# V700 + RTD Setup

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Greetings all.

First round setting up a RTD and, after reviewing forum posts here, I believe I am more confused than before... but likely an easy answer for most around here as I'm much more of a software guy.

Our setup: V700 Model with an RTD going to a IO-ATC8.  I do not have the RTD model# on me at the moment (typing this after hours but can provide tomorrow) but I know that the RTD contains a Prosense XTH-0300F-PT1 transmitter (0-300 degree F even though the RTD itself is capable of a much larger but unneeded range)

I understand that the IO-ATC8 is 14bit.

The confusion: I've seen forum posts where the linearization block references the "resolution @ 4-20mA" stated by the RTD/transmitter (I cannot locate this on the spec sheet for the model listed above, or maybe it's called something other than "resolution" but means the same thing?).  I've also seen where the linearization is based off the 14bit module (3277 - 16383) - which is in line with what we did with level sensors... aaaand I've seen where, in other cases, an RTD does not require additional linearization (other than dividing by 10, but I assume this is for 10bit setups?).

Any advice would be appreciated with a layman's outline of rationale.  Cheers!

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• MVP 2023

There are a couple of I/O module that can you can directly connect a 100 ohm RTD to - the V200-18-E3XB snaps onto the back of the V700 and the IO-PT400 expansion module is similar to the IO-ATC8.  Both of these modules return the temperature directly in 0.1 degree units, which probably what's confusing you.

But this is not what you have in the Prosense (translation: Evil Empire) transmitter.  You need to understand what you'll be feeding the IO-ATC8 electrically from the transmitter.  It's not ohms.  Let's lay it out:

0 F = 4 mA

300 F = 20 mA

The IO-ATC8 is an A/D (analog to digital) converter that produces a number to the PLC based on it's input.  You said you understand it's 14 bit, but do you know what that means?  Let's lay that out, too:

0 mA = 0 counts to PLC

20 mA = all 14 bits on = 2^14 = 16384 counts to the PLC

So your transmitter starts at 4 mA.  This means at 0 F the IO module will return (4/20*16384) = 3276.8 counts, which it will round to 3277.   At 300 F it will return 16384 counts.

Read the Help on the LINEAR block, which includes some fun diagrams under the subtopic "Linearizing Analog I/O Values"

So you set your linear block up like this:

X1 = 3277  (point 1 input)

Y1 = 0  (point 1 output)

X2 = 16384 (point 2 input)

Y2 = 300 (point 2 output)

Map X to the register assigned to the channel on your IO module and Y to whatever MI you want to use for the output.

If you still need help post your program.

Joe T.

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• MVP 2023

As I was writing my answer, Joe's has popped up and says it all.

One thing I might add is to go this topic and get the calculator, which might help you understand some of the principles a little easier.

cheers, Aus

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My god... I actually understood that.  Much thanks!  This is how I had it set up when I left today but was unsure of it (aside from using 16383 vs 16384 for whatever reason), maybe just a typo.

In my defense, the RTD and transmitter wasn't my choice, I'm' just the one trying to make the PLC like it.

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• MVP 2023
11 hours ago, Pendalar said:

aside from using 16383 vs 16384 for whatever reason

It's 16383. There are 2^14 counts, but it starts at zero, so the highest number is 2^14-1 (all 14 bits on would be 2^14-1, or 16383). This is what goes into your linearization block, though you would likely never notice it in the real world if your linearization was off by 1 part in 16384.

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Thanks for the explanation / makes sense.

As you said, probably never notice it in the real world, but then shouldn't the lower bound be 3276.8 - 1 = 3275.8, rounding to 3276 rather than 3277?

Asking more for learning purposes than concern over nth level of degree accuracy of the RTD.

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2 hours ago, Flex727 said:

but it starts at zero

I always tell my guys that get confused, "to a computer, zero is a much a number as one is, only people think zero means nothing"

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• MVP 2023
4 hours ago, Pendalar said:

but then shouldn't the lower bound be 3276.8 - 1 = 3275.8

Where are you getting a decimal point from? Everything in the context of this discussion involves integers only. Don't confuse the 4-20mA analog input with thermocouple or RTD inputs that provide temperature directly. Even with those, the input is still an integer with an implied decimal point.

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From Joe's explanation, the 4/20 with a 14 bit would round to 3277, but I'm asking if the value in the linear function should be -1 from that (3276)

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• MVP 2023
2 minutes ago, Pendalar said:

From Joe's explanation, the 4/20 with a 14 bit would round to 3277, but I'm asking if the value in the linear function should be -1 from that (3276)

Ah, sorry, I misunderstood.

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• MVP 2023

And the final follow-on to all of this number talk.  At first glance the calculator is a useful thing just to get your head around how Linearisation works.  But it does have a "higher purpose" and that's why I did it.  In the theoretical world, all 4-20 sensors work perfectly on those numbers mentioned.  But in the real world, sensors, the plc's input or both can be slightly off. If it is a simple linear difference over the entire span, then a + or - amount to the result  is all that is needed to get complete accuracy.  However, if the span itself has some innate error, the calculator lets you play with numbers very easily to get the best match to adjust things.  In all the sensors I use, many of my linearisations are NOT set at the numbers above, they have small variations to correct for this error.  My "need no adjustment at all" percentage is probably 25% of the sensors in use.  How critical your reading needs to be is an important factor in whether to play with this or not.  In my case I have to have all temperature sensors perfectly matched to a master device, otherwise I can get control conflicts.  In other situations this is not strictly necessary.

cheers, Aus

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