Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Rosemary Ainslie]

Q2 never turns OFF completely throughout the oscillation cycle. In the case where Q1 is present (and hence its body diode), some of the current path is through Q2 and some through Q1's body diode. When Q1 is removed, all the current path is through Q2. There is also the body diode of Q2; if it is ever forward biased (it never is in this case), it too provides a current path from Source to Drain.

.99

Not sure what you mean by 'forward biased'.  If the body diode is positioned that the anode is against the anode of the battery - as it IS and as it's shown in your schematics in your 'replacement' of Q1 - then it does indeed provide a path from the source to the drain.  Correctly you should be putting that body diode across Q2 with that same 'forward bias' as you put it.  And it will certainly PREVENT a discharge of positive current from the anode to the cathode of the battery.  The question is how does it allow a current flow from the drain to the source?  Because there's evidently a flow in this direction.

Now.  Back to my question.  What in the circuit enables that 'positive' current flow through Q2?  You really just need to give a simple answer here Poynty.  One assumes that there is a path because the current is most decidedly positive.  WHERE DOES IT FIND THAT PATH? 

It goes from the + at the battery terminals - then it flows to the Gate of Q2? or WHAT? on it's way back to the negative terminal of the battery.

Regards again,
Rosie

[poynt99]

Not sure what you mean by 'forward biased'.

With reference to a diode, "forward biased" means that the anode of the diode is at least 0.7V higher in potential than the cathode. When forward biased, a current path is formed in the direction from anode to cathode.

Now.  Back to my question.  What in the circuit enables that 'positive' current flow through Q2?  You really just need to give a simple answer here Poynty.  One assumes that there is a path because the current is most decidedly positive.  WHERE DOES IT FIND THAT PATH? 

It goes from the + at the battery terminals - then it flows to the Gate of Q2? or WHAT? on it's way back to the negative terminal of the battery.

Regards again,
Rosie

The normal flow of current through a N-channel MOSFET is from Drain to Source, as long as the Gate is sufficiently potentialized to turn the MOSFET on to a degree. The Drain to Source current path is created when the MOSFET is "ON". The degree to which the MOSFET is "ON" is determined in general by the Gate to Source voltage.

.99

[Rosemary Ainslie]

With reference to a diode, "forward biased" means that the anode of the diode is at least 0.7V higher in potential than the cathode. When forward biased, a current path is formed in the direction from anode to cathode.
The normal flow of current through a N-channel MOSFET is from Drain to Source, as long as the Gate is sufficiently potentialized to turn the MOSFET on to a degree. The Drain to Source current path is created when the MOSFET is "ON". The degree to which the MOSFET is "ON" is determined in general by the Gate to Source voltage.

.99

OK.  SO.  Assume that the drain to source is clockwise - positive as it relates to ground.  Correspondingly source to drain is negative - counter clockwise - negative as it relates to ground.  Now.  Look at the body diode.  That's only polarised to allowed a counter cockwise flow.  And the signal at the gate of Q2 is what?  Negative?  As you've shown it? -5v's? 

So.  WHAT path does the current flow when it's generated from the CEMF when it flows clockwise? Because it really DOES flow clockwise for 50% of each of those sinusoidal waveforms.

Thanks
R

[poynt99]

And the signal at the gate of Q2 is what?  Negative?  As you've shown it? -5v's? 

The voltage at the Gate wrt the Source (called VGS), has about a +5V potential (minus some drop across the 2 Ohm resistor). The VGS is also modulated up and down by the oscillation.

So.  WHAT path does the current flow when it's generated from the CEMF when it flows clockwise? Because it really DOES flow clockwise for 50% of each of those sinusoidal waveforms.

The current paths are a combination of the Q2 D-S channel, and through the Q1 body diode and wire down to ground.

.99

[Rosemary Ainslie]

The voltage at the Gate wrt the Source (called VGS), has about a +5V potential (minus some drop across the 2 Ohm resistor).

That's not what you show in your schematic.  I'll post it again.  You show -5 volts.  Are you saying that you're actually applying a positive or +5 volt signal at the gate of Q2?

The VGS is also modulated up and down by the oscillation.

I've looked at those voltages very carefully.  When you say 'up and down' are you implying that the voltage across the gate ALSO crosses ZERO during the oscillation phase?

The current paths are a combination of the Q2 D-S channel, and through the Q1 body diode and wire down to ground.

This explains NOTHING.  The question is this.  WHERE EXACTLY is the path in your Q2 - sans Q1 schematic - that allows a POSITIVE cycle of each half of that sinusoidal waveform??

R
here's that schematic again.
added.