Allen Bradley CPT Instruction

Ron Beaufort · Mar 24, 2017

there are several ways to set up the scaling for your input module ... we're not sure which of these setups you're using ... and (no offense) you're not really sure of what values are being used ...

now for the classic question:

are you allowed to post the ENTIRE project file (ACD extension) ... we can move along a LOT faster once we see what's behind the curtain ...

you might also put a milliamp meter in the circuit - and tell us how many mA are actually flowing through your module's input ... (that would probably be easier than working with the transmitter - at least at this point of the game) ...

and by all means - confirm that the transmitter is putting out the correct value ... (maybe it has various settings and adjustments that aren't quite right ...

ELake20 · Mar 24, 2017

Ones_Zeros said:
here is what i know & what normally we have scaled the setting on the
1769-IF8 analog card
the transmitter is scaled +75 to -10
so this should be my scaled max = +75 and min = -10
the input Max = 16000 & Min = 4000 (based on the scaling setting on the 1769-IF8 analog input card) I might have these wrong
the Raw input 6546 is the actual 4-20ma raw value coming into the card from the transmitter

the actual read out on the transmitter is somewhere around 0.2 to 0.3
so based on the scaling on the transmitter i know this is what my final value needs to be after working out the expression

I may not have these values correct since the expression is not working out correctly
what ever i need to change on the expression, please let me know

thanks alot i appreciate you guys looking over this

This may help you....

Split up your equations for sanity sake.

y=mx+b

Solve for X otherwise known as RATE:
X:=(rScaled_Max - rScaled_Min) / (rInput_Max - rInput_Min)

Solve for B otherwise known as OFFSET:
B:= rScaled_Min - (rInput_Min * rRate)

Solve y=mx+b using previously calculated X and B values:
Scaled_Output:=(iRaw * rRate) + rOffset

So you are using Engineering Units, respanning 4-20mA to 4000 to 16000? Why is this? I would do this with 4000 to 20000 for 4-20mA signals.

Ron Beaufort · Mar 24, 2017

here are some of the ranges available for your 1769-IF8 module ...

we can't do much more than guess until we know which of these setups you're using - and how much current (mA) is flowing through the input channel ...

good luck - I've got to go cut the grass now ... (spring has sprung) ...

.

Allen Bradley CPT Instruction

Ron Beaufort

Lifetime Supporting Member

ELake20

Lifetime Supporting Member

Ron Beaufort

Lifetime Supporting Member

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