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Overunity Machines Forum



another small breakthrough on our NERD technology.

Started by Rosemary Ainslie, November 08, 2011, 09:15:50 PM

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Rosemary Ainslie

Quote from: TinselKoala on April 05, 2012, 10:33:00 PM
Rosemary, the power dissipated in a resistance is given by P=I^2 R. In this case we have 800 mA flowing through 11.11 ohms at the load. By Ohm's Law, we then have P=(. 8) (. 8) (11.11) = a little over seven Watts, not 50.

And you know nothing of impedance, obviously.

TK - I was referring the output from the battery.  Wattage delivered is most certainly closer to 7 watts.  And you know this perfectly well.  You're just hoping that our readers will overlook this.  And how you can argue that the impedance does not effect the resistance value of the shunt is beyond me.  Clearly I know considerably more than you do.

Rosie Posie

picowatt

Rosemary,

Channel 3, trace D is indeed the voltage applied to the Q1 gate, correct?

This voltage, when positive, indicates to a fair degree, the actual ouput of the function generator, as the loading on the generator is minimal when its output is positive (Q2 is biased off and the gate of Q1 is a very high impedance once Ciss is charged).

Things change a bit when the FG goes negative, as the bias current for Q2 now causes a voltage drop across the function generator's internal 50 ohm resistor.  If we knew the open circuit voltage versus the closed circuit voltage of the function generator's output when its output is negative, we could calculate the bias current for Q2.  Without those figures, all replicators will have to approximate or guess at the bias current for Q2 based on IRFPG50 data sheets and trial and error of the negative going setting.

It is important at the least, that all the scope shots are correct and reflect the actual test described so that FG settings can be set as close as possible to your operating conditions.

PW

 

Rosemary Ainslie

Quote from: TinselKoala on April 05, 2012, 10:36:15 PM
The dates of the photo and the scope shots do not support your argument. And we know already how many times you have argued strongly for a wrong position -- like the "50 watts" above -- , and later had to retract it in public. "Trust me on this".... and then you had to apologise to me AGAIN for being wrong and pig-headed about it.
I don't know what is the case here, I'm just going by Fuzzy's posts... but I certainly find him to be much more reliable than you.

Guys let me make this absolutely clear.  Since I have learned so thoroughly how TK's mission is deny EVERYTHING he can, I have NEVER apologised to TK for anything at all.  Nor will I.  And I cannot dismantle the existing state of the apparatus if I wanted to.  I do not have the competence nor the eyesight.  That photograph was taken by me before it was modified to this Q-array circuit.  Why would I lie about that?

Rosemary

Rosemary Ainslie

Quote from: picowatt on April 05, 2012, 10:39:13 PM
Rosemary,

Channel 3, trace D is indeed the voltage applied to the Q1 gate, correct?

This voltage, when positive, indicates to a fair degree, the actual ouput of the function generator, as the loading on the generator is minimal when its output is positive (Q2 is biased off and the gate of Q1 is a very high impedance once Ciss is charged).

Things change a bit when the FG goes negative, as the bias current for Q2 now causes a voltage drop across the function generator's internal 50 ohm resistor.  If we knew the open circuit voltage versus the closed circuit voltage of the function generator's output when its output is negative, we could calculate the bias current for Q2.  Without those figures, all replicators will have to approximate or guess at the bias current for Q2 based on IRFPG50 data sheets and trial and error of the negative going setting.

It is important at the least, that all the scope shots are correct and reflect the actual test described so that FG settings can be set as close as possible to your operating conditions.

PW

Picowatt - are you saying that you're trying to replicate this?  In which case are you using a function generator?  Then if so, you'll be able to apply that 'offset' as required and you will see what it is that you'll get.  And the beauty of that little LeCroy is that there is no way it can do anything other than reflect the actual waveforms.  Well within it's competence.  And I certainly don't have the competence to alter any of the data that it shows.

If you read the text you'll see that our emphasis is absolutely to allow experts to evaluate the evidence.  And that evidence is easily replicated.  And I'm entirely satisfied we're not talking about vagaries in the function generator.  But again.  That's for you experts to determine.  And I certainly do not include TK in that list.

Regards again
Rosemary


MileHigh

Okay!  It's time for the really big shew!!!

Here is what I figured out:  The circuit started out with one MOSFET, Q1.  Then they added the big aluminum heatsink set with the Q2-Q5 MOSFETs.  When they did that they did the miswiring.

Look at the original diagram in the clip.  Q1's source connects to the shunt, all is seemingly normal.  But then when you add the wiring mistake when they added the Q2-Q5 setup you end up getting the Pegboard of Doom!

In negative oscillation mode you end up bypassing the current sensing resister!  Muhahahaha....

Muhahahahaaaaa! Muh-ha-ha-ha-haaaaaa....