such that they are in the same direction as the current that is considered positive in that circuit, that is the current in the direction of the diode. On the currents diagram, red is the current at the gate, and blue is the current in the circuit. The 400 units there correspond to 400 microamperes for current at the gate, and 40 microamperes for the current in the circuit. The current in the circuit is multiplied by 10 and the current at the gate is reversed, to enable to compare these currents better.

First TheComet was right in that this circuit really works similar to a boost converter. Yet he was not right that voltage spikes start from the end of the pulses, what can be considered voltage spikes there start from the beginning of the pulses. Because during the spike the resistance of the circuit is higher, which enables the voltage to go high. There is another "spike" in the opposite direction, but because of more current in the circuit, it is so flat that it cannot be called spike. The TheComet's simulation is questionable also for other reasons. Like it doesn't correspond to the circuit whith which the experiments were done and the diode is not similar to the diode used in the experiments. The biggest problem was though his screw you style.

Compared to a boost converter though, this circuit has some advantages, in that it needs only one input from the signal generator, it doesn't need a separate voltage source as the boost converter does. So it may be better like for testing overunity of some different types of coils.

What shows the best what happens there, is the currents diagram. As you see at the beginning of the pulse, the current goes first negative. The reason for that is that when the input capacitances of the mosfet (gate-source and gate-drain) are charged at the beginning of the pulse, the gate-drain capacitance charges faster, because its capacitance is smaller. This leakage is small because the mosfet opens quickly. When the mosfet is open, there is no leakage, because the resistance of an open mosfet is small, may be less than an ohm.

The only other time when the mosfet leaks, is when it closes at the end of the pulse. Then the current is positive, because concerning the discharging, its input capacitances are in series for the circuit, and the gate-drain capacitance discharges faster. The input capacitance discharges then, but it takes some time. And during that time the current from that capacitance goes to the circuit.

Now when you look at the currents on the currents diagram, it can be seen that the current in the circuit after the end of the pulse, can come from the mosfet's input capacitance. Considering its capacitance (some 500pf) this can cause the effect in the experiments. I have not calculated though, but it looks like so. I don't exclude though the possibility of overunity in that circuit at certain frequencies or duty cycles. You may research it more, and maybe i also do.

What makes me wonder a bit though, is that there seems to be no correlation between the discharge current of the input capacitance, and the current in the circuit. The input capacitance seems to discharge the same as a separate capacitor, by a perfect exponential curve, not disturbed by anything. Does this indicate that the input capacitance only discharges to the output of the signal generator, and doesn't leak into the circuit at all? I don't know, because i have not analyzed or calculated that.

What concerns the induction in the coil, quick rise of the current causes voltage to go higher, but because of the lenz law it also goes faster down, so the pulses are narrower, and thus create less current. But as much as i remember, TinselKoala said that things are not ideal. Exactly because they are not ideal, there can be overunity when the lenz law doesn't work directly against the current.

This may also be the case in the DePalma's n-machine, which is based on the Faraday's homopolar generator, where the direction of the current relative to the magnetic field is 90 degrees different from that angle in a coil. The coils where the wire is winded at different angles, i don't know. maybe there too the negative emf is at different angle and thus also not directly against the current, especially maybe when to switch different windings in such coils.
Title: Re: Negative discharge effect
Post by: TheComet on September 29, 2015, 12:52:37 PM
I still don't understand why you would suspect overunity.

Clearly over 99% of the energy from the signal generator is being "short-circuited" through the mosfet directly to ground and dissipates over the first resistor. Only a tiny fraction of the input energy ends up on the output.

Can someone please tell me why they suspect overunity in a circuit that is less than 1% efficient? Two people have proven this with simulations already.
Title: Re: Negative discharge effect
Post by: ayeaye on September 29, 2015, 02:35:11 PM
TheComet, what matters is research, like possible overunity in the coil.

Maybe the mosfet leaks through resistance, and not through internal capacitances, as it is often said. Look at the horizontal region of the gate current on the currents diagram. What is this? This may as well be a continuous leakage from the input capaciance to the circuit, through a 42 k resistance, well, calculate. It does not show any correlation whatsoever with the circuit though. Is it a power necessary to attract the electrons in the mosfet? But whatever it is, it cannot be seen where it really goes, and that a part of it doesn't go to the circuit. These are the problems, we cannot see inside everything, thus also not possible to see for certain, what the real leakage is.

We can know what the maximum leakage is, this though is almost certainly a lot more than the real leakage. The only real leakage i think there certainly is, is leakage through the input capacitance when the mosfet switces on, which is very small though and lasts for a very short time. But it provides the intitial voltage to the coil, and starts that oscillation. The maximum leakage though can also be useful to know. One may try the simulation with coils with higher inductance, for example, and different frequencies and duty cycles. I don't exclude the possibility that with some of these there can be overunity even considering the maximum leakage.

I don't assume the overunity, i just don't exclude the possibility. This is what the research is all about, we assume nothing, just find out what happens. Just as Faraday did, he did it to find out how god made the world, because he was very religious. Well and, without him we didn't have the electricity today. And overunity, it looks like, is only some small effect, at least when using conventional ways of doing things. Later the small effect can be increased, like it has happened a lot in technology.

Like Faraday first saw only a slight declination of a compass needle, for a very short time. Something which some may not even notice, or consider some interference. To exclude that possibility he, btw, invented his Faraday cage, to protect his galvanometer from interference. But this slight effect is what all the electric motors and generators are based on today.
Title: Re: Negative discharge effect
Post by: TheComet on September 29, 2015, 02:56:12 PM
Quote
Look at the horizontal region of the gate current on the currents diagram. What is this?
I'm not sure what you mean, can you please be more specific? Where is this diagram? What current doesn't make sense? Please point it out by downloading the image, drawing on it with an image editing tool and re-uploading it and explaining what doesn't make sense.

I can help you understand.

Quote
It does not show any correlation whatsoever with the circuit though.
Why? What is your reasoning behind this statement?

Quote
But whatever it is, it cannot be seen where it really goes, and that a part of it doesn't go to the circuit.

Baseless statement. Download the simulation and show me with your own measurements what you mean.
Title: Re: Negative discharge effect
Post by: ayeaye on September 29, 2015, 03:25:25 PM
I'm not sure what you mean, can you please be more specific? Where is this diagram? What current doesn't make sense?

The diagram below, from the simulation results, showing the currents at the gate and in the circuit. You see the current drawn with red there, this is the current at the gate from the beginning of the input pulse, up to the half of the cycle. The pulse ends when the gate current goes to zero.

Look at the current drawn with red there, this is the gate current. It first goes down (it is drawn reversed), then up again, and then there is a horizontal region, it stays at the same value for some time, and then goes up again to zero. This horizontal region means continuous consumption of the gate current, not charging the input capacitance, which happens by an exponential curve. This means a continuous power which goes to the mosfet. We cannot see inside the mosfet, so we cannot see where it goes, whether it is consumed by the mosfet, or it goes to the circuit.

Correlation, the two things that are somehow connected, should show some dependence on each other. Like the gate current, when it leaks to the circuit, should somehow reflect the current oscillations in the circuit. It does not though, there is a perfectly straight horizontal line. I cannot help you to understand, what concerns things like that. This needs an analytical general thinking, the people who lack that ability, are never able to understand some things. And there are many such people.
Title: Re: Negative discharge effect
Post by: ayeaye on October 03, 2015, 08:27:49 PM
I now also did a simulation of the same circuit, but with the inductance of the coil 10H. The energy going to the capacitor in a single cycle was only 0.7pJ, which is much less than the mosfet's leaking energy with the circuit with an equivalent impedance. Which by the thread here on how much the mosfet leaks http://overunity.com/16062/how-mosfet-leaks/#.VhBEDG943iY (http://overunity.com/16062/how-mosfet-leaks/#.VhBEDG943iY) , is 0.7nJ.

This doesn't necessarily mean though that the output energy cannot be greater in a real circuit. As simulators don't always simulate everything so as it happens in the real world.

I also added the circuit diagram as it should be, below. One thing to notice there, the resistance of R3 is written there as 100000, not 10M. This is very important, because ngspice recognizes all suffixes such as k, m, u, n and p, but doesn't seem to differentiate between M and m, or doesn't recognize M. So the values with the suffix M will have a wrong value, and thus instead of using M, the value should be written either with a full number, or then with a suffix that is recognized, such as k.