another small breakthrough on our NERD technology.

[Rosemary Ainslie]

No need to apologize Rosemary.

My only hope is that if you wish to have a productive discussion about your circuit, we can "talk the same language" and agree in terms of how the circuit is connected, what the various points in the circuit are "named", and what the polarity is across the FG.

Until we do, it would be extremely difficult for me to answer any of YOUR questions, because I would not understand exactly what you are asking.

So, can we agree to properly use the nomenclature as denoted on YOUR schematic?

Good.  Thanks for that.  Now.  We've agreed that the terms related to source and drain are this.  If I refer to Source Rail or Drain Rail then I am referring to the circuit connection to the negative and positive respectively.  By the same token if I refer to the source or drain on one of either legs of the transistors - then it is referred to as Q(1 or 2)source or drain.

I may need to go back to one of my posts.  I'll edit it in the light of these terms.  And then I'll repost. 

BRB
Rosie Pose

[poynt99]

If I refer to Source Rail or Drain Rail then I am referring to the circuit connection to the negative and positive respectively.

I can work with "Drain Rail" if I have to (I would prefer Q1-D or Q2-D), but it would be impossible to work with "Source Rail" because the two MOSFET Sources are not connected together. Again, the preference in order to avoid confusion, would be to refer to the MOSFET leg directly, such as Q1-S, which means the Source pin of Q1.

[Rosemary Ainslie]

Poynty - this is just a repeat of my previous post. I've highlighted those places that needed qualification - but otherwise it's exactly the same post.  Would you care to comment here?

Kindest regards,
Rosemary

Here's that argument
Actually guys - this may be a better way to explain the anomalies and it may also get to the heart of Bubba's objection.  The oscilloscope probes are placed directly across the batteries that ground is at the source rail and the probe is at the drain rail.  Which is standard convention.  Then. During the period when the oscillation is greater than zero - in other words - when the battery is DISCHARGING - then it's voltage falls.  And it SERIOUSLY falls.  It goes from + 12 volts to + 0.5.  Given a  supply source of 6 batteries for example, then it goes from + 72 volts to + 3 volts.  At which point the oscillation reaches its peak positive voltage.  And this voltage increase is during the period when the applied signal at the gate of Q1, is negative.  WE KNOW that this FAR EXCEEDS THE BATTERY RATING.  In order for that battery to drop its voltage from + 12V to + 0.5V then it must have discharged A SERIOUS AMOUNT OF CURRENT.  Effectively it would have had to discharge virtually it's ENTIRE potential as this relates to its watt hour rating.  We EXPECT the battery voltage to fall during the discharge cycle.  But we CERTAINLY DO NOT expect it to fall to such a ridiculous level in such a small fraction of a moment AND SO REPEATEDLY - WITH EACH OSCILLATION.

Now.  If we take in the amount of energy that it has discharged during this moment - bearing in mind that it has virtually discharged ALL its potential - in a single fraction of a second.  And then let's assume that we have your average - say 20 watt hour battery.  For it to discharge it's entire potential then that means that in that small fraction of second -  during this 'discharge' phase of the oscillation it would have to deliver a current measured at 20 amps x 60 seconds x 60 minutes giving a total potential energy delivery capacity - given in AMPS - of 72 000 AMPS.  IN A MOMENT?  That's hardly likely.  And what then must that battery discharge if it's rating is even more than 60 watt hours?  As are ours?  And we use banks of them - up to and including 6 - at any one time.  DO THE MATH.  It beggars belief.  In fact it's positively ABSURD to even try and argue this.

NOW.  You'll recall that Poynty went to some considerable lengths to explain that the battery voltage DID NOT discharge that much voltage.  Effectively he was saying 'IGNORE THE FACT THAT THE BATTERY VOLTAGE ALSO MEASURES THAT RATHER EXTREME VOLTAGE COLLAPSE'. JUST ASSUME THAT IT STAYS AT ITS AVERAGE 12 VOLTS.  Well.  It's CRITICAL - that he asks you all to co-operate on this.  And in a way he's right.  There is NO WAY that the battery can discharge that much energy. SO?  What gives?  Our oscilloscope measures that battery voltage collapse.  His own simulation software measures it.  Yet the actual amount of current that is being DISCHARGED at that moment is PATENTLY - NOT IN SYNCH. 

But science is science.  And if we're going to ignore measurements - then we're on a hiding to nowhere.  So.  How to explain it?  How does that voltage at the battery DROP to +0.5V from +12.0V?  Very obviously the only way that we can COMPUTE a voltage that corresponds to that voltage measured across the battery - is by ASSUMING that there is some voltage at the probe of that oscilloscope -  that OPPOSES the voltage measured across the battery supply.  Therefore, for example, IF that probe at the drain rail - was reading a voltage of +12 V from the battery  - and SIMULTANEOUSLY it was reading a negative or -11.5 volts from a voltage potential measured on the 'other side' of that probe - STILL ON THE DRAIN RAIL - then it would compute the available potential difference on that rail of +0.5V.  Therefore, the only REASONABLE explanation is to assume that while the battery was discharging its energy, then simultaneously it was transposing an opposite potential difference over the circuit material.  WHICH IS REASONABLE.  Because, essentially, this conforms to the measured waveforms. And it most certainly conforms to the laws of induction.

OR DOES IT?  If, under standard applications, I apply a load in series with a battery supply - then I can safely predict that the battery voltage will still apply that opposing potential difference - that opposite voltage across the load.  Over time.  In fact over the duration.  It most certainly will NOT reduce its own measured voltage other than in line with its capacity related to its watt hour rating.  It will NOT drop to that 0.5V level EVER.  Not even under fully discharged conditions.  So?  Again.  WHAT GIVES?  Clearly something else is coming into the equation.  Because here, during this phase of the oscillation, during the period when the current is apparently flowing from the battery - then the battery voltage LITERALLY drops to something that FAR exceeds it's limit to discharge anything at all.  And we can discount measurement errors because we're ASSURED - actually WE'RE GUARANTEED - that those oscilloscopes are MEASURING CORRECTLY.  Well within their capabilities. 

SO.  BACK TO THE QUESTION?  WHAT GIVES?  We know that the probe from the oscilloscope is placed ACROSS the battery supply.  BUT.  By the same token it is ALSO placed across the LOAD and across the switches.  It's at the Drain rail.  And its ground is on the negative or Source rail.  And we've got all those complicated switches and inductive load resistors between IT and its ground.   Could it be that the probe is NOT ABLE to read the battery voltage UNLESS IT'S DISCHARGING?  UNLESS it's CONNECTED to the circuit?  Unless the switch is CLOSED.  IF there's a NEGATIVE signal applied to the GATE by the signal generator then it effectively becomes DISCONNECTED?  In which case?  Would it, that oscilloscope, not then pick up the reading of that potential difference that IS available and connected in series - in that circuit?  IF so.  Then it would be giving the value of the voltage potential that is still applicable to that circuit.  It may not be able to read the voltage potential at the battery because the battery is DISCONNECTED.  It would, however, be able to read the DYNAMIC voltage that is available across those circuit components that are STILL CONNECTED to the circuit?  In which case?  We now have a COMPLETE explanation for that voltage reading during that period of the cycle when the voltage apparently RAMPS UP.  What it is actually recording is the measure of a voltage in the process of DISCHARGING its potential difference from those circuit components.  Which ONLY makes sense IF that material has now become an energy supply source. 

It is this that is argued in the second part of that 2 part paper - as I keep reminding you.  Sorry this took so long.  It needs all those words to explain this.  The worst of it is that there's more to come.   

[Rosemary Ainslie]

I can work with "Drain Rail" if I have to (I would prefer Q1-D or Q2-D), but it would be impossible to work with "Source Rail" because the two MOSFET Sources are not connected together. Again, the preference in order to avoid confusion, would be to refer to the MOSFET leg directly, such as Q1-S, which means the Source pin of Q1.

NO POYNTY.  That's not the meaning of the term RAIL.  I was given to understand that any reference to RAIL refers to the wire that is connected directly to the supply source being the positive, DRAIN RAIL or the negative SOURCE RAIL of the supply's terminals.  In every context the word source must be referenced either as it relates to the transistor legs - in which case it is Q(1 or 2)S or as Source rail or Drain rail.  That way there are no ambiguities and it allows full circuit reference as required.

Surely?  In any event that's how I've referenced it in the above post.  Take a look in there and see if you can or can't understand it.

Regards,
Rosie

[poynt99]

OK, I see now the Drain rail is actually the battery + (B+). "Drain Rail" is confusing when actually referring to the battery + (B+), but I can work with it if necessary.

"Source Rail" is actually the FG-, agreed? At any rate, I can work with that if it makes it easier for you.