Forgive me for chipping in, but isn't this just a process instrument with a process output of 4 to 20mA? - (0mA used to signal line break, like early telegraphs, I was led to believe).
The idea of 4 to 20mA output is that you can daisy chain a number of sensors and and analogue inputs on the same loop, as long as the device supplying the current has enough voltage (compliance) to supply all the other devices. Most process stuff works on 24-V, so that limits the number of devices on the loop.
If you want a voltage signal to read rather than current, the standard method is to put the 4 to 20mA through a precision resistor (500R for 0 to 10V, 250R for 0 to 5V - V=Ri, 0.020A through 500R is 10V)
The problem is that 500R or 250R at 0.1% tolerance are specialised beasts and cost a lot more than a bog standard 240R or 510R resistor.
If this is going into a "process", I would be using the precision resistors.
If the Labjack has a 0 to 5V input, use 250R, bearing in mind that your scaling is 1 to 5V corresponding to 0 to 100% process, with a 1V offset

 

It's a lot more helpful to have a schematic, if you could take the time to draw one it would be great. That would be shielding wire to shield the signal from any stray noise.

 

This may be a rhetoric and dumb question, but since the signal output is in mA, can I treat it as a current source?

Then I would used ohms law to calculate the voltage though a resistor.

 

Yes, that's exactly what you do.

 

If you want a voltage signal to read rather than current, the standard method is to put the 4 to 20mA through a precision resistor (500R for 0 to 10V, 250R for 0 to 5V - V=Ri, 0.020A through 500R is 10V)

It's taken a while, but mug up on 4 to 20mA current loops, used on just about every instrument in every factory, power station, car production line, brewery etc. that I've ever seen. SCADA is worth a look as well. In some ways, you've thrown yourself into the deep end.
Get a 510 Ohm or 240 Ohm resistor, connect it across the 4 to 20 mA output then put your voltmeter across the resistor.
Once that works and you have understood that 4 to 20 is an offset signal, get the correct value resistor (no self-respecting process electrician would "make up" the precision values) and you are good to go.
The whole idea of 4 to 20 is that a remote instrument on the factory floor can send an output to one or two control panels that can "pick off" the mA signal, convert to a voltage etc. all provided the transmitter (pressure gauge) has the oomph (compliance) to drive the current.
One transmitter will generally be able to provide compliance to drive at least two receivers - your spec will give the compliance voltage, say 24-V, the rest is Ohm's Law.

 

Ok, I am really really stump again now. I thought the signal output (mA) changes depending on the pressure??? The signal output changes when I change the voltage source.

 

If it us under 5 volts, you should be good.

Also, one meter has a full scale of 500 and one has full scale of 300 units so any non-zero input will have different value.

Any idea on my current question?

 

I don't think you're grasping the basic principles here.
Take some time to really make sure which model and operating specification you have.
The data sheet or whatever will tell you what the operating voltage is. Get that right to start with.
Then make sure you know exactly what the output is - it will be a current output or voltage output.
Within the types, there could be different ranges. 4 to 20mA is standard, 0 to 20 is also sometimes seen - voltage outputs are generally 0 to 5-volts, 0 to 10-volts, 1 to 5-volts
Sometimes the instrument will allow you to choose voltage or current output in the setup.
If it's current, you can just use a multimeter in mA mode to see what's going on - if it's volts, just use a meter in V mode.
I suspect you aren't grasping the offset bit - 4 to 20-mA represents (usually) 0 to 100% of the pressure scale, so 0 to 20-mA is -25% to 100%.
The negative numbers generally signal a fault.
You need to do the legwork, I don't have time to read the data sheet, and if it is an offset output, the maths is trivial - 16mA represents 0 to 100%.
If all else fails, you'll have to contact the manufacturer for guidance.
You will get there in the end and all will be clear. Just let us know the outcome.

 

Any idea on my current question?

Again, you should only have two wires connected to your LabJack. Is that how you are connected? No wires should be connected to the Vs port on the LabJack.

 

Looking at your sensor the sensor has a 4 to 20 mA output which is popular. The way this works is when the sensor is powered with an adequate voltage the sensor will output 4 to 20 mA proportional to the sensor's span. For example a 0 to 1000 PSI gauge will output 4 mA at 0 PSI and 20 mA at 1000 psi. Now looking at that we have a span in mA of 16 mA. Now if I run that output through a 250 Ohm resistor the voltage will be 4mA = .004 * 250 = 1 Volt and 20 mA = .020 * 250 = 5 Violts. If we change the resistor to 500 Ohms we would get 2 to 10 Volts. The trick here is the voltage applied to the sensor needs obviously to exceed the voltage we expect to see. Additionally the resistors used with sensors like this are generally 1% precision resistors but that can be worked around.

Now back to your very first post. I have never used a labjack module but many similar for basic data acquisition. Using your link: https://labjack.com/support/app-notes/measuring-current Figure 1. Current Measurement With Arbitrary Load or 2-Wire 4-20 mA Sensor is what you want. Well similar to figure 1 in that you want your resistor across the output of your sensor. The resistor I mentioned as 250 or 500 ohms (1 to 5 volts or 2 to 10 Volts) is labeled SHUNT in figure 1. They mention 240 Ohms where I used 250 Ohms the maximum difference being 4.8 or 5.0 volts full scale.

I am guessing you have the LabJack U3 LV model which is a USB DAQ device with 16 flexible I/O, and 4 dedicated digital I/O. The flexible I/O can be configured as either digital or analog, thus providing up to 16 analog inputs, or up to 20 digital I/O. It also has two 10-bit analog outputs, up to 2 counters, and up to 2 timers. All we really care about at this point is a single analog signal input.

• Up to 16 12-bit Analog Inputs (0-2.4 V or 0-3.6 V, SE or Diff.)

SE or Diff refers to Single Ended or Differential Input, for your purposes you can just use SE since we won't be seeing a negative going input. The better choice here would be the HV (High-Voltage) Version:

• 4 HV Inputs have ±10 Volt or -10/+20 Volt Range.

Either way the unit is a 12 bit Analog to Digital converter. So full scale in bits is 2 ^ 12 or 4,096 bits full scale. Does all of this so far make sense to you? Eventually the number of bits will be converted to an engineering unit like Volts or PSI in your case. While I am not familiar with the module you have they are all pretty similar. I wrote most of my code for what I used in old Visual Basic 6.0 and I see where they offer code samples for your module. What do you plan to do?

My apologies for just crashing into the thread. I have done quite a bit of data acquisition with similar modules to what you have.

Ron

 

Is this homework or a lab project of some kind? We have a couple of threads involving this LabJack that originate from the same university. If it IS a lab project, that's fine, but it should be in Homework Help, which should work more to your educational advantage.

 

That's it then.
Always nice to know how people get on.