How to connect a SIKA pressure sensor to the cDAQ-9174 with module NI 9219 and read the signal?

[WM_John_Weeks]

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@DP123 wrote:
Hey John

First I want to say thank you for taking your time to help me.

I am pretty sure that the module I got can receive inputs of 4-20 mA. The thing is that I am really new to programming and with wiring modules and hardware. Therefore simple instructions are very needed. The reason why I asked if it was a full bridge, half bridge or quarter bridge is because labview gives you an example (connection diagram) based on which type you choose. If I choose the full bridge option it shows me that all 4 wires should be connected to the module which kinda makes me think something is wrong since the one wire is for grounding and atleast one of the wires should be for powering the sensor.

DP

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My previous reply, where I said "You do not need to worry about the bridge- it is internal to your transducer, and you don't interact with it at all. The transducer has electronics inside it to read the bridge and generate the current output." was intended to directly address your question about the bridge type. You don't need to know that, and the examples you found that tell you how to connect a bridge are not relevant to your pressure transducer.

Your transducer produces a current- you need to look up connections for current measurement. Unfortunately, the Sika web site is not easy to navigate and I was not able to find any information on which wires on your transducer are to be connected. You will need to ask them for that information.

Meanwhile, I also outlined how you would wire the transducer for current measurent: connect a suitable resistor across the V+ to V- inputs (differential mode) or from V+ to Ground (RSE). Connect your transducer output to V+ and the transducer common (or ground or earth, etc.) to the V- or Ground on your device.

Be aware that the resistor will need to be a good quality resistor, or it will introduce temperature drift.

I see on the Sika web site that your transducer also comes in a voltage output option. That might be easier to use.

If you don't know about Ohm's law, and can't compute the resistor value to use, perhaps you need to take a lab course in instrumentation, or get a good book about the subject.

[Henrik_Volkers]

And hip hip huraa , all information marked on the sensor...

Spoiler

That was my guess, wasn't it?

Nice sensors they have in the school labs today 😄 😄

[DP123]

Hi everyone

Is this the scenario I am looking at making (attached file)? I am currently studying marine engineering but I just started. So I got the basics set but I still have alot to learn. I am trying to learn how to do this in my spare time so I am sorry if I sound like a complete idiot. 

DP

[WM_John_Weeks]

That looks right. I will confess I don't have experience with 4-20 mA systems, but I know quite a bit about instrumentation and some about electronics. That diagram squares with my electrical knowledge. Perhaps someone with actual experience with current loop transmitters will chime in here.

Your DAQ device will be the "Control Room Panel Mounted Instrument".

Sorry if I came off a bit peavish in my last post- it was meant as friendly advice. When I was an undergraduate in geology, I took a non-majors course in electronics. It was perhaps the most useful course I ever took! And I spent a number of years in my graduate and post-graduate life reading books and learning the hard way.

Welcome! It can be both a lot of fun tinkering with these things, and a lot of frustration.

[DP123]

Hi John

You were not too harsh. No worries :). I just wonder if I have a module that is able to read current signals why would I need to connect a resistor between the two inputs? 

DP

[WM_John_Weeks]

That's how you measure current. Ohm's Law: E = iR, voltage = current * resistance. The 4-20 mA current flows through the resistor, and that makes a voltage appear across the resistor that is proportional to current. The 250 ohm resistor in the picture you posted would make a range of voltage of 0.004*250 = 1 Volt to 0.020*250 = 5 Volts. By connecting the resistor across the V+ to V- inputs (differential) or from V+ to ground (RSE) you generate that voltage where your device can measure it.

Find that electronics course!

[DP123]

Hi John

Yes I know that I would get a voltage across the resistor that my device would be able to read. But what I am wondering is if its necessary when the hardware should be able to read the current signal and use that directly. 4mA = 0% and 20mA = 100%. I thought that it was only necessary if the module could not read current signals? I

DP

[WM_John_Weeks]

OK- I have never used a 9219 device and forgot that part of your posting title. I looked up the specs for the 9219 and you are right- it has an option to measure current. That should do the trick just fine.

You still have to remember to complete the current loop through the 9219 and the transducer, but you don't need the resistor. Effectively, the resistor is inside the 9219 when you choose to measure current.

[DP123]

Awesome John! 

Next part I have to understand is when to use Ex+ and Ex- and when to use HI and LO terminals on my module. I have tried looking up information about it but it seems hard to find anything useful.

I know that EX+ = Positive sensor excitation connection, EX- = Negative sensor excitation connection, HI = Positive input signal connection and  LO= Negative input signal connection. Anyway you could tell me the difference? 

I guess that I my case I have to use the HI and LO since I am working with current signals? Second the sensor output should probably go to the HI or am I wrong?  

DP