What I find quite curious, is if you performed the battery voltage test as I requested, why you did not post an pictures or video of the result AND the test setup?

As a first step, I need to see precisely where you placed the scope probe to measure the battery voltage. I'll draw something up.

OK Rosemary,

Please identify using two of the six "P" numbers on the diagram, where the scope probe and probe reference were placed for your battery voltage measurement.

Hello PhiChaser,

You've based your argument on the fact that a MOSFET transistor is a capacitor because it shares some material properties with a standard capacitor.  Which is interesting.  However, the only thing that it does not share is the ability to store charge.  It is designed to deliver charge.  If it had the ability to store charge then indeed your argument would hold.  But it is physically IMPOSSIBLE to disconnect a MOSFET and then apply it to the same or to an alternate circuit and expect it to then discharge what was first delivered.  That - essentially - would be required for one to argue its definition a capacitor.  Quite apart from which that capacity in that transistor then needs must be ENORMOUS. 

I see where you're going.  You're proposing that during the time that the battery is applying a positive charge to the gate of Q2 - then it's in the process of storing that charge.  Then the switch at the Gate of Q1 changes to become positive.  And simultaneously the switch at the Gate of Q2 changes to become negative.  And somehow, during this transition then all that energy STORED in Q2 is then discharged. On the face of it - it could perhaps be plausible.  Assuming always that it can even find a path through it's source leg Q2S to the Drain rail or battery positive - as Poynty refers to it.  Again.  Bear in mind that IF this energy is being returned it is still showing a voltage that is greater than zero.  And also.  Bear in mind that in the process of returning this energy it is also discharging about 72 000 amps of stored energy.  That's to account for the battery voltage reducing from 12 volts to 0.5.  Which not only begs a storage capacity somewhat larger than a standard capacitor.  But for some reason - rather confusingly - this returning energy BACK to the Drain rail of the battery - that positive terminal - somehow manages to then REDUCE that battery voltage from 12 volts to 0.5 volts,  Under normal circumstances IF energy is returned - one would expect it to recharge that battery.  And one would also expect the voltage to then be less than zero.

But that slew of improbable events is actually not even relevant.  Because, in point of fact, the signal at the Gate of Q2 does NOT change from a positive to a negative during this oscillation.  It stays negative.  For the duration. So.  I'm not sure that this proposal can be resolved by proposing that the MOSFET is acting as a capacitor.  Unless I've missed something.

Regarding your questions. The drawing of the paralleled MOSFET's is doable.  Poynty has some in his own schematics.  And I am not about to alter that artifact - nor am I interested in doing other experiments.  But there's nothing to prevent you from doing this.  And thank you for this proposal.  It is SO much more palatable than Poynty's rather repetitive dialgue while he tries to duck the issue.  At LEAST it's arguing the case.  Most appreciative PhiChaser.

Kindest regards,
Rosemary

Rosemary,

So the charge isn't held in the MOSFETs, I can see that so scratch that off the list of possibilities. I remember reading somewhere that voltage spikes can go reversed bias through a diode. Has this possiblility been considered?
Oscillation means waves meeting waves in some sort of harmony (?) to me (just because I'm a musician maybe?)... 4/4 is a pretty common time signature... (Well, there ARE 4 MOSFETs on one side?)
Have you tried to get the same (similar) results using just 2 MOSFETs for Q2 and adjusting the frequency from your generator to double (or half?)?
And lastly (heh heh, for now) do you think the type of MOS you're using is responsible for your (anomalous) results Rosemary?
I swear I see a mobius loop between Q1 and Q2...

PC

OK Rosemary,

Please identify using two of the six "P" numbers on the diagram, where the scope probe and probe reference were placed for your battery voltage measurement.

P2 AND P4 WITH THE USE OF ONLY ONE BATTERY.  I trust that answers your question Poynty.  But having said that I need to add the caveat that we do NOT use that circuit you've drawn.  IT'S SIMPLY WRONG.

Regards,
Rosemary

P2 AND P4 WITH THE USE OF ONLY ONE BATTERY.

Not sure what you mean. You had only one battery powering the circuit?

But having said that I need to add the caveat that we do NOT use that circuit you've drawn.  IT'S SIMPLY WRONG.

That kind of comment does not move this argument forward, does it? What is wrong exactly?

Poynty.  If I had the time or the interest I'd be able to refer you to as least 4 times in the past 24 hours  that I've told you this.  In fact we have NEVER powered the circuit with one battery.  For some reason we need 2 - on our element resistor - to generate that oscillation.  BUT WE HAVE used JUST 1 BATTERY on other circuits.  AND WE GET PRECISELY THE SAME WAVEFORM WITH PRECISELY THE SAME VOLTAGE SWING ACROSS THE BATTERY AND WITH NO MORE THAN 18 inches or thereby of CIRCUIT WIRING.  That oscillation is most certainly NOT dependent on the length of wire that is used connecting the batteries in series or even connecting the batteries to the circuit apparatus.

What is wrong with that circuit schematic is that the arrangements of the transistors is not our own arrangement.  I only mention this lest in the distant future you then state that I had ENDORSED that schematic. 

What arrangement of the MOSFETs are you referring to? I've NOT SHOWN ANY arrangement...ON PURPOSE! It's not required. It's called a "block diagram" FYI, and that is why it is labeled "Q1-Q5". You don't think I know the circuit? Do you really think it is necessary to show the MOSFET arrangement in order to further this part of the discussion?

This is why it is near impossible to have a meaningful discussion with you, because you take everything said and either misunderstand it, misinterpret it, or read into it something that is of your own imagination.

And on the battery measurement, sorry, I just don't buy it. Until I see what you've actually done, I can't be assured you did it correctly.

I don't think this discussion will end in any agreement or conclusions, so until and if I ever build this thing, only then will I be able to convince you of your measurement errors. And even then, I'm quite certain you won't accept my argument.

Until then, I bid you adieu. Life is too short for this, and I've got plenty of productive projects on the go that require my deserved attention.