Quote from: poynt99 on April 25, 2012, 02:27:12 PM
The sim would not work properly with 50 Ohms.

Of course the FG has a 50 Ohm output. 

.99,

Do you have any idea as to why the sim would not work with 50R?

PW

Only two things come to mind, although there may be other better reasons:

1) There are no RF effects in the simulation
2) The MOSFET model is not perfect.

Quote from: poynt99 on April 25, 2012, 09:28:40 AM
GL,

Your second paragraph in the diagram is not correct.

There will be no power added to the circuit in the scenario shown. What you have there is simply an isolated VGS bias that turns ON the right MOSFET. There isn't even a return path for the FG source, so it can't provide any power to the circuit.

.99,

Yes it is. When the FG goes positive pulse the RIGHT MOSFET will switch on. The current output from the FG
will go through the internal diode in the LEFT MOSFET and will be burned as heat in the FG resistance.
And that is what I wrote in the text in the drawing.

GL.

Quote from: Groundloop on April 25, 2012, 04:50:11 PM
.99,

Yes it is. When the FG goes positive pulse the RIGHT MOSFET will switch on. The current output from the FG
will go through the internal diode in the LEFT MOSFET and will be burned as heat in the FG resistance.
And that is what I wrote in the text in the drawing.

GL.

GL,

Even when fully on, the right MOSFET will only pull the drain down to around 11 volts (assumes 72 volt supply battery, 11R load resistor, and 2R as MOSFET on resistance).  The left body diode will not conduct until the gate voltage of the right MOSFET is about 2 volts (body diode fwd voltage) above the drain, or in this case approx. 13volts.  So, yes, if the FG was outputting +15volts open circuit, for example, there may be approx 2 volts drop across the 50R due to the left body diode conducting.  The supply battery voltage, the level of the FG drive, and the MOSFET Rdson, will determine if sufficient voltage pull down is acheived at the drain to allow the left body diode to conduct.

PW

Quote from: picowatt on April 25, 2012, 12:22:17 PM
.99,

I think GL's simplified schematics provide a better tool to help Rosemary understand what the FG is doing.  Particularly the first drawing posted in reply #732 as it regards to her Q2 array.

As for GL's second paragraph, I spotted that too but did not at the moment want to complicate things.  The body diode on the left will, however, conduct and clamp the right side gate voltage to a couple volts or so above the drain voltage.  When the right side MOSFET is on, the drain voltage will be dependent on the battery voltage, the load resistor drop, and the MOSFET on resistance.  With a 72volt supply battery, the gate voltage would be clamped at around 14 volts and with a 48 volt supply battery, the gate voltage would be clamped at closer to 9 volts.  Any voltage applied to the gate by the FG (or a pwr supply) that exceeds this "gate clamping voltage" will be dropped across and dissipated by the 50R.  At the supply battery voltages typically used by Rosemary, and with a typical FG, this dissipation from gate voltage clamping would be minimal. (if any at all, depending on the FG settings)

Again, I really like GL's first drawing and was hoping it would help Rosemary understand how her Q2 array is being turned on and off.  I was then going to attempt to help her understand the current limiting/regulation that occurs regarding the voltage applied to the 50R and the threshold voltage of the MOSFET, that limits the Q2 DC current to around 200ma or less (thought I would discuss only DC conditions for a while).

Then, in response to her wondering why current flow at the CSR is not observed, hope to make her see that the 200ma of bias current will make the CSR only 50 millivolts positive, which would be barely visible at her scope settings.

It sounds like you have covered all this ground in the past.  Apparently it has yet to "sink in".

PW

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.