Testing the TK Tar Baby

picowatt · April 25, 2012, 05:22:45 PM

Quote from: Groundloop on April 25, 2012, 05:14:09 PM
PW,

You are right, but let us keep this simple. When the FG is positive then the current from the FG
will be burned as heat in the 50 Ohm and not in the LOAD. That was the point of the drawing.

GL.

When the FG output is positive there will likely be no current flow through the FG or 50R.... unless, the FG drive and drain pull down is sufficient to allow the left body diode to conduct. Of course there will be a short burst of current to charge the right MOSFET Ciss (and discharge the left), but once charged, and at say +9 to +10 FG drive level, the left body diode will likely not conduct and there will only be right MOSFET gate leakage current being drawn from the FG for the bulk of the FG's positive cycle (and the gate leakage is a very small amount of current).

PW

Groundloop · April 25, 2012, 05:33:39 PM

Quote from: picowatt on April 25, 2012, 05:22:45 PM
When the FG output is positive there will likely be no current flow through the FG or 50R.... unless, the FG drive and drain pull down is sufficient to allow the left body diode to conduct. Of course there will be a short burst of current to charge the right MOSFET Ciss, but once charged, and at say +9 to +10 FG drive level, the left body diode will likely not conduct and there will only be right MOSFET gate leakage current being drawn from the FG for the bulk of the FG's positive cycle (and the gate leakage is a very small amount of currentl).

PW

PW,

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.

Agree?

GL.

picowatt · April 25, 2012, 05:35:29 PM

Quote from: Groundloop on April 25, 2012, 05:33:39 PM
PW,

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.

Agree?

GL.

GL,

Agreed!!

PW

Groundloop · April 25, 2012, 05:46:32 PM

Quote from: picowatt on April 25, 2012, 05:35:29 PM
GL,

Agreed!!

PW

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.

picowatt · April 25, 2012, 05:50:36 PM

GL,

If we want to be even more precise, if the drain pull down and FG drive is sufficient to turn on the left body diode, the current through the load will actually decrease slightly. And, the bulk of the FG drive above the left side body diode clamping voltage 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. If 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.

But, 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.

So again, agreed...

PW