voltage data acquisition in fuse testing

[anita]

I need to find an appropriate Data acquisition board to obtain data on an analog voltage signal coming from a current transformer used in a TCC fuse tester

[markwysong]

Anita,

The important things to know when picking an acquisition board are:

- how accurate (12-bit or 16-bit board)?
- how fast (samples per second)?
- how big is the signal? (volts)
- Do you need any signal conditioning? (filters, amplifiers, straing guage, etc)

This will help very much when selecting a board. I can help you select an appropriate board with that information.

Mark

[anita]

This is to Mark Wysong,
Thank you for your comment. If you have time, maybe you could consider my project application further.
I need to develop LabView software to use on a TCC tester. These are some of the requirements:
Target user: technician

Inputs: Current signal from a Current transformer

Calculation: Calculate RMS Current, and amps squared seconds to melt

Screen output: time to melt, RMS current,amps squared seconds, visual graphic representation of waveform (long time to melt dislay the peak voltage for each half cycle, fast times show breakdown of wave form for up to 20 cycles with 50 points per cycle)

Statistics: mean, variance, etc. on data

DAQ: not too accurate, I suppose
50 samples per second for fast melt times
Si
gnal voltage up to 2.5 V
I don't think that there is a need for signal conditioning, unless you think that there is a need for signal amplification

thanks, anita

[markwysong]

Anita,

I believe that for your application, a 12-bit DAQ will probably suffice. If you have a computer already with an open slot, this would work just fine.

Therefore, I think you'll need:

AT-MIO-16E-10 Data acquisition board

(100 kS/s, 8DI/16 SE analog inputs, 2 DACs, 2 counters, 8 digital lines)

SH6868EP cable

SCB-68 Shielded I/O connector block

And, of course, a copy of LabVIEW.

If you are truly reading a current, not a voltage, you will need some external CVR (current viewing resistor) to read a voltage. This is just a very low resistance put in series with the transformer, and a voltage reading across the resistor is taken. Current can then be calculated.

That should do it!

Mark