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Author Topic: Graham Gunderson's Energy conference presentation Most impressive and mysterious  (Read 193078 times)

TinselKoala

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TK,

I have no issue with Gunderson measuring his Pin after the H-bridge. I think we can cut him some slack here, as obviously the transformer gadget is the DUT. It just requires a particular input wave form.

If the transformer itself is measured to have a COP>1 that is all that counts, right?

If you had a DUT that required a square wave input, you would measure the output power of your FG, not its AC input, agreed?

Look at it this way. The measurement point is measuring the input to the transformer. But this same point is also the _output_ of the H-bridge. If the H-bridge is putting out "0.000 Watts" but is still able to power something downstream, like a 10 watt brake light bulb.... wouldn't you want to know what the _input power_ to this magical H-bridge is? And where its input power is going?

Maybe it is the _H-bridge_ that is the special DUT and the transformer is just another Red Herring.  (insert tongue in cheek emoticon here.)


TinselKoala

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So why was the information that MH posted on GGs past removed from this thread?-->is this thread now being censored ?.


Brad

What's the matter? If "Open Source" now costs two million dollars (USD, that is 2.68 million AUD), and some incomplete information about it is being sold as DVDs, videos, downloads, etc., then what's surprising about information being censored?

tinman

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« Last Edit: July 24, 2016, 12:45:32 PM by tinman »

poynt99

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It depends. If the FG was only providing a clock input then it might be acceptable to measure only its power output. But in this case the H-bridge is providing the power to the transformer, and as such is a critical part of the device. OK, I'll accept that you can ignore the timing input to the H-bridge, but you cannot ignore the input of power that is being switched by the bridge.
 
I've obviously encountered this issue before, and I've built oscillator-driver circuits that are powered by the overall power supply to the DUT, or even from the bank of batteries that the DUT is supposed to be charging (hence "output")  so that there is no question about where the power is coming from.
 
I say again, if a component or subsystem is necessary for the device to operate, the power supplied to those components or subsystems must be included in the "input power" of the device. For a simple square wave clock signal, obviously what is generating this signal could be a mains powered FG, or it could be replaced with an oscillator-driver that is powered from the DUT's own power supply or even its output power. But in this case the H-bridge is supplying power to the transformer over the link where a measurement of "0.000 Watts" has been quoted, yet the device keeps running, and this link cannot be broken if the device is to keep on running. Hence the H-bridge is a critical component of the DUT and its input power must be included in the COP calculation.

OK, I will allow the power to the _clock_ signal generator for the H-bridge to be ignored, but not the power that is being switched by the H-bridge, this must be included.

As far as I can tell, the special transformer (DUT) itself requires no auxiliary power source or switching input of its own. (yes I know about the active diode network on the output, but I'm not referring to that) Would you agree?

Therefore you can think of the h-bridge as the input power supply, just like the battery is in a variety of other dc-powered devices such as the Joule Thief. As such, the Pin measurement should be captured at the output of the h-bridge, as it is the power source. In other words, Gunderson is measuring at the proper point for input power.

TinselKoala

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Obviously I disagree.

The input to the transformer is indeed switched or cycled, that is, the H-bridge is being clocked at a definite mix of frequencies. It is not simply providing a +/- voltage input to the transformer. As Smudge has determined, the H-bridge waveform is a sinusoid at 75 kHz but is being chopped or switched or gated at 50 kHz so that the "input" to the transformer is one full cycle of a near-perfect sinusoid (from an H-bridge!) followed by a dead time equal to one half-period of the sinusoid. See the attached scopeshot below.

Furthermore, what do you make of this statement:
Quote from: Spokane1
Also, Gram said that by adding a 100 pf Silver-Mica capacitor between the Source and Drain of one of the Back End FETS would completely destroy the OU effect.
If by "back end FETs" he means the H-bridge, then it becomes very clear that the exact specifics of the H-bridge and its drive parameters are critical to the performance of the entire unit. Therefore the power supplied to the H-bridge is part of the power necessary to produce the effect and must be included in the "input power" for the COP calculation. Ditto for the FETs in the "synchronous diode".

poynt99

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Obviously I disagree.

The input to the transformer is indeed switched or cycled, that is, the H-bridge is being clocked at a definite mix of frequencies. It is not simply providing a +/- voltage input to the transformer. As Smudge has determined, the H-bridge waveform is a sinusoid at 75 kHz but is being chopped or switched or gated at 50 kHz so that the "input" to the transformer is one full cycle of a near-perfect sinusoid (from an H-bridge!) followed by a dead time equal to one half-period of the sinusoid. See the attached scopeshot below.
I didn't say the input to the transformer was a +/- voltage. What I did say was that the h-bridge is the power source for the input to the transformer. In terms of the input power measurement, the wave form going into the transformer is irrelevant (the scope setup will handle anything thrown at it). The H-bridge is simply the power source (with a custom wave form), nothing more.

"Back end" would be referring to the synchronous diode circuitry I think.

TinselKoala

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In terms of input power measurement the waveforms are irrelevant because the scope can handle anything thrown at it... but in terms of the device operating as claimed, the waveforms are _very_ relevant. Supply it with DC for example.... the scope measures this power just fine, but the device will not work. It must be supplied with the correct waveforms in order to work, and in order for the "input power" to be "zero" the waveforms must have a specific relationship.

As Smudge has pointed out, the "input power" as seen by the scope is a result of multiplying two waveforms that are almost exactly 90 degrees out of phase. This is why they result in a "zero" average power measurement.

If this is considered as the "input power" to the device, then I am happy to report that I have a handful of devices here in my lab that also count as massively OU, because the "input power" measured at the primary coil is also made up of current and voltage waveforms that are 90 degrees out of phase, hence resulting in a "zero" average input power. Yet the secondary outputs of the air-core transformers will light up bulbs, charge capacitors to high voltage, even run motors.

So by the same logic that seems to be applied here, I have had these "OU" high COP devices sitting around for years already. And they are simple and cheap to build, require no "unobtanium" parts, have been fully described and demonstrated and I am still willing to discuss them.

So wherz my 2 million cheezburgerz?    ;D

lancaIV

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http://www.google.com/patents/US7830065 (whom belongs  the patent rights ? Chava ( Energy ) Llc ?)
The mechanical action of an electrical generator is thereby synthesized without use of moving parts.


ttps://patentscope.wipo.int/search/en/detail.jsf?docId=WO2016070292
VOLTAGE SYNTHESIZER BASIC BUILDING BLOCK CIRCUIT DIAGRAMS

minnie

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  Chet K, you ought to be moderated yourself.
       John.

ramset

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Minnie
Start a forum of your own and hang your name on the door and start calling out known persons by their full name and associate them with things which they had nothing to do with.


It has no place here ,and is unethical to say the least .
And criminal in most countries .

MileHiigh did this at OUR and got the owner in trouble with lawsuit threats.
And to be clear the threats were well on their way to realities...


If you have issues about how Stefan runs his forum perhaps you should ask HIM about these things ?
Or start your own...


Respectfully
Chet K


PS
I see Minnie's request below ,I am traveling and working out of town .
I will Gladly Discuss this upon return .
And to be PERFECTLY CLEAR ,I have never asked for Miles to be banned [and Specifically stated that when I was Speaking with Stefan]. ,I too appreciate his technical contributions .


However If his contributions become a personal liability to Stefan .
That's a no Brainer ...















« Last Edit: July 24, 2016, 04:12:40 PM by ramset »

minnie

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  Chet K,
         PM. me and explain what was wrong.
         I've learned a lot from MH.
               John.

poynt99

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Considering TK's described scenario, a question to all:

With a given black box power source producing a pure sine wave, and when measured the voltage and current are precisely 90º out of phase, could there be real power delivered and measured at the load?

See attached representation of this hypothetical setup.

poynt99

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If this is considered as the "input power" to the device, then I am happy to report that I have a handful of devices here in my lab that also count as massively OU, because the "input power" measured at the primary coil is also made up of current and voltage waveforms that are 90 degrees out of phase, hence resulting in a "zero" average input power. Yet the secondary outputs of the air-core transformers will light up bulbs, charge capacitors to high voltage, even run motors.
I didn't say the input power is to be measured at the primary coil.

Input power must always be measured at the source itself. In the case of Gunderson's H-bridge, that would be right at its output. Just as in the case of Ainslie's battery, where the measurement should have been taken right at the battery terminals, not at the other end of 30 feet of wire where the "phase" relationship becomes grossly skewed.

TinselKoala

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I didn't say the input power is to be measured at the primary coil.

Input power must always be measured at the source itself. In the case of Gunderson's H-bridge, that would be right at its output. Just as in the case of Ainslie's battery, where the measurement should have been taken right at the battery terminals, not at the other end of 30 feet of wire where the "phase" relationship becomes grossly skewed.

Well, then, my measurements of my devices definitely qualify, because there is very little distance between the thick PCB traces that connect the mosfet Drains to the capacitors of the tank circuit and the primary coil, and I am attaching my probes quite near the board or even sometimes directly to the mosfet drains, whereas the coil itself has perhaps 6 cm (x2) of straight wire standing it off of the board. The 0.25 ohm non-inductive Ayrton-Perry-wound current-viewing resistor pair is at the board end of one of the coil legs.

However as you have poynted out, Gunderson's measurements are apparently taken nearer to the transformer rather than directly at the H-bridge output. The actual distance is unknown (at least by me).

TinselKoala

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Considering TK's described scenario, a question to all:

With a given black box power source producing a pure sine wave, and when measured the voltage and current are precisely 90º out of phase, could there be real power delivered and measured at the load?

See attached representation of this hypothetical setup.

Even though the current and voltage are 90 degrees out of phase, the _current_ is very real and in the devices we are discussing can be pretty large, tens or even hundreds of amps. This current produces a very real, changing magnetic field, which in turn can induce a voltage in another magnetically coupled coil, even at some distance. Can this induced secondary voltage then provide real power to a load?