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query on Area measurement

Question asked by 90804ucl on Sep 16, 2009
Latest reply on Dec 17, 2009 by ko2054

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Dear supporter,
In my experiment, I measured the area of signals.
I observed the unit of the measure result is pVs.
Does this means volt* picosecond？Actually, I want to
get charge of signals, so I divided the result by 50ohm
(since the oscilloscope is DC coupling and 50ohm input impedance).
In the view of unit, I got it, volt*picosecond/ohm=amp*picosecond,
is just the unit of charge. But I still wondered my calculation is correct or
not? If it's not correct, how can I get the output signal's charge via our
Agilent 90804A?

Thanks a lot!

Best regards,
J.H.L.

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90804ucl Sep 21, 2009 2:59 AM
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Hi, ulf.jakobsson and other reader,
Thanks for your reply!
I am sorry but have to say, do you know how to get an "official" confirmed answer
on this case?
On the other hand, if the area measurement does not represent charge measurement.
why Agilent design such a measuring item, I mean area.

Thanks again!
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algoss Nov 3, 2009 5:11 AM
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I'm sorry I don't have my scope in front of me right now, but it is possible to make the measurement you want, using the 'Probe Setup' dialog.

The scope has no way of knowing that you have a 50 Ohm terminated system, and that you want to measure charge. Using the 'Probe Setup', you can tell the scope that you are measuring Watts, or Amps, and that there's a scaling factor. When you do the Area function, then the scope will display the correct units.
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ko2054 Dec 17, 2009 1:54 PM
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Hello 90804ucl, I checked with our lab and they say that the equation below should be correct. Area is Volts * Seconds but if the range of interest is not very long you will get pVs or mVs. Dividing by 50 ohms works, but scope does not know what you are really trying to accomplish so units may not be what you expect. I asked our lab to add intelligence to the math functions or allow user's to better specify units as needed. This happens in a variety of math function. You are not the only customer to bump into this issue. But your forumla is correct.
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