Quote from: Rosemary Ainslie on May 04, 2012, 02:58:40 AM
And with regard to this post.I have stated this PUBLICLY - right here on this forum.  WE PUT THE PROBE OF THE FUNCTION GENERATOR DIRECTLY ON THE GATE OF Q1.  WE PUT THE TERMINAL OF THAT PROBE DIRECTLY ON THE GATE AT Q2.  When the probe voltage applies a positive voltage at Q1 then the voltage at Q2 is in anti polarity.  There most CERTAINLY is an applied voltage at the gate of Q2.  And Q1.  ALWAYS. They're in anti phase or anti polarity.  It is NOT.Not sure what you mean to achieve by taking out the CSR. It makes no difference either way.   Q1 voltage is precisely in anti phase to Q2 polarity  whichever way the function generator has applied that voltage.  This is IMPOSSIBLE.  We apply that terminal directly to the Gate at Q2.  If the Gate at Q1 is negative - then correspondingly the Gate at Q2 is positive.  Q2 is turned on by the positive signal applied DIRECTLY to that gate.  The Q2 source leg is NOT negative during this phase of the switching cycle.  It is POSITIVE. Besides which,  Q2's source leg is NOWHERE connected to the negative rail of the battery supply source.  (And MileHigh - IF you don't like that term - you'll need to live with it.  It's entirely acceptable terminology.) Effectively it FLOATS.  One would hope that you'd be dealing with the actual circuit and not some idea that you may have of its configuration.

Do you actually believe the gate of Q2 is not always held at the potential of the non-battery end of the CSR?

There is no way in your circuit that the FG can apply a positive signal directly to the Q2 gate.  The Q2 gate is always at or very near ground.  Q2 is turned on by the FG when the FG makes the source of Q2 more negative than ground.

The operation of your circuit is fully understood.  It is a common configuration used everyday, afterall, there are only three possible configurations for all FET's from which all circuit's are designed.  You, apparently, are the only one, or amongst the very few, that does not understand how this circuit operates. 

You further demonstrate your lack of understanding regarding your circuit's operation in this post.  If you would listen instead of always arguing, you might have had a better understanding of its operation by now.  I'm sure most readers would agree.  Indeed, go prove something.

'tis a pity,
PW

   

Is there yet a different schematic, other than the one below, being used??

Or is Rosemary simply talking out of her head again, making things up and contradicting herself yet once more?

And what about that IsoTech GFG324.... the magic invisible function generator that makes all this possible. What about that?

QuoteAnd with regard to this post.I have stated this PUBLICLY - right here on this forum.  WE PUT THE PROBE OF THE FUNCTION GENERATOR DIRECTLY ON THE GATE OF Q1.  WE PUT THE TERMINAL OF THAT PROBE DIRECTLY ON THE GATE AT Q2.  When the probe voltage applies a positive voltage at Q1 then the voltage at Q2 is in anti polarity.  There most CERTAINLY is an applied voltage at the gate of Q2.  And Q1.  ALWAYS. They're in anti phase or anti polarity.  It is NOT.Not sure what you mean to achieve by taking out the CSR. It makes no difference either way.   Q1 voltage is precisely in anti phase to Q2 polarity  whichever way the function generator has applied that voltage.  This is IMPOSSIBLE.  We apply that terminal directly to the Gate at Q2.  If the Gate at Q1 is negative - then correspondingly the Gate at Q2 is positive.  Q2 is turned on by the positive signal applied DIRECTLY to that gate.  The Q2 source leg is NOT negative during this phase of the switching cycle.  It is POSITIVE.  Besides which,  Q2's source leg is NOWHERE connected to the negative rail of the battery supply source.  (And MileHigh - IF you don't like that term - you'll need to live with it.  It's entirely acceptable terminology.) Effectively it FLOATS.  One would hope that you'd be dealing with the actual circuit and not some idea that you may have of its configuration.

Oh really? Both the Gate and the Source of Q2 are POSITIVE at the same time, even though they are hooked to either end of the FG's output? Are you sure about that?

And the FG's internal 50R impedance means that the sources of the Q2s are indeed connected to the negative rail, and the FG is acting in series with that main battery. This has been shown to you in circuit diagrams, explained in publications from the manufacturers of function generators, and has been demonstrated in hardware and simulations. Yet you refuse to understand or believe it.... because it would mean that you would have to admit that you are wrong. But ANY TEST that could show that we are correct and that you are wrong... you refuse to do yourself. JUST DO IT and show the results, LIKE STEFAN ASKED YOU TO.

Quote from: TinselKoala on May 04, 2012, 03:37:41 AM
Is there yet a different schematic, other than the one below, being used??

Or is Rosemary simply talking out of her head again, making things up and contradicting herself yet once more?

And what about that IsoTech GFG324.... the magic invisible function generator that makes all this possible. What about that?

TK,

More importantly, the gate of Q2 is connected directly to the CSR and therefore the voltage at the gate of Q2 can never be any voltage other than that of the non-battery end of the CSR, which is very near ground.  I did not know the obvious needed stating.

I seem to be correct in my assumption from that read of the blog link and subsequent responses that RA does not understand that an FG has a bipolar supply and that she believes the FG is somehow applying a positive voltage directly to the gate of Q2.  Apparently, based on her own statements, she cannot even read her own schematic. 

How can the FG signal ground or the Q2 gate EVER be any voltage other than that of the CSR?  It can't.

Again I am flabbergasted,

PW

TK,

Have you ever done a search online that directed you to a page of totally disconnected words only there for the purpose of attracting a search engine response?

PW