another small breakthrough on our NERD technology.

[poynt99]

Here is the output path for the Instek GFG-8216A, courtesy Kenenth Ho at Instek America. I received this Jan 30th.

The switches are for 20dB attenuation, but are shown in the normal unattenuated position. I don't believe Rosemary was using the attenuation setting (which would be a pulled AMPLITUDE knob).

.99

[TinselKoala]

Thanks for posting that from the INSTEK circuit, .99. Of course, nobody that can read that diagram pretends that Rosemary is correct in her claims.... and Rosemary can't read that diagram or interpret it at all.

What's the green trace Rosemary?

For someone who accuses me of trying to silence you.... you are remaining awfully silent on this issue.... except of course to say that I'm wrong. But you WON'T EXPLAIN why the diagram and the presenter and the actual physical circuit show that the GREEN TRACE is the mosfet common drains... unless of course, as at 0:42 of the video... something ELSE is also connected there which is NOT SHOWN on the circuit diagram.

You have ample opportunity to ANSWER MY DIRECT QUESTIONS about these contradictions and impossibilties but you REFUSE to do so time and time again.

YOU ARE THE ONE IMPOSING A CONE OF SILENCE ABOUT THE ACTUAL PARAMETERS AND BEHAVIOUR OF YOUR INDUCTIVE CLAMP TEST CIRCUIT. Open Source Project? Again.... a redefinition of words by Rosemary Ains-lie.

The GREEN TRACE. What does the GREEN TRACE represent?

[TinselKoala]

Now... eatenbyagrue, you think you aren't sophisticated enough in electronics to be able to evaluate her circuit.
But actually... you are.

Have you ever experienced a loose or corroded positive battery cable in your automobile? Most any driver has, and knows that any loose connection here will interfere with the charging system and also will not be able to carry the current necessary for the starter. Now... the Source and Drain connections of a power MOSFET are kind of like the battery cables in your car. They must be securely attached to whatever they are attached to, and the connections must be of low resistance themselves. The internal "ON" state resistance of the IRFPG50 mosfet is 2.0 Ohms. Have you ever touched and felt the temperature of a loose battery  cable after you've tried to start your car and run the battery down doing so? Feel that heat? That is (P=I squared R) heat, aka Joule Heating. The resistance of the loose connection is high and with high current flow, much power is dissipated at the loose connection in the form of heat... and this power isn't getting to your starter.

Now... take a look at the construction of Rosemary's demonstration circuit. See all those ALLIGATOR CLIP LEADS, clipped to long pieces of ALLTHREAD ROD or long screws to make the parallel connections and further on through the circuit. ALLIGATOR CLIPLEADS on long bits of wire, connected by spring tension only to something that isn't even designed to be an electrical conductor.

Now... take her claim of supplying 26.5 million Joules to something in 100 minutes (6000 seconds.) Since we know (some of us) that a WATT is a JOULE PER SECOND, we can take the 25.6 MILLION JOULE figure and DIVIDE BY the number of seconds to arrive at a WATTAGE figure that must have been applied CONTINUALLY for the entire 6000 seconds in order to transfer that much energy. That wattage figure is over 4200 Watts, continually. Since this power was supplied at 60 or 62 Volts we can determine the AMPERAGE necessary, since P=V x A, so we have A (or I for current) = P/V or.... wait for it..... over 70 AMPS, continuously, for the entire 600 seconds without interruption. If her system is only supplying power during the ON part of a 50 percent duty cycle... then it must supply 140 AMPS during those periods, since that means that the power is actually only on for half the 100 minute timespan.

Now... can those cliplead connections carry SEVENTY AMPERES of current? If the MOSFET's internal ON state resistance is 2.0 Ohms, how much power MUST be dissipated within the mosfet with 70 amps (or even 1/4 of 70 amps in the stack of 4) flowing through it? This is given by the relation P= I squared x R, or, P = (70)(70)(2), or a whopping 9800 Watts heating up that single mosfet, or a quarter of that for each in the stack of 4. That is the power that must be dissipated IN THE MOSFET ALONE at a current of 70 amps through a 2 ohm mosfet. Of course the transistor itself can only carry about 4 amps maximum before it blows up. So to claim the performance that she has claimed and defended over and over requires that you swallow such a linked chain of absurdities that it is entirely IMPOSSIBLE, even if her circuit actually COULD work as claimed. There is no way that the apparatus itself could sustain the power levels required.

The true current levels in Rosemary's apparatus will be limited by the ability of the mosfets to handle the current. The single MOSFET can handle a bit more than 4 amps when COLD and less when at operating temperatures. Multiply by 4 for the behaviour of the stack, so at BEST, her circuit might be able to handle 16 amps for a short time. And her claim equates to a claim of 70 amps for 100 minutes continuously, or 140 amps at a 50 percent duty cycle.

And she says "Do The Math", in her overweening arrogance and willful ignorance.

Note that she can't claim that the power is made "in the load" somehow and dissipated there without affecting the mosfets ... .since she's used the ELECTRICAL PARAMETERS of the rest of her circuit to derive her bogus 25.6 megaJoule figure and her other "evidence" of OU performance.

ETA: attached the IRFPG50 mosfet data sheet.

[TinselKoala]

I said earlier,

The green trace is set at 100 volts per division and is AC coupled in order to display on the screen and not shoot up above it. Since this is the drain signal, it should be HIGH when the mosfets are off... and it should be LOW when they are on... so one should see the same kind of up and down jump, with oscillations on the on portion, as we see in the gate drive signal, I think. Only this jump's magnitude should be near the battery's voltage. I don't know if the scope's AC coupling is flattening this out or not. The scope is telling us that the oscillations have a 44 volt p-p amplitude. No surprise there.

And Rosemary Ains-lie replied,

Yet more of those egregious violations.  LOL.  This trace has absolutely NOTHING to do with the drain signal.  Not even close.  It's a shame that so much presumption is also based on all that pretension.

And further,

Guys, I think this is the reference TK's asking me about.  If so, then I've said all I intend to say about this.  I certainly won't indulge him a free lesson in the art of waveform analysis - albeit much required.  In fact I'm not sure that there's any point in answering any of his posts ever.  I think he's guilty of egregious violations - all over the place.  Puts me in mind of Hitler.  Or Savonarola.  They were both rather self-righteous - and it's tediously inappropriate to a science forum.

Kindest regards,
Rosemary

There... you see... she lies about the trace and then compares me to Hitler and Savonarola for asking her to explain her assertions.

Everything I've said is backed up with external references. Where are the references to support what she says? Take a look... they ALL come from her.

[TinselKoala]

I saw this heating element in the surplus store today so I picked it up to try. Note that it is a Camco 02142, rated 1500 Watts at 120 V. This is a water-heater element, one of a pair that's normally installed in a 40-gallon home water heater. It has a resistance measured on my Simpson of just over 10 Ohms. At 120 V and 10 Ohms, Ohm's law says V=IR, so I=V/R or 120/10.1 = 11.88, call it 12 amps. So the power dissipated in the element at a supply of 120 Volts must be P = I^2R, or (12)(12)(10.1) or... 1454 Watts, close enough... that is, within the accuracy limit of my resistance measurement.

I haven't measured its inductance or tried it with the Ains-lie circuit.... yet.... but I'm sure it will be interesting when I do.