measuring resistance but value is not constant

[Zancis]

Im measuring resistance using the program attached in the image below, but the value  of the resistance is not constant.  Also I cant understand what excitation  source and value are. Also attached there is an image of the objects Im using .

[crossrulz]

A DAQ cannot just measure resistance.  You have to supply a constant current through the system.  The DAQ will read the voltage and convert to resistance based on the current you told it you are supplying.

[johnsold]

According to your image, you specified an External current source.  No such source appears in the other image, but clearly if it external, you need to set the value.

Please tell us exactly what equipment/devices you are using. Also tell us what is inside the foam block you are measuring.

Is this the same project as in your other thread where you are using a PCI-6251? You have already been told that you cannot directly measure resistance with that device.

Also, it is very unlikely that you will get consistent or meaningful results by sticking small wires into blocks of conductive foam.  But that is a separate discussion.

Lynn

[Zancis]

Thank You for reply. I am using PCI-6251 and CB-68LP.

1) Could You please explain, what is excitation value and do I have to use another wire to  add the external source?

2)  if i  add the external source and set the excitation value will i find the resistance?

3) is there another way to measure resistance if this version is wrong?

[johnsold]

1) Excitation is the voltage or current applied by the instrument to the unknown resistance. The instrument then measures the current or voltage and uses Ohm's Law to calculate the resistance.  The PCI-6251 can generate a voltage and it can measure a voltage, but it cannot do the voltage and current combination required for Ohm's Law.

2) If you apply an external current source and measure the voltage, you can then calculate the resistance.  That block of conductive foam you showed in one of the images you posted probably has a fairly high resistance ~10E5 ~10E7 ohms. That would mean that you would need a current in the microampere range. If it really is in the 10E1 to 10E2 range as you suggest from the settings you show in Channel Parameters, then you would need a larger current.

3) You can use the voltage divider technique. Select a fixed resistor of known value. Its value should be comparable to the values expected for the unknown. Connect the fixed resistor in series with the unknown resistance. Connect an Analog Output to the series string. Set the output voltage to a value which is within the input range of the Analog Inputs of the DAQ device and which will provide a suitable current through the resistors. Use tow Analog Input channels to measure the voltages across each resistance. Calculate the unknown resistance using Ohm's Law and Kichoff's Law.

Lynn

[Zancis]

Thanks for reply. 

1) I found some information in google about excitation and from Your answer I understood, what it is. 

2) I didn't know that resistance will be so high. Ok, I can change parameters, but the main question - how could I add external source? Do I have to use another wire/s? How I need to connect them into my schema? What I have to do in LabView to generate the current?

Sorry for my small knowledge... I am total begginer, but I deffinetely must do this experiment.

[crossrulz]

Go with Lynn's suggestion about using another fixed resistor.  Your DAQ should have an Analog Output.  Set the AO to 10V.  Measure the voltage across the known resistor to calculate the current and measure the voltage across the test resistance and divide by the current measured earlier to get the resistance.  You can setup both reads with a single task.

[Zancis]

Thank you for your reply 🙂 

We made this structure to find unknown resistance as you see from the image attached. Also we made 2 tasks(1-voltage out task for pin 22 and the 2nd a  voltage task for pin 68/34 and pin 66/32).Is it possible if you can help us with the code  as we didnt code anything because we dont know how to start the program. We know that  we have to generate and measure the voltage. 

[johnsold]

That wiring looks OK.

Code: Start by looking at the examples for AO and AI. You should be able to run them independently and get some preliminary results.

Lynn

[Zancis]

Thanks, I will try!