ratio for analog outputs

4-20mA again.

After my post yesterday a person at work gave me this ratio;

It is x = strokes/min * mA

-----------------
20 mA

So if I want to know the strokes for 15 mA, I do 360 * 15, then
divide that answer by 20.

I am a bit confused becase this ratio gives different results
then the answer I received yesterday. Which is correct ?

Thanks

This would be true if it wer a 0-20 ma signal. But it's a 4-20 ma signal.

Greetings Tom,

the answers to the "scaling" questions you've been posting are included here ... it might take some time to wade all of the way through it ... but if you're really interested in the "nuts-and-bolts" of this subject, this would probably be a handy resource ... it's helped quite a few people in the past ...

and if you want an easy to print .PDF format, you can find it here as part 1 and part 2 ...

Greetings Tom,
I don't know about your previous post, But here is the formula:

Engineering value =
(Your binary counts - min. binary counts)/(Binary counts span - Engineering value span) + min engineering value.

See the attachment for better view!

Where, ​

Your binary counts are the analog value in counts.​

Min. binary counts is the low limit of the binary counts ​

Binary counts span = (max. binary counts – min. binary counts)​

Engineering value span = (max. engineering value – min. engineering value)​

Note: read the PLC manufacturer's manual to know the Min. binary counts and Max. binary counts, as it depends on the analog channels resolution and whether the offset (i.e. 4mA) is compensated to zero counts or not.​

​

​

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Remeber that the formula above is the most generic one, It work in all cases and if you find any different formual please understand that it's a special case from the generic one above (I tried this my self for any formula I find).