RSLogix 500 Scale Function Test

hahaliao · Feb 22, 2017

Hi,

First time using scale function and not quite sure whether it is right or wrong. Please advise and below is what I have
1. Using Allen Bradley 1766-L32AWAA
2. Temperature Transmitter is Prosense TTD-20-N40160F-H
3. PT100 is Prosense RTD0100-10-010-H
4. Write a simple Scale function. Input number is changing but I see no change on the output number

Please let me know why or what I should do to make sure I do it right.
Thank you!

Dravik · Feb 22, 2017

Is your PLC in run? Is the ladder being scanned? Can you post a screenshot?

hahaliao · Feb 22, 2017

Here is the image
http://imgur.com/a/nHIsq

I just wrote a simple scale function.

hahaliao · Feb 22, 2017

hahaliao · Feb 22, 2017

Since I am new to this scale function, if someone can provide me some simple quote and method of testing, I would very appreciate that.

Then I can move to next step where I need to do some control according to the temperature change.

Mike_RH · Feb 22, 2017

You used ladder file 17, is there a JSR to it?

hahaliao · Feb 22, 2017

Hi Mike,

Since I am testing on this scale function, I wrote this in ladder file 17 and plan to use it later in my main when writing the some lesser than or greater than function to control my output.

Is that because I do not write any JSR in my main and it cause the output file got no number change when the input number is changing?

Dravik · Feb 22, 2017

Yes, If there isn't a JSR, the logic isn't being scanned and outputs will not update.

hahaliao · Feb 22, 2017

You guy are awesome!
After adding some simple command in my main, now I saw the changing on the output. I know I am in the right path now and continue working on the temperature control part.

By the way, I have a simple question to ask. Here is what I found when I doing the wiring connection
1. without a resistor between analog output and analog com, the result seem not correct
2. with a resistor between analog output and analog com, the result seem right

I will do the #1 method. Can someone tell me whether I am making the right choice and tell me why?

Thank you!

ASF · Feb 22, 2017

Most likely your sensor is a 4-20mA type sensor. The analog input on the Micrologix PLC's is a 0-10V analog input. If you put a 500ohm resistor across the analog input, then your 4-20mA will flow through the resistor, and ohms law will tell you that you now have 2-10V across the analog input.

So, without the resistor, you have a 4-20mA signal and a 0-10V input. This won't work. By putting the resistor there, you convert the 4-20mA signal to a 2-10V signal, and this will work on your 0-10V input.

hahaliao · Feb 22, 2017

Thanks for the detailed explanation.

Does it matter if I am not using 500 ohm? Is it critical if not using exact 500 ohm?
I only have 470 ohm or 680 ohm. Unless I solder it and put them in series.

James Mcquade · Feb 22, 2017

Yes,

AB says that you need a 500 +/- 0.25 ohm resistor.
if you don't have one, the closer to 500, the better results you get.
you will also need to scale the input for best results.

We bought several from our electronics dealer and we use them for calibration and other things.

james

ASF · Feb 22, 2017

It doesn't matter if you account for it in the scaling. It's simple ohms law: V=IR.

Lets say you have a 470ohm resistor. We need to know the voltage at 4mA and 20mA.

For 4mA:
V=I*R
V=0.004*470
V=1.88

For 20mA:
V=I*R
V=0.020*470
V=9.4

So if you use a 470 ohm resistor, your 4-20mA will become 1.88-9.4V. This is within the range of the 0-10V input, so it will work just fine - you will just have to work out what values 1.88V and 9.4V correspond to on your analog input so that you can scale correctly.

If you were to use the 680ohm resistor, your 20mA would calculate out to be 13.6V which is above the measurable range of the analog input, so that will not work.

Long story short; use as close to 500ohms as possible, but if you must vary, go lower rather than higher and calculate accordingly.

hahaliao · Feb 22, 2017

Thanks for the suggestion, James.

Thank you, ASF for the detailed information with the calculation.

Conclusion, it needs in the range of 0-10V.

Mickey · Feb 22, 2017

hahaliao said:
Conclusion, it needs in the range of 0-10V.

No it does not, you could use a 250 ohm giving you 1 to 5 v.
You will just lose some resolution with a smaller resistor.

Also keep in mind if you are using the embedded analog inputs the raw data
is 0-4095 for an input of 0-10v.

Lets say you have a 470ohm resistor. We need to know the voltage at 4mA and 20mA.

For 4mA:
V=I*R
V=0.004*470
V=1.88

For 20mA:
V=I*R
V=0.020*470
V=9.4

Using ASF's example..

1.88 v = 770 raw data value
9.4v = 3849 raw data value

RSLogix 500 Scale Function Test

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