temperature measurement pt100 daq

[cebailey]

Good find, smercurio. I followed beyond the link and opened a pdf for that product and noticed the note at the bottom that looked like it meant "Contact INOR", and guessed that INOR is a company name. A web search on INOR rtd PT100 gave me this link:: http://www.ecnasiamag.com/article-19199-inorsbrandpt100temperaturesensortransmitter-Asia.html

Now I suspect we may be dealing with a linearized signal already, coming from a transmitter made by INOR - could that be?

[smercurio_fc]

I don't think so since the OP is measuring a voltage, not a current, which is what the INOR outputs. Also, the INOR product is labeled as a "transmitter", rather than a sensor, and the OP indicated he's working with a sensor.

[sergymax]

That is the company I called but they didn´t know where to find these parameters, weird. And the pdf doesn´t say anything about Iex

The sensor Pt100 measure temperature and send a current signal according to a slope y=2000x - 1E-14 from 4 to 20 mA. Thats the only documentation I have from the sensors.

Could I calculate the Iex someway?

[smercurio_fc]

<Saverio just sits at his desk shaking his head>

Message Edited by smercurio_fc on 02-20-2009 01:30 PM

[sergymax]

What does it mean smercurio?

[cebailey]

>The sensor Pt100 measure temperature and send a current signal according to a slope y=2000x - 1E-14 from 4 to 20 mA. Thats the only documentation I have from the sensors.

We have been solving the wrong problem. You **don't need** to read a Pt100 signal. You need to read a 4 to 20 mA current signal. Therefore you need to measure current. Maybe you also need to provide a DC power loop to run through the transmitter. I forget what hardware you are using (was this the thread with the USB6009 or was that another one?), but you must either have a resistor built in to the DAQ device, or you must run your 4-20 mA through a resistor of your choosing and measure the voltage drop across it. From the resistance and voltage you know the current. THAT's what you want, the current. Between 4 and 20 mA.

OK, here's something we can do. Cut and paste the text from the documentation into a reply here, either the whole thing if it is tiny or else, at least, the lines that discuss this temperature and current and slope and y and x and 4 to 20 ma. Maybe that will clear things up.

Or if that doesn't work, if you can get two temperature readings off of some other thermometer and get the 4 to 20 mA signal that this device gives you when it is seeing those same temperatures, we can calculate a slope.

[cebailey]

I think maybe smercurio just means he wishes he read Swedish better.

You did say you have a USB 6009. That does not read currents directly. You have to run your 4-20 mA signal through a resistor and use the 6009 to measure the voltage across the resistor, using a differential input channel. The 6009 has a maximum 10 V input, or 20 V if you are careful to keep whatever ground it sees midway between the inputs. You can use a 1000 ohm resistor and get a 20 V signal at the top of the scale. More people use 250 ohms, though, which will give you a 0.1 to 5 V signal you can easily measure with the 6009. It also is more likely to be within the voltages the transmitter can handle. And, if the transmitter is using the current loop as its source of power (which I would guess it probably does), you have to be creating the loop voltage with your own power supply, maybe 10 or 24 volts or so.

What I am explaining is how to set up a current loop measurement with your transmitter and your 6009. I don't do this much, so if you or somebody there already knows how to do this, then go ahead.

Give me more info and I can help figure out temperature from your current reading, as I described in the last post.

[cebailey]

I said that wrong. The 250 ohm resistor will give you a 1 to 5 V signal, not 0.1 to 5.

[Ben]

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sergymax wrote:
What does it mean smercurio?

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It means that you are working with a device that outputs a 4 to 20 milliamp signal. Milliamp is a measurement of current and not voltage.

Ben

PS: Boy am I glad I only read this thread after Y'all did all of the work.

Sergymax,

You could partially recover from this fiasco if you very quickly go back and give Kudos (thank yous) to all of those people that have worked with you.

Message Edited by Ben on 02-20-2009 03:04 PM

[sergymax]

Sorry guys. I first tried to make a VI to measure the current and convert it into temperature according to the slope, but the responsible of the project told me I should measure voltage as the device just receives voltage signals.

I dont understand so much what are you asking to do now, cebailey. Should I create a current imput or how?

I am afraid I can not give you more information. The only thing I have is the name of the sensors and the loop current-temperature.

Thank you and sorry for the missunderstandings