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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 741408 times)

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #960 on: April 29, 2011, 04:33:17 PM »
Correct.

However, power mosfets, in general use with switched power circuits are usually biased at the gate with a nominal 10 volts or more.

Thanks for posting hoptoad.

In this circuit, the MOSFETs are being applied in more of a "linear" sense, and VGS never (or rarely) approaches anything more that 5V, one polarity or the other.

.99

neptune

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #961 on: April 29, 2011, 04:48:27 PM »
@poynt99 . Thankyou for answering my questions.

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #962 on: April 29, 2011, 05:29:30 PM »
Rose,

Unless the demonstration video footage is not authentic, I have already painstakingly well established what the actual "AS-BUILT" circuit connections are and posted them. Have you seen those posts?

It is quite clear from the video footage that the actual schematic is per the one below, with the exception for the function generator, shown here as a fixed DC source. In particular, the CSR is most definitely connected to the Q1 source and Q2 Gate as shown.

Poynty - it does not make a blind bit of difference to the waveform or the results whether the ground from the signal generator is directly on the shared negative rail or if it's in series with the shunt.  The fact that it was set up that way at the demo was happenstance. The fact is that it's more convenient to keep the ground on that shared rail because we were running two oscilloscopes and 4 channels - concurrently.  That made the junction at D rather crowded.  In point of fact it is normally at the pin marked at D on the video.  But that's because I usually only ever use the LeCroy.  I've just checked the video and the board.  Right now and for those shots I took to argue your 'undersampling' quibble - it is and usually is positioned at D - in series with the shunt.  I wonder what difference it would make to your waveforms if you placed it at the negative rail.  I'd be interested to see.  I suspect very little.  But in any event it's wrong.  It is properly in series with the CSR.

Why are you changing the pin designation on "Q2-5"? The "g" means "gate", and "s" means "source".

I know this Poynty.  That's exactly why I needed to transpose it.  Think it through.  We swap the Gate and the Source.  We do not change the drain.  Am I talking to myself here?  Let me try this again.  Q1 takes a positive current from the battery via a postive charge applied to the gate.  That's during the 'on' time.  Then.  Let's just take it that we've got a battery signal source because that gets to the meat of the issue best.  Q2 takes a positive charge from the signal supplying battery via a postive charge applied at the gate.  The same thing.  I'ts only NEGATIVE relative to Q1.  That transposition effectlvely generates a POSITIVE CHARGE at the gate.  Which is precisely what the MOSFET wants.  It's been built that way.  I do hope the penny drops.  Effectively the MOSFET is reading the source as the drain and the drain as the source relative, that is, to Q1.

Regarding that small capacitor in parallel with the load, it is of course not a discrete component of the load, but all inductors have some capacitance associated with them, and it was included only to allow for a more accurate simulation. The consequence however of removing it is minimal. It is of no concern in actuality.


Good.  Let's get rid of it. 

Kindest regards,
Rosemary

Added.
And some more.

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #963 on: April 29, 2011, 05:49:54 PM »
Now that the "as-built" circuit connections have been firmly re-established (reference diagram depicted in Simplification01_schema01.png), would anyone like to try and predict what changes might occur in the circuit operation if Q1 is completely removed from the circuit, and no capacitor or diode connected in its place?

.99

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #964 on: April 29, 2011, 06:05:17 PM »
Now that the "as-built" circuit connections have been firmly re-established (reference diagram depicted in Simplification01_schema01.png), would anyone like to try and predict what changes might occur in the circuit operation if Q1 is completely removed from the circuit, and no capacitor or diode connected in its place?

.99

Yes.  I would.  I already know.  PROVIDED you keep the bias of the diode at the drain as per Q2 as you've SHOWN it and not as our circuit actually has it - then it won't make a blind bit of difference.  The circuit will oscillate.  But to what end?

This is where I get more than a little exasperated Poynty.  Are we simply to attempt more and more schematics to generate an oscillation?  Is that all the merit you see here?  The POINT of the study of that oscillation was not HOW TO GENERATE IT - but to consider that it's there at all.  I am reasonably satisfied that it will persist certainly as long as a charge is applied at the Gate.  And when we do get that oscillation then we have current moving to and from the battery that has all kinds of conservation benefits.  But we also need to 'up the ante' and that means that all this discourse could have been spent in studying how to get the 'extreme' values in those oscillations.  For instance - at certain settings the voltage peak to peak across the CSR is 4 volts or thereby.  But at other settings - usually with adjustments to the offset and the duty cycle - it's possible to get voltages that are 10 times that value and oscillations that are greater by a factor of 3.  And so it goes. 

We need to move on.

Rosemary

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #965 on: April 29, 2011, 06:13:26 PM »
Yes.  I would.  I already know.  PROVIDED you keep the bias of the diode at the drain as per Q2 as you've SHOWN it and not as our circuit actually has it - then it won't make a blind bit of difference.  The circuit will oscillate.  But to what end?

Could you explain in some detail for myself and the readers here, what you mean exactly?

The circuit is precisely as I have depicted it. Please indicate where it is that you are having difficulty seeing it, and I will try to make it more clear.

.99

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #966 on: April 29, 2011, 06:19:59 PM »
Could you explain in some detail for myself and the readers here, what you mean exactly?

The circuit is precisely as I have depicted it. Please indicate where it is that you are having difficulty seeing it, and I will try to make it more clear.

.99

Poynty - look at Q2 on that schematic that I modified of yours.  You'll notice that the source and gate have been transposed.  But there has been no change made to the drain.  It is still biased to enable a 'negative' current and block a 'postive' current.

Rosemary

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #967 on: April 29, 2011, 06:43:24 PM »
Poynty - here's the thing.  I am very aware that I have no skills at circuit design.  And frankly that's what's needed.  You asked - some time back - where to from here?  Well.  I would dearly love to leave this thread to someone of your exceptional competence that you guys can brainstrorm and apply your collective skills and your knowledge to precisely this reach.  And I know it'll be to good effect.  But I'm concerned that you're chasing red herrings and I'm not sure why.  Unless - of course - it's to further test the principles applied.  But they've been done to death now.  Surely?

I would dearly love to spend some constructive time in getting a paper prepared and in generally showing things to our learned and revered.  And I can't do anything other than obsessively chase posts that may mitigate against the progress of this technology rather than otherwise.

Please let us know where you stand on this.  If you can't take it over - is anyone else prepared to?  I really have so much work I need to do.  And I can't contribute any further here.  It needs your kind of expertise and your kind of lateral thinking.  Will you PLEASE help us. 

Kindest regards,
Rosie

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #968 on: April 29, 2011, 07:04:53 PM »
We can not arbitrarily at will relabel the pins on an electronic device such as the MOSFET. The industry creates the symbols for these components so that everyone is "on the same page" regarding the conveyance of information via circuit diagrams.

Look at the attached capture from the IRF document locate here:
http://www.irf.com/technical-info/appnotes/mosfet.pdf

It clearly indicates which pins on the symbol are the Drain (D), Gate (G), and the Source (S).

Also shown is that characteristic ID vs. VGS curve I emphasized earlier.

Note also the indication of current flow in the Drain...it is in the down direction.

.99

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #969 on: April 29, 2011, 07:23:44 PM »
We can not arbitrarily at will relabel the pins on an electronic device such as the MOSFET. The industry creates the symbols for these components so that everyone is "on the same page" regarding the conveyance of information via circuit diagrams.

I HAVE NOT RELABELLED ANYTHING.  What is wrong with you?  What we've done is PHYSICALLY SWAPPED the SOURCE WITH THE GATE.  And what I've then done is relabelled the schematic accordingly. 

Look at the attached capture from the IRF document locate here:
http://www.irf.com/technical-info/appnotes/mosfet.pdf

It clearly indicates which pins on the symbol are the Drain (D), Gate (G), and the Source (S).

Also shown is that characteristic ID vs. VGS curve I emphasized earlier.

Note also the indication of current flow in the Drain...it is in the down direction.


I there anyone out here who can explain it better?  POINTY PLEASE LISTEN UP.  The MOSFET IRFPG50 is designed that it can take a postive charge at the gate and then conduct a current through the device.  We've positioned that IRFPG50 Q2 that it can take a POSITIVE CURRENT FROM THE BATTERY SIGNAL SUPPLIER and conduct a current through the device.  That NEGATIVE CURRENT at the source is now a POSITIVE CURRENT with respect to the new positioning of the source and the gate.  That MOSFET can still do what it does best.  It's conducting a positive current.  Proof is precisely when you place a battery in series with Q2.  It takes that as a positive current.  But it's negative with respect to the actual circuit and it's primary supply being the battery supply source.  In other words Q2 sees the actual circuit DRAIN as the SOURCE.  And there is no restriction to the current it can enable.  Therefore it enables that current.  But - relative to the primary battery supply - IT'S NEGATIVE.

Please just think it through.  I can't explain it any better.  Just think it through.

Kindest
Rosemary

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #970 on: April 29, 2011, 07:32:32 PM »
I HAVE NOT RELABELLED ANYTHING.  What is wrong with you?  What we've done is PHYSICALLY SWAPPED the SOURCE WITH THE GATE.  And what I've then done is relabelled the schematic accordingly. 

If it is not clear, I have been, and am referencing the apparatus (and how it was connected) in the video demonstration.

If you are now discussing a different circuit connection from the one used in the video demonstration, then I believe it would be prudent to advise the readership to that effect.

.99

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #971 on: April 29, 2011, 07:39:58 PM »
If it is not clear, I have been, and am referencing the apparatus (and how it was connected) in the video demonstration.

If you are now discussing a different circuit connection from the one used in the video demonstration, then I believe it would be prudent to advise the readership to that effect.

.99
No.  I am not referring to anything other than the 'modified schematic' that I posted - here today.  I'll repost it for clarity.

I see you're getting angry there Poynty.  And I also see that you're missing this all by a mile.  Just look again at the positioning of the source and the gate.  And then consider what it sees when it sees a postive charge from the signal generating battery.  Just look at it upside down if you have to.

Here it is again.  And, with the exception of the bias of that zener - it's precisely how our circuit is designed - up to and including the position of the CSR.

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #972 on: April 29, 2011, 07:54:01 PM »
No.  I am not referring to anything other than the 'modified schematic' that I posted - here today.  I'll repost it for clarity.

Here it is again.  And, with the exception of the bias of that zener - it's precisely how our circuit is designed - up to and including the position of the CSR.

Please then clarify for the readers; does your above diagram accurately depict the circuit connections for the apparatus in the video demonstration?

.99

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #973 on: April 29, 2011, 07:59:46 PM »
Please then clarify for the readers; does your above diagram accurately depict the circuit connections for the apparatus in the video demonstration?

.99

Yes.  With the exception of the position of the ground from the signal generator.  And there has to be a connection between those two (or more) MOSFETS that's omitted.  And - correctly - the zener at the drain on Q2 needs to be reversed biased.   Otherwise I think it's right.

Rosemary

ADDED.  And that we're using a functions generator - obviously.

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #974 on: April 29, 2011, 08:17:07 PM »
Yes.  With the exception of the position of the ground from the signal generator.  And there has to be a connection between those two (or more) MOSFETS that's omitted.  And - correctly - the zener at the drain on Q2 needs to be reversed biased.   Otherwise I think it's right.

Rosemary

ADDED.  And that we're using a functions generator - obviously.

At a point not too long ago, the attached was the circuit connections diagram for that video demonstration.

Clearly the diagram you are now presenting, is substantially different from the original in the video.

Could you please, for the benefit of the readership here, explain how the CSR has been relocated to an entirely different position? Also, PLEASE provide a complete schematic of the apparatus in the video demonstration, if the attached is not already one that is correct.

This is all not making much sense Rose, and I think you're losing not only readership here because of this, but losing everyone in general (including me), in terms of their understanding of what exactly you are trying to discuss here.

.99