Yes, through the switching of the function generator
additional energy can be flown into the circuit via the
Gate Source and Gate Drain capacitances.

So it will be wise to just use a negative DC power
supply on the Gates to start and keep the oscillations.

Yes, this will eliminate the giant spikes but still a Kelvin sensing ultra-low inductance shunt should be placed on the battery side of the common ground point to eliminate the currents in the gate-source circuit loop resulting from the oscillations at 1.5MHz (as well as any gate drive spikes, which will go away if the circuit can be made to oscillate with just a DC bias).

Stefan, you must remember that it is not only energy that comes from the sig-gen flowing in the gate-source current loop (and showing up in the shunt even though it is not flowing in the larger circuit loop that includes the batteries). 

Ok.  This is more comforting.  We put a .5 Ohm resistor at the gate to measure the energy.  We established that there was something in excess of 5 watts - but that this was being returned to the functions generator.  One of the guys there said that there was enough capacitance associated with the generator to absorb this energy.  I am not qualified to comment.

Regards.
Rosemary

So Rosemary's team acknowledges that 5W of 1.5MHz RF is circulating in the gate circuit when there is no dynamic transition signal coming from the signal generator.

We see this clearly when the 1,5MHz oscillations are going and there is AC voltage at the gate having to exist across the 14nF of gate capacitance.  That represents significant power that is neither coming from the sig-gen nor flowing through the larger outer circuit loop and battery.  But it appears on the shunt because the shunt is inside both the smaller and larger current loops at present.  This why the shunt should be moved to the battery side of ground and thus taken out of the gate-source current loop.  If the goal is to measure just the battery current flow by itself, this must be done.

If this is done, please note that the polarity reverses on the shunt.  Think about that carefully and you will see that when current flows in the normal direction out of the battery (draining the battery) a shunt on the left (battery) side of the ground point will show a negative voltage below ground while a shunt on the right side (MOSFET source) of the common ground will show a positive voltage above ground.  Remember the two shunts are in series with ground at the center tap.

Regarding that non-driven approach (DC bias without input pulses) Rosemary might find it won't readily begin oscillating.  In my simulation, I used a DC bias and got continuous oscillation but I needed to have one single sharp pulse to get it to start going.  I'm not predicting one way or the other, just noting a minor point that I found using the simulator.  The actual hardware may be different.

Bryan

Have you ever ran your circuit for a longer period of time than would be possible on a battery (or bank of batteries) than without your circuit to the heat resistive element?

Seems like your circuit and your goal as intended to generate heat without depleting the source battery (or bank of batteries) to pre-heat water or other uses could run this circuit once started and tuned indefinitely. If you don't mind me asking, how long have you ran this circuit without interruption?  I know you mentioned pushing the circuit and that the battery was still charged, but can you keep it running endlessly if not pushing it too much and still go past the C20 capacity rating for the battery?

Sorry for all the questions, I know you're busy but if you could answer the above I don't see what all the fuss about measuring is if you can keep the battery powering extra heat indefinitely  

Regards,
Paul

It's a good question Paul - and I've sort of answered it all over the place.  It seems to be a preferred way of proving things because it's so logical.  I have NOT managed to find any loss on our own batteries - used pretty well continuously for 5 or so months.  But the batteries are HUGE.  I'll try and find the post that refers to this and then repost it.  If you're asking this still then there are others with the same question. 

Hang ten - I'll post it hereunder - when I find it?  Not the quickest around this internet thing.  LOL

Kindest regards,
Rosemary

It's a good question Paul - and I've sort of answered it all over the place.  It seems to be a preferred way of proving things because it's so logical.  I have NOT managed to find any loss on our own batteries - used pretty well continuously for 5 or so months.  But the batteries are HUGE.  I'll try and find the post that refers to this and then repost it.  If you're asking this still then there are others with the same question. 

Hang ten - I'll post it hereunder - when I find it?  Not the quickest around this internet thing.  LOL

Kindest regards,
Rosemary

It's a simple question, what is the longest time you've run the circuit off the batteries? Why don't you know this offhand? Seems fairly important. May I suggest trying tinselkoala's requests on the bottom of page 1 as a starting point to answering questions which seem to be going around in circles.

Thanks.

happyfunball.  Another gross misnomer.  LOL Your posts read like the prophets of doom. I've just trailed through a page of them.  Your denial of OU is somewhat brutal.  They're about as inspirational as as a tall glass of tepid tap water.   And you're wrong of course.   Measurements are given all over the place.  It's just when the stack up to contradict what you clearly require - then they're ignored - or considered fallacious.

Just to fill you in here I'll say this again.  When BP (SA)  evaluated these results - some decade ago - they insisted that it would ONLY be proved on batteries.  I was involved in a series of the most boring tests that I have ever been involved with.  All the more arduous as I am - absolutely not - an experimentalist.  I won't here go into the protocols.  But it required close testing of controls against the experiment and run concurrently.    The timing of those batteries was determined like this.  When either one of those supply banks depeleted their PD from 24 v's to 20 volts or when each battery depeleted from 12 to 10  - then the tests were terminated.  That constituted the 'test period'.  What was evidenced is that the controls were entirely 'flat' when the test had barely lost a fraction of a volt.  On the strength of these results PB (SA) allowed us to use their names as accreditors of that early test.  Those early tests are on record as showing a COP>17.  In effect we proved that the test batteries outlasted its watt hour rating against the control.

Now.  When it came to giving a published report on those definitive tests - the PUBLISHER refused to allow ANY REFERENCE TO THE CONTROL.  The publication was a technical journal.  The editor was advised by an electrical engineering academic.  They determined - regardless of my protests - that any reference to battery duration was entirely IRRELEVANT to the argument.  Therefore was I not allowed to reference batteries.  I ASSURE YOU - that as often as you guys state that the battery needs to be tested to it's full duration - just as often will that evidence be ignored.   Batteries vary - one from another.  Some batteries retain their charge and then collapse in moments - to nothing.  Others distribute their charge more gradually.  Others require small currents to match their ratings.  Others don't.  The electrolytes vary - one from another.  So.  If I was to test one then - for conclusive results - I'd need to test them all.

Then.  We have hooked up as many as 7 of those very large batteries in one single test -  apparently discharging nothing.  Now.  The artefact matters.  When this experiment finally gets to our academies, then equivalent and nonequivalent capacities will need to be tested.  In these tests we only used that same bank.  And we could measure absolutely zero loss over a 5 month period.  Exactly how long would it be required to run those tests?  Would it take 2 years to prove it?  10? 6 months?  What?  What exactly would satisfy you?  And how then does one run a control?  Must we SHOW that under normal operating conditions a battery will discharge?  I would have thought that that much could be relied on.  And even then.  I am ready to put money on it that while the most of you engineers require it - our learneds will, to a man, insist that the battery duration is irrelevant.    I wonder if I can state this more plainly.  They're right.  The minute you start evaluating the battery performance - then you are trying to resolve a result in line with specific commodity with a market supply that has varieties that are probably counted in their thousands if not their hundreds of thousands.  That's an awful lot of testing. 

What is intersting is this.  We have an energy returned to the supply that is far greater than the energy delivered from that supply.  Now.  Here's the thing.  If, as is widely assumed by mainstream - that energy is lost to a battery when it discharges current flow - then - by the same token one would expect the energy to be increased in line with a recharge cycle.  In point of fact the batteries voltages varied under test conditions.  The stronger the current discharge the quicker the decline.  But OF INTEREST - is that immediately thereafter it systematically climbs - within minutes - to it's previous high.  Not higher.  Perhaps there are those subsequent tests that may take it higher.  In previous tests we have certainly found a climb to a higher 'start condition'.  But in these tests we did not.  It never exceeded its 'kick off' voltage level. 

I would modestly propose  therefore, that there is a fixed amount of energy that is available from that potential difference - and that no new material - electrons or whatever classical assumption requires - has been introduced to the system.  That's interesting.  That implies that this may be a closed system. It also implies a whole lot of other things.  But for now - just consider that.  That is, if you are not 'happily' out to throw more of that tepid tap water on this research.

Rosemary

Got there.  Strangely this was in answer to Happyfunball.  Seems like he either missed the post or the sense in that post.  I will not again get embroiled in evaluating the battery draw downs.  Anyone feel strongly about this feel free to do your own tests.  Meanwhile - as a reminder - if I had to rely on battery performance then it would eliminate one half of a very strong argument.  My intention is to show that these gains apply to AC or DC supplies.

Thanks for the defense here Paul.  I'm always grateful for this.  It seems that either my personality or the facts of these experiments tends to polarise opinions. 

Kindest regards,
Rosemary

Yes, add some thick wires there and measure with the scope
head directly at the positive terminal
and with a thick diameter wire connected directly
connect the ground line of the scope to the neative pole of the battery.

Hmm, but it could also be, that the green area is bigger all in all.
As the scope says it is a negative nanoVolts ,
the negative area seems to be only very minuscule bigger...



Regards, Stefan.

Remember that there are long wires in between each of the batteries too, so just moving the probes to the +/- terminals of the end batteries will not get rid of the large inductive voltage swings. 

The reality is that the battery voltage as measured by the DMM seems to always exactly or very closely match the mean battery voltage as reported by the scopes.  We all know the battery voltage itself is DC and has only a very small AC ripple due to its internal impedance.  So the only problem is that the wild 100V+ AC voltage swings that Rosemary is feeding into the scope math as the battery voltage plus the current waveform due to the shunt inductance are giving bad numbers when multiplied on an instantaneous basis.

It's really simple to measure the input power.  The battery voltage is a DC quantity with negligible AC ripple.  All we need now is a good low inductance shunt placed properly to sense ONLY the battery current.  Then get the average of that either by using the mean function on the scope or by a simple low-pass filter consisting of a single resistor (10K) and a single capacitor (1uF) on the shunt with a simple DMM measurement.  Calculate the current considering the shunt value. Then multiply.  Power!  The end.

Bryan

Thanks NP.  Always a pleasure to read an upbeat post.  I have a problem with the ac supply source.  For reasons which will entirely exahust everyone's patience - the circuit would probably need to carry two loads - negative to one load and positive to another.  I can't see it working through a supply with a rectified current.

But I've undertaken to test this which, hopefully, will not be too far away.  Otherwise, I agree.  With the caveat that the circuit is configured with two entirely separate loads - then indeed - the numbers should stack up.

Kindest regards,
Rosemary

Thanks NP.  Always a pleasure to read an upbeat post.

 I have a problem with the ac supply source.
...
 - the circuit would probably need to carry two loads - negative to one load and positive to another
...
 

upbeat is good! 

there shouldn't be a problem replacing your battery bank - which is currently connected as a unipolar supply, i believe - with a unipolar output power supply** unit fed from the AC supply

one benefit of this scheme is that you can transfer your experiment without needing to change any of the wiring

(** need to use a simpler, non switched-mode type supply, so that your circuit can return energy to the output ballast capacitor - which must be directly across output, with no 'non-return' diodes)

if you mean that you want to use a split-supply (ie., Pos-Gnd-Neg), then that will still work with the Kill-o-Watt type test scheme outlined above - just replace the unipolar supply with a split-supply unit, and connect to your circuit as you wish)

looking forward to hearing about any 'AC supplied' results!

all the best
np


http://docsfreelunch.blogspot.com
 

Hi Rosemary .Rearrange your batteries into a U shaped line , so that the input and output terminals are physically close together . Now your scope probes will reach .If you lived closer I would get on my bike and come and help you lift them! .Upbeat as ever , Neptune .

LOL.  Yet again Neptune.  I clearly have that dysfunctional IQ that Poynty keeps pointing to.  Of course.  But that too will need to wait until tomorrow.  I will, God willing, and assuming that I survive the night - solemnly undertake to show those waveforms directly across the battery. 

Golly.  Clearly in need of more intellect.    

Regards,
Rosemary