Quote from: Groundloop on April 25, 2012, 05:46:32 PM
PW,

Great :-)

So now we know that only the negative pulse from the FG will inject power to RLOAD.
I know RA does not agree to this but this a fact.

It is also a fact that when the FG is negative then the circuit runs as an Colpits oscillator.
The 180 degrees feedback needed to run is through the MOSFET internal D to S capacitance.
The MOSFET runs in the linear region.

I know RA has a question about the negative measurement over
the SHUNT resistor when the oscillator runs, but I will take that later.

GL.

GL,

I think it might be best for you to discuss DC conditions with RA for a time as if there were no inductances involved, and then move on to AC conditions.

I am unsure if RA fully understands or accepts the DC conditions indicated by your first set of drawings (your reply #732).

PW

What on earth are you trying to say picowat?

Quote from: picowatt on April 25, 2012, 05:50:36 PM
GL,

If we want to be even more precise, if the drain pull down...

What drain pull down?  We're talking MOSFETs not IC circuitry.   MOSFETS are by definition 'solid state' switching devices.  So. There is no 'drain pull down'.  What there is, is a positive or negative signal applied by the function generator or by 555 circuitry - to the gate of Q1 in anti phase to Q2.  That's it.  And the only thing that then happens is that the current from the battery is either enabled or not - depending on the MOSFET type and the level of potential difference from the supply.  We're using N channel 'types' therefore the gate oxide is designed to enable a flow of current from the battery when that applied signal is positive.

Quote from: picowatt on April 24, 2012, 06:03:05 PMand FG drive is sufficient to turn on the left body diode, the current through the load will actually decrease slightly.

What?  When will this 'decrease' take place?   The current flow will be enabled to the extent that the gate is open.  And that rate of current flow is determined by the resistance in the path of the current from the supply.  that's it. 

Quote from: picowatt on April 24, 2012, 06:03:05 PMAnd, the bulk of the FG drive above the left side body diode clamping voltage ...

What on EARTH is 'the left side body diode clamping voltage'?  The ONLY clamper in that circuit is from the function generator's 'off set'.  And if there's a 'left side body diode' during one phase of each switching cycle there's also a 'right side body diode' at the other.  They both have IDENTICAL functions.  The only difference is their connection to the the battery source rail which in the one is enabled and in the other it is NOT enabled. 

Quote from: picowatt on April 24, 2012, 06:03:05 PM'...will be dissipated in the 50R and to a lesser degree in the left side body diode and in the right side drain to source resistance.

Right side drain?  Source resistance?  What are you talking about?  If you mean the drain or source legs of Q1 or Q2 then say it.  If you are referring to the drain or source rail of the battery then say it.  This entire phrase is entirely undefined.  It is the simple 'rule' of science that terms must be defined and clearly expressed.  Anything less and we're NOT talking science. 

Quote from: picowatt on April 24, 2012, 06:03:05 PMIf drain pull down and FG drive are insufficient to turn on the left body diode, then no power will be drawn from the FG or dissipated in the 50R.

Again.  WHAT 'drain pull down'?  And what left and what right?  Do you mean Q1 or Q2?  When the offset of the function generator is adjusted that its signal applied to the gate of Q1 is positive yet has been 'pulled down' below that positive setting - then it will not allow the flow of current from the battery supply as the net value of that applied signal is NOT positive but negative.  Is that what you mean? 

Quote from: picowatt on April 24, 2012, 06:03:05 PMBut, as you say, if there is any current dissipated by the FG during the positive output portion of the FG cycle, it will not be dissipated in the load.

the function generator does not typically 'dissipate' any energy on the circuit at all.  If you are referring to the 'heating' of the MOSFET TRANSISTORS then that is the result of the 'applied voltage' from the function generator.  And the heating of the MOSFET is then as a result of the extent to which that applied voltage is then able to deliver current from the battery source supply - OR NOT. 

I wonder if it would be as well to try and keep ones terms as clear as possible.  I think we're all rather tired of obfuscation.  It simply confuses the argument.  With respect. 

And Groundloop - just a small point.  You would need to justify your argument when there is a steady DC application by the function generator of a negative voltage at the Gate of Q1.  No switching at all.  Because that's what is evident.

Kindest regards,
Rosie

Groundloop,

Quote from: Groundloop on April 25, 2012, 05:46:32 PM

So now we know that only the negative pulse from the FG will inject power to RLOAD.  I know RA does not agree to this but this a fact.

Actually what you've pointed to is that the signal from the supply source and the batteries are 'in series' during the application of a negative signal at Q1.  There is no-one would argue.  I agree.  Wholeheartedly.  And, as I pointed out in my previous post - this is continual during an extended switching phase of the function generator.  We literally apply this negative signal for upwards of 2 minutes.

Quote from: Groundloop on April 25, 2012, 05:46:32 PMIt is also a fact that when the FG is negative then the circuit runs as an Colpits oscillator.  The 180 degrees feedback needed to run is through the MOSFET internal D to S capacitance.
The MOSFET runs in the linear region.

I am not sure how the Colpitt's oscillator runs.  But I believe it does not use MOSFETS - and therefore does not 'typically' have that internal body diode and the 3 pin connection to the circuitry and signal generator - simultaneously.  Are you saying that one MOSFET - on its own - would then enable that 'Colpitt's oscillation'?  What I DO know is that we can, indeed, just use the one transistor - as Poynty Point's already confirmed.  And that we then only need to apply a continual positive signal to the source rail of the battery supply to generate that oscillation.  Which is one of the tests that we've got lined up to show you all.  So.  I can't comment regarding that Colpitt's oscillation analogy.  But you're CERTAINLY correct in that continual oscillating waveform as a consequence.

Quote from: Groundloop on April 25, 2012, 05:46:32 PMI know RA has a question about the negative measurement over the SHUNT resistor when the oscillator runs, but I will take that later.

It's not the only thing Groundloop. Quite apart from the evidence of a negative voltage at the 'onset' of that oscillation - is the fact that we measure that negative wattage.  That's pivotal.  But I like your 'style' of argument.

Kindest regards,
Rosie

Sorry Groundloop.  I'm working these posts reading 'latest' first.  Just a point here.

Quote from: Groundloop on April 25, 2012, 05:33:39 PM
My point was that the current (if any) from the FG at the positive pulse will NOT go to the RLOAD but will be burned as heat in the 50 Ohm instead.

Not so much.  On both the 555 switching circuits AND on our NERD Q-array - we're able to heat the load to significant levels purely from that oscillation.  No current drawn from the battery at all during the 'on' phase - or when the signal at Q1 is positive.    Not sure where you're going with this - but bear that in mind.

Kindest again
Rosie

Quote from: Groundloop on April 24, 2012, 07:37:07 PM
.

Rosemary,

I was having a discussion with GL regarding one of his drawings.  He did not label the MOSFETs in the drawing we were discussing, so we were referring to them as "left" and "right".  As to the rest, I am not surprised that you were not able to follow along.  There was no "obfuscation", it was a discussion between people with a common technical background in electronics using terms that one with such a background would easily be able to follow.  It would only appear as "obfuscation" to someone without that background.


To continue our discussion, if you so desire:

In the drawing that GL graciously provided in reply #732 on page 49 of this thread, there is a "supply battery" and a "bias battery".  The bias battery is the battery connected to the 50 ohm resistor at the MOSFET source.

Do you understand and agree that the MOSFET is turned on when the bias battery negative terminal is towards the 50 ohm resistor?  (This is the instance on the left above)

Do you also understand and agree that when the MOSFET is turned on, current flows from the supply battery, thru the load resistor, thru the MOSFET, thru the 50 ohm resistor, through the bias battery, and back to the supply battery?

Do you also understand and agree that when the bias battery positive is towards the 50 ohm resistor the MOSFET is turned off and no current flows?

These are simple yes or no questions and a response with just yes or no answers would greatly facilitate further discussion.

PW

ADDED location of GL's drawing discussed