Testing the TK Tar Baby

[TinselKoala]

Sorry.  I should have referenced this separately.If the function generator applies a negative to Q1 - are you conceding that it then applies a positive to Q2?  Because that is NOT what picowatt is claiming.

And this...
Everything that needs explaining that relates to current is FULLY explained in our second paper.  Intrigues me you guys go to some considerable lengths to avoid mentioning it.

Rosie Pose

Ainslie, if one terminal of a BATTERY is positive, what is the other terminal?
If one terminal of a POWER SUPPLY is positive, what is the other terminal?

If one terminal of a FUNCTION GENERATOR is positive.... what, then, is the other terminal?

You are once again misrepresenting what PW said and what I say, and you are ONCE AGAIN displaying your abysmal ignorance and unwillingness to learn.


And THIS is what your "second paper" presents as an "explanation": a set of incoherent, if entertaining, cartoons that have no connection to any reality. Why does nobody mention it? Because it's meaningless nonsense, that's why.

[Rosemary Ainslie]

Ainslie, if one terminal of a BATTERY is positive, what is the other terminal?
If one terminal of a POWER SUPPLY is positive, what is the other terminal?

If one terminal of a FUNCTION GENERATOR is positive.... what, then, is the other terminal?

You are once again misrepresenting what PW said and what I say, and you are ONCE AGAIN displaying your abysmal ignorance and unwillingness to learn.

You're arguing MY argument.  NOT picowatts.

Rosie Pose

[TinselKoala]

Not actually TK. The Source leg of Q2 FLOATS.  It is NOT connected to the circuit.  HOW CAN YOU CLAIM THAT IT IS?  JUST CHECK OUT THE SCHEMATIC.

R

I have notated the connections of Q2's source leg to the rest of the circuit, on YOUR OWN SCHEMATIC.
IT DOES NOT "FLOAT".

[Rosemary Ainslie]

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

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.  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

Let me add this while I can still modify this post.  I'm referring to the circuit that he uses with just one MOSFET.  Hope that's clearer.

Again
R

[picowatt]

ABSOLUTELY NOT.  Q2 is NEVER on during the oscillation. 

R

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.