current testing

[kenypatel]

Hi,

I am trying to test/measure analog dc current. I am using a 9v battery and a 3k resistor. I am also using cDAQ-9181 chassis and a NI-9219 analog input adaptor. I have the wiring correct (Pretty Sure). I put the analog input adapter wires across the resistor to measure the current in my small circuit. 

So, when I try to use the labview express vi DAQ Assistant, in the pop up where it shows you a graph or a chart of the input before you turn it into an express vi I cannot get the correct current reading. (9v/3k) = 3mA, but the express vi reads a constant 25mA.

The express vi's settings is that the max input is 25m and the min input is -25m. I have it set on shunt resistor internal. When I try to change it to shunt resistor external and choose the external resistance to be 3k, it gives me an DAQ assistant error where it tells me "Requested value is not a supported value for this property. The property value may be invalid because it conflicts with another property." I have no idea on how to trouble shoot this problem. I want the current to read 3mA not 25mA. 

Any help in this would greatly be appreciated. If I was unclear on any of this let me know and I will try to clarify better.

Thank you!

[rgwagner888]

If you put the analog inputs "across" your resistor you will not be able to measure the current flowing through the resistor. You will need to put your instrument in series with your resistor.

-RGW

[billko]

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@rgwagner888 wrote:

If you put the analog inputs "across" your resistor you will not be able to measure the current flowing through the resistor. You will need to put your instrument in series with your resistor.

-RGW

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You can calculate it with reasonable accuracy using Ohm's law and reading the voltage across the resistor.

[kenypatel]

I have attached a (poor drawing) picture of how I wired it. Please let me know if it is incorrect

I have also attached pictures of where I am having trouble in case it will help clarify.

I know I can calculate the current with ohms law, but the point is to have labview do it. My application will eventually be to monitor current

I appreciate all the help

[Dennis_Knutson]

Try setting It for just a voltage reading. You can use a custom scale to convert to amps.

[mikeporter]

All you have to do is measure the voltage and divide that value by the resistance -- in LabVIEW. Ate you sure of the resistance? What are the color bands on the resistor?

Mike ..

[natasftw]

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@kenypatel wrote:

I know I can calculate the current with ohms law, but the point is to have labview do it. My application will eventually be to monitor current

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LabVIEW is doing exactly what you told it to do.  You can't wire your sensor in a way that can't measure current directly and program the VI to measure current.  Think about the measurements.  When you want to measure current, you want your device to have the least resistance possible.  If the measurement introduces resistance to the circuit, it'd read a lower current than it should.  Without any resistance, it's a short.  If you short the resistor, as you have in your design to measure current, you won't be reading the value you expect.  Your circuit now has 0 resistance and 9 V.  I = V/R = inf.  You're getting a result that is the maximum you can read.  That's in line with what you've setup. 

You have two options:

1) Measure voltage as you are currently and use Ohm's law to determine the current.  You can do this with scales or by dividing your result by resistance all on your own.

2) Wire your circuit correctly (assuming your module measures current.  I didn't double-check.)

In no case do you blame the tools you're using because you're using them incorrectly and tell others that are helping that you expect the tools to magically know you've set them up improperly and correct things for you.