Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[poynt99]

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]

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]

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]

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 I_D vs. V_GS curve I emphasized earlier.

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

.99

[Rosemary Ainslie]

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. V_GS 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