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Problems using DAQmx Channel Property Node in Labview

I'm updating some of my programs to use DAQmx. I need to find the units allocated in each Global channel. The Channel Property Node seems to need a Task as input and not a Channel. I don't want to create a Task as there will be several measurement types in the Task ie Voltage, Thermocouple, PRT's and it's not straightforward building the task when there will be over 100 channels (I've seen the DAQmx Task Example for mixed measurement types).

The only way I've managed to achieve this so far is to create a task for each channel then remove it (see attached picture). This works but it takes too long.

There must be a better way - Any ideas?

Thanks.
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Hi Ian,

You can wire a DAQmx Channel directly to the channel property node. For that matter, you can wire a DAQmx Channel to any DAQmx node/input that takes a DAQmx Task. LV does the conversion to a Task for you.

Deborah
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Hi Deborah

I tried that first without success. I could only get the information when I created a tsk.

Maybe you can spot any silly mistake in my program. The default values on the panel show what I'm trying to achieve.

Thanks
Ian
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Hi Ian,

I can run your VI with success when wiring the chanel names in directly. What is the incorrect result that you are seeing? What property node is failing you? What is the error and result from that node?

Deborah
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Thanks Deborah

It's good to know that the code does work. Mine gives the following error when I go straight into the channel property :

code -200088
Property Node DAQmx Channel (arg 1) in add_units[1].vi

Encouraged by your success I'm going to go back into the MAX explorer and check through my channel set up to see if that's where my problem is.

Thanks again for your help, I'll let you know how I get on.

Ian
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Hi there Deborah

After finding no faults in my set up I re-installed my version of Labview and MAX Explorer. My program now works fine.

Thanks for your help, and all the others who contribute to the group, my jobs been made much easier because of it.

Ian
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