Partnered Output Coils - Free Energy

[EMJunkie]

Ahem... I am 'lite' on the pop culture these days.   I looked at a trailer and I see the good old Hoover dam blows again.  You could be typing a message into your computer and all of a sudd

MileHigh - You should read this: Joe Brandt

   Chris Sykes
       hyiq.org

[picowatt]

PW,

Agreed, but you did not answer my question.

All measurements will still have a degree of In-accuracy! No matter how expensive the equipment! This is unavoidable.

   Chris Sykes
       hyiq.org

Ultimately, the required accuracy depends on what it is you are measuring and what you are going to do with those measurements.

If low cost gear is all you can afford, then by all means, do what you can with what you can afford, but be aware of the limitations of your equipment and the measurements made.  If greater accuracy is required, more accurate/expensive equipment can be rented to verify measurements.

But back to that 20KHz BW Hantek, I would not suggest it for measurement of pulsed wavefoms with anything resembling vertical edges...  A properly utilized low inductance CVR would provide better results.

PW   

[TinselKoala]

OK ... so ... how any other way is it supposed to be? (I really want to learn the right method).
I actually used 1/6 for the positive waveforms and 5/6 for the negative.
The energy calculated in the first 1/6 of the period was 0.02678569mJ and in the last 5/6 of the period was 0.00690478mJ.

I think the Vishay calculator gives you the average across the entire waveform. That is, the "1.8 W" "average" value is the value of the On pulse, averaged over the entire duration of the whole period. But we want the "average" of the ON portion separate from the "average" of the OFF portion for comparison. So looking at just the ON portion we have something around 9 Watts 'average', and then the duration of that is about 25 microseconds, so the energy is around 9 x 0.000025 = 0.000225 Joules or 0.225 mJ. And the "average" of the OFF portion alone is something around  0.5 W and it lasts for around 115 microseconds, so the energy there is around 0.5 x 0.000115 = 0.0000575 J or 0.0575 mJ. So there's about 4 times the energy in the ON portion as in the OFF portion.
My figures are rough, of course.

[EMJunkie]

Ultimately, the required accuracy depends on what it is you are measuring and what you are going to do with those measurements.

If low cost gear is all you can afford, then by all means, do what you can with what you can afford, but be aware of the limitations of your equipment and the measurements made.  If greater accuracy is required, more accurate/expensive equipment can be rented to verify measurements.

But back to that 20KHz BW Hantek, I would not suggest it for measurement of pulsed wavefoms with anything resembling vertical edges...  A properly utilized low inductance CVR would provide better results.

PW   

PW,

I like your post, its decent, achievable for all here.

Still you avoided my question - No matter.

   Chris Sykes
       hyiq.org

P.S: So far so good on the San Andreas, no one hurt! I hope!

[TinselKoala]

OK,you lost me. How do you get a positive from two negative values?.

That's just simple math! Multiply any two negative numbers and you get a positive number as an answer.

If we look at AC,and consider the positive above the ) volt line,then a negative value would mean a value below the 0 volt line. In this case,we wish to know which way the current is flowing in which direction,and when the above and below the 0 volt line are averaged out over each cycle,we see a higher average below the zero volt line per cycle-both voltage and current.

Yes, but there seems to be something wrong with the "average" that's being reported by your scope. By now you've seen my video on the subject:
http://www.youtube.com/watch?v=Q_fOJw8rePg
Now I've just read the manual for the Atten 1000 series scopes and I can't tell from that whether the scope uses the entire memory or just the displayed waveform for calculations. I think, from the description in the FFT section, that it just uses the display. I did see that the "average" just uses the "first period" to compute the average, whereas the "mean" uses the "entire waveform", according to the manual. Whether this means the "entire displayed waveform" or not I'm not sure of. But the fact that the "average" just uses the first period rather than the whole dataset probably accounts for your result here:

No,they remain the same-->is this good or bad,as you know im still learning about scope's.

That is, changing the horizontal timebase on your scope when you have "average" set and it's reading just the first period means that the value computed won't change. On my scope demonstrated above, the "average" I used is working on the whole displayed waveform, which may contain more or fewer complete cycles or portions of a cycle, so the measurement changes as I change the timebase; I do have a "period average" that does the same thing as your "average", I think, but I made the video before I read the Atten manual....

There are good reasons why measurements should only apply to what's actually on the screen rather than using the whole memory buffer. Such as a pulse train with gaps in between groups of pulses -- if you have the horizontal timebase set to display the individual pulses and can't see the gaps, and you want frequency or duty cycle of these pulses, if the measurements are using the whole buffer including the gaps, the result will not be what you expect or are asking for.

Also, according to my read of the manual you can select "Math" as the measurement source and you should then be able to display all the usual measurements from the Math trace just as from any other trace. Maybe we can figure this out over on the other forum or by PMs.

(Reading the manual was very informative. The Atten 1000 series scopes have some nice features that my Rigol doesn't have, like alternate triggering so that two signals that aren't synched can be displayed stably. The Atten manual also explains some features that we both have, better than the Rigol manual does.)