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Author Topic: Negative discharge effect  (Read 31566 times)

Offline ayeaye

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Re: Negative discharge effect
« Reply #60 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.

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Re: Negative discharge effect
« Reply #60 on: September 23, 2015, 12:24:19 AM »

Offline ayeaye

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Re: Negative discharge effect
« Reply #61 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.

Offline TheComet

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Re: Negative discharge effect
« Reply #62 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?

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Re: Negative discharge effect
« Reply #62 on: September 23, 2015, 09:25:46 AM »
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Offline ayeaye

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Re: Negative discharge effect
« Reply #63 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 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?


Offline TheComet

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Re: Negative discharge effect
« Reply #64 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.

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Re: Negative discharge effect
« Reply #64 on: September 24, 2015, 01:04:41 AM »
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Offline ayeaye

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Re: Negative discharge effect
« Reply #65 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.


Offline TinselKoala

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Re: Negative discharge effect
« Reply #66 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.

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Re: Negative discharge effect
« Reply #66 on: September 24, 2015, 07:10:56 AM »
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Offline ayeaye

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Re: Negative discharge effect
« Reply #67 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.


Offline ayeaye

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Re: Negative discharge effect
« Reply #68 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.
« Last Edit: September 29, 2015, 04:09:02 AM by ayeaye »

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Re: Negative discharge effect
« Reply #68 on: September 29, 2015, 01:51:28 AM »
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Offline TheComet

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Re: Negative discharge effect
« Reply #69 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.

Offline ayeaye

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Re: Negative discharge effect
« Reply #70 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.

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Re: Negative discharge effect
« Reply #70 on: September 29, 2015, 02:35:11 PM »
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Offline TheComet

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Re: Negative discharge effect
« Reply #71 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.

Offline ayeaye

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Re: Negative discharge effect
« Reply #72 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.

Offline ayeaye

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Re: Negative discharge effect
« Reply #73 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 , 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.

 

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