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Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

Started by hartiberlin, February 20, 2011, 06:14:05 PM

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0 Members and 9 Guests are viewing this topic.

poynt99

Quote from: hartiberlin on March 22, 2011, 12:25:24 AM
Hi Poynt,
please try again your simulation with this setup please
and show burst waveforms (many cycles) and 3 to 4 cycles
on one scopeshot.

Many thanks.

Here you go Stefan. Notice the difference across the shunt, with and without the 110nH series inductance...

.99
question everything, double check the facts, THEN decide your path...

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Rosemary Ainslie

Harti,
Regarding your question related to the 'default' mean average voltage to positive during the 'high' wattage dissipation on the load - here's the reference in the report under RESULTS

3.2 Second test
The mean average and cycle mean average voltage across Rshunt indicates that some current has been discharged by the battery to the source rail. However, instantaneous wattage analysis applied to the voltage measured across the battery and Rshunt indicate, here too, that the battery supply source has had more energy returned to recharge it than was first applied to the circuit.  When this is applied to each sample from a spreadsheet analysis across the 500 000 to 1 million samples supplied by the digital storage oscilloscopes, then the product of this and the battery voltage represents the instantaneous wattage. The sum of these values, divided by the number of samples, represents the average wattage delivered over the entire sample range. This results in a negative value indicating that more energy is still being returned to the battery than was delivered. This is in line with the math function of the DSOs where it, too, indicates an increase of wattage back to the battery supply over the amount of wattage initially delivered from that supply.

More wattage returned to the battery than was delivered
Wattage dissipated at RL1= 44 watts
Switching results in the generation of extreme spiking at the transitional phases of the switch. 


The following is the video @ 6.33 minutes highlights the neg mean over the shunt - compared to 7.03 minutes that highlights the default to positive - notwithstanding the continued negative mean average.  This is in the annotations at the start of the video.  In the report this was listed under 'anomalies' PRECISELY as it's required to get an expert evaluation here.

http://www.youtube.com/watch?v=fyOmoGluMCc

There was an extensive debate held at OUR.com related to the correct wattage measurements.  The consensus was that wattage is correctly computed as the instantaneous product of vi over time.  If you look closely at the antiphase condition of the voltages across the shunt and the battery you will see that when the battery voltage is trending high - then the voltage across the shunt is trending low.  And conversely when the battery voltage is trending low - then the voltage across the shunt is trending high.  In effect, the returning current flow from the circuit 'trumps' the discharge from the battery that there is a zero loss to the battery - and, according to the math - results in a gain.  THAT IS THE SIGNIFICANCE OF THE ANTIPHASE CONDITION OF THOSE VOLTAGES.  I have referenced this in this thread - extensively - and on the blog.  I have, in fact, been trying to draw your attention to this from the get go.  This is one of the many aspects of this 'evidence' that requires research and expert analysis.  This is also PRECISELY why we included two tests - to highlight this very point.  If the measurement of instantaneous wattage is a correct reflection of the energy delivered by the supply then here there is an inevitable net gain to the supply.  Interestingly - the voltage measured across the battery - defaults to less than its 12 volt each 'start' voltage and then, within minutes, recovers to its previous value. 

While I appreciate that this is now being used as a 'last ditch' effect by Humbugger as he reaches deep into that barrel in his efforts to refute the claim - it is, in point of fact, the ENTIRE theme of that demonstration.  That, and the fact that there is a negative mean average over the resistor at all.  Good heavens.  We could not have been more transparent in our evidence if we had recorded the demo in 3D.

There is NO simple explanation here guys.  There are challenges to conventional protocols, conventional predictions - conventional assumptions - ALL OVER THE PLACE.

Regards,
Rosemary

Rosemary Ainslie

Quote from: poynt99 on March 22, 2011, 09:32:38 PM
Here you go Stefan. Notice the difference across the shunt, with and without the 110nH series inductance...

.99

Poynty - I can no longer see your waveforms and i would LOVE to see them.  There are others of us who also cannot open those files.  WOULD YOU PLEASE POST A PICTURE and just size it that it fits this thread.

Kindest regards,
Rosie

ADDED
This is copied over from your forum - written by you Poynty Point.

'Indeed, I am having difficulty figuring out why her shunt trace is at zero. Mine is showing about 1.5V or so, and hey, 0.25/11 x 72 = about 1.6V. It would seem what I'm showing is about right.'

I think what you meant is 0.25/11 x 72 = about 1.6 watts, NOT VOLTS.  In which case can you then explain the temperature over the load which, typically, is at 6 watts or greater at 72 volts applied.

Rosemary Ainslie

Quote from: cHeeseburger on March 22, 2011, 04:49:20 PM
Rosemary,

At 8:40 into the video, while your colleague is demonstrating 190C on the load with a 50VDC battery voltage, there is a good closeup of the LeCroy which shows that there is +243mV (about 1 Ampere on a 0.25 Ohm shunt) flowing out of the battery.

Could you please explain to everyone why the scope math that is showing us the product of the +243 mV trace and the +50.3 VDC battery voltage is telling us that the product is -5.43 VV?  How does the scope get a negative small number by multiplying two positive numbers?  By my figuring (even without using a calculator) 1A x 50VDC = 50 Watts.  All positive numbers flowing out of the battery.

Please clear this up.  It's rather confusing.  Thank you.

cHeeseburger

I trust that my previous post has now addressed this Humbugger.  It is the result of the phase shifts that the advantage goes to the charge condition of the battery.  Just bear in mind that the negative product of the voltages done by the math trace on both the LeCroy and the Tektronix is born out in the spreadsheet analysis.  AGAIN.  We have asked for EXPERT opinion on this as we are applying conventional power measurement protocols - based as they are - on wattage analysis.

Rosemary

Rosemary Ainslie

Guys - I feel I'm trying to walk in a bath of treacle.  I thought that these facts had already been understood.  Harti.  You keep asking for 4 traces.  Please look at the frequency of that 'burst oscillation mode'.  That occurs at a rate that is determined - not by the switch - but by some resonating condition that is imposed on the circuit as a result of inductance.  Then.  Please note that the oscillation is self sustaining whether it is dissipating high or low energy on the load.  We ARE, indeed, showing you four or six cycles.  We then zoom into that oscillation.  There is no other way that this can be represented.

Kindest,
Rosemary