Level Pro ITC 450, Руководство пользователя

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Let the input mode = 4-20 mA, and the user selected the 10 segment characteristic. To do this it is necessary
to enter X and Y coordinates of 11 points (see
Menu ”inPt”
). The calculations will be done for three different
input currents from example 2, so in calculations some of the segments will be used only.
Let the following points will be given:
Additionally all other points must to be defined and stored in the device memory.
The segment defined by X6 = “30.0.” and X7 = “40.0.” for this In will be selected. Accordingly to expressions
given for user defined characteristic (see page 42) X6(PL) = 30, Y6(PL) = 30, X7(PH) = 40, Y7(PH) = 80 and Ip =
0,3 , the displayed value :
b) I
in
= 2,5 mA and In = -0,0938, because of the normalized In value is lower than 0, the segment defined by X1
and X2 will be selected. X1(PL) = 0, Y1(PL) = -50, X2(PH) = 10, Y2(PH) = -30 and Ip = 0. For these values the
displayed value W
≅
-69.
c) I
in
= 20,5 mA and In = 1,0313. , because of the normalized In value is higher than 1, the segment defined by
X10 and X11 will be selected, and X10(PL) = 90, Y10(PL) = 900, X11(PH) = 100, Y11(PH) = 820 and Ip = 0,9 for
these values the displayed value W
≅
795.
Let the current output parameters be:
“modE” = “4-20” , “ OUtL” = 100, “OUtH” = 200, “Lo r” = 5.0, “Hi r” = 5.0
Parameters “Lo r” and “Hi r” define working range of current output to 3.8 ÷ 21 mA.
Output current will be calculated for three displayed values „D”:
a) D = „17.5”
According to formula from page 34:
I
out
= (17.5-10.0) / (20.0-10.0) × 16 mA + 4 mA = 0.75 ·16 + 4 = 16 mA
a) Iin =10 mA and In = 0,375
X1 = “00.0.” , Y1 = “-50.0” ,
X2= “10.0.” , Y2 = “-30.0” ,
....
X6 = “30.0.” , Y6 = “30.0” ,
X7 = “40.0.” , Y7 = “80.0” ,
....
X10 = “90.0.” , Y10 = “900.0” ,
X11 = “100.0.” , Y11 = “820.0” ,
Example 6: The user defined characteristic
Example 7: Current output value calculation
Y Y
X X
Y = I
n p
80-30
40-30
30 67
=(0.375-0.3)
≅
43
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