COP 17 Heater | Rosemary Ainslie

[witsend]

The problem that seems to be occurring here is folks are associating "voltage" with "ON", and "no voltage" with "OFF"....

So, monitoring the voltage is confusing folks here. If the current was being monitored (i.e. place a bulb as the load), then it would become clear what is going on.

.99

.99 with respect. Who is confused? Even I, with my limited knowledge, understand this. Is TK giving us a public lecture? Is he asking the public the obvious significance of the voltage? Is he using this new point to throw doubt as to whether we are delivering energy while the switch is off? What is his point? I cannot work it out. EDIT - nor, I may add can anyone I've spoken to. And I've referred many to that video of his.

[Aaron]

magnetic field strength

Aaron,

IF we can bend the "rules" regarding the accepted behaviour of electron flow and all associated (and possibly hidden) fields, then as far as I know you will not find the "method" or exposé in any classical text book.

.99

Thanks and I appreciate the honest answer.

Like Luc and so many others here, I do not have any training in physics or electronics. When I need to calculate something, I use use online calculators or I ask my friends and most of us can be resourceful enough to learn what we need to learn for whatever project we happen to be working on in the moment.

Anyway, there appears to be quite a few examples of the current:voltage relationship being violated like operating certain transistors in their negative zone where current drops while the voltage is increasing.

There are various electromagnetic circuits that produce a magnetic field with no voltage - in a way, they imitate a permanent magnet that has current but no voltage.

Capacitors can be alternately discharged into opposite ends of a coil in a novel way that charges the coil with all benefits of the amperage but there is no voltage - violating ohms law, lenz law, etc...

Anyway, I'm glad to see you're open to the possibility.

[Aaron]

Mosfet question

I have tested in PSpice the 555 timer circuit as per the Quantum article, and my findings are that the 555 timer output is indeed at about a 96% duty cycle when adjusting for about 2.4kHz frequency. The full range of the 2 potentiometers were swept and the lowest duty cycle achieved was about 73%. There was no setting that could achieve 3.7% duty cycle. Obviously the flyback diode was missed in the article, but it has no bearing on the 555 duty cycle output.

It seems quite clear from your posts and from the EIT paper that the desired duty cycle for driving the MOSFET is 3.7%, and I can accept that the circuit in the Quantum article achieves quite the opposite, because as you say you had no involvement with the design or testing of that circuit. No problem there for me.

It appears the fellow that did the 555 timer design for the article was doing so to achieve 3.7% duty cycle at the MOSFET Drain rather than the MOSFET Gate . Errors and mis-communications do happen, so I think it's safe to say that the 555 timer circuit shown in the Quantum article can be dismissed as valid for your experiment, and that those wanting to replicate your experiment do so with either a function generator or some other oscillator capable of achieving the desired 2.4kHz/3.7% duty cycle wave form, and driving the MOSFET.

"It appears the fellow that did the 555 timer design for the article was doing so to achieve 3.7% duty cycle at the MOSFET Drain rather than the MOSFET Gate"

I'm just wondering about something and anyone feel free to jump in.

Q1) The bottom line is that for each pulse/cycle, the battery should only be delivering current for 3.7% of the time. Is this correct? That is the point of having the duty cycle to begin with is that power is delivered for x% of the time per cycle.

Q2) If the Quantum article circuit has a 3.7% duty cycle at the Mosfet DRAIN - doesn't that mean that the Mosfet is allowing current to pass through from the battery at 3.7% of the time per pulse?

Q3) In the below diagram, if the Mosfet drain is conducting 3.7% of the time per pulse will the LED be on 3.7% of the time?

------------------------------------------------------------------



Example application of an N-Channel MOSFET. When the switch is pushed the LED lights up.

The LED only lights when the switch is closed to send power to the gate, which allows voltage and current to move through the mosfet (source/drain). If the mosfet wasn't off when the switch is disconnected, the led would stay lit at a 100% duty cycle making the mosfet obsolete meaning that the semiconductor industry has tricked everyone into thinking it is a switch. lol

[witsend]

Aaron, Loved the example. I'm now going to sleep. It's been a great evening on the thread. But I feel my AGE. I need to walk the dogs. OH HOW I LOVE THIS SUBJECT.

[RAMSET]

An engineers opinion

Aaron with reference to your question and diagram: imagine a voltmeter or oscilloscope connected at the mosfet drain. Or at the top of the resistor where it connects to the positive rail. When the gate signal is ON and the mosfet is conducting, what is the voltage at these points? When the gate signal is OFF and the mosfet is NOT conducting, what is the voltage at these points

When the mosfet is conducting the voltage at those points is LOW, not high. When the voltage at those points is HIGH the mosfet is not conducting, it is off. Joit's trace shows the voltage going HIGH at the mosfet drain for short periods. The transistor is OFF at these times. just hook up a bulb like I have done. Try it!


.
Slow the freqs down, and think about what you are seeing. Carefully.

Aaron the drawing is fine, and yes, as shown the mosfet STATE (on or off) clearly exactly follows the gate drive state: ON is ON, for sure. But that's not the issue: the issue is HOW LONG it's on, and how that on time is measured. If you are looking at the Point A in Ainslie's circuit or the drain of the mosfet in your diagram, the VOLTAGE that the scope is measuring--what it uses to give the duty cycle figure...that voltage is HIGH when the MOSFET ( and the light) IS OFF. Use that exact circuit and put a meter in the exact place Rosemary does. Push that button and tell us what the voltage is on your meter.

[Hoppy]

@ all

I would like to return to Luc's last video demo where he pointed out that the input lamp was very dimly lit. The output lamp on the other hand was brightly lit. It had been strongly suggested by some prior to this test that the input lamp was a good indication of the total power being consumed in the circuit. Assuming this is the case and keeping things simple without EE equations involved, where is the additional energy coming from to light the lamp connected as the load? Logically, I think we have two possibilities, either from the supply battery or direct from the environment.

If we accept that the energy is being tapped from the environment and in some way finding its way to the load without affecting the input lamp, then surely we are in an OU situation big time! If we accept that all energy is being derived from the battery, then logically this energy must be present in the input circuit prior to being transferred to the load. If the energy is being derived from the battery, why is the lamp so dimly lit if it is supposed to be indicating the total power input to the circuit?

I apologise if this sounds rather confusing but I feel it is very important to apply a bit of logic to what we are observing.

Hoppy

[poynt99]

"It appears the fellow that did the 555 timer design for the article was doing so to achieve 3.7% duty cycle at the MOSFET Drain rather than the MOSFET Gate"

I'm just wondering about something and anyone feel free to jump in.

Q1) The bottom line is that for each pulse/cycle, the battery should only be delivering current for 3.7% of the time. Is this correct? That is the point of having the duty cycle to begin with is that power is delivered for x% of the time per cycle.

Q2) If the Quantum article circuit has a 3.7% duty cycle at the Mosfet DRAIN - doesn't that mean that the Mosfet is allowing current to pass through from the battery at 3.7% of the time per pulse?

Q3) In the below diagram, if the Mosfet drain is conducting 3.7% of the time per pulse will the LED be on 3.7% of the time?

------------------------------------------------------------------



Example application of an N-Channel MOSFET. When the switch is pushed the LED lights up.

The LED only lights when the switch is closed to send power to the gate, which allows voltage and current to move through the mosfet (source/drain). If the mosfet wasn't off when the switch is disconnected, the led would stay lit at a 100% duty cycle making the mosfet obsolete meaning that the semiconductor industry has tricked everyone into thinking it is a switch. lol

Aaron, some very good questions ...ones that could finally put an end to the confusion here.

1) YES! "Duty Cycle" in general refers to the percentage of a given time period that something is delivering power to a load. In this case Rosemary has specified that the MOSFET should be driven at the GATE with 3.7% duty cycle wave form such that the MOSFET conducts current from the battery for only 3.7% of the full 2.4kHz cycle.

2) NO. Remember my question above about the ideal switch and what the voltage is between its two terminals when the contacts are closed? If the contacts are OPEN, i.e. the switch is "OFF", then you WILL measure a voltage between its terminals. It is the same with the MOSFET. When the MOSFET is "OFF", you can measure a voltage at its DRAIN. So in this case if we are measuring a voltage for only 3.7% of the time (the rest of the time we measure almost zero volts), what percentage of the total cycle is the MOSFET "OFF"?

If that is still unclear, then imagine removing the MOSFET from the circuit. You now have an open space where the MOSFET was and there is no possible connection between the coil and ground. If you place your DC voltage meter between the coil - terminal (the coil + terminal is connected to the battery +) and the circuit ground, you will measure a DC voltage equal to the supply voltage, correct? Let's say your supply voltage is 12VDC. So you will measure 12VDC on your meter. Now, while your meter is still connected to the same points, take a length of wire and connect the coil - terminal to one end, and the circuit ground to the other end of the wire. If you use a heavy enough wire, your meter will now read a very low voltage, probably very close to zero (and your wire may become very hot, so be careful LOL).

3) YES! If the MOSFET Drain is conducting for 3.7% of the total cycle PERIOD, then yes the LED will be ON also for 3.7% of the time. This is of course assuming that you are using a very low frequency so you can actually see the LED turn discretely ON and OFF.

If you can sort out answer #2 in your mind, you will have mastered how to read a scope with these types of circuits and wave forms, and you'll know what the true duty cycle is.

Hope that was helpful.

.99

[gyula]

Hi Hoppy,

You can consider the input lamp and the rest of the circuit as a voltage divider, fed from the input DC source. The upper member of the voltage divider is the input lamp, the lower member of the divider is all the rest of the circuit.

What constitutes the individual dissipation of these two members?

Obviously the voltage across their own two legs and the common current flowing in them, I say common current because these two members are connected in series.
So the input lamp is hardly lit because the voltage drop across its two legs is small due to the divided DC source voltage, the bigger voltage part gets across the two legs of the rest of the circuit (in which the the output lamp is too).

And because the current is the same in the two members, the voltage drop is very different, this explains the hardly lit input lamp's case.

Simply put it. Study voltage dividers a little to understand more, if needed.

rgds, Gyula

[Aaron]

simple question

Ramset,

That is right but that does not answer the specific questions.

A closed switch is like taking a voltmeter and putting both leads next together on the same wire, there will be no potential difference even though current is flowing and there is voltage moving. That much is common sense. Actually, there will be small milivolt reading because there is a small potential difference but for all practical purposes, there is no voltage.

If the DRAIN has a 3.7% duty cycle, IS IT or IS IT NOT conducting current from a power source for 3.7% of the time per pulse?

It really is a simple yes or no question.

[Hoppy]

Hi Hoppy,

You can consider the input lamp and the rest of the circuit as a voltage divider, fed from the input DC source. The upper member of the voltage divider is the input lamp, the lower member of the divider is all the rest of the circuit.

What constitutes the individual dissipation of these two members?

Obviously the voltage across their own two legs and the common current flowing in them, I say common current because these two members are connected in series.
So the input lamp is hardly lit because the voltage drop across its two legs is small due to the divided DC source voltage, the bigger voltage part gets across the two legs of the rest of the circuit (in which the the output lamp is too).

And because the current is the same in the two members, the voltage drop is very different, this explains the hardly lit input lamp's case.

Simply put it. Study voltage dividers a little to understand more, if needed.

rgds, Gyula

Hi Gyula

If you read my earlier posts you will see that I have explained exactly what you are saying. I'm trying to get others to understand what we are both saying and also what Poynt99 is trying to convey in his report. The alternative is to accept that this circuit is running at rampant OU and I don't think many of us really believe that this is the case with this circuit. This is why I asked Luc to run this lamp demo as it shows an extreme difference between the brightness of the lamps, which some have strongly suggested represents true power level in a circuit.

Hoppy

[gyula]

Hi Hoppy,

Ok, and sorry for not being aware of your full contributions ( I was also blocked out from the energeticforum.com since last Saturday like some other innocent member here (Jetijs, Michael Nunnerley) from Europe..., for unkown reason for me.)

I would be happy to read Peter's experiences with this circuit, for his version of schematics appeared on the very first page of this thread, and since then nobody has addressed his circuit.

Thanks, Gyula

[poynt99]

.99 with respect. Who is confused?

Rosemary,

Where there is one, there are likely many more.

Even I, with my limited knowledge, understand this. Is TK giving us a public lecture? Is he asking the public the obvious significance of the voltage?

No.

Is he using this new point to throw doubt as to whether we are delivering energy while the switch is off? What is his point? I cannot work it out. EDIT - nor, I may add can anyone I've spoken to. And I've referred many to that video of his.

Actually the point he and I and others have been trying to make is unchanged from the beginning. I don't think there is a new point at all in his new video. I'm sure TK will correct me if I'm wrong.

I think what you may be getting from the videos and portions of some posts is that TK is solely attempting to debunk your claims of energy gains from your circuit.

That may be partly true, but as far as I can tell the main point lately has been to establish that the 555 circuit everyone is building in an attempt to replicate your circuit, will produce the wrong drive output from the get-go.

If there is any chance of replicating your results the oscillator driving the MOSFET switch needs to be set to exhibit a 3.7% ON duty cycle, and from what I, TK and others have seen, folks are building the 555 circuit from the Quantum article, which is not producing the drive you have specified.

So again, the point lately has been to let everyone know that they should be using a function generator or a corrected 555 circuit, and I am quite certain that if asked, TK would design one so that folks can get on with testing the circuit as specified by you.

Folks are also asking questions and trying to understand WHY the published circuit won't give the right result's and that is a big positive step forward for everyone here.

If TK is unwilling to design a corrected 555 circuit (he probably already has), then anyone else that wishes to, feel free to step forward. Luc? In fact it wouldn't be a bad idea for all interested to take a crack at it, perhaps in a new thread so as not to throw this one any further off track than it already is. There are several folks around (myself included) that could help out with the design as questions come up. Ramset, you want to head this 555 project up perhaps?

Regards,
.99

[RAMSET]

Reply to Aaron

NO.
When the drain signal is at battery voltage (high) the mosfet is OFF and the load is non-conducting. Clearly. You can wire up that circuit, using a real bulb instead of an LED, and if your voltmeter is sensitive enough to pick up the small voltage drop, you will see it. What looks like a big signal on the scope traces is really a very small fluctuation sitting on top of a large DC offset---the battery voltage. This is why 'AC coupling' must be used to resolve it. If you use DC coupling, if the screen shows a line at battery voltage the fluctuation will be too small to resolve--the voltage doesn't drop very much when the mosfet is ON. But it drops, for sure. If an oscilloscope is relatively dumb, like the Fluke 199, it needs to be TOLD whether you are calling the "high" signal "OFF" or "ON" to give a duty cycle output. There are 2 separate controls in the FLuke that must be set properly: the trace invert function AND the duty cycle definition. Only ONE of the Four possible combinations of these controls is correct in this experiment.

PS

If you want to call the drain signal a 3.7 percent duty cycle, that's OK as long as you know the MOSFET is OFF at that time. Not ON.

PPS
Jolt how is that light bulb test going

[Aaron]

mosfet switch

Aaron, some very good questions ...ones that could finally put an end to the confusion here.

1) YES! "Duty Cycle" in general refers to the percentage of a given time period that something is delivering power to a load. In this case Rosemary has specified that the MOSFET should be driven at the GATE with 3.7% duty cycle wave form such that the MOSFET conducts current from the battery for only 3.7% of the full 2.4kHz cycle.

2) NO. Remember my question above about the ideal switch and what the voltage is between its two terminals when the contacts are closed? If the contacts are OPEN, i.e. the switch is "OFF", then you WILL measure a voltage between its terminals. It is the same with the MOSFET. When the MOSFET is "OFF", you can measure a voltage at its DRAIN. So in this case if we are measuring a voltage for only 3.7% of the time (the rest of the time we measure almost zero volts), what percentage of the total cycle is the MOSFET "OFF"?

If that is still unclear, then imagine removing the MOSFET from the circuit. You now have an open space where the MOSFET was and there is no possible connection between the coil and ground. If you place your DC voltage meter between the coil - terminal (the coil + terminal is connected to the battery +) and the circuit ground, you will measure a DC voltage equal to the supply voltage, correct? Let's say your supply voltage is 12VDC. So you will measure 12VDC on your meter. Now, while your meter is still connected to the same points, take a length of wire and connect the coil - terminal to one end, and the circuit ground to the other end of the wire. If you use a heavy enough wire, your meter will now read a very low voltage, probably very close to zero (and your wire may become very hot, so be careful LOL).

3) YES! If the MOSFET Drain is conducting for 3.7% of the total cycle PERIOD, then yes the LED will be ON also for 3.7% of the time. This is of course assuming that you are using a very low frequency so you can actually see the LED turn discretely ON and OFF.

If you can sort out answer #2 in your mind, you will have mastered how to read a scope with these types of circuits and wave forms, and you'll know what the true duty cycle is.

Hope that was helpful.

.99

Yes, very helpful and I understand it perfectly because any closed switch has no potential difference on it if the ground and positive lead of a voltmeter is at both ends of the switch. I give the example in response to Ramset before I saw this post.

But my question 2 is not in reference to a scope shot? Just asking about an authentic 3.7% conducting time.

[FuzzyTomCat]

Ground - Grounded - Grounding

Hi everyone,

There has been something that I have noticed looking at many older documents and illustrations of the term "Ground" as in Earth or "Terra firma" Terra firma - Wikipedia, the free encyclopedia . As you all know common voltages from countries vary, the UK, Europe, Africa parts of Asia many more not to be named for example use 240 volt ..... but the United States and other North and South America countries use 120/240 volt ..... the big difference is how the "Ground" is connected and referenced.

Where 240 volt is and the only voltage available meaning "NO" 120 volt, a ground conductor is actually Grounded to "Ground", Earth or "Terra firma".

Where 120/240 volt is a ground conductor is bonded to the Neutral wire in a Electrical Service Panel (circuit breaker or fuse) whether in a residence, commercial or industrial application. This would also include "Bonding" of any ground rods, water pipes ( if metal ) and natural gas lines all bonded to the Service neutral conductor. SO A GROUND WIRE AND EVERYTHING CONNECTED TO IT ( neutral wire, pipes ) IS NOT A TRUE GROUND and anything connected to it can be subject to unwanted frequency's or harmonics induced into the grounding system through the neutral conductor.

How To Fix -

A separate "Ground" Earth or "Terra firma" connection must be used generally called a "Isolated Grounding System" using one 8'-0" ground rod a minimum of 6 (six) feet from any other ground rod system or underground water and gas lines. It must be totally isolated using a minimum of a #8 AWG insulated green conductor and must not be connected in any way to you existing grounding system ..... any questions you should contact a qualified person (disclaimer).

Testing equipment causing harmonics can be somewhat isolated during operation using a "UPS" power battery back up supply ( AC to DC to AC ) but that does not solve the ground reference problem.

An electronic circuit to free energy

Not need an oscilloscope to demonstrate on the unit: it is enough to measure currents ex 2: between the earth and the resistance R2, and between the land and the resistance R1 "

Best Regards,
Glen