03-01-2010 04:09 AM
HI all,
I want to know about DAQmx Create Channel for Accelerometer hoe it will do the convertion any one reply me
03-02-2010 07:42 AM
From the detailed LabVIEW help:
"Accelerometers typically convert acceleration measured in g's to voltage. For example, a sensor with a rated output of 10 mV/g should produce 50 mV when subjected to 5 g of acceleration.
... sensitivity is the sensitivity of the sensor. This value is in the units you specify with the sensitivity units input. Refer to the sensor documentation to determine this value"
Also, if you do not want linear scaling, you can use the "custom scale" input - "custom scale name specifies the name of a custom scale for the channel. If you want the channel to use a custom scale, wire the name of the custom scale to this input and set units to From Custom Scale. "
03-15-2010 04:30 AM
Hi Vaibhav,
Thanks for reply,
What ever you have conveyed is right but, if I have connected the sensor (10mV/g) to PXI-4472 and I am simulating 5g from the Vibration calibrator and I have created a task for Accelerometer then I am getting 5g in my system but if I connect the same to Oscilloscope then I am not getting 50mV I am reading something else so I want to know the difference b/w these two
03-19-2010 11:23 AM
03-21-2010 11:55 PM
thanks for reply, I want know about this ICP can you just explain it clearly
03-22-2010 09:45 AM
Most of the accelerometers that i have used till now are ICP devices.
They need about 5mA current to power them selves up. Other wise they give inaccurate output.
I guess you are measuring a raw time domain vibration signal.So you just need to
I have attached my own code of how i acquires a vibration signal usin PXI-4462 and 4461 modules. Keep it secret till i do my thesis :).
You have to specify channel name ,device name, and remember to create the channel using the AI-accelerometer -accelerometer VI ,if you use any other form of this polymorphic vi then ICP source options will not be there and your sensors will not have the current to work properly.