Allen Bradley SLC503 PLC Analogue

New to analogue I/O's. I have seen 4 to 20 mA inputs and the ladder programme that this represents (I think). The 4 ma is 3424 and the 20 mA is
16528 (from memory) can check these numbers later. Can anyone tell me how we arrive at three thousand and odd and the sixteen thousand and odd. Is this some equivalent for the 4 to 20? What would this number be for say 12 mA? TIA
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You are in the ballpark but it looks like your zero might be slightly elevated. When the analog module and the transmitter are properly calibrated, the PLC counts at 4ma should read 3276 counts, at 20ma the counts should read 16384. 3276 is the PLC count 20% offset. 9830 will be the 12ma reading At 12ma or 50% (9830 - 3276) / 13108 = .5 TW

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Thanks for the info. Just what I was looking for.

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