Simex SPT-86L, Руководство пользователя

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User manual for RTD,TC,mV/4-20mA converter module SPT-86L
b) U
inp.
= -9.38 mV and U
n
= -0.0938.
Normalized result is negative, so the returned value is equal to ”
InputLow
” parameter:
W
=
-300.
c) U
inp.
= 103.13 mV and U
n
= 1.0313.
W
≅
1223.
Example 6: The user defined characteristic
Let assume the input mode ”0-100” is chosen, and the user selected the 10 segment
characteristic. To do this, it is necessary to enter X and Y coordinates of 11 points (see
registers 70h÷97h).
The calculations will be done for three different input voltages from example 2, so in
calculations some of the segments will be used only.
Let the following points will be given:
X1 = ”
00.0. ”, Y1 = ” -50.0 ”,
X 2= ”
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 ”,
Additionally all other points must to be defined and stored in the device memory.
a) U
inp.
= 37.5 mV and U
n
= 0.375
The segment defined by X6 = ”
30.0.
” and X7 = ”
40.0.
” for this U
n
will be selected. Accordingly
to expressions given for user defined characteristic (see page 14) X6(PL) = 30, Y6(PL) = 30,
X7(PH) = 40, Y7(PH) = 80 and U
p
= 0,3 , the returned value:
W
=
U
n
−
U
p
 ×
[
Y

PH
−
Y

PL
]
[
X

PH
−
X

PL
]
×
100

Y

PL
=
=
0.375
−
0.3
 ×
[
80
−
30
]
[
40
−
30
]
×
100

30
≃
67
b) U
inp.
= -9.38 mV and U
n
= -0.0938.
Because of the normalized U
n
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 U
p
= 0. For these values the returned value W
≅
-69.
c) U
inp.
= 103.13 mV and U
n
= 1.0313.
Because of the normalized U
n
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 U
p
= 0.9 for these values the returned value W
≅
795.
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