08-05-2009 03:57 AM
Hi,
I have done a number of projects using labview in the past measuring DC voltage and current using shunt resistors and such.
I have a new project in an area I am not too familiar with. I understand health and safety issues working with 230v etc, and this is not a problem.
My project involves measuring and plotting inrush current of some 230v products. My question is thus; How do I measure AC current using a NI USB6008?
Thanks in advance
Martin.
08-05-2009 07:24 AM - edited 08-05-2009 07:25 AM
Hello,
I would suggest trying to find a clamp meter or power quality analyzer that can measure current and output the value over serial(rs232) or GPIB, you can then use NI-VISA/GPIB to read this and bypass your USB-6008.
I suggest this over using your USB-6008 for safety and ease.
08-05-2009 03:31 PM - edited 08-05-2009 03:38 PM
Hmm... While not technically impossible, measuring a 230V current source with a 10V non isolated data acquisition device sounds like a "thermal event" waiting to happen 😉 I'd strongly urge you to consider a product designed to handle such high voltages (for your own safety!).
That said, if you happened to have discrete points of an AC waveform (say, +/- 1V across a shunt), it is possible to convert them to an RMS value fairly quickly: http://en.wikipedia.org/wiki/Root_mean_square . You might also look into the various RMS VIs in labview (I like the one called RMS PtByPt). After appropriate scaling you might end up with a reasonable picture of current through your shunt.
Since the 6008 is only 12bits / 10ksps, I worry that you might not get a very good reading depending on how fast your inrush is and the frequency of your source.
Quickie simulation using a *slowly* ramped up sine wave:
This all assumes that you're still alive after setting your system up 😉 Sometimes it's best not to leave these things to chance! NI does make high resolution, high voltage, isolated multimeters capable of current measurements that would give you a much better picture without all the hassle of math, shunts and possible electrocution 🙂
Hugs,
memoryleak
08-06-2009 05:00 AM
08-06-2009 05:29 AM
Use a current transformer for AC. A core with one turn primary and 200 - 1000 turns secondary and a 10 to 100 Ohm resisitor.
Better units use active compensation methods with a hall sensor LEM(.com) is a good source.
Both provide isolation 🙂