16 bit module to power a 24 bit module

[nssid]

Hi,

I have a cDAQ and I am powering my sensor with 0.5V DC using a 16bit analog output module (NI 9263) . I am then measuring the signal using a 24bit analog input module (NI 9219). Is this ok ? 

I see a lot of noise in my signal when I do this. When I connect the output module directly (9263) to the input module (9219), the voltage values change from 0.4998 to 0.4992 ( I ask the 9263 to give me 0.5V DC). I suspect this is causing the unwanted noise in my signal. Any thoughts on this ?

Thanks,

Sid

[mudson]

Hello Sid,

Please confirm that you are you generating 0.5V with a NI-9263.  You won't damage anything with that setup, but you may want to use a module that is intended for smaller voltage ranges.

Looking over the specs for that device, the behavior you report is surprisingly good!  The minimum voltage range for that device is -10V to 10V, with a minimum voltage accuracy of 0.11V.

NI 9263 Product Page

From what kind of sensor are you reading (make, model)?  How is your experiment setup?

[nssid]

Hi Matthew,

Yes I am generating 0.5V DC with the 9263. Is there any other cDAQ output module that has a smaller range  (I couldnt find any when I searched) ?

My sensor (TR 10) is a linear potentiometer (1000 ohm). Since the 9263 has a max current drive of 1mA, I could use upto 1V across the 1000 ohm pot. However the specs of the potentiometer say that I need to have a wiper current less than 1uA. So, just to be on the safe side I am using 0.5 V ( Technically I should be able to use 1 V too as the 9219 has an input impedance of  greater than 1giga ohm in the 1 and 0.125 V ranges).

I am trying to measure small changes of displacement with my pot and hence I need as much accuracy as possible.

If there is any other output module which you think works better in the smaller range please let me know.

Sid

[chuggins1434]

I'm curious as to why you would power a pot with the output from a voltage signal generator?  ...I'm assuming you have some type of voltage divider circuit for measuring the position?  Why not just use a cheap power supply?  I'm looking at the spec sheet you linked to and from what I'm reading you can throw up to 24 volts (DC I'd assume) at this thing with no issues.  A computer power supply will give you several different voltages to play with for little to no cost.  Not criticizing, just thinking of options! 

Enjoy!

Chad

[nssid]

Chad,

Although I can use up to 24 V, I need to choose a appropriate voltage such that my wiper current is less than 1micro amp (recommended operating wiper current < 1uA). Since I am using the 9219 to measure my signal and its impedance in the 4V and above range is 1mega ohm, the wiper will draw more than 1uA. So, I need to keep my voltage less than 4V (limited due to the 1 mega ohm impedance of the 9219)

Cheers,

Sid

[mudson]

Hello Sid,

I think I spoke out of turn back there.  -10V to + 10V looks to be the range available for all of our cDAQ modules with analog outputs.  The NI 9269 has a max voltage accuracy of 6.3mV, so that may be a more trustworthy than the device you are currently working with.  At the time of my last post, I was thinking of the specs of our PXI DC power supplies and SMUs.  Sorry about that.

A thought:  you may be able to output a higher voltage, use a voltage regulator to bring the output down to 1.25 V, and then divide it down from there to 0.5V.  I know this kind of the "Frankenstein" solution — and EE professors the world over would cringe at the thought of using a divider there.  However, it doesn't sound like you'll be sourcing much current or dissipating a whole lot of power in this setup, so it may still work out.

Best regards.

[nssid]

Matthew,

Thanks for your reply. I think I figured out a way to remove a lot of noise in my signal. I am now measuring the voltage the output module is delivering to the pot and doing a ratiometric to these values instead of the value I specified (0.5 V)  my noise is significantly lower. May be I should consider the 9269 in the future. It looks like a much better card than the 9263 as far as accuracy is concerned.

Thanks,

Sid 

[kellam]

Hi Sid,

I had a quick question for you. Your sensor is a 1k resistance sensor.  The 9219 is designed to measure resistance directly by powering the sensor itself, and will take very good resistance measurements, especially in four wire mode.  Is there something about your system that requires you to externally power the sensor?

-Jeff

[nssid]

Jeff,

My sensor has a power,signal and common (3 wires) and I beleive it is supposed to be used in a ratiometric to voltage form. Hence, I never thought about the 4 wire resistance mode. That being said the wiper current in the sensor (potentiometer) must be less than 1micro amp and it might be a limiting factor to be used in the resistance mode. I am not sure how much current the 9219 would use when I connect my 1kohm pot. I did not try.

Sid

[kellam]

You should be OK for current.  The excitation for a 1k load is 430mV, and in resistance mode the 9219 has an input impedance of over 1GOhm.  Anyway, I was mostly curious, since if our products aren't meeting the needs of our customers, that's always good feedback.

-Jeff