Temperature with 0.01° accuracy

Hi all,
I have following problem. I work with a Schneider M241 PLC and have the TM3TI4 module. This module can hold 4 PT-100 or PT-1000 sensors and can I configure these indeed as one of these 2 sensors with a direct readout with a 0.1°C accuracy but also have the opportunity to have an INT value between -32.768 and +32.767.

The problem I have is that I need a 2 decimal value (0.01°C).
Can somebody tell me if there is a solution for this please? Is there a way to get more accuracy?

Thank you all, Vincent

The RTDs themselves are not that accurate.

Thank you for the answer Steve. I used the wrong word and should have been resolution instead of accuracy. We can always calibrate the sensor and such but I need the 0.01°C resolution as I steer a 3-way valve with a 0-10VDC signal and that means the 3-way open and closes with 10% at the time.

http://www.pyromation.com/Downloads/Doc/Training_RTD_Theory.pdf
The most accurate will be RTD.
DIN 43760 class B = ± .12% @ 0°C
DIN 43760 class A = ± .06% @ 0°C
ASTM E-1137 grade A = ± .05% @ 0°C
ASTM E-1137 grade B = ± .10% @ 0°C
You are asking for something ± .005 @ 0°C. I have not worked with anything like that.
Regards,

That level of resolution without corresponding accuracy doesn't accomplish much, other than give someone a warm fuzzy feeling. What are you trying to do?

Let me sketch the issue at hand

As being from Belgium do we work in °C. We use the M241 for climate control by means of an Ani Handling Unit equipped with a water heating and water cooling coil. Each of the coil has a 3-way valve controlled by a 0-10VDC signal.

By having a temperature with a 0.1°C resolution and the 3-way valve controlled by a P controller that has a 1°C bandwidth. That means the 3-way valve goes open in 10% steps what is too much for an accurate temperature control. Yes, a PI would be more accurate but it replaces a Siemens RLU controller and this far less advanced controller does work with the same PT-100 (or PT-1000) sensors with a 0.01°C accuracy.

Cheers,
Vincent

You can scale the raw analog input using the line equation y=mx + b where x is the input variable, m is the line slope, and b is the y intercept coordinate.

To compute the value of the slope we need to know the scaled span divided by the inputs span, or

m = (Scaled_Maximum - Scaled_Minimum) / (Input_Maximum - Input_Minimum).


and

b = Scaled_Minimum - (m*input_Minimum)

Check your input card manual to see what the RTD input range is. It will be something like -200°C to 600°C

The input maximum and input minimum will be 32767 and -32768 if what you gave us was correct.

Plug those values into the equations I gave you for m and b. Then take the raw integer analog input, multiply it by m and add b. That will give you a scaled value that will appear to you to have more resolution. Whether that apparent resolution will be real or not depends on your actual hardware, so check the manual for that information.

The spans of the RTD input card can be rather wide, -200 to 600 Deg C for a Pt1000, -200 to 800 Deg C for Pt100.

An alternative is to use a temperature trnasmitter and range it for its minimum range, sometimes as low as 0-25 Deg C.

With a 4-20mA analog input card, 25 deg C over 16mA is 1.375 Deg/mA. Getting a mA signal to resolve 1.xx mA or .XX Deg is common. But all those asterisks and notes on accuracy come home to roost when dealing with hundredths of a degree.

Thank you all already.

TConnely, will try indeed your formula.
DanW, a transmitter is too expensive I'm afraid.

s you can see in the specs the Pt100 setting has 10500 points, and yes it is a 16 bit card, however the resistance is measured from 0 to 400 ohms.
this is divided in this 65232. the 10500 is from -200 to 850 Celsius.
No it is not possible in this setting to get higher resolution.
However there is a solution for this.
Most probably you only have a limited scale needed.

assume it is -50 to 50C
this is 100 Kelvin, instead of 1000 so the resolution will be 10 times better so you will meet your 0.01 C in resolution.
what you need is a converter that makes a 3 wire Pt100 signal into a 4-20 mA signal, costing around 25-150 euro (yes i can supply them.)
They contain a XTR105 circuit. you will have a resolution of 100/65232*4/5, however the accuracy will be around 0.1 due to the price and stability.
more info send mail to [email protected]

While writing the message a lot happened, connely is not correct as the line is not straight, but is curved, and yes i do have this formula.
Your theory is correct, however when you work it out, you will see the Integrator is the better control.

have a look at Oscat.de (yes working on schneider/codesys)(it has a good PID controller, and in building it has a valve control.
dont worry to much, the 0.1 resolution is good enough for control.
lets assume you want 23 C, the P band should be around 5 C
the I about 10 minutes.
try the oscillation keep it below 0.5 C.
etc.

Yes, it does!

Tom Jenkins said:

That level of resolution without corresponding accuracy doesn't accomplish much, other than give someone a warm fuzzy feeling. What are you trying to do?

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It reduces quantizing error ( digital non-linearity ). Higher resolution allows the use of derivative gain if necessary.

Do some math on it in floating point and then you can get all of the false accuracy you want.

Vinnie_vdb said:

I need 0.01°C resolution...

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Vinnie_vdb said:

...a transmitter is too expensive I'm afraid.

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ASF said:

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My two cents