Negative discharge effect

[ayeaye]

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.

[TinselKoala]

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.

[ayeaye]

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.

[ayeaye]

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.

[TheComet]

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.