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Old 08-03-2009, 09:56 AM
witsend witsend is offline
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Join Date: May 2009
Posts: 1,881
TK - I'm hoping I can bend your mind around this problem.

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value) . Then how do you work out the product of the energy available at that moment? In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

Now. If you do not do the integration simultaneously how is this advantage or gain made evident? And more alarmingly, if you simply do the product of both values and add it to the general loss - then your methodology is wrong.

If, as you say the LeCroy is doing the math - then I'm afraid I need to see it. I'm still concerned that it is not capable of doing that DC offset. To my primitive way of thinking this means that it cannot gauge zero in order to make a comparison as to where the waveform was found in relation to zero.

Please explain this. Sorry to impose.

You are right. Its's the best I can do to explain my concern. I use primative example and analogy. But it serves its purpose.

Last edited by witsend : 08-03-2009 at 09:59 AM.