Claimed OU circuit of Rosemary Ainslie

[TinselKoala]

Hmm. OK, I believe I have reproduced what Hoppy reports. And I see why Aaron has put a heatsink on his mosfet.
The 555 timer circuit is definitely behaving in a "non-linear" way, and it gets worse when you run it to the mosfet gate.
From what I'm seeing it looks like this: As the 555 timer is adjusted to that critical point where the frequency stops increasing and drops abruptly, keeping a very short duty cycle: It looks like the mosfet shouldn't be turning on. But: in my case, what happens is that the Baseline of the 555 pulses rises from zero to about 5 volts. If you are on AC coupling you might not notice this. Of course this, with the short duty cycle, is enough to turn the mosfet almost completely--but not quite full on 100 percent of the time. The Simpson analog meter in series with the battery shows about 1.8 amps, and I see 500-600 mV drop continuous DC, no pulses but with a little ripple, on the Ainslie point B, across my 0.25 ohm shunt. And the mosfet gets very hot very fast and so does the load. If I use a higher inductance, like Aaron is using, I get things that look very similar to what Aaron shows in his latest stills.
I'd like to see, at this point, on Aaron's digital scope, a shot of the oscillations "across the battery", that is, Ainslie point "A", on one trace and the Ainslie point B (voltage drop across the shunt) on the other, during these "oscillations." Since that's a 4-trace scope we might also be able to see the raw output from the timer too.

Let me emphasize here that what I am seeing is the mosfet responding to what it's being told; and it's being told to stay just on the edge of being full on 100 percent of the time.
I think.
So that is probably why Aaron is now using a heatsink on his mosfet.

[TinselKoala]

Like I said in my little MosfetMotel example, the apparent excess energy comes from the same place the extra dollar went. Into mistaken problem setup and calculation, caused by the garden-path patter of the initial story combined with some correct numbers to lull the complacent mind.

Not only that, but here's the mistake:

Next we see an immediate 468 Watts of Power at the inductor. How much energy is that? And what will happen to it? If that power were a straight DC wattage, then the energy would be easy to calculate, W x t = J. But it's not, it is sinusoidal, and therefore we must take its RMS value if we intend to flatline the energy. But this is not a consistent thing, that 468W only represents the first peak in the first cycle. Each cycle is less and less. And over a 200ms period it decays enough that we can just say its back to zero for easy calcs. If you draw a line through all the tops of the peaks you'll have a slope and relative to the baseline it makes a triangle. A little simple geometry tells us if we drop the left side of the line and raise the right side of the line, then we will get a level line and our triangle will be a rectangle. The area of both is the same. Therefore, if I divide my initial power by 2 I get the same area under my line. Now I need to apply the RMS factor of .707 and this gives us the flatline power value. So now we have the flatline watts, we multiply it by the time period and we get the Joules dissipated in the inductor.

You are adding up the same energy, over and over, by using your mistaken calculation method. RMS Power does not represent any useful real quantity in this case. See the pdf file linked below. In addition, you are conflating energy, which has the units watt-seconds, with your calculation which has the same units but does not represent an energy--because you got to it wrong.
ALL the energy available from the ringdown is in the area of the first peak. All of it. It moves out of the inductor into the circuit's capacitance, and then the current changes sign and it comes back. Lather rinse and repeat. It loses a bit each cycle. THIS LOSS IS THE POWER DISSIPATION. If there were no loss, the ring would not ring "down" but would continue forever. If there were any gain, the ring WOULD INCREASE IN AMPLITUDE.
Over the entire time length of the ringdown, all the energy, which is contained in that first peak ONLY and has been sloshing back and forth, is lost to heat. You cannot compute the energy the way you are doing it. The proper comparison is to take the area of that FIRST BIG SPIKE. That's the energy that was stored; that's all the energy there is. Compare the value of that spike, in Watt-seconds, with your input energy value.
If you want to know the energy balance in the ringdown, you need to integrate the waveform, not "rms" it and multiply by the time. You can even tell, by the shape of the decay envelope, whether more power is being dissipated in inductances or capacitances.

And You know all this full well, Harvey, so quit screwing around.

So, you take that 468 Watt initial spike, and since you can't solve that nasty shape analytically you use the tried and true Grungey-Kuttoff method of numerical approximation --which can overestimate by as much as a factor of 2 or 3 -- and call it a rectangle of height 468 Watts and width, er, um, 1/2f seconds. At 1.6 MHz which is typical for my ringdowns with my load, that's, let's see, my gosh, nearly a hundred and fifty MICROJOULES. Sorry, my caps key got stuck I was so excited by the amazing amount of energy stored in a small inductance.
But that's the real number to be compared against the number you cite as input energy. And if you go along the ringdown and add up all the little pieces by which the consecutive peaks decrease, you will be back to the original 150 microJoules.

Unless I misplaced a decimal again.

[poynt99]

There is very little power dissipated in the resistor while ringdown occurs (I and V are 90º apart), and much more when the inductor is being energized. The power in the inductor is huge in comparison to that in the resistor because of the relatively large Inductance to resistance ratio. This huge (mostly apparent) power will drain slowly due to the long ringdown.

.99

[TinselKoala]

Uh huh, that's another way of looking at it. Ultimately, a tiny bit might get back into the battery during the first couple cycles, and that contributes to the decay. But ultimately the stored energy is all dissipated as Joule heating of one kind or another. I think.

Now, I'm really getting paranoid that Harvey is setting me up for a big embarrassing fall. OK, at this point I don't care too much, I have as much respect for a good fisherman as anybody. I flunked diffy-Q's anyway.

OK, what's behind door number three?

(EDIT: and of course, I mean the power is being dissipated in the real resistances associated with real inductors and real capacitors. In the real situation, you can tell from the envelope in which _components_ the power is being dissipated, I should have said. I know that ideal inductors and caps don't dissipate power through Joule heating. They may radiate it as RF or something...OK, tear me apart.)

[poynt99]

I don't think there's a big setup.

I could be wrong but I think my last couple posts point out the little trick Harvey was playing to see if anyone would catch it. He 'apparently' even gave a pretty big clue.

.99