I just hooked my recently aquired FG directly to an LED and guess what?!?
Yeah, it lit up. Hmmm...
Easy test for RA to see that a FG passes current  right? Not sure how that could be argued...
Sorry to take up valuable space, just wanted to share that.
Imagine that, Rosemary is WRONG AGAIN!

PC

EDIT: Added AGAIN and a glow...
2nd EDIT: This is added to make my post asinine Rosemary: WHERE IS YOUR NEW MATH?

My dear PhiChaser

Try and catch up with the facts.  I have NEVER claimed that a function generator does not have current flowing through its probes.  I have ONLY claimed that the current from the function generator does not move away from those probes.  The current that is INDUCED by the function generator is from its applied voltage at the GATES of the those MOSFETS.  It is NOT FROM that function generator as YOU, PICOWATT, MILEHIGH and TK erroneously suppose.

Quote from: PhiChaser on May 08, 2012, 10:50:42 AM
I just hooked my recently aquired FG directly to an LED and guess what?!?
Yeah, it lit up. Hmmm...
Easy test for RA to see that a FG passes current  right? Not sure how that could be argued...
Sorry to take up valuable space, just wanted to share that.
Imagine that, Rosemary is WRONG AGAIN!

PC

EDIT: Added AGAIN and a glow...
2nd EDIT: This is added to make my post asinine Rosemary: WHERE IS YOUR NEW MATH?

And WHAT new math do you keep referring to - in your rather weak efforts at joining in this 'tar Rosemary' thread?  We all know you're anxious to 'join in' with what you 'assume' is a majority opinion.  In FACT you are sharing a minority opinion with 4 heavy weight OU disclaimants.  And that STILL makes you nothing but a bully while they're seasoned anti OU campaigners.

Rosie Pose

Quote from: Rosemary Ainslie on May 08, 2012, 11:06:17 AM
My dear PhiChaser

Try and catch up with the facts.  I have NEVER claimed that a function generator does not have current flowing through its probes.  I have ONLY claimed that the current from the function generator does not move away from those probes.  The current that is INDUCED by the function generator is from its applied voltage at the GATES of the those MOSFETS.  It is NOT FROM that function generator as YOU, PICOWATT, MILEHIGH and TK erroneously suppose.

And WHAT new math do you keep referring to - in your rather weak efforts at joining in this 'tar Rosemary' thread?  We all know you're anxious to 'join in' with what you 'assume' is a majority opinion.  In FACT you are sharing a minority opinion with 4 heavy weight OU disclaimants.  And that STILL makes you nothing but a bully while they're seasoned anti OU campaigners.

Rosie Pose

You have over the past several months given all readers the impression that you believed a FG cannot pass, sink, or source current.  From your thousands upon thousands of ohms comments to implicit denial.  I am not sure what you mean by "move away from those probes", and likely most others do not as well, but it does seem some progress is being made by you in accepting that an FG can indeed pass, sink, or source current.

As for the rest:

The gate of Q2, the source of Q1, and the FG signal common are all tied together and connected to the battery negative via the CSR.  Therefore, the gate of Q2, the source of Q1, and the FG signal common can only be and will always be at the same voltage, which is the voltage observed at the CSR.  Look at the schematic, this is both obvious and irrefutable.

When the FG output is a positive voltage, i.e., a voltage above its signal ground, the gate of Q1 is made positive and Q1 turns on.  Current flows thru Q1 and the CSR to the battery negative.  Q2 remains off.

When the FG output is a negative voltage, i.e., a voltage below its signal ground, the source of Q2 is made negative with respect to the Q2 gate and Q2 is partially turned on, being biased into a region of linear operation.  Current flows thru Q2, the FG, and the CSR to the battery negative.  Q1 remains off.

The amount of DC current that flows thru Q2 and the FG when the FG output is negative is dependent upon the turn on voltage of Q2, the open circuit negative voltage of the FG, and the FG's internal Rgen of 50 ohms.

Print this out and show it to your experts.  No "expert" would disagree with any of the above.

PW

PhiChaser

Quote from: picowatt on May 08, 2012, 11:51:30 AM
You have over the past several months given all readers the impression that you believed a FG cannot pass, sink, or source current.  From your thousands upon thousands of ohms comments to implicit denial.  I am not sure what you mean by "move away from those probes", and likely most others do not as well, but it does seem some progress is being made by you in accepting that an FG can indeed pas, sink, or source current.

SHOW ME ONE - JUST ONE POST - WHERE I HAVE STATED THIS?  I have ALWAYS KNOWN that current flows in the function generator probes.  IF it did not then there would be no VOLTAGE applied at the MOSFET gates.  BUT current does NOT pass from the function generator probes to the the circuit.  TK and picowatt DEPEND on this argument.  Their argument is that current from the function generator passes TO THE CIRCUIT.  I have DENIED this.  THEN.  They have then 'spun' the fact that I am CONFUSED and claim that there is no current flow in the function generator.  AND.  Because you're basically a bully and because you're trying to get onto the 'team' so to speak - and because you're not that perspicacious - you BOUGHT into their argument.  You have, effectively, been duped. 
As for the rest:

Quote from: picowatt on May 08, 2012, 11:51:30 AMThe gate of Q2, the source of Q1, and the FG signal common are all tied together and connected to the battery negative via the CSR.  Therefore, the gate of Q2, the source of Q1, and the FG signal common can only be and will always be at the same voltage, which is the voltage observed at the CSR.  Look at the schematic, this is both obvious and irrefutable.

IF the argument were irrefutable - then we would not be able to refute it.  Since we can - then the argument is NOT irrefutable.  We can REFUTE it simply by applying the source of Q2 directly to the negative rail of the battery supply and then the voltage across the  CSR will be POSITIVE?  HOW then would you explain that?

Quote from: picowatt on May 08, 2012, 11:51:30 AMWhen the FG output is a positive voltage, i.e., a voltage above its signal ground, the gate of Q1 is made positive and Q1 turns on.  Current flows thru Q1 and the CSR to the battery negative.  Q2 remains off.

THIS is correct.

Quote from: picowatt on May 08, 2012, 11:51:30 AMWhen the FG output is a negative voltage, i.e., a voltage below its signal ground, the source of Q2 is made more negative than its gate and Q2 is partially turned on, being biased into a region of linear operation.  Current flows thru Q2, the FG, and the CSR to the battery negative.  Q1 remains off.

THIS is not correct. 

Quote from: picowatt on May 08, 2012, 11:51:30 AMThe amount of DC current that flows thru Q2 and the FG when the FG output is negative is dependent upon the turn on voltage of Q2, the open circuit negative voltage of the FG, and the FG's internal Rgen of 50 ohms.

This is also NOT correct.  It is also disprovable - very easily - by applying the source of Q2 directly to the Source of the supply battery. 

Quote from: picowatt on May 08, 2012, 11:51:30 AMPrint this out and show it to your experts.  No "expert" would disagree with any of the above.

WHY would I bother?  It is nonsense.  We'll demonstrate this together with a whole lot of little side tests to refute every single one of TK's 'allegations'.  LOL.

Rosie Pose

Quote from: Rosemary Ainslie on May 08, 2012, 11:06:17 AM
My dear PhiChaser

Try and catch up with the facts.  I have NEVER claimed that a function generator does not have current flowing through its probes.  I have ONLY claimed that the current from the function generator does not move away from those probes.  The current that is INDUCED by the function generator is from its applied voltage at the GATES of the those MOSFETS.  It is NOT FROM that function generator as YOU, PICOWATT, MILEHIGH and TK erroneously suppose.

Yeah, okayyyy... (Facts, LMAO!!!) The current does not move away from the probes? So if I hook a really long wire to the LED it won't work? Er... I'm obviously confused then...

Quote
And WHAT new math do you keep referring to - in your rather weak efforts at joining in this 'tar Rosemary' thread?  We all know you're anxious to 'join in' with what you 'assume' is a majority opinion.  In FACT you are sharing a minority opinion with 4 heavy weight OU disclaimants.  And that STILL makes you nothing but a bully while they're seasoned anti OU campaigners.

Rosie Pose

You're kidding right? The (bad) math referred to has been reposted several times. Your 'results'  were badly in error. How many watts in a Joule in an hour in the sunshine Little Miss Mosfet? Do you remember doing those calculations? I'm sure someone can dig them up again...

I will admit I am not qualified to engage in their technical discussions, I TRY to learn from them. And yes, I am anxious to join in because they know more than I do about electronics. That is how I learn, by listening to more educated people on the subject I want to learn. If you feel like I'm just getting on the bandwagon to bash Rosie, you would be wrong about that. Reading your posts just annoys me sometimes because I can see that these people know WAY more than you do about electronics. Because they tell you that you are wrong, that makes them the 'enemy of OU' and out to 'tar' you. You're like a broken record, you know that?
They have been patiently TRYING to explain what is happening while the circuit operates (more clearly than ANYTHING I have read from you) and you continue to DENY them because they don't agree with your paper. Either you know what you're talking about or you don't. A good researcher tries to DISPROVE their anomalous results. You haven't done that. Not that I can see anyways... Your scientific methodology involves you pointing at a 'paper', and claiming you have proper and accurate measurements because you used an expensive scope with a math function...
Put up or shut up Rosemary. The Dim Bulb test suggested to you (and done several times in time lapse by TK) PROVES that your batteries will drain. Do the test! Prove him wrong! No special tools required, just some batteries and light bulbs... Still waiting for that BASIC TEST.
Your endless excuses are pathetic. Quit typing, make a video that proves ANYTHING related to your 'claim' that is COP>17 one day, COP=Infinity the next, you never 'claimed' OU, only 'measured' it (improperly) the next. Talk about word salad... Start your own thread, build your circuit from scratch, make videos, do tests, find out for yourself. Hell, build TKs Tar Baby if you want to try and disprove his results. You haven't done that either...

It doesn't seem like anyone can tell you that you are wrong about anything. As long as someone believes your 'claim', you will continue to deny ALL EVIDENCE TO THE CONTRARY. Amazing...

As far as OU goes, why would they waste their time and money if they didn't want OU to be a reality? TK built a better circuit than yours that does the same thing. You keep denying this!
Build something!!! Make a video! TEST something! Re work your math. We're still waiting for those new OU numbers...

One last thing in regards to the Tar Baby comments: At first I thought it was a racist remark until I looked at it as it was intended; A sticky situation that you become more entangled in the more you try to fight it. You are tarring yourself lady, they're just giving you something to stick to. (Funny pun there since you can't answer simple questions without convuluted dissertations leading to nowhere.)

More asinine me: Where is that NEW MATH??

And I'm sorry if you think I'm a bully; I think you're delusional and refuse to accept reality because it is contrary to your (un) 'truth' that you desperately cling to. It's okay, really...
Again, sorry for crapping up your thread TK. I won't post on here again but I will keep reading in case RA ever gets around to doing those new calculations...

PC