COP 17 Heater | Rosemary Ainslie

[poynt99]

Hi .99
Always read your posts,very instructive,but I can't get my head around current going in the opposite direction to voltage,is there an easy way to understand it? Thanks
peter

Yes, Peter. This is very easy, and I think puts in to question the zippon theory of reversed current when the flyback diode is considered. Please refer to the attached diagram of the RA circuit.

1) When the switch (MOSFET) is on and conducting, I think we all agree that: a) the current direction through the coil is from terminal 1 to terminal 2; b) the voltage potential across the coil is 1+ and 2-; c) the diode is reversed-biased with no forward current through it; and d) the MOSFET conducts the same current as the coil.

2) When the switch turns OFF, and due to the collapsing magnetic field of the coil, the voltage potential across the coil reverses such that: a) the voltage potential across the coil is now 1- and 2+; b) the switch is OFF and effectively isolated from the coil and diode; c) the coil, diode, and coil capacitance effectively forms its own isolated circuit because the lower end is now effectively floating; d) with the coil voltage as it now is, this forward-biases the flyback diode and it forms the major current path around the coil; e) the current through the flyback diode must be from bottom to top (anode to cathode) in order for it to conduct, and I think we all agree that it does indeed conduct while the field is collapsing; f) as the current exits the top of the diode, it can not go into the 24V supply because there is no electrical path to the negative side of the supply, so it conducts around down through the coil (and capacitor) again in the same direction it was when the switch was ON.

If RA concedes that the flyback diode does indeed conduct during the collapsing field of the coil, how can the current be reversed as she proposes?

.99

[MileHigh]

Rosemary:

> Medium size and small buckets, both with holes would pertain to the first analogy? Why were they omitted?

They were intentionally omitted to keep it simple and easier to understand but you are correct they still are there. Energy is also lost in the resistive part of the coil and the diode's effective resistance as well as going into the battery. The voltage drop across each device will give you the relative amounts of energy lost or stored.

> In other words the current generated by the resistor during the 'off' period of the duty cycle must go back to the battery via the body diode in the MOSFET?

Yes, with some clarifications. The current is generated by the coil part of the resistor-coil during the "off" period. The "body diode" in the MOSFET is reverse biased in this case and is not conducting any current. It is a "protection" diode with a high reverse-breakdown voltage to protect the MOSFET from inductive spikes. So MOSFET itself is switched off and the protection diode is switched off. So the energy discharge from the coil will choose the path of least resistane and go through the diode that's across the coil.

.99:

> MH and I are in slight disagreement on this issue, but perhaps there is a happy medium?

More a question of relative scale, no? If the energy returned to the battery is less than 1/1000th of the energy discharged as heat then it is insignificant. I feel that it is less than one-millionth.

> I believe that makes it's way into the battery (due to this temporary path) as I showed in my scope shots, and I believe TK did as well.

Watch out for when two scope cables are carring fast rise time square waves. It is easy for one cable to influence the other cable and create small exponential voltage spikes on the other cable, especially if what the other channel is looking at is already at high impedance. I am not saying that's what is happening, but you should be aware.

All: The current in the coil does not reverse direction when the MOSFET switches off. This is a phenomenon associated with observing the voltage polarity switch and making the assumption that the current must have changed direction to create this effect. To understand this you have to put yourself in the place of the coil. If current is flowing in your feet and out your head and suddenly the flow is cut off, you want the current to continue flowing "up through your body" and out the top (higher voltage), through some sort of resistance, and then back into your feet (lower voltage). To keep this happening you "push out the top" and you keep your "head voltage" higher than your "foot voltage" to keep the current flowing. Everybody with a house observes this effect. If you go into the basement and you shut off a water valve suddenly, you hear the "watter hammer" effect and there is a big "thunk". The flowing water has mass and stores energy by virtue of it's mass and velocity. This is like an inductor storing energy in the flowing current. When you shut off the water valve suddenly, the water pressure on one side of the valve instantly shoots up to a spike in pressure, and the energy in the flowing water is dissipated mechanically in the pipe system and it generates heat. So that's a mechanical inductor discharge.

[witsend]

MileHigh and .99 Such an interesting discussion. It is honestly the first time I've been given an explanation as to how you guys, as classicists, see current flow. You should look at wiki explanation. I found it very confusing. It's the only one I know except for Heniecks. Many thanks. Now I've got to go and think about it and it'll take me some time. Am somewhat 'slow of thought'. But to both - many thanks for the explanation.

[poynt99]

.99:

> MH and I are in slight disagreement on this issue, but perhaps there is a happy medium?

More a question of relative scale, no? If the energy returned to the battery is less than 1/1000th of the energy discharged as heat then it is insignificant. I feel that it is less than one-millionth.

The bottom line is they are there, very real, and they are quite visible/evident, both in the sim and with the real scope shot. I trust TK's build and scoping techniques. As such, they need to be dealt with to appease any doubts and account for all energies that evident in the circuit. I am merely conceding that they are there, and noting their relative feeble (i.e. 20ns width) nature.

> I believe that makes it's way into the battery (due to this temporary path) as I showed in my scope shots, and I believe TK did as well.

Watch out for when two scope cables are carring fast rise time square waves. It is easy for one cable to influence the other cable and create small exponential voltage spikes on the other cable, especially if what the other channel is looking at is already at high impedance. I am not saying that's what is happening, but you should be aware.

Indeed this can happen, but not in a sim. These spikes are visible in the sim.

.99

[MileHigh]

A tip about the build: <techie corner> The MOSFET switching ringing could also be because there is ringing on the gate input. Putting your scope probe ground on the MOSFET source pin (or as close as possible) and then your probe on the gate pin you can check for ringing on the square wave signal. With a signal generator set to 50 ohms output and a coax feed, grounded close to the MOSFET drain and a 50 ohm resistor across the MOSFET Gate-Drain, you will get a rock-solid pulse wave with no ringing. It would be better to move the shunt resistor to the battery positive to keep the MOSFET source at the true circuit ground. With the 555, you should keep the signal wire as short as possible and check for ringing also. A single 25-50 ohm series resistor at the 555 output may clean up any ringing. The 555 ground should also have a good solid connection to the MOSFET source, like a thick copper wire and it's local supply should have a pair of decoupling capacitors. Square wave signals need special attention. Ok, enough teckie teckie... lol

[witsend]

If you're still there .99 - a quick question. If supply is from AC supply, say utility source - what happens to the waveform across a load, such as a light?

I think I'm asking this. Does the current change on a resistive load if the voltage is AC?

[RAMSET]

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[poynt99]

Following quotes from .99 post 876
I agree with both, MH and RA here.
Both right? Find this difficult to understand. Our analysis is diametrically opposed.

From the classical perspective, you are both partially correct. At the initial transient, there IS a very brief (20ns or so) spike of current going into the battery. AFTER this initial spike, there is no path to the negative side of the 24V supply, so no current goes back into the battery. See? it's not a matter of IF you are both correct, it's a matter of WHEN you are correct.

What actually happens here I believe is that when the MOSFET is OFF, the remaining energy in the coil really can only circulate round it's own closed loop so to speak.
What remaining energy? Does this energy come from the battery or from the coil and has nothing to do with the collapsing fields?

All the energy initially comes from the battery. THEN it is stored in the coil, and upon switch-OFF, the energy in the coil de-energizes into all dissipative elements it finds in its way. Yes inductive kickback is the de-energize phase.

BUT at the very instant the MOSFET turns OFF, there is a fast reversal in voltage across the coil and this very fast transient will briefly conduct through the reversed-biased body diode of the MOSFET, as this does represent a real but small capacitance to ground.
Yet more electric energy and still nothing to do with inductive laws?

The reversed potential of the coil due to inductive laws IS what allows this short pulse of current to enter the battery. If there was no conductive path to ground due to stray (and parasitic) capacitances in and around the MOSFET and coil, this reverse spike into the battery would not occur.

It is this initial transient spike I believe that makes it's way into the battery (due to this temporary path) as I showed in my scope shots, and I believe TK did as well.
What causes the spike is actually the question. Could you elaborate on this?

What causes the spike is what I have been saying above. There are always minute stray (and parasitic) capacitances around a circuit. Here are some in your circuit, as minute as they may be, but THEY ARE THERE AND WILL CONDUCT SIGNIFICANT CURRENT FOR A VERY SHORT DURATION:

1) MOSFET D-S
2) MOSFET D-G
3) MOSFET G-S
4) MOSFET reversed-bias body diode capacitance
5) Coil (and leads) proximity to circuit ground
6) Breadboard socket-socket capacitance

Hit these capacitances with a very fast-rising/falling edge and they will quite happily conduct current and provide a path. Once the transient edges are passed they no longer provide a significant current path. The biggest and fastest edges occur when the coil is first energized, and when it is first de-energized (inductive kickback), so there will be two spikes per input pulse.


I trust that covered your questions. If not, let me know.

.99

[poynt99]

If you're still there .99 - a quick question. If supply is from AC supply, say utility source - what happens to the waveform across a load, such as a light?

I think I'm asking this. Does the current change on a resistive load if the voltage is AC?

I'm not 100% sure what you are asking. Does the current change? Change how?

For 50/60 Hz AC power into a resistive load, the current alternates direction yes.

Inductive loads are quite different as the current lags the voltage in phase (90º I believe) with sine wave inputs. With pulse inputs we have already covered that.

.99

[witsend]

.99 - I got it. You've more or less covered your thoughts on this this in the previous. I wonder if you could explain your thoughts on current flow from an AC supply. Does this reverse?

[MileHigh]

Rodemary: This may help you understand how a coil works:

Suppose you have a 1 Henry coil with 1 amp of current flowing through it across a zero ohm resistance. It is an idealized coil and there is no resistance. The current will flow in the coil forever. By the formula E = 1/2 L i-squared, there is 1/2 Joule of energy stored in the coil. Wether you say it is stored in the magnetic fileds, or in the amount of current flowing, they are one in the same.

Now if you change the resistance across the coil to 1 ohm, what happens? The answer is that the coil will generate 1 volt across it's terminals to "push" the one amp of current through the resistor. This causes the exponentially decaying voltage waveform across the coil as the energy in the coil discharges and gets burned off in the resistor.

2 ohms = 2 volts
10 ohms = 10 volts
1 kohm = 1000 volts
1 Mohm = 1 million volts
open circuit = infinity

As the voltage gets higher, the time constant or "width" of the pulse gets narrower and narrower. In each case the amount of energy discharged will be 1/2 Joule.

So the voltage spike 0.99 is talking about occurs because for about 20 nanoseconds the resistance across the coil is very high until the diode switches on and then everything changes. When the diode switches on the resistance gets very low and the inductor generates 0.6 volts.

Ideal capacitors do exactly the same thing but for current instead of voltage and thus get the short end of the stick because voltage spikes are sexy but current spikes aren't.

MileHigh

[witsend]

I'm not 100% sure what you are asking. Does the current change? Change how?

For 50/60 Hz AC power into a resistive load, the current alternates direction yes.

Inductive loads are quite different as the current lags the voltage in phase (90º I believe) with sine wave inputs. With pulse inputs we have already covered that.

.99

Ok - sorry I missed this. Are you then saying that lower frequencies allow current to flow with voltage. Higher frequencies not?

[poynt99]

following quotes from .99 post 868

The only difference I see between your zippon current model, and the classical model is the direction of this current when the coil voltage reverses.
In the classical model, the current does not reverse, it maintains the same direction as when it was being energized by the battery. The voltage reverses yes, but the current continues in the same direction.
I agree. This is the difference. But I can prove it by showing the battery recharge using the second battery example as well as the extra energy heating the load that much more.

Excellent. Then you SHOULD prove it. Please post some data and scope shots to that effect.

If the current DID reverse direction as you propose in your model, wouldn't we be able to get more than just a brief (20ns) spike back into the battery? I would think so.
We would and do. You must remember your result is based on your simulator.

Excellent. Then you really should prove it. Please post some data and scope shots to that effect.

Also, I wonder how the modern world could have overlooked this reversed current phenomenon (if in fact it does exist) when folks have been designing with this circuit for at least 30 to 40 years. What designs? Switched-mode power supplies, Buck-Boost converters etc.
If current reversal were allowed then there would be an acknowledgement of OU. Jibbguy goes to some lengths to explain how these effects were eliminated. And I can quote you many more such comments from highly skilled experimentalists.

That is a very weak argument. 99.9% of engineers and technologists working on and designing these well-known common circuits aren't even aware that Free Energy Forums exist and that there are 1000's of folks trying to obtain overunity. How could there be an "allowance" or "dis-allowance" (if there even is such a word) of something when the designers are oblivious to the existence of free energy in the first place? If one of these individuals stumbled upon a reversed-current effect or overunity, you can bet in the last 40 years we would have heard of at least one instance.

The "allowance" of how this circuit operates is largely dictated by nature, not by the "intervention" of some clandestine group of energy-controlling conspirators. You'll need to do better than that to answer that question.

Which post of Jibbguy's are you referring to? Not post 243 I hope? I do not see anything in that post that would suggest the "suppression" of this reversed-current effect. Please elaborate.

.99

[witsend]

Following quotes from .99

Excellent. Then you really should prove it. Please post some data and scope shots to that effect.
Data on the paper. Scope shots unfortunately omitted. We're hoping to get something up in the near future.

That is a very weak argument. 99.9% of engineers and technologists working on and designing these well-known common circuits aren't even aware that Free Energy Forums exist and that there are 1000's of folks trying to obtain overunity. How could there be an "allowance" or "dis-allowance" (if there even is such a word) of something when the designers are oblivious to the existence of free energy in the first place? If one of these individuals stumbled upon a reversed-current effect or overunity, you can bet in the last 40 years we would have heard of at least one instance.
I know. It puzzles me too. One would have thought? And 'disallowance' perfectly good word.

The "allowance" of how this circuit operates is largely dictated by nature, not by the "intervention" of some clandestine group of energy-controlling conspirators. You'll need to do better than that to answer that question.
I entirely agree with you. I have never thought there was a clandestine group. Why do you say this? However, I do think that there's a reluctance by academics to evaluate OU claims based on personal experience.

Which post of Jibbguy's are you referring to? Not post 243 I hope? I do not see anything in that post that would suggest the "suppression" of this reversed-current effect. Please elaborate.
Not sure. It wasn't that long ago - 243 is way back. But I'll check and get back to you.

[poynt99]

.99 - I got it. You've more or less covered your thoughts on this this in the previous. I wonder if you could explain your thoughts on current flow from an AC supply. Does this reverse?

Yes.

Ok - sorry I missed this. Are you then saying that lower frequencies allow current to flow with voltage. Higher frequencies not?

That's a loaded question because it does not contain enough information. Please elaborate with more details and specifics.

.99