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A pressure transducer is connected to input 0 and we want to read the
value in engineering units. The pressure transducer measures pressures from
0–1000 psi and provides a 0–10V signal to the analog module. For a 0–10V signal,
the analog module provides a range between 0–32,767. The following program
rung places a number between 0–1000 into N7:20 based on the input signal coming
from the pressure transducer into the analog module.
Rung 2:0| +SCP––––––––––––––––––––+ ||––––––––––––––––––––––––––––––––––––––––––––––––––––+SCALE W/PARAMETERS +–|| |Input I:1.0| || | 0| || |Input Min. 0| || | | || |Input Max. 32767| || | | || |Scaled Min. 0| || | | || |Scaled Max. 1000| || | | || |Scaled Output N7:20| || | 0| || +–––––––––––––––––––––––+ |
PrefaceInstruction Set Reference Manual
3–16
Example 2
In the second example, an analog I/O combination module (1746-NIO4I) is in slot 1
of the chassis. We want to control the proportional valve connected to output 0.
The valve takes a 4–20mA signal to control how far it opens (0–100%). (Assume
that additional logic is present in the program that calculates how far to open the
valve in percent and places a number between 0–100 into N7:21.) The analog
module provides a 4–20mA output signal for a number between 6242–31,208. The
following program rung directs an analog output to provide a 4–20 mA signal to the
proportional valve (N7:21), based on a number between 0–100.
Rung 2:1| +SCP––––––––––––––––––––+ ||––––––––––––––––––––––––––––––––––––––––––––––––––––+SCALE W/PARAMETERS +–|| |Input N7:21| || | 0| || |Input Min. 0| || | | || |Input Max. 100| || | | || |Scaled Min. 6242| || | | || |Scaled Max. 31208| || | | || |Scaled Output O:1.0| || | 0| || +–––––––––––––––––––––––+ |
Math Instructions
3–17
Scale Data (SCL)
When this instruction is true, the value at the source address is multiplied by the rate
value. The rounded result is added to the offset value and placed in the destination.
Example
SCL
SCALE
Source N7:0
100
Rate [/10000] 25000
Offset 127
Dest N7:1
377
The source 100 is multiplied by
25000 and divided by 10000 and
added to 127. The result 377 is
placed in the destination.
Note Anytime an underflow or overflow occurs in the destination file, minor error bit
S:5/0 must be reset by the program. This must occur before the end of the current
scan to prevent major error code 0020 from being declared. This instruction can
overflow before the offset is added.
Note that the term rate is sometimes referred to as slope. The rate function is
limited to the range –3.2768 to 3.2767. For example, –32768/10000 to
+32767/10000.
Entering Parameters
The value for the following parameters is between –32,768 to 32,767.
• Source can be either a constant or a word address.
• Rate (or slope) is the positive or negative value you enter divided by 10,000. It
can be either a constant or a word address.
• Offset can be either a constant or a word address.
33 333
SCL
SCALE
Source
Rate [/10000]
Offset
Dest
Output Instruction
PrefaceInstruction Set Reference Manual
3–18
Updates to Arithmetic Status Bits
With this Bit: The Processor:
Carry (C) is reserved.
Overflow (V)
sets if an overflow is detected; otherwise resets. On overflow, minor
error bit S:5/0 is also set and the value –32,768 or 32,767 is placed in
the destination. The presence of an overflow is checked before and
after the offset value is applied. À
Zero (Z) sets when destination value is zero.
Sign (S) sets if the destination value is negative; otherwise resets.
À If the result of the Source times the Rate, divided by 10000, is greater than 32767, the SCL instruction overflows,
causing error 0020 (Minor Error Bit), and places 32767 in the Destination. This occurs regardless of the current
offset.
Application Example 1 – Converting 4mA–20mA Analog Input Signal to PID
Process Variable
Scaled Value
Input Value
3,277
(Input Min.)
16,384
(Input Max.)
16,383
0
(Scaled Max.)
(Scaled Min.)
Math Instructions
3–19
Calculating the Linear Relationship
Use the following equations to express the linear relationship between the input
value and the resulting scaled value:
Scaled value = (input value x rate) + offset
Rate = (scaled max. – scaled min.) / (input max. – input min.)
 (16,383 − 0) / (16,384 – 3277) = 1.249 (or 12,490/10000)
Offset = scaled min. – (input min. x rate)
 0 – (3277 × 1.249) = –4093
Application Example 2 – Scaling an Analog Input to Control an Analog
Output
Scaled Value
Input Value
3,277 4mA 16,384 20mA
32,764 10V
0 0V
(Scaled Max.)
(Scaled Min.)
(Input Min.) (Input Max.)
PrefaceInstruction Set Reference Manual
3–20
Calculating the Linear Relationship
Use the following equations to calculate the scaled units:
Scaled value = (input value x rate) + offset
Rate = (scaled max. – scaled min.) / (input max. – input min.)
 (32,764 – 0) / (16,384 – 3277) = 2.4997 (or 24,997/10000)
Offset = scaled min. – (input min. x rate)
 0 − (3277 × 2.4997) = – 8192
The above offset and rate values are correct for the SCL instruction. However, if
the input exceeds 13,107, the instruction overflows. For example:
17mA = 13,926 × 2.4997 = 34,810 (actual overflow)
34,810 – 8192 = 26,618
Notice that an overflow occurred even though the final value was correct. This
happens because the overflow condition occurred during the rate calculation.
To avoid an overflow, we recommend shifting the linear relationship along the input
value axis and reduce the values.
The following graph shows the shifted linear relationship. The input minimum
value of 3,277 is subtracted from the input maximum value of 16,384, resulting in
the value of 13,107.
Math Instructions
3–21
Scaled Value
Input Value
0 4mA 13,107