As an alternative to replacing the MOSFET Rose, you can solder in a diode with high VBr (reverse breakdown voltage) across the Source and Drain leads.
This will tell you right away if the internal body diode is faulty or not. If the wave forms change significantly, then the body diode is damaged. If there is no change, then the diode is ok, and I am wrong.
Pay attention to the diode polarity if you try this.
.99
I'm going to take the 10% bet on this, Poynt. You're 90% sure it's the body diode. I'm not at all convinced. Your challenge, above, is a tiny bit of a heads-I-win, tails-you-lose situation. If the body diode is good and she adds another diode in parallel with different recovery characteristics and added capacitance, there will surely be some change in the waveform, however small, and possibly significant change.
If it's bad, and she puts a weak little diode across there, it could pop open (rare, but your theory says highly possible) quickly and she'd see no change after the first few moments of operation.
I just finished doing your no-Mosfet sim, exactly as you showed yours, with the switch and all the latest inductance changes, etc.
With the diode gone, the oscillation still dies out rather quickly...quicker than my MOSFETwith diode intact sim with the quite reasonable and highly possible gate drive impedance mismatch. Yes, it gets absolutely whacked down with the diode in place in your switch circuit, for sure. No argument there.
Why doesn't it get equally quickly whacked down with my MOSFET-installed model? My theory is because the circuit is operating under normal Nyquist feedback common source mode. Gate and drain are already 180 degrees out of phase; add another 90 and you have the verge of instability.
The switch (non-Mosfet) model has no gate to accept feedback, no parasitic Crss or Ciss capacitances, source lead inductance, etc. That's why the ringing tank dies out pretty quick (due to Rload losses) even without the diode and really quick with it. There is no possible gain/feedback mechanism with the switch simulation.
As far as the proof being in the pudding, I don't see where you have posted any continuous oscillating circuit at any time, switch or MOSFET. I looked, but did not find. Did I miss it? At least I did post one configuration (with impractically large gate inductance added) that rang on at a non-decaying level.
Finally, I am still at a loss to understand why you say that unless the diode is open, the Vbatt (at the point you suggest probing) can't ever go below the nominal DC battery voltage. The only thing the diode clamps is when the actual drain itself tries to go south of the source. Between the actual positive battery terminal and the drain itself, there are two inductors (2.5nH and 10nH) plus the load resistance. Those nodes are relatively free to go anywhere between the peak of the first flyback spike all the way to the source voltage and a volt or two below (source "ground"). Of course they will swing below the battery positive node during the oscillations. The body diode doesn't turn on until drain is south of source. Nothing to do with battery positive voltage.
That's my position and I'm taking the 10% odds against your 90% confidence level the other way. Until I'm re-educated, which is entirely possible.
Unless my position is so unreasonable or poorly stated or so hopelessly stupidly flawed that it deserves to be dismissed with appropriate ridicule, I await your corrections to my thinking at your leisure, Sir!
Humbugger
Author
Topic: The Rosemary Ainslie Circuit (Read 477192 times)
