Analogue IO

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Analogue IO
May 16, 2008 2:22 am, by Chris
Subject : PLCs
from the Automation List dept.
Text :
Can someone please explain mV/cnt or uA/cnt?

If I have an input that has 5.13uA/cnt, how do you calculate 20mA?

The input is 12 bit left justified.
Reply

Re: Analogue IO
May 17, 2008 12:31 pm, by zaki
Hi Chris,

1000mV = 1 volt
1000uA = 1mA

5.13uA is your count per resolution.Says,(4-20mA) 16000uA/5.13uA=3118.9 counts ,(0-20mA)20000/5.13uA=3898.6 counts nearly to 12 bit resolution.Input devices having 12bit=4096 counts.

I hope can add some

rborn
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Re: Analogue IO
May 16, 2008 9:57 pm, by Leonid
12 bit is 4096 counts, and because that max input is 5.13uA*4096 counts = 20000uA or 20mA.
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Re: Analogue IO
May 17, 2008 12:32 pm, by Roy Matson
Leonid,
The highest integer number we used to see in 12 bit Modicons was 4095 (started at 0), I may be wrong though.

5.13uA*4095 counts = 21,012 or ~21mA.
The extra mA must be for over-range assuming it starts at zero and not some other value below 4 ma (Under-range)
5.13 uA / count seems a bit odd to me, I wonder why they chose that?
Hopefully Chris will fill us in to what module it is.

Regards
Roy
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Re: Analogue IO
May 16, 2008 9:54 pm, by mike
I presume the "cnt" means count? If so then it means that the device is using integer values for passing information. Most microprocessor equipment loves to talk with whole numbers (1,2,3,4,5...). Microprocessors don't like to talk with real numbers (1.2, 1.3, 1.4, 1.5...). So, when trying to pass real numbers between devices, there is a little ritual that takes place. The sending device will convert everything to whole numbers (integers). In the first case, the sending device does the math using a conversion factor of 1 count per u-amp. The sending device sends the resultant math integer.

The receiver device does the inverse math. It'll convert the counts back to real numbers using the inverse conversion factor 1 u-amp per count.

But remember both devices must know the conversion factor well before the data flows.

Example:
(I can't remember, is u mean micro? X 10E-6? For the sake of this example, I'm going to assume 10E-6)

The sending device measures 0.0215 amps. Using the conversion factor of 1 u-amp/count, that would mean the sending device would send the value of 21500 counts. The receiving device knows the conversion factor so it does the inverse math and converts the number back to a real number of 0.0215 amps.
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Re: Analogue IO
May 16, 2008 9:44 pm, by Roy Matson
5.13 uA per count, that means each time you increase the signal by 5.13 uA the count goes up by 1:
10.26=2, 15.39=3 and so on.
Assuming your input starts at zero,
4 mA would be 4,000/5.13 or ~780 counts
20mA would be 20,000/5.13 or 3,899 counts
for 12 bits the highest count is 4095
4095 x 5.13 = 21,007 uA or 21 mA

What is your system? Someone who works with it will be able to elaborate on my explanation.

Regards,
Roy
Reply

Re: Analogue IO
May 17, 2008 12:30 pm, by Chris
Thanks Roy,
That is the way i have always calculated values, but i recently purchased a process calibrator and was doing some emperimenting and found my 20mA scaling to be incorrect.

21mA = 32768 or 4095 x 2^3
20mA = 30840 or 3855 x 2^3

i had been using 31184 or 3898 x 2^3.

This is a AB flex i/o module 1794-ie8.

Regards
Chris
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