another small breakthrough on our NERD technology.

[Rosemary Ainslie]

Nope, didn't propose that. I propose just taking current FROM one battery to supply the 'connected' part of your circuit and pOUT to ground, or another battery, or anything else BUT the other side of your battery.

I can't follow this at all.  Can you also perhaps draw a circuit?  If there is no connection to the battery supply then that battery can't supply.  If there IS a connection then there has to be some commonality between batteries.  If they're paralleled then all terminals would be connected and we'd have the same conditions that we see.  If they're not - then either one or the other would be supplying - which by default means that one or the other is discharging. 

This same questions persists through all the following quotes PhiChaser.  And I think they're argued in that earlier post.  If not - then let me know.

.   You said it powered itself, not me...
.   Right on.Shazam!!! There is where things get interesting.
.   Okay, I get that.
.   What I see is a function generator connected to an interesting mosfet circuit with some resistors and powered by batteries

Regarding this question

(Just curious: Are your results the same if you use a .25ohm resistor instead of four 1ohm resistors in parallel?)

Not on our circuit.  On others where we generate considerably less voltage.  Then it's within their tolerance levels.  We anticipate that the use of ours will introduce a margin of error which is factored in. But as we're not dealing with marginal evidence, in fact we've got huge energies being dissipated - then that potential margin of error is indeed marginal.  We only use those resistors because we need to accommodate the high current flow.

Now THAT is worthy of study, no doubt about that. HEAT = WORK. Free heat = free work.

I'm not sure that it's 'free'.  What we find is that it's ridiculously cheap.  Certainly far, far cheaper than our paradigms allow for.

Again, kindest
Rosemary

[Rosemary Ainslie]

Actually - back to this question and my reply...

I can't follow this at all.  Can you also perhaps draw a circuit?  If there is no connection to the battery supply then that battery can't supply.  If there IS a connection then there has to be some commonality between batteries.  If they're paralleled then all terminals would be connected and we'd have the same conditions that we see.  If they're not - then either one or the other would be supplying - which by default means that one or the other is discharging. 

I'm taking the trouble to post over Poynty's schematic.  Not sure if it's what you had in mind.  But if it is, then, as mentioned - we've tested this.  Except that we used LED's in place of lights.  What we found was that the one rail stays lit.  The other not.  Poynty asked which one stayed lit.  I can't for the life of me - remember.

[PhiChaser]

Rosemary,
Thanks for taking the time re-word it and put it more plainly for me. It does make more sense when worded that way maybe... I really do try to keep up, I just don't know much about transistors/MOSFETS (although I'm trying to learn, honest I am!).
Not sure your light switch analogy holds since you can still get oscillation from an open circuit. A cheap AC tester will beep near 'hot' wires (current oscillation) whether the light switch is 'on' or 'off'.
But I think I really DO get it now, and please correct me if I'm wrong (again)...
To use your switch analogy you really need two switches correct? So... Let's see if PC really does 'get it' or needs another 'explanation'!

Let's say that Q1 is the breaker (switch) in the panel that opens and closes the circuit from the power company (battery source). When the breaker (Q1) is turned on, power can go (from source through Q1) to the light switch (Q2) which is turned off. As soon as Q2 gets power from Q1 it turns on completing (closing) the circuit to the light bulb (turning it on) AND trips the breaker (turning Q1 off) at the same time. Even though the circuit breaker (Q1) is in the off position the bulb stays lit and you can see the bulb/light switch circuit oscillating but the supply (current) running from source to Q1 is ZERO??? Do I get it now?

PC

[Rosemary Ainslie]

Rosemary,
Let's say that Q1 is the breaker (switch) in the panel that opens and closes the circuit from the power company (battery source). When the breaker (Q1) is turned on, power can go (from source through Q1) to the light switch (Q2) which is turned off. As soon as Q2 gets power from Q1 it turns on completing (closing) the circuit to the light bulb (turning it on) AND trips the breaker (turning Q1 off) at the same time. Even though the circuit breaker (Q1) is in the off position the bulb stays lit and you can see the bulb/light switch circuit oscillating but the supply (current) running from source to Q1 is ZERO??? Do I get it now?

PC

PC?  Is that what you're prefer to be called? Anyway.  Regarding this explanation. It's NEARLY right.

Let's say that Q1 is the breaker (switch) in the panel that opens and closes the circuit from the power company (battery source). When the breaker (Q1) is turned on, power can go (from source through Q1)...

This is right.

BUT this, not so much...

to the light switch (Q2) which is turned off. As soon as Q2 gets power from Q1 it turns on completing (closing) the circuit to the light bulb (turning it on) AND trips the breaker (turning Q1 off) at the same time. Even though the circuit breaker (Q1) is in the off position the bulb stays lit and you can see the bulb/light switch circuit oscillating but the supply (current) running from source to Q1 is ZERO?

Here's an even easier explanation. Current is dynamic.  It always moves.  And it always moves from its source, wherever that is - back to its source - wherever that is.  IF it CAN'T get back to its source - then it simply CAN'T flow.  There would be NO CURRENT.  That's a GIVEN.  No-one would presume to argue.  Various forms of Flux can flow away from its source.  CURRENT CAN'T.  So.  If you use a breaker, or whatever you want - if you OPEN the circuit - you're preventing the current flowing FROM its source BACK to it's source.  Which means that there's simply no current.  Now - we also KNOW that if current is NOT flowing - then there's NO ENERGY BEING DELIVERED.  Which means that a disconnected power supply - is simply NOT able to deliver any energy at all. 

Which is why the energy delivered is measured in voltage - x - the amount of current flow - x - the period of time over which that current flowed.  The voltage or potential difference at the supply source - CAN DO NOTHING - unless it can deliver current.  And it can't deliver current through a circuit that is OPEN - or DISCONNECTED. 

PC.  I am DELIGHTED to explain this.  If you're asking - and you're really bright - then how many others are asking the same thing?  This is the problem with these forums and the beauty of Open Source.  We never know if we're entirely understood.  In any event.  You see this now?  When  Q1 has a negative signal applied to the gate - then that circuit is OPEN.  The voltage potential at the batteries can do NOTHING.  They're passive.  The same applies to Q2.  Unless it's connected to that 'battery negative' - it also can't deliver any current.  Simple really.  And there's no connection to the battery from Q2S to the negative terminal of the battery.  Therefore it can't deliver current.  Which means that regardless, when Q1 has a negative applied signal then there's NO OPEN PATH FOR THE TRANSFER OF ENERGY FROM THOSE BATTERIES.

What's intriguing about those MOSFET transistors - or switches - is that they have what is called a body diode.  This is biased to allow current flow to move in an OPPOSITE direction.  It's dielectrics are designed to take an 'opposing' current to the current that is first applied.  We use that.  Because - here's the thing.  When current flows its also induces an IDENTICAL AMOUNT OF POTENTIAL DIFFERENCE OVER THOSE CIRCUIT COMPONENTS.  But.  Most importantly.  That potential difference is PRECISELY opposed to the supply.  Now.  If the circuit is OPEN and the battery can't deliver a current from a positive potential difference (that battery voltage) - but there's a body diode that's pointing in the right direction - to be loaded with a whole lot of negative voltage - then it CAN find a path - THROUGH THOSE BODY DIODES.  So.  The circuit can take current from one direction.  But it can't take current from another.  The traditional supply source - being the batteries - have been taken out of the equation.  There's very clear evidence of current flow.  So.  It MUST therefore, be coming from the circuit material. 

Are you there yet?  Let me know.  This is really nice.  I'm feeling rather smug that I can explain this.  It's a first that anyone has asked me.  LOL.

Kindest as ever,
Rosemary

ADDED - for clarity
and changed 'delivered' to 'transferred'

[Rosemary Ainslie]

Here's an even easier explanation. Current is dynamic.  It always moves.  And it always moves from its source, wherever that is - back to its source - wherever that is.  IF it CAN'T get back to its source - then it simply CAN'T flow.  There would be NO CURRENT.  That's a GIVEN.  No-one would presume to argue.  Various forms of Flux can flow away from its source.  CURRENT CAN'T.  So.  If you use a breaker, or whatever you want - if you OPEN the circuit - you're preventing the current flowing FROM its source BACK to it's source.  Which means that there's simply no current.  Now - we also KNOW that if current is NOT flowing - then there's NO ENERGY BEING DELIVERED.  Which means that a disconnected power supply - is simply NOT able to deliver any energy at all. 

Just another quick point.  If you think of the flow of current like the flow of a river - then that's good.  Except.  Don't presume that Q2 can act like a dam wall.  It can't.  It can't store the flow of current.  Not even a capacitor can store charge unless it is also first charged by the 'moving current' flow.  Which means that it too - needs to provide some path for that current to get back to the supply source that's charging the cap.

And if you've refreshed the page to read this - then read back over the previous - because I've added some highlights.
Regards
Rosemary