another small breakthrough on our NERD technology.

[Rosemary Ainslie]

As an example; If one was to take the FG leads that are connected to the G and S of a MOSFET (normally + to the Gate, and - to the Source), and swap them around, THAT would be an inversion.

OK.  Then the answer is simple.  We DO NOT connect the G (gate) and S (source) of the MOSFET like this.  Ever.  But I think I'm getting into your argument.  I'll give it another shot.

BRB (be right back)
R

[Rosemary Ainslie]

OK.  We now understand that 'inversion' is defined as follows 'the physical transposition of the signal generator's probe from the gate of Q1 TO the gate of Q2.  In which case - in order to function - then the ground of the signal generator would still be applied to the Source.   And the assumption is then made that the Q2 is the 'functioning' transistor.  In which event Q2 has simply REPLACED Q1.  This would result in a voltage that is steady - greater than zero.  Until - again that switch was opened by the application of a negative signal applied to that gate.  There would be no evident inversion of the waveform.  It would remain greater than zero.  There would simply be an alternate path opened.

So.  When you state, as you do here ...

The first inversion is caused by connecting the FG- to the Q2-Gate, and the FG+ to the Q2-Source. It would normally be connected in reverse of this.

...then I'm not sure that it's right.  The ground of the signal generator (FG-) is NOT connected to Q2-Gate.  It's connected to Q2 source.  It's only link to the drain is via that Gate - through the drain rail.  The logical and unarguable assumption then is as you've stated it.  The applied signal from the signal generator (FG) has simply changed from Q1 to Q2.  The one has replaced the other.  In which case?  We SHOULD, by rights, have a continual DC current flow discharged by the battery with, at the most, some spiking at the transitional phases of the duty cycle.  That's NOT evident.

The second inversion is caused by introducing negative offset to the FG (as shown in the video), such that the FG output, if measured with a + probe on the + terminal and - probe on the - terminal, would measure a negative voltage when the FG output is in a LOW state, and a 0V or slightly positive voltage when the FG output is in a HI state.

Ok.  I think I'm beginning to see it.  The applied positive signal now changes to a negative signal.  You're suggesting that during that transitional phase then the actual applied voltage from the signal itself, gradually diminishes - over time - from say - +5 volts to -5 volts - as the signal changes and the new voltage level kicks in.  Which is fair comment.  BUT.  While the signal at Q2 is changing back to an 'open' or 'negative' signal - then simultaneously the signal at the gate of Q2 is changing in anti phase.  The sum of both those changes would allow precisely the same amount of delivery of current from the battery.  What it would do is possibly show a small 'drop' from say a high of 12 volts to a zero - then back to 12 volts.  In a partial oscillation that would still be entirely above ground. There would be no oscillation as seen in our waveforms as the current flow from that applied battery voltage would always be maintained.

The resulting Q2 V_GS voltage is such that Q2 will turn partially ON (the bias) and completely OFF.

During which time there is some moment when the applied signal is crossing zero and there's no voltage at all. Agreed.  But that would still not account for the zero crossing as evident in the oscillation. See the above point.

As Q1 G-S is connected in reverse of Q2, Q1 is always OFF with this setting.

Q1 ground source is not connected in reverse.  It is never disconnected from the circuit. 

Regards,
Rosemary

added - two points - an another - so.  3 points added.  I hope they're highlighted.
and took out 'to' - repetitive
Sorry and a whole lot more corrections.  I really need to check the 'preview' more often.  Apologies.

[poynt99]

So.  When you state, as you do here ......then I'm not sure that it's right.  The ground of the signal generator (FG-) is NOT connected to Q2-Gate.  It's connected to Q2 source.

I'm going to try another slightly different approach.

I'm going to take your diagram from your paper, Fig. 1, and focus ONLY on the connections between the FG and Q2 for the moment. I am going to erase all of the components and connections (wires) except the FG and Q2, and the electrical connections between them. I am not going to draw any new lines in. See the changes below.

1) From the cleaned up diagram, which FG lead is connected to the Q2 Gate lead?

2) From the cleaned up diagram, which FG lead is connected to the Q2 Source lead?

[Rosemary Ainslie]

I'm going to try another slightly different approach.

I'm going to take your diagram from your paper, Fig. 1, and focus ONLY on the connections between the FG and Q2 for the moment. I am going to erase all of the components and connections (wires) except the FG and Q2, and the electrical connections between them. I am not going to draw any new lines in. See the changes below.

1) From the cleaned up diagram, which FG lead is connected to the Q2 Gate lead?

2) From the cleaned up diagram, which FG lead is connected to the Q2 Source lead?

There is only one FG or signal generator.  The probe is signed  '+'  the ground of the signal generator is signed '-'.  This does not vary.  What varies is the applied signal at the probe.

R

[poynt99]

There is only one FG or signal generator.  The probe is signed  '+'  the ground of the signal generator is signed '-'.  This does not vary.  What varies is the applied signal at the probe.

R

My focus for the moment is ONLY on the wired connections between the FG and Q2. The actual signal from the FG isn't the issue at this point. Let's establish FIRST which FG lead is connected to which lead of Q2 please.

Now, with reference to the two questions above, please post your two answers.