Quote from: picowatt on May 04, 2012, 04:27:18 AM
That's pretty confusing.  You lost me at -12 wrt the battery positive...

PW

Yes, it is confusing, that I freely admit.

If you hook the negative lead -- the reference lead -- of the DMM voltmeter to the battery positive pole and the other lead to the Q2 gate as the circuit is drawn ... what does the DMM voltmeter read? Is the voltmeter wrong?

(ETA: AH... I see the 12 is the problem, not the -. Sorry, that's right, I was thinking just the one battery. The number will be the battery voltage of course, more properly 48, 60 or 72 volts. )

A Rosie Poseo:

QuoteThe CSR is NOT tied to ground.

What a joke.  The lunatics have taken over the asylum.

Rosemary if you only knew how ridiculous a spectacle you are making of yourself.  Electronics is not "your bag" and you don't stand a chance.  Just do the dim bulb test and find out the truth for yourself.

I will try to catch up over the weekend but that comment caught my eye.  You really are suffering from that syndrome.  You are clueless with respect to electronics yet you forcefully reject the truth as told to you by the experts.  It's nuts.

The Bizarro Universe is unfolding as it should.

MileHigh

Quote from: Rosemary Ainslie on May 04, 2012, 01:47:01 AM
I claim NOTHING related to the NERD circuit.  We have presented two papers that need expert evaluation and those papers only POINT AT QUESTIONS. We are asking those questions.  And there is no mere mortal who will EVER know the truth in any absolute sense.

Regards
Rosie Pose

Rosemary,

I'll take this as a retraction then of your application for the OUR Award.

Quote from: picowatt on May 04, 2012, 03:50:10 AM

More importantly, the gate of Q2 is connected directly to the CSR and therefore the voltage at the gate of Q2 can never be any voltage other than that of the non-battery end of the CSR, which is very near ground.  I did not know the obvious needed stating.

Picowatt - I assure you.  The Gate leg of Q2 is connected to the Source leg of Q1.  The Source leg of Q1 is connected to the battery negative rail.  It is ONLY the source leg of Q1 that has the current sensing resistor in series with that negative terminal.  Q2 source leg goes NOWHERE.  It floats.  Q1's source leg is most assuredly negative - as you say - but it is negative ONLY when the Gate at Q2 is positive.  Then.  The Source leg of Q2 is connected to the Gate of Q1.  The source leg of Q2 is absolutely not connected to anything other than the Gate at Q1.  Please look again at the NERD schematic and then look again at my diagram of the connection points.  Because that configuration is a fact.  And that is most certainly how those legs are LITERALLY connected to each other.  And that the Q2 Source leg is floating is also evident on a view of those photos of the circuit board.  And it is and will be easily evidenced when we do our demos.  I do not know if that's how TK has configured his circuit.  It most certainly is how we've configured ours.

For the battery to be able to discharge any current through Q2 during the 'off' period of each switched cycle it would not be able to discharge that positive current flow through its source leg and onto the Gate at Q1.  The blocking diode at Q1 opposes a positive flow of current.  It would effectively leave the circuit open.  The resulting positive current flow - from the battery - assuming that there was any - would need to move through the function generator's terminal to it's probe and then onto the Source leg of Q1 and then only onto the source rail of the battery supply - literally THROUGH the function generator probe and terminal.  Then it would need to resolve as a positive current flow.  Above zero.  And typically - the current from the battery supply needs must pass from the Drain leg of Q2 to the Source leg of Q2 before it can go anywhere at all.  The voltage at this point in the switching cycle is NEGATIVE.  So clearly it is NOT coming from the battery supply.  Quite apart from which that battery supply would also then need to be discharging a negative current flow evident in that negative voltage during that 'off' period of each duty cycle.  Which is impossible.  One could argue that the voltage is weakening during that switching cycle - and that capacitance from the MOSFETs kick in.  Then there would be a reversal of that voltage.  But there is way too much current to be accounted for by capacitance.  So where does that really robust current flow come from?  If that current flow is indeed able to pass through the function generator's probe and terminal - then the strength of that current flow - as mentioned - is far greater than the capacitance that is available at the MOSFETs and far greater than the level of applied voltage that is measured across those switches.

continued/...
And the resulting current flow is assuredly AC because the resulting oscillation is AC.  But in our first test - as described in our paper - at no stage during a period of upwards of 2 minutes between each switched cycle - does the function generator apply anything other than a DC signal to those MOSFETs.  The positive applied directly to the gate of Q2 while the negative is applied directly to the gate of Q1.  I am well aware of the fact that the applied signal from the function generator CAN be AC and therefore sinusoidal.  But the applied setting during that first test of ours it is only AC if you take in each switch change at every 2.8 minutes or thereby.  Between switches it's effectively a DC current.   

Quote from: picowatt on May 04, 2012, 03:50:10 AMI seem to be correct in my assumption from that read of the blog link and subsequent responses that RA does not understand that an FG has a bipolar supply and that she believes the FG is somehow applying a positive voltage directly to the gate of Q2.  Apparently, based on her own statements, she cannot even read her own schematic.

You are liberal with your criticisms of my lack of comprehension.  And then you seem to take offense when I point out that it is YOUR OWN erroneous assumptions related to both the circuit and my understanding.  How can I not know - after all these years - that a function generator is capable of putting out a variety of waveforms?  I would need to have less than a functional intelligence.  It is NOT a difficult thing to understand. It is ABSURDLY simple.  I know very well what signals a function generator is capable of generating.  I've said this before. This kind of reckless reliance on insulting my intelligence is positively slanderous.  And as ever - it is patently evident that you have NOT yet seen that the Source leg of Q2 is floating and is absolutely NOT connected to the source rail of the battery supply.  So who then is the 'fool' that you're so anxious to 'spin'? 

Quote from: picowatt on May 04, 2012, 03:50:10 AMHow can the FG signal ground or the Q2 gate EVER be any voltage other than that of the CSR?  It can't. Again I am flabbergasted.

And so am I. And I'm also looking forward to showing you this 'lack of connection' between the Q2 source and that Q1 source - that you're depending on.  Because I see now that it has entirely eluded you.

Rosemary