Testing the TK Tar Baby

[Rosemary Ainslie]

So guys.  There you have it.  Another slew of pure 'spin' based on anything BUT the 'real truth'.  It's laughable - but nonetheless seems to be enough to dupe some of you.  I am NOT an expert at anything at all.  Nor is my math dependable.  Nor am I always perfectly articulate - albeit that I do my best.  But these representations of my work - what I think - what I 'believe' are UTTERLY ERRONEOUS.  Please be advised.  We're dealing here with a group of vigilantes - led by TK and his very appropriately termed TAR BABY and its historical associations.  Which all is intended to cast absolute doubt on our test results based on the misinterpretation of our work.   And bear in mind.  There's a mission here executed with all the zeal of any vigilantes - anywhere.  And way beyond acceptable.

Kindest regards,
Rosemary

[Rosemary Ainslie]

And as for THIS

Rosemary,

I assume this to mean that you now understand and agree with Groundloop's first drawing wherein the FG is depicted as a battery in series with a 50 ohm resistor.

Understanding how, when the FG output is negative, Q2 is biased "on" into a linear region of operation and that the DC bias current consequently flows thru Rload, Q2, the FG, and the CSR, is the first step toward understanding the how and why of the oscillaton.

Golly, progress, good for you!

PW

Spare us all your continuing inanities.  Or answer those questions.

Rosemary

[happyfunball]

Another slew of pure 'spin' based on anything BUT the 'real truth'.  It's laughable - but nonetheless seems to be enough to dupe some of you.  I am NOT an expert at anything at all.  Nor is my math dependable.

You can't claim to know next to nothing and also claim to know the truth.

[happyfunball]

delete

[picowatt]

And as for this number...LOL  It seems picowatt has still not mastered that copy paste number.  And yet he's trying to teach us anything at all about a MOSFET.  What he needs to show us is how an applied positive voltage at the GATE of Q2 can then conduct upwards of 4 amps of negative current flow generated from capacitance across the MOSFET  > through the function generator probe and its terminal > back to the Source Rail of the battery supply > and then inject this upwards of 4 amps of NEGATIVE current flow > across the current sensing resistors > move this through upwards of 26 volts from a battery supply source > and then through the load > back to the DRAIN of Q2 while it remains - unarguably and unequivocally BELOW that zero crossing line.  And he argues that all is to be expected?  IF that energy is from capacitance at the MOSFET - then that's an ENORMOUS amount of capacitance.  Somehow able to generate in the order of 4 amps x 26 volts and UPWARDS.  Which capacitance - ATYPICALLY - and contrary to ALL standard prediction - is then also ABLE TO DISSIPATE HEAT AT THE LOAD.  Clearly it's a load on nonsense.  Which makes the balance of his statements - ?  Whatever.

Kindest regards,
Rosemary

There is no applied voltage at the gate of Q2 from the FG.  Again, more of your your words put into the mouths of others.

The gate of Q2 is always at or very near ground potential.  If the CSR were temporarily replaced by a wire, then the gate of Q2 would ALWAYS be EXACTLY at ground potential, which is the negative terminal of the battery.  When the FG output is negative, the FG makes the voltage at the source terminal of Q2 negative with respect to ground to which the Q2 gate is connected.  Pulling the source of Q2 negative with respect to its gate turns on Q2.

Look at your own schematic, the gate of Q2 is connected to the CSR.  The FG can do nothing to change that, the gate of Q2 will always be at or very near to ground potential (the voltage indicated at the CSR).  Ignoring the small voltages seen across the CSR, the gate can, for discussion of DC conditions, be considered as tied to ground.  It is the source leg of Q2 that sees the FG output voltage, not the Q2 gate.  Q2 is turned on by having a negative voltage applied to its source by the FG.  The voltage at the gate is effectively unchanged.

From your written statements and drawings I personally do not believe you comprehend this.   You apparently seem to think that the FG is somehow changing the voltage at the gate.  It is not.  The gate of Q2 is ALWAYS at the voltage of the non-battery end of the CSR.

As to the rest, and more of your "death by a thousand cuts" insults, digs and jabs...

Golly, if you could comprehend the DC conditions of your circuit, maybe you could understand how all that AC current flows.

Believe what you want, but you further erode your credibility with each of your posts,

PW