Claimed OU circuit of Rosemary Ainslie

[Hoppy]

That's the  Drain voltage, point P2 in my test plan (green trace). The Red trace is point P3, the Source/shunt voltage. Yep, about 750V peak.  Very odd that my power supply or filter caps haven't blown up yet, LOL. 

Varying the grounding by length and position etc., alters the shunt wave form a great deal, and hence can have a huge effect on the power calculation, even though that spike only lasts for 150ns. Overall for the frequency (T=400us) I am using, that represents an average power over once cycle of about -1W.

In Glen's case at 500kHz, it is a completely different story. We go down from about 400us period to about a 2us period, yet the power spike is approximately the same width. In glen's case it represents about -100W over one cycle, and accounts for the highly skewed results once crunched in Excel.

.99

Your improved star earthing should give more reliable results now. As you have found, the grounding length makes a big difference to results. I had been trying in vain to get the importance of this earthing across to Aaron before I was ejected from his forum for making critical remarks. As you say Glen's results will be even more crazy running at 500KHz!

A good way I have found to 'star earth' is to use a strip of thin copper and solder all earth leads onto this strip as closely as possible. The earth leads to different parts of the circuit should be as short as possible and preferably all the same length. The copper strip is then conected directly to the neg battery terminal or connected to a PSU ground using a very short and thick lead, again soldered close to the other leads. Another way is to use ring crimps on the ground leads and solder as well as crimp the rings to the leads. These can then be tightly bolted together, first ensuring the the crimp surfaces are very clean.

Hoppy

[PaulLowrance]

I quite doubt it would indicate COP>1 by ignoring the spikes. I'm not sure how one would go about doing that anyway?

If we replace all the spikes with the value it would be as a pure resistance, then most likely we would be left only with Ohmic calculations, and they should come out as per theory.

The premise of Rose's theory is that the spike itself generates (or re-generates) excess power in the return portion of the cycle. Killing the spikes from the data dump would preclude this theory entirely.

.99

That makes sense if that's her theory, but it's very easy to be fooled by spikes, thinking it's current flowing through the resistive component, when it's usually the current pulses through the reactive components.

That's why from the start I suggested the sure method of measuring how fast each component will heat up in a given time period, and then compare that to the control experiment. That's very easy to do that.

Paul

[poynt99]

Glen,

Take your pick, all the photos look substantially the same. Here are two of yours, one from the OTS resistor, and one from test #5. You apparently missed my point anyway. Yes the peak voltage is higher in the second shot and will result in a higher skewed power.  I have 750V peaks, so I guess I'm obtaining even more "gains" then you with my lowly OTS resistor  

The focus of my post was and has been on the shunt voltage.

.99

[MileHigh]

Rosemary:

Just need to draw your attention to the 'second part of the cycle'.  If there is enough energy to both dissipate heat and recharge the battery then it must - theoretically - be more energy stored than was first delivered?  Surely?

I don't know how you see that.  Suppose 100 units of energy are delivered by the battery in the short ON pulse.  50 of those units become heat in the load resistor, MOSFET and shunt resistor.  The other 50 units end up sitting in the MOSFET drain-source capacitance and stray capacitance.  This very small capacitance is at a very high voltage.  The current flow then reverses direction and 25 of these units become heat in the various resistive elements and the other 25 units go back into the battery.

Just to touch on how the battery handles these 25 units of return energy again:  The safest statement is that there is an unknown split between this energy becoming heat and recharging the battery.  Some will argue that most of the energy recharges the battery and my argument is that most of the energy is lost as heat.

You seem to be having trouble with current flowing back into the battery becoming heat.  For starters any time you charge a battery in whatever charge state it may initially be in some of the charging energy becomes heat.   It's simply impossible for 100% of the electrical energy that you put into the battery to become chemical energy.  Supposing that you have a fully charged battery and you start putting current spikes into it.  What do you think happens in this case?  There is only one thing that can happen, the battery gets hot.  There are no more available molecules in the battery for the chemical reactions to take place, so the supplied electrical energy has to become heat energy.  I am simplifying things a bit here because there are some molecues available, but very few.  Some molecules will eventually "come out of hiding" and "swim onto the playing field,"  so to speak.

And the measurements here comply to the apparent battery draw down.  That we need to account for some extraordinary wattage measured across the resistor without any apparent work - is a given.  But even in measured heat dissipated over the resistor - there is clear evidence of excess energy over the supplied.

Well .99 has highlighted the fact that the "extraordinary wattage" measured across the load resistor is in fact an illusion because the wattage derived from the thermal measurements is at least an order of magnitude smaller than the DSO data (excluding .99's recent measurements, but they are suspect also).  Your paper indicates that you measured real thermal heat showing a COP of 17, it was based on real world measurements.  The real world measurements are not showing that for Glen or .99's setups.  If the load resistor measurements done with the DSO are suspect, then we have to conclude that the battery or power supply and MOSFET DSO measurements are suspect.  It is really unfortunate that this is the case, I was not execting that to happen.  However, .99 stated that it is not the equipment, and I agree with him.

Just to briefly touch on measuring the average power for the load resistor.  When the current in it is increasing, you are charging the inductor.  When the current is decreasing the inductor is discharging.  That implies that if you monitor the load resistor for power you may be doubling the measurement for the stored inductive energy because of the charge-discharge cycle.  Then there may be transmission line/delay line/stray capacitive effects also.  At any instant in time the load resistor is dissipating energy proportional to the current through it and it also has a baseline amount of stored energy, and the stored energy is either increasing or decreasing.  Then on top of all that the voltage across it is a combination of the voltage drop due to the resistive element and the rate of change of current with respect to time times the inductive element.

Even with all of those dynamics going on, I think that you can make a good clean measurement of the energy being dissipated in the load resistor with the DSO if you are an expert at doing these kinds of things.

I think that the reason for the very high wattages measured across the load resistor have to do with the reactive power creating high voltages across the load resistor when the current through it is in fact decreasing down to zero.   The key is that the voltage you are measuring is always two voltages added together, the reactive inductive voltage and the dissipative resistive voltage.  The real power being dissipated in the load resistor is the product of the dissipative resistive voltage and the current only.

I still must make note that .99 showed much better and more realistic load resistor power measurements recently.  I suppose that with a proper set up the DSO can record the dissipative + reactive power flowing into the resistor and then as things turn around you are recording the dissipative power less the reactive power as the reactive energy leaves the load resistor and charges the MOSFET capacitor.

I am going to be totally honest and qualify everything that I am saying above.  My confidence in what I am saying is only fair to good, I am not speaking authoritatively.  I really am not an analog design engineer that specializes in this kind of stuff by any means.  However, if I was given a few days with the DSO and a setup and just focused on understanding the dynamics of the load resistor power measurement then I would stand a decent chance of figuring it all out.

Another way to think of the "waveform" across the load resistor is to give it a few extra energy waveforms to get a clearer picture of what it going on.  There is a resistive power dissipation waveform vs. time and a stored reactive energy waveforms vs. time and a reactive power out/in waveform vs. time along with the voltage and current waveforms.   You simply have to average the the resistive power dissipation waveform over one cycle to get the true load resistor dissipative power.

MileHigh

[fuzzytomcat]

Another improvement.

I only had one 6" croc lead there Hoppy, but have now eliminated that as well. I noticed a marked improvement by doubling up on the croc lead back to the shunt, so I've soldered a 3" piece of 12 AWG to the shunt -'ve and brought it out for connection to all 3 scope probe gnd leads.

As a side note, I believe Glen is using long croc leads all over the place, not to mention the long leads to his pots...nice for noise coupling indeed. 

.99

Here is a photo of The TDS 3054C oscilloscope probe connections  .... as you can see there is "NO" long croc leads all over the place ..... false allegation or misdirection