Claimed OU circuit of Rosemary Ainslie

[MileHigh]

Aaron:

It has CLEARLY been explained by the skeptics that the entire oscillation effect is a red herring and therefore, they have disqualified themselves (on many occasions) from even having enough experience to discuss this circuit.

Without even looking at your new clips (which I will later) here are my first two comments/recommendations:

1.  Check the 555 output to see if it is stable or not.  If it is running at a high frequency, is it the same frequency as your spikes?

2.  If the 555 output is stable, then wet the tip of your finger and start touching the circuit to see what happens.  The first place that you want to touch with your fingertip is the MOSFET gate input.

And I have some questions:  How come your battery voltage seems to have about one volt peak-to-peak of noise on it?  Can you trigger on it to see if it is regular?  How about seeing the 555 output displayed at the same time as your battery voltage?

MileHigh

P.S.:  I just watched the clips.  The unknown is that you did not look at the 555 waveform, so the jury is still out on that.  There is the distinct possibility that the 555 circuit is too suspectable to outside influence as Hoppy stated and the oscillation loop is between the 555 circuit and the MOSFET/coil-inductor circuit.

Certainly the oscillation can be due to the MOSFET itself going into a metastable on/off positive-feedback defibrillation.  It all looks very impressive, but all that means is that a lot of energy is being burned off in the MOSFET.  Yes indeed there are spikes going back into the battery, but as previously stated, it is simply a revolving energy door.

It appears that the current consumption may be going up as Hoppy stated.  In previous postings I said that the current consumption would go down because the inductor in the coil would choke the current off.

These two statements are seemingly contradictory but they are not.  The precise on/off duty cycle for the MOSFET in defibrillation will determine how much current ends up flowing through the coil-resistor.  The amount of inductance in the coil-resistor is a big determining factor also.

Anyway you have a nice piece of equipment to work with and it should be possible to figure out what is going on.

A little tip about the dispay when the MOSFET is in defib:  There is probably a "one-shot" display mode where you just show a single captured frame after a trigger happens.  It is worth trying that out so that you can look carefully at a single snapshot in time.  This removes the distractions of the display updating all the time.

MileHigh

[TinselKoala]

Well, meanwhile I have been playing with the circuit of Aaron's that Hoppy recommended earlier.
The only way I can get it to make a short ON duty cycle is to adjust it so that the output is actually clipped a bit and the circuit is definitely in a non-linear mode.
There's a point where adjusting the freq pot in the increasing freq direction, then suddenly the freq decreases, with a very short on duty cycle, and the freq is quite unstable at that point. I have no doubt that the mosfet will amplify this craziness which is coming from the 555 all on its own.
I'm just testing the clock alone; later I'll hook it into the rest of the circuit and see if the interaction with the mosfet does anything interesting.
But first I need to refresh my beverage. This is thirsty work.

(And Aaron: MH's advice is great. But you can forget the finger, just use your tongue.)

[MileHigh]

Harvey:

However, the battery may not be accepting any current during that pulse duration. Remember, the battery has an internal inductance, so it will take time for the current to start flowing into it after the voltage is applied.

That is an excellent point.  The coil acts as a current source for the brief time it discharges.  Let's assume in simplified terms, the current source hits the wall of the switched-off MOSFET, and this generates a very high voltage.  Some of the energy in the voltage spike is dissipated in the MOSFET itself, and some is reflected back and goes back to the battery.

Normally the battery would muffle the reflected high voltage spike down to a very low voltage if it was directly absorbed into the battery.  However, like you said, there can be another "stray inductance" or "battery inductance" that is in fact preventing the battery from absorbing the bulk of the energy from the reflected spike and you instead get a ringdown.

Everything I am stating is just a preliminary guess so I can't be held to it.

In essence we are talking about some sort of transient analysis associated with the 555/MOSFET in defib, or the MOSFET itself in defib.  It is not trivial but can be figured out with some focused investigation.

The downer for all of the believers is that all of this is "normal" and just represents energy from the battery being dissipated in the circuit.  That can be confirmed with the DSO in conjunction with the thermal analysis.

MileHigh

[Harvey]

TK still seems to be wrestling (or Wrastlin' as they say in Springfield) with the power dissipation portion of the 400ms Energy Cycle I posted in the other forum. There is no dispute that all of the energy represented is stored in the inductor during the first 200ms. What happens to this energy now that it is stored? And how much energy is it? To simplify this, I took the current slope of the sensing resistor and averaged it. The slope goes from near nothing to about 942ma - so being generous I said 800ma flat. We take the battery voltage, times the current to calculate the power. We then multiply that times the time interval to determine the energy in Watt Seconds or Joules (3.84J). That's what we put into the circuit and then the flow of energy from the battery ceases. 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. If we knew the frequency, then we could use TK's approach and arrive at the same value. Just think of the ringing as available AC power running through a heating element in your house for 200ms. How many KWH did your little wheel turn?

Now, to confuserize things a bit (dropped my fuse in my beverage so now its soakin' fused) - power is never consumed by inductive or capacitive reactance. This means that an ideal inductor and capacitor in resonance will never stop ringing. The consumption of power is in the resistance. So the ringing you see represents available power and when it fully dissipates then the power is considered to be consumed (even though we all know its just converted into something else).

So where did the apparent extra energy come from?
Show your work. 

[poynt99]

So where did the apparent extra energy come from?
Show your work. 

Tried twice and no one got it ? (or did they?)

It's apparent power, therefore not real. Volt-Amps, not Watts.

.99