COP 17 Heater | Rosemary Ainslie

[witsend]

OK - have had a chance now to go through all the comments. Aaron - that ringing - have never eliminated it - but nor have I used a cap. But there's nothing wrong with it. It's returning while the battery is disconnected. Just adds to the general efficiencies.

My own summation - as we've not yet got the required heat - is this. If the batteries or the cap recharge and there is heat - then that is overunity. But it's only going to get better from here. You'll see both.

The real value of the video is that it shows that oscillation is achievable. I've always said it. I cannot understand why this wasn't evident elsewhere? Maybe it needed your 'break through'. It'll be interesting to watch. And interesting to see classical argument to refute it.

[Aaron]

scope

The battery voltage insreasing is a clear sign that the MOSFET is in self-oscillation. The coil is simply choking the current flow here because of the high frequencies involved.

Can you power-down and check the resistance across your pot when you trigger the self-ocsillation? I have a feeling that the pot is going open-circuit at the end of the turn. If that's the case, the MOSFET input is left floating, and I explained the self-oscillation process for that a day or two ago. If I am correct about the pot going open-circuit, then your observation confirms what I said, which is great for everybody. <EDIT: I am of course referring to the gate resistance, not sure if you were playing with the gate resistance or a 555 trimpot. If it was a 555 trimpot, then it would indicate true self-oscillation>

When you see the self-oscillation, you shoud try to get the scope to trigger on it anyways because it still looks like it lost trigger.

Finally, in certain points of the clip, some of the alleged self-oscillations could simply be the scope loosing trigger. That's exactly what it looks like, something we have all seen thousands of times before. It looks like you are AC coupled, you can see a "jump" in the waveform every now and then, and that could cause you to temporarilly loose trigger. I doulbe-checked and you do indeed loose trigger when the waveform jumps down, and the trigger comes back when the waveform moves back up. Depending on where your trigger threshold is set, you are almost certainly loosing trigger in these "jump down" cases.

So we both may be right, the glass is more than half full.

Just a final comment about your clip: The cap charging example returns to the battery ground, forming a full circuit path, a loop that allows the caps to charge. Territory that was prevoiously discussed, it does not show any charging can go back to the source battery in the real circuit because the MOSFET switch is off and breaks the loop.

MileHigh

Right, higher frequency will be less flow from the battery.

If the signal pot is right at the verge of oscillation but not at it. Decreasing duty cycle will cause it to self-oscillate and/or increasing frequency will cause it to do the same. All of those are predicted by the gas pressure analogy of voltage. In either case, with 99% duty cycle and the slowest frequency I can get with my circuit, I can increase gate resistance (at 99% duty cycle and slowest frequency) and it will do the self oscillation.

My scope is connected to a 400w inverter running on a 12v car battery. That is the fan sound in the background (inverter). That was the only way I could isolate the scope from my Gray circuit and left it like that. Isolation transformer I used didn't even isolate it.

With scope disconnected, I can turn the pots, see the voltage on battery climb and then hook the scope to it and 100% of the time it shows the self oscillation. I can remove the scope, turn the pots until the battery voltage drops and 100% of the time I hook the scope back up, I see the wave form. It coincides 100% of the time with what the battery shows. I don't know if this answers anything you're asking.

The pot on the gate is 10k. The oscillation is around 10% max (up to 1k) depending on what the duty cycle and frequency is set at. The lowest it is at when triggered in self oscillation is probably around 300 or so and has stayed out of self oscillation up to probably 800~1000. It is actually difficult to keep it out of oscillation with that much resistance.

The ohmeter in my multimeter is blown and I blew up 3 other multimeters in the last couple weeks on my Gray circuits. Will get another.

[poynt99]

Aaron,

I know you have been at this FE thing for a relatively long time, and obviously you have picked up quite a bit of information along the way (kudos to ya). Also, I assume that electronics is not your formal background, so may I suggest that although you currently have some electronics knowledge, you may not necessarily have ALL the information you need to make blanket statements about how these circuits operate?

For example from your document:

Some 'experts' claim the spike is so puny that it
can't do anything but I was able to charge the cap
to 30+ volts in a few seconds and that is a high
capacitance 60uf. Relatively high capacitance for
the small spikes and it still pushed that cap.

Those spikes DO charge the front battery and can be
sent to a separate battery or capacitor. Capacitor
charges higher than the battery.

You are dealing with two very different circuits here. In case A you have the flyback diode directly across the coil and you are claiming that it is charging the source battery. In case B, you take the flyback diode off the positive end and feed it to a capacitor for charging (cap+ to diode and cap- to +24V rail).

Do you see that in case A there is no path (not in-circuit) for the spike to go through the battery (aside from the capacitive one I already identified), and that in case B the capacitor is directly IN the circuit with the coil? Charging a cap or battery this way is normal.

Do you realize that when a coil is energized it is analogous to charging a capacitor, and that when the coil de-energizes, that too is analogous to a capacitor dis-charging? Why then do you feel that a de-energizing coil is somehow OU? The inductive kickback of voltage in an inductor is analogous to the current kick you get when shorting out a charged capacitor, yet we don't call that OU.

Utilizing the inductive kickback of the inductor is NOT getting energy from nowhere, it IS the original energy that you put into the coil from the battery in the first place. If we place a flyback diode across this coil it allows only a single (for the most part) half-cycle de-energization into itself as the load. If we remove the diode, again the coil de-energizes into itself but this time over several ring-down cycles. If we re-route the flyback diode into an external load such as a battery or capacitor or bulb, depending on the load impedance there will be as little as one half cycle or several half cycles present to totally de-energize.

Do you fully understand how the inductive kickback behaves and how to manipulate it into giving you mostly current or mostly voltage?

In general what you describe in the above quote are "normal" characteristics. Yes you can drive a load from the flyback diode but all that is happening is you are de-energizing the coil's already-paid-for-energy into that load.

Here is one possible and easy test for you or anyone to try, in order to see how much effective charging really is taking place in the source battery: Insert a fast heavy diode (such as a MUR1520 or equiv.) in series with the battery so that any currents going backwards into the battery will be blocked. Now test to see if the battery voltage falls, rises, or stays the same.

I think it will be the same as without the diode, but you never know. I'm keeping and open mind.

Are we still friends

.99

[Aaron]

future reporting

Aaron and everyone: It was fun discussing the circuit and the clips. I don't know if I will have much more to say unless people start crunching some power-in power-out numbers from real tests.

I'll post whatever results I measure. Most of all, I wanted to take the mystery out of the mosfet oscillation just to get the ball rolling. I still need lower shunt resistance and a 555 circuit that gives me under 50% duty cycle. I'm sure there (I hope) will be other replications before I post again now that I have shown the mosfet will oscillate as claimed.

I will also get a 100w 6 inch long 10 ohm wire wound resistor and will get one the diameter or Rosemary's. Someone here in town has some that I can use.

I'll also use the 20ah batts...I used my 7ah gels just to do something fast.

[witsend]

Aaron and everyone: It was fun discussing the circuit and the clips. I don't know if I will have much more to say unless people start crunching some power-in power-out numbers from real tests.

For Rosemary: No need to take a swipe at the "classical" types. Your comments impugning their abilities and understanding are way off base and they are not worth rebutting. I hope that some of what you learned will be beneficial to you. You dropped a hint about wanting to replace the diode across the coil-resistor with a LED because it makes sense to you to see it flashing and it likely would never have occured to you unless you were here on this forum to learn and I did not state it. The bottom line is results. Somebody has to run the setup and make make some credible power measurements, that is a fact that we can all agree on. Good luck!

MileHigh

MileHigh - why the aggression? No swipes made I assure you. I am absolutely unable to understand these comments. MileHigh, I am an AMATEUR. I cannot put a circuit together. And I can only draw very simple circuits. You guys - all - have forgotten what I know. Not only that - but nor am I into conventional power applications. So - not only do I not know - but nor am I ever likely to learn. I obsess about physics. You guys are practical. I have every respect.

What I really do need to understand is how you all apply classical understanding of current flow to any circuitry at all. I am constantly surprised at the difference in our view points. But everyone acknowledges that it is the classicist who has taken electrodynamics to levels unequalled in any other branch of physics. And that deserves my respect and lots of it. Please, if it seems that I have insulted you then I am truly sorry. Never meant. Really never meant.

What I now need to wrap my mind around is how you argue the result of the recharging. To me that is unarguably over 1. How can I counter your arguments unless I understand what it is. Sorry MileHigh - really - from every level. I did not mean to imply any disrespect for your knowledge. On the contrary. I only want to discuss.

[Aaron]

measurements

Aaron - was glued to the computer all day - then fell asleep a the critical moment111 Have just watched the video. What a pleasure. That self-oscillation - AT LAST. There's something wrong with my Fluke. I'm going to get it fixed and will then post it to you. You need some good power measurements now. And I would love you to use it. It'll be in more capable hands than my own.

WOW. I am really buzzing. Well done.

And what a relief to see that oscillation. That's always amazing and for those that replicate this test that's the point that benefit is inevitable.

Anyway - am just so chuffed. Many thanks for the trouble in doing that video. And from here on - enjoy the trip. It only gets better.

Hi Rosemary, I'd be honored! Everything gets put to good use around here.

[Aaron]

comments

MileHigh - why the aggression?

That was meant for me I believe. Bugged me on the scope comments about the oscillation and I commented very bluntly about it. Considering what TK has put forth, it seemed like another road bump to slow down people's enthusiasm for the circuit.

[MileHigh]

Rosemary:

>
Quote:
Originally Posted by witsend
MileHigh - why the aggression?
That was meant for me I believe. Bugged me on the scope comments about the oscillation and I commented very bluntly about it. Considering what TK has put forth, it seemed like another road bump to slow down people's enthusiasm for the circuit.
>

I was a mistunderstanding, please go back and read my edited posting and good luck again!

MileHigh

[poynt99]

99, you appear to be proficient in EE and related. However, when you question the circuit and concept contained it as it relates to operating COP and Efficiency (I agree who cares about efficiency if cop is over 1.0), I find it most interesting that your very definition of efficiency doesn't include environmental input, which is required. This, to me, is a red flag and is very questionable.

Well, with all due respect Aaron, your credentials are also in question, in how qualified you are to question not only my knowledge and professional opinions but those of thousands more like me.

I do not want to get into a debate about semantics and frivolous definitions that are really irrelevant to the goal of achieving OU. It's also not at the heart of the matters we are discussing here nor of any great importance. If you create an overunity device great. What you call it after that (overunity, >100% efficiency, COP>1) whatever, who cares?

.99

[witsend]

MileHigh - I know only too well - what tiredness does. But for the record - unless we had you and .99 and any others with critical input - we'd never be able to hone an argument as required.

If we're going to claim OU - we need to get our ducks in a row - and they have to be pretty orderly. It's thanks to you guys that set the standards, so to speak. It's critical.

So please. Don't stop discussing. Your input is far more important than I think you realise.

[witsend]

Well, with all due respect Aaron, your credentials are also in question, in how qualified you are to question not only my knowledge and professional opinions but those of thousands more like me.

I do not want to get into a debate about semantics and frivolous definitions that are really irrelevant to the goal of achieving OU. It's also not at the heart of the matters we are discussing here nor of any great importance. If you create an overunity device great. What you call it after that (overunity, >100% efficiency, COP>1) whatever, who cares?

.99

.99 If I may comment here. I think Aaron's qualification to comment is that he's looking at experimental evidence. That's got to be up there in any scientific evaluation of anything at all. And I don't think qualifications are at question. The fact is that qualified or not Aaron is experimenting. That's got to be pretty high on the credential list. Our hope is that you, hoppy or .99 will also comment on the evidence at hand - eventually.

And I would have thought that the definition of overunity needs to established. For the record, my own is simple 'more out than in'. How's that to start the debate?

[poynt99]

.99 If I may comment here. I think Aaron's qualification to comment is that he's looking at experimental evidence. That's got to be up there in any scientific evaluation of anything at all. And I don't think qualifications are at question. The fact is that qualified or not Aaron is experimenting. That's got to be pretty high on the credential list. Our hope is that you, hoppy or .99 will also comment on the evidence at hand - eventually.

Experimental evidence is worthless if the experimenter is not qualified to fully understand the evidence, nor the ramifications of it. It is evident that incorrect assumptions are being made about the evidence at hand and even the operation of the experiment itself. Indeed if you do not have a sufficient enough understanding of something, how can you then be qualified to properly comment on it?

Experimenting is fine and should be encouraged, but what gets me is when folks become authoritarian and downright bold about their results and conclusions when in fact they are sorely lacking the qualifications to do so, especially, disrespectfully and defiantly in the faces of those that are.

.99

[Aaron]

spikes

You are dealing with two very different circuits here. In case A you have the flyback diode directly across the coil and you are claiming that it is charging the source battery. In case B, you take the flyback diode off the positive end and feed it to a capacitor for charging (cap+ to diode and cap- to +24V rail).

Do you see that in case A there is no path (not in-circuit) for the spike to go through the battery (aside from the capacitive one I already identified), and that in case B the capacitor is directly IN the circuit with the coil? Charging a cap or battery this way is normal.

Do you realize that when a coil is energized it is analogous to charging a capacitor, and that when the coil de-energizes, that too is analogous to a capacitor dis-charging? Why then do you feel that a de-energizing coil is somehow OU? The inductive kickback of voltage in an inductor is analogous to the current kick you get when shorting out a charged capacitor, yet we don't call that OU.

Utilizing the inductive kickback of the inductor is NOT getting energy from nowhere, it IS the original energy that you put into the coil from the battery in the first place. If we place a flyback diode across this coil it allows only a single (for the most part) half-cycle de-energization into itself as the load. If we remove the diode, again the coil de-energizes into itself but this time over several ring-down cycles. If we re-route the flyback diode into an external load such as a battery or capacitor or bulb, depending on the load impedance there will be as little as one half cycle or several half cycles present to totally de-energize.

Do you fully understand how the inductive kickback behaves and how to manipulate it into giving you mostly current or mostly voltage?

In general what you describe in the above quote are "normal" characteristics. Yes you can drive a load from the flyback diode but all that is happening is you are de-energizing the coil's already-paid-for-energy into that load.

Here is one possible and easy test for you or anyone to try, in order to see how much effective charging really is taking place in the source battery: Insert a fast heavy diode (such as a MUR1520 or equiv.) in series with the battery so that any currents going backwards into the battery will be blocked. Now test to see if the battery voltage falls, rises, or stays the same.

I think it will be the same as without the diode, but you never know. I'm keeping and open mind.

Are we still friends

.99

Since you're keeping an open mind

In case A with the diode across the resistor, my battery resting voltage was 24.40v. I connected it and the timer and started it. When it was NOT in oscillation, the voltage dropped to 24.31 or so. After in self-oscillation, the battery continued to climb to 24.65 volts. That is 0.25 volts increase over resting voltage. How is that not a battery receiving the potential? If what you say is true, the voltage can only go down.

I have done enough battery tests to know what the draw down voltage looks like over time and it never consistently goes up like that ever. It is usually a little up and down with an overall down trend. But steadily increasing form 24.31 up to resting voltage of 24.40 and then steadily up to 24.65. That was over 10 something hours from around 10pm when starting - or midnight? Still a little tired... to 859am when I saw it at 24.65. That is steady up, I didn't see any down... and early this afternoon, the battery voltage started to drop below 24.65 and still didn't hit 24.40.

That voltage on the battery is performing work on the circuit and is above the 24.40. Even if zero heat is made, there is extra potential that wound up in the battery. Any increase in the voltage is increase in potential difference - no matter if it is "fluffy" or rock solid, potential is potential.

In case B, I'm just happy as a clam that you see this as what is supposed to happen.

I think your analogy of the coil charged and capacitor are similar but the capacitor doesn't have the assistance of a magnetic field collapsing the way a coil does.

I'm not saying de-energizing a coil is OU. OU I accept as a term to me over 1.0 COP. I'm saying the inductive spike makes it possible for very simple reasons.

----------------
Based on this belief:
Originally Posted by MileHigh
This circuit burns off 99.9999% of the energy stored in the coil-resistor in the resistive part of the coil-resistor and in the diode.
------------------

If the battery delives A volts at B amps over C time, that is a known amount of potential used up by the battery. If 99.9999% of that is burned off in the resistive part of the inductive resistor, then there should be nothing that is able to be recovered.

Lets say 10 joules were expended by the battery. If 99.9999% is burned off in the resistor for the most part, that is 10 joules of battery potential that went bye bye in the resistor...if it is dissipated, it doesn't matter how much detectible heat there is, 99.9999% is 99.9999% of that in dissipation.

If 10 joules goes into the resistor and it is all supposed to be burned up and I wind up with a capacitor (the B concept of disconnecting diode and putting to cap and putting cap back to circuit) 5 joule worth of charge. That 5 joule can do work with its own losses. That plus our input of 10 joules worth of work in the resistor is MORE work that can be done than we paid for. That is over 1.0 COP and still under 100% efficient.

According to the belief (Milehigh quoted it - not you - the 99.9999% burn off in resistor - but seems to be a strong belief by conventional thinking) - that all of the watts put into the resistor should be burned up. If it is all burned up, I shouldn't be able to recover anything in a cap even in situation B. For the fact that I can means AFTER all the energy is burned up, I still wind up with potetential that can do work. That free potential coming back in the form of a spike + the potential that left the battery is of course MORE than what left the battery.


Implying that I am saying that the spike is automatically ou is taking everything out of context.

It is very simple. If x watt/seconds/minutes/hours/whatever is burned in the resistor that left the battery and it is all burned up, are you saying that somehow some wattage snuck through the resistor undetected by the resistor, never dissipated and that is what I'm winding up with?

The resistive wire allows the spike to generate "current" just like a resistive load on a battery with the inductive kickback charge generates "current" in measurable watts of power.

You say the inductive kickback is the original power we put into the coil. I'm not arguing that point necessarily but if it is the original energy wasn't that original energy supposed to be burned off in the resistor? If the claim is supposed to be held that the energy is burned in the resistor then any extra energy in the kickback must be from elsewhere.

If it is not from elsewhere, then it must be admitted that energy doesn't need to be dissipated in the resistor in order to generate heat as evidenced by the fact that we still wind up with energy in the spike coming back. It is self-evident in many cases that conservation of energy is completely erroneous.

If the inductive kickback de-energizes itself in the coil, it will simply make more heat since that is what restive wire does to a voltage potential spike and this will happen regardless of what direction current is going. That is simply MORE heat generated than can be accounted for from the battery.

You say: "es you can drive a load from the flyback diode but all that is happening is you are de-energizing the coil's already-paid-for-energy into that load."

Yes, I agree. We already paid for the energy but it was supposed to have went bye bye in the resistor since the resistor burned it all off. Therefore, if we get back what we already paid for, then the heat really was generated for a fraction of what we paid for meaning the math is wrong.

Your test with the battery and diode won't work. With the positive terminal (only terminal the spike can enter) having a diode leaving it so the cathode is in contact with the resistor, the diode is of course fully letting voltage leave the battery to the resistor. During the OFF cycle, the diode is still OPEN since the 12v potential on the annode of the diode is more than the voltage potential of the cathode. Then when the spike moves to it, it will simply move through the cathode out the diode and to the battery terminal. When the diode meets its reverse voltage rating, it will then shut off but does not shut off until AFTER the reverse voltage has already moved through it.

I may have posted more documentation and videos on the internet than everyone else combined showing the nature of diodes and how a positive potential can move backwards through the diode from another voltage source.

However, if the diode can sense that voltage is coming to it before the voltage gets there and shuts off first, then it will turn it away, but I have never seen this ability in a diode.

Here is one video showing voltage moving backwards through a string of 15 diodes. YouTube - Positive Jumps to Positive with Common Ground

I'm open to trying this and if it blocks anything (spike is pretty quick), I believe it may only reduce what the battery takes while still letting it takes some and simply making the resistor have more.

I have a small variety of diodes to try this with. 1n914, 1n4001/2/4/7 I think??, 6a100's and a few microwave diodes.

[Aaron]

not semantics

Well, with all due respect Aaron, your credentials are also in question, in how qualified you are to question not only my knowledge and professional opinions but those of thousands more like me.

I do not want to get into a debate about semantics and frivolous definitions that are really irrelevant to the goal of achieving OU. It's also not at the heart of the matters we are discussing here nor of any great importance. If you create an overunity device great. What you call it after that (overunity, >100% efficiency, COP>1) whatever, who cares?

.99

I don't know thousands but I know dozens of highly qualified EE's that simply do not understand and cannot replicate what I have been demonstrating for a long time. I do know a few highly qualified EE's that agree with me as well as some friends with degrees in physics that also agree with me. The majority opinion is only an opinion and just because McDonalds sells more burgers than anyone doesn't mean its a good burger.

And it really isn't about semantics. If conservation of energy, 2nd law and 3rd law are wrong, which they are, it isn't about semantics. Mis-defining the very definition of efficiency leaving out environmental input is not semantics, it is mathematically a whole different ball game and if it is about semantics, then 1 and 2 are equal.

[witsend]

Following quotes from .99. I know they were addressed to Aaron but would be glad if you could consider my comments here.

Do you see that in case A there is no path (not in-circuit) for the spike to go through the battery.
I would argue this on the basis that a second battery - not supplying the circuit - can be charged via the flyback. Provided that the second battery has a higher voltage than the supply battery yet is recharged from that second battery then this point may remain contended.

Why then do you feel that a de-energizing coil is somehow OU? The inductive kickback of voltage in an inductor is analogous to the current kick you get when shorting out a charged capacitor, yet we don't call that OU.
I would argue that the kick from counter electromotive force is a response to inductive laws. The collapsing fields become changing magnetic fields. Changing magnetic fields induce electric fields. This, in turn, induces current flow. An actual regeneration of energy from the system away from the supply.

Utilizing the inductive kickback of the inductor is NOT getting energy from nowhere, it IS the original energy that you put into the coil from the battery in the first place.
If what you claim is correct then one would expect rather less energy to be returned to the system than evident. Your own sim. indicated a small somewhat irrelevant spike level. The experimental evidence contradicts this.

If we remove the diode, again the coil de-energizes into itself but this time over several ring-down cycles. If we re-route the flyback diode into an external load such as a battery or capacitor or bulb, depending on the load impedance there will be as little as one half cycle or several half cycles present to totally de-energize.
Again - this is assuming that the energy is not generated during the 'off cycle' when the fields collapse. I'd be glad if you could comment regarding the result of these changing magnetic fields in terms of inductive laws.

Do you fully understand how the inductive kickback behaves and how to manipulate it into giving you mostly current or mostly voltage?
.99 and with the utmost respect - if there is a path enabled for current to flow from potential difference - it will flow. If there were no path - then your point is valid. Is this what you're claiming?

In general what you describe in the above quote are "normal" characteristics. Yes you can drive a load from the flyback diode but all that is happening is you are de-energizing the coil's already-paid-for-energy into that load.
Again. What about the regeneration of energy from collapsing fields? If it did not generate a second cycle of current flow - then it would be conflicting with inductive laws.

Here is one possible and easy test for you or anyone to try, in order to see how much effective charging really is taking place in the source battery: Insert a fast heavy diode (such as a MUR1520 or equiv.) in series with the battery so that any currents going backwards into the battery will be blocked. Now test to see if the battery voltage falls, rises, or stays the same.
We have done this test. It would be nice for Aaron to show this.