Cable Position Transducer with NI 9237

[lalala1234]

Hello,

I have a NI 9237 Half/Full-bridge analog input module. Recently I aquired a cable driven position transducer, which is basically a potentiometer. You can find the circuit diagram from the link below or in the attachment.

http://www.celesco.com/_datasheets/sp2.pdf

I realized the NI 9237 requires a half bridge or full bridge circuit in the transducer but apparently the potentiometer is not a bridge circuit, and when I connect the output and excitation terminals to the module, I'm not able to measure the change of voltage or resistance. Do you know any way I can use NI 9237 to measure the potentiometer?

Thank you!

[J-Barticus · Accepted solution]

Hi larryklee,

The 9237 was not designed to measure voltage and resistance, but rather a ratiometric measurement based upon the excitation voltage. There would be a round-about way to measuring the voltage and/or resistance but it would require you to scale or post process the data.

Additionally, you would have to supply an excitation voltage that would allow you to read the voltage coming from the transducer which is dependent on the input voltage you are supplying to the transducer. The input range of the NI 9237 is limited by the excitation voltage, at the max internal excitation setting the range is +-250mv.

-Jake B.

[biodegraded]

I had a similar question.  Can you put the potentiometer as one arm of a wheatstone bridge and excite the bridge with the Vex and then read the Vo from the bridge with the 9237.  If properly balanced, the change in the potentiometer would result in a change in the Vo.  My only concern is if it is possible to use a conventional potentiometer and choose appropriate resistors in the bridge to still stay within the limits of the Vex and Vo of the 9237.  Is this possible?

Thanks,

Corbin

[Henrik_Volkers]

Here are two ways to do it:

Build a half bridge with two R ( R>=350 ohm),  say 470 ohm 🙂
 in parallel put your pot (10k? ) and between the wiper of the pot and mid of the half bridge add another resistor R_p valued R*(1-S)/S  with S as (half the) range 25mV/V  equal to 39 times R (for R=470Ohm this leads to ~18.3k  so pick a 20k) 🙂

BUT now the pot position is not linear to the readout , some more homework to do 😄

BUT you have a low impedance coarsly matching a bridge.

Given the same pot (10k) you can add two resistors at top and bottom valued >39/2 times the pot value ~> 200k  , linear readout, but high impedance.