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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 711659 times)

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #930 on: April 27, 2011, 03:47:40 PM »
Which all brings me back to this answer to nul-points. 

The positive peak excursion across the CSR resistor, is a result of the sharply cut off Drain voltage of Q2 when it turns ON. Prior to this when Q2 is OFF, the CSR's inductance is energized by the up-swinging Drain voltage (via the 'Q1' and Q2 capacitance), and so this sudden cut-off causes the CSR's inductance to reverse it's voltage across itself (Lenz's law),...
- when I would have thought that the collapsing fields across the CSR result from Faraday's Inductive Laws.

And then

and this positive peak voltage is then "limited" by the forward-biased diode,...
when far from limiting this it would enable this.

And then

which is why the positive peaks appear "squashed" and widened. This latter part happens when Q2 is ON, so the diode is forward-biased into a 0-volt potential, which is why it can limit the positive peak, even though it may only be a couple volts in amplitude.

What 'forward-biased' diode is this?  The zener body diode?  Or the diode that you've now put across the cap to replace the Q1 MOSFET?  Because if you left the Q1 MOSFET as was earlier shown - then you actually don't get any 'squashing' or anything other than that clean sinewave shape. 

So.  Again.  I'm not sure that it's actually desirable to get rid of Q1 MOSFET.  I can't see how you'll be able to vary the input voltage from the source to up the power dissipated at the load and on the circuit generally.  So I'm not sure that it's a good idea to get rid of it.  And I'm not sure that it's correct to claim that any of these effects are due to anything short of Inductive Laws.  Unless of course Lenz Law allows for a complete breach of thermodynamic constraints.  And while your simulators can most certainly mimic our results - I'm also not sure that the simulations will be able to compensate for the input power at higher levels.  And I'm absolutely not sure what you actually think is going on here.  Perhaps you can advise us.

Rosemary

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #931 on: April 27, 2011, 04:06:18 PM »
Here's the kind of example that I'm trying to point to.  At this level we're dissipating in excess of 80 watts.  Just check out the math trace.   And I absolutely do not know how we could get here without the Q1.  I seriously propose that this is required.  But I'm open to correction.  Perhaps there's a way of doing it without that MOSFET.

Kindest regards,
Rosemary

Added

Sorry.  This is not the one I meant to post.  Here is only upwards of 40 watts.  But I'm hoping it's pointing to the issue at hand.

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #932 on: April 27, 2011, 04:25:26 PM »
Here's a better example where we're into wattage dissipation at something close to 80 watts.

Added.  I've been going through the hundreds of samples that we've got.  But the real 'runaway' numbers just happen too quickly to get a down load of the data.  I could however get a screen shot and shall try and do so this week.  The sample I've shown here is at a high frequency.  We can get this at slower frequencies - but I need to be nimble.  It often ignores the setting and then just keeps going north.  And I'm not that keen on spoiling any more of those FETs.


Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #933 on: April 28, 2011, 03:17:52 AM »
All,

With reference to Q1's function in the circuit, I hope this helps:

How a diode can effectively convert a wave form with a mean value of zero, to one with a mean value that is non-zero. This is one of the two functions Q1 is performing in the circuit via it's internal body diode.

(The other function is to provide additional D-G feedback capacitance in parallel with Q2's own internal Cg-d).

This is a simple example that illustrates one of the effects Q1 has on the circuit. This ties in with my response to np's question about the positive portion of the "CSR" voltage. This hopefully gives you some insight into what can cause the MEAN negative voltage across the "CSR" resistor.

Let me know if anything does not seem clear and easy to understand.

.99

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #934 on: April 28, 2011, 03:35:16 AM »
I encourage anyone following along and wanting to get the most out of the information being presented, they save and print out this diagram, which is the simplified single-MOSFET equivalent of the original 5-MOSFET version.

Please pay careful attention to the labeling and the connection/location of the "CSR" resistor (Lcsr1 and Rcsr1), and what used to be the function generator, and is now replaced with the fixed DC voltage source (Vbat2).

There are still a couple changes that can be made to simplify it even more, and I am still hoping one of the readers here might spot these.  :)

.99

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #935 on: April 28, 2011, 04:54:40 AM »
Which all brings me back to this answer to nul-points. 
 - when I would have thought that the collapsing fields across the CSR result from Faraday's Inductive Laws.
Faraday's law is of course responsible for the voltage across any inductor as a result of the current through it changing, but Lenz's law determines the polarity of that voltage, and it is the polarity I was trying to emphasize.

Quote
What 'forward-biased' diode is this?  The zener body diode?  Or the diode that you've now put across the cap to replace the Q1 MOSFET?
The MOSFET body diode (if the Q1 MOSFET is used), and the diode I've replaced it with.

Quote
Because if you left the Q1 MOSFET as was earlier shown - then you actually don't get any 'squashing' or anything other than that clean sinewave shape. 
I am of course referring to the "CSR" wave form "shape" we are quite familiar with now. I would encourage you to view my recent post illustrating how the said diode limits the positive voltage excursion (in both circuits). In this sense, it "squashes" the otherwise narrower, higher amplitude, positive peak.

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #936 on: April 28, 2011, 06:23:28 AM »
I encourage anyone following along and wanting to get the most out of the information being presented, they save and print out this diagram, which is the simplified single-MOSFET equivalent of the original 5-MOSFET version.

Please pay careful attention to the labeling and the connection/location of the "CSR" resistor (Lcsr1 and Rcsr1), and what used to be the function generator, and is now replaced with the fixed DC voltage source (Vbat2).

There are still a couple changes that can be made to simplify it even more, and I am still hoping one of the readers here might spot these.  :)

.99

My dear Poynty.  You're trying so hard.  Golly.  You've shown us a schematic with two shunts - one at the negative and one at the positive rail of the battery.  Could I impose on you to show us those voltages?  That may go some way to answering MileHigh's obsessive concern that somehow the functions generator is 'fudging' the results.  Because there is absolutely NO DIFFERENCE where you put that CSR.

Regards,
Rosemary

I'll get back to your other points in due course.  Lots of errors there Poynty Point.

ADDED
BTW - if there is a variation here then it is not representative of what is happening on our circuit.


and another point.  This is the schematic I'm referring to.
« Reply #932 on: Today at 03:35:16 AM »
« Last Edit: April 28, 2011, 06:54:48 AM by Rosemary Ainslie »

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #937 on: April 28, 2011, 06:42:22 AM »
Well Poynty.  It seems that you are entirely ignoring most of my questions and those that you do acknowledge are answered incorrectly.  Pity.

So.  I'll start with the one that I thought was clearest.  Would you please oblige us by explaining how you plan to adjust the power from the supply without the use of what you refer to as Q1 - being the MOSFET that is positioned as traditionally applied?  While I'd be delighted to see variations of this circuit - as I'm sure we all would - I earnestly recommend that they're still able to perform some reasonable level of work. 

Then you say that Lenz Law determines the 'polarity"?  Since when did that happen?  Lenz law has nothing whatsoever to do with polarity.  That polarity was determined by Faraday and defined in inductive laws.  What Lenz law proposes is that the strength of those collapsing fields depends on the level of retained charge in those inductive components.  In fact his arguments go some way towards establishing those dreaded Laws that were imposed on Mr Faraday's discoveries.  Here's a Wiki link to this.

http://en.wikipedia.org/wiki/Lenz%27s_law

So.  With respect.  I rather suspect that your reference to this was to endorse those very laws that your results appear to defy.

Kindest regards,
Rosemary


Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #938 on: April 28, 2011, 07:30:08 AM »
All,

With reference to Q1's function in the circuit, I hope this helps:

How a diode can effectively convert a wave form with a mean value of zero, to one with a mean value that is non-zero. This is one of the two functions Q1 is performing in the circuit via it's internal body diode.

This is a little confusing.  You show a CSR between two positive terminals of two separate supplies.  For any kind of current to flow you presumably have those batteries with different voltages.  So.  Let's assume that you can get a current to flow from the one to the other - then?  You apply a switch? is that it?  Or you change that signal to a negative?  I would have thought that if you can induce that tiny bit of inductance in the CSR to 'reverse' it's voltage (Faraday's Law) then it is unlikely to return to its source.  But.  If you used something with a little more inductance in it then you would, theoretically - not only have generated a sine wave but you would also be able to re-energise that supply.

So.  Far from 'converting a wave form with a mean value of zero' it's waveform would then first be greater than zero  and then less than zero and theoretically a pure sine wave would be the consequence.  This the more so as the voltage drop across the diode would apply in both directions. 

I'm probably missing your point.  But your recipe or explanation of 'How a diode can effectively convert a wave form with a mean value of zero, to one with a mean value that is non-zero' is posibly not the best example.  And I certainly do NOT  see that this is what Q1 is doing.  On the contrary.  Q1 is enabling the current from the battery supply.  For some reason there is a small period where this conduction is prevented.  That's where my own interest lies.  Personally I think it is to do with the amplitude of the setting.  But our amplitude is always 'full on' so that's still not the whole of it.  Hopefully the Tektronix will get to the route cause better.  At least we can experiment with shorter 'on' periods.

So.  And again.  This is absolutely NOT one of the two functions that Q1 is performing in the circuit via its internal body diode.  If only it were that simple.

(The other function is to provide additional D-G feedback capacitance in parallel with Q2's own internal Cg-d).

You really need to explain your terms.  What is a D-G feedback capacitance?  And why is this in parenthesis?  Is it your fall back argument?

This is a simple example that illustrates one of the effects Q1 has on the circuit. This ties in with my response to np's question about the positive portion of the "CSR" voltage. This hopefully gives you some insight into what can cause the MEAN negative voltage across the "CSR" resistor.

Let me know if anything does not seem clear and easy to understand.

So again Poynty Point.  I'm sorry but you're dealing with a gross amateur here.  If you can't be bothered to answer these points I won't blame you.  I'm sure that everyone reading here fully understands you.  But I don't.  And I certainly don't agree with you.  While I am delighted to see simplification of anything at all - I absolutely cannot follow this reasoning.  The more so as Q1 which you're so ANXIOUS to dispose of - is CRITICALLY required to conduct at higher energies.

Kindest regards,
Rosemary 

edited spelling.
« Last Edit: April 28, 2011, 11:54:16 AM by Rosemary Ainslie »

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #939 on: April 28, 2011, 07:57:44 AM »
I encourage anyone following along and wanting to get the most out of the information being presented, they save and print out this diagram, which is the simplified single-MOSFET equivalent of the original 5-MOSFET version.

Please pay careful attention to the labeling and the connection/location of the "CSR" resistor (Lcsr1 and Rcsr1), and what used to be the function generator, and is now replaced with the fixed DC voltage source (Vbat2).

Which makes me rather alarmed Poynty Point - that you are now recommending this schematic to 'replace' our own.  Modify our own - by all means.  But I'm afraid if you go this route there is likely NO advantage whatsoever in this circuit.

I can't reference the diagram without seeing it but will do so when I've got that 'print out' that Poynty's recommending.  But there are so many obvious glitches and I'd be rather concerned that anyone go to too much trouble to build this proposed circuit when the schematic will most certainly not do what is required.  Unless I'm wrong.  I do hope so.  Doubly troubling - lol - when those 'pencilled in' comments all need such close review.  I'll deal with it when I've managed to get a copy of it for reference. 

 ;D

Kindest regards,
Rosemary

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #940 on: April 28, 2011, 02:55:12 PM »
As I previously mentioned, Faraday's law states that a voltage develops across an inductor any time the current (flux) through it is changing, and it is Lenz's incorporated law which determines the polarity of that voltage.

From Hyperphysics page:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #941 on: April 28, 2011, 03:59:29 PM »
So guys.  It seems that the real danger is nothing to do with the actual circuit variations - but that the variations are then so 'off the mark' that they no longer have relevance to the original circuit objectives.  That would be a pity.  Because - in essence - what we've got here is a method whereby we can dissipate some hefty wattage at the load with apparently no energy expended from the supply. 

I need to stress this point.  There is nothing that this circuit does that cannot be done with the MOSFETs in the traditional setting.  And I would re-iterate that the object of our demonstration was NOT to guage those high wattage outputs - but simply to resolve that extraordinary waveform.  You see this I trust.  The fact that we can exceed those equivalence principles then implies that the 'sky's the limit' when it comes to applying more energy.  Or.  Possibly it would be better to say that the amount of energy that is then dissipated is actually the next most challenging aspect to unfold.  And THAT is when Q1 most certainly comes into play and - possibly - where we need to keep to some semblance of those early first principles used in that original design.

In any event.  There is a gross tendency to 'wax obscure' especially when it comes to determining the properties and the functions of sundry transistors.  The explanations becomes somewhat more confusing than the rather dark art that is applied to the manufacture of these transistors in the first instance.  In the interests of keeping this explanation as simple as possible I'm therefore proposing my own 'take' which has the real merit of focusing on the measurable effects rather than proposals as to what or why these are happening.  And more to the point.  It's also all that's needed.  Just follow the logic.  It's straight forward.

We know that the circuit can be fully connected to the supply and yet - if the functions generator is not 'turned on' then the circuit remains open.  I know this well.  It's the easiest way for me to disconect things at the end of our tests and it's all I ever do.  So.  When we turn it on - we're closing the circuit.  And we're first applying a positive signal to the gate of the FET.  This provides that 'bridge' across a gap that then also allows current to to flow through the circuit.  Under usual conditions the current then flows to the negative terminal and this persists until the end of that 'on' period. In effect, the applied voltage at the postiive terminal of the battery is able to bridge that open condition courtesy the positive signal applied at the gate.

Then the signal defaults to negative.  Immediately the battery stops conducting and there is a return of negative voltage induced from the collapsing fields of all those circuit components.  (Faraday's Law.)  Now.  Just as the positive charge was presented at the drain courtesy the MOSFET Q1 - now a negative charge is presented at the source courtesy of the MOSFET Q2 (and upwards as required).  And in the same way that the Gate provided a positive charge during the 'on' time - it now provides a negative charge during the 'off' time.  And appropriately, this is provided at the source which then allows for the 'negative' voltages throughout the circuit to flow freely.  Prior to this - under usual conditions - only the reverse body diode allows this path.  And while there is no 'problem' with using this path - it is also NEVER enough to fully allow that continuous resonating condition.  We know this.  We've tried it. 

So.  The actual focus is that there is this much negative voltage that then flows this freely as negative current.  Of especial interest to me is that it also never seems to fully discharge without re-establishing the potential difference at the battery - which then needs to discharge.  And on and on.  What we have here is a continual current flow.  And it's not a ripple - as was early argued all over the place.  It's evidently energetic and it's most certainly sustaining a temperature at the load and, indeed, over some of the circuit components. 

I'd be very sorry to find that we delve into an obscure explanation related to a lot of simulated schematics and consequent waveforms that also then took the focus off this point.  There is a relationship established between the supply and the circuit components that seems to mutually re-inforce each other that the current can persist over some considerable time and to some considerable advantage to the energy supply source.   What I personally, would like to resolve is how it is that there is this small 'window of opportunity' so to speak, in the setting of the 'offset' that also allows that oscillation at a zero discharge from the battery during the 'on' time and yet it fully enables the negative voltage and all the consequent oscillation.  It is almost as if the circuit is able to get energised without a flow of current - and I can't see how that can be a realistic explanation.  Nor, MileHigh, is it energy that is being delivered to or from the functions generator because, self-evidently, it can all be achieved without using one. 

But having said all that.  Please remember that we're not out to 'solve' the puzzle of that oscillation in our replications - for those of you who wnat to go this route.  We're trying to maximise the energy output at the least possible cost to the supply.  And for that you need that Q1 because you also need that initialising 'on' period with its consequent positive voltage during that time.  And here there's a wealth of potential discovery.  Because we've hardly opened the door a fraction.

Kindest regards
Rosemary

Offline i_ron

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #942 on: April 28, 2011, 05:14:40 PM »
So guys.  It seems that the real danger is nothing to do with the actual circuit variations - but that the variations are then so 'off the mark' that they no longer have relevance to the original circuit objectives.  That would be a pity.  Because - in essence - what we've got here is a method whereby we can dissipate some hefty wattage at the load with apparently no energy expended from the supply. 

snip
Kindest regards
Rosemary

Rosemary,

Excellent post as usual.

Now for a builder, the logical schema at this time seems to be the two FET model, one forward, one backwards?

So what is the ratio of on time to off time for FET 1? I know you have said but if you could repeat once more please? And the voltage of the Function generator, is it plus minus 5 volts or 10?

What I was considering was doing away with the FG and using just a battery and a timed change over relay. Would this work? Has the group approached this from this angle?

Warm regards

Ron

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #943 on: April 28, 2011, 05:32:55 PM »
Rosemary,

Excellent post as usual.

Now for a builder, the logical schema at this time seems to be the two FET model, one forward, one backwards?

So what is the ratio of on time to off time for FET 1? I know you have said but if you could repeat once more please? And the voltage of the Function generator, is it plus minus 5 volts or 10?

What I was considering was doing away with the FG and using just a battery and a timed change over relay. Would this work? Has the group approached this from this angle?

Warm regards

Ron

Hello Ron.  What a pleasure to see that you're going to try this.  I don' know how to do away with the functions generator.  But Harti gave a schematic using a battery to supply the required signal.  I'll see if I can find it.  The guys have only got an oscillation at this stage.  And it's not stable and it certainly is not able to show a gain.  Still very much a work in progress.  I'm a clutz at design - unfortunately.  I was rather hoping that Poynty's schematics would evolve to something 'doable'.  Perhaps that's still on the cards. 

But if you can explore ways of doing this all with a switch and those 'back to front' FET's as you describe them - then that would be a good thing.  Once you get testing you'll see the point of what I'm going to say here.  It is absolutely IMMATERIAL what frequency you apply.  It finds its own resonating frequency.  What you WILL need is a reasonably good pot to explore all this potential.  And it would be WONDERFUL if you could try this with batteries that are not quite as long lasting as our own.  It would certainly answer some questions that keep coming up. 

But let me see if I can find that schematic that Harti gave us.   I'll get back here.  Delighted to see that you're up for this. 

Kindest regards
Rosie

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #944 on: April 28, 2011, 06:00:34 PM »
Hi,
what about this circuit ?

Just use a 9 Volts battery and a pot to supply the
negative bias voltage at the gates.
To get it to oscillate you might need to switch the
9 Volts battery on and off a few times.

Then also as Humbugger said the shunt will only
have the battery current and not the 9 Volts battery current.


Well to measure also the battery voltage with a dual channel scope that
has a common ground you need to do this circuit then.

See attached picture.

Many thanks.

Regards, Stefan.

I can't do anything about posting over the schematic Ron.  But this is the post
« Reply #347 on: March 26, 2011, 04:09:20 AM »

I think the same schematic is applicable - just add in that MOSFET. 

I'll be able to get hold of the guys early next week and I'll ask them for that schematic they're trying out.  Then I'll post it here.

Hope that helps.  I'm afraid you'll need to be very creative as I absolutely can't help here.

Again, kindest and very best of regards,
Rosie