Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[cHeeseburger]

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

[woopy]

Hi Magluvin

thank's very much for your reply 208 and included schematic

i did a small test of your idea and hope this is not too much off topic here

perhaps you should open a specific thread for this specific idea  ?

http://www.youtube.com/watch?v=XrwgEb5ac_w

And of course my BRAVO to Rosemary and her team for sharing 

good luck at all

laurent

[hartiberlin]

Good question Cheesiebugger,

it probably comes from the Minus 70 Volts offset in the channel 2
which is not substracted.
So channel 2 had about -20 Volts x 245 mV= about - 5 VV...

So the multiplication settings of the scope was not set right at this time.

But much more interesting is this attached scopeshot,
where you can see that the mean current at the shunt is really negative.
Not only from the numbers that show Minus 25 MilliVolts
but also from the display of the yellow burst showing more amplitude
below the ground line.

The ground line is the left yellow line at the number 1.

So in this condition is  really seems to charge the batteries.

Depends probably all on the working points it is running on.

So the first shown test seems to recharge the batteries but the
second one seems to discharge them at the higher temperature...

Regards, Stefan.
P.S. Well done video Rosemary !
Thanks a lot.
Clears up many questions.

[cHeeseburger]

So you agree that he second test shows 44W of heat costing 50W of battery drain, right?  And thank you for clearing my post.  It is an honor to be allowed to post here once in a while.  I'll not abuse the privilege,

Regarding the first test, then, we all agree that it would have been nice if Rosemary had given us a shot or two showing the actual waveform of the oscillation, rather than exclusively showing low-sweep-speed shots of the 100Hz duty cycle where no one can see the cycle by cycle shape of the actual oscillations.

SInce we know that the actual power into or out of the battery depends on the areas under the curves above and below zero and not on the peak voltage excursions there, and we acknowledge never having been showed those areas at any time, how can we conclude anything realistic about the first test based on only those peak excursions and the math trace which we agree was faulty and in error on the second test?

What if the cyclic oscillation waveform looks like this picture?

Sincerely,

Cheesebreath

[hartiberlin]

Then the yellow burst would probably have looked different,
but I agree, that we need better zoomed in waveforms,
showing only 3 or 4 cycles and not this burst only.