Testing the TK Tar Baby

[Magluvin]

Hi Poynty
The Gate and Source LEGS of a MOSFET are physically tied to the same structure. 
Our Source leg of Q2 FLOATS and is NOT therefore connected to the circuit.  As per the schematic.

I think that's right
Regards
Rosie
added

I would like to see the circuit that the source leg of Q2 is not connected to the circuit. 

If you think thats right. 


Mags

[Rosemary Ainslie]

Oh picowatt

How then do you explain the oscillation with the use of ONLY one MOSFET?  NO FUNCTION GENERATOR.  NO POSTIVE SIGNAL ANYWHERE AT ALL.  Anything from the battery BLOCKED. 

You are wrong.  Q2 is ONLY on during the oscillations.

When you apply a negative voltage to the source of a MOSFET that is connected as in your schematic, it will turn on.

Just as it is impossible for Q1 to not turn on when its gate is made more positive than Vth, it is impossible for Q2 to not turn on when a negative voltage less than -Vth is applied to Q2's source.

This is predicted as per what YOU would call "the standard model", it has been verified in .99's sims, and it has been measured in TK's replication.  The amount of current Q2 passes is what we refer to as "bias current" or "Ibias" for short

The Q2 bias current was predicted to be between 50 and 200milliamps, .99's sims confirmed it, TK measured it.

Your own 'scope captures prove it as well.  When you set the FG to full negative offest, there should be -12 volts or so coming out of the FG.  However, in all captures, your FG output trace only shows around -4 volts, also as predicted.  This is due to the voltage drop across the FG's internal 50 ohm output resistor caused by the current flow thru Q2 and that 50 ohm  resistor.

Q2 is never turned fully on, it is merely biased into a region of linear operation.  The Q2 bias current is, again, 50-200ma. depending on the FG setting.  The observed voltage at the source of Q2, when the FG output is a negative voltage, is always going to be the -Vgs for the current flowing thru it, which in the case of your circuit, is around -4volts.

All very standard, well understood electronics.  Any first year EE would agree.

Rosie Pose

Anyway guys - this is FUTILE.  They're DESPERATELY trying to minimise something that's HUGELY significant.  And there are MANY of them contributing to their own argument.  That was courtesy Stefan's belated membership allowed to MileHigh and to FTC.  They're colluding. 

And right now I'm off to get some sleep.

Kindest regards,
Rosemary

[TinselKoala]

You're arguing MY argument.  NOT picowatts.

Rosie Pose

No, Ainslie. You are misrepresenting and misunderstanding what PW is telling you.

Read his posts again. He is telling you that the Potential at Q2's GATE is always at the same potential as the negative pole of the battery. This does not change no matter what the FG is putting out. The "black" lead of the FG is also connected to this point: therefore IT DOES NOT CHANGE POTENTIAL EITHER.
In fact it is the other "red" lead of the FG that swings to voltages that are either positive or negative with respect to this UNCHANGING POTENTIAL at the negative pole of the battery/the system ground/the black FG output lead.
PW is telling you this and I am telling you this as well. Now, when the FG's RED lead is putting out a voltage that is sufficiently NEGATIVE WITH RESPECT TO THE BLACK LEAD, which is at ground potential, then the Q2 turns on. However in your case, the Q2 does not _fully_ turn on, it is rather functioning in the "linear response" region of its performance-- in other words, it's not acting as a simple switch but more like a rheostat. This is because it cannot get more than -4 volts to the source (equivalent to +4 at the gate) due to the reason described by pw earlier, and even this voltage fluctuates due to the oscillations. It then oscillates in and out of that region, partially conducting and nonconducting at the oscillation frequency.

[picowatt]

Oh picowatt

How then do you explain the oscillation with the use of ONLY one MOSFET?  NO FUNCTION GENERATOR.  NO POSTIVE SIGNAL ANYWHERE AT ALL.  Anything from the battery BLOCKED. 
Rosie Pose

Anyway guys - this is FUTILE.  They're DESPERATELY trying to minimise something that's HUGELY significant.  And there are MANY of them contributing to their own argument.  That was courtesy Stefan's belated membership allowed to MileHigh and to FTC.  They're colluding. 

And right now I'm off to get some sleep.

Kindest regards,
Rosemary

Which circuit are you referring to in this post?

It sounds more desperate than sensical.

ADDED:  And keep in mind, I never said you needed a POSITIVE  voltage from the FG in your NERD circuit to make it oscillate.  So again, this post makes no sense.



 

[TinselKoala]

Guys - I've yet again been advised to stay away from this thread.  It's really achieving absolutely NOTHING - and I know this.

So you keep promising....
But it is achieving something. It is making you look stupid, for sure.

In any event just let me explain this much.  TK is WELL able to get that oscillation from Q2 with a CONTINUAL NEGATIVE SIGNAL applied to the gate.

YOU GROSS AND BALDFACED LIAR.

I am NOT applying a continuous NEGATIVE SIGNAL TO THE GATE OF Q2. LOOK AT MY SCHEMATICS. And neither are you. Oh.. .that's right, you can't read schematics. Well, get your advisors to explain it to you.

As are we  And that's only with the use of Q2.  IF there's a continual negative charge - and YET there's an oscillation - then he's yet to explain how the battery discharges current NOTWITHSTANDING the restriction imposed at the gate of the transistor - from that negative voltage signal applied.  Remember that each positive half of each of those oscillations represents the current flow from the battery supply.  So. Where is the path for that?  Certainly not across that gate.

It's absurd that they pretend to have the answers.  They're nowhere near.  Nor are we.  But we at least have a consistent argument proposed for consideration.

Regards,
Rosemary

And the rest is nonsense as usual. You have no consistent argument, either, just a bunch of handwaving conjectures, bad math and mendacious descriptions of your experiment.

If you look at the schematic below, you will see that the continuous NEGATIVE bias current is applied to the SOURCE of Q2 at the point marked "+" (from your old schematic) and that the positive pole of the bias supply is connected to the negative pole of the main battery, thus THEY ARE AT THE SAME POTENTIAL, system ground. Definitely no "negative signal" bias current is being applied to the GATE as you claim and misrepresent.