User manual - METER SUR-457
Example 6: The user defined characteristic
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:
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”
,
Additionally all other points must to be defined and stored in the device memory.
a) I
in
=10 mA and I
n
= 0,375
The segment defined by X6 =
“30.0.”
and X7 =
“40.0.”
for this I
n
will be selected. Accordingly
to expressions given for user defined characteristic (see page 45) X6(PL) = 30,
Y6(PL) = 30,
X7(PH) = 40
,
Y7(PH) = 80 and I
p
= 0,3 , the displayed value :
W
=
I
n
−
I
p
×
[
Y
PH
−
Y
PL
]
[
X
PH
−
X
PL
]
×
100
Y
PL
=
=
0,375
−
0,3
×
[
80
−
30
]
[
40
−
30
]
×
100
30
≃
67
b) I
in
= 2,5 mA and I
n
= -0,0938, because of the normalized I
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 I
p
= 0. For these values the displayed value W
@
-69.
c) I
in
= 20,5 mA and I
n
= 1,0313. , because of the normalized I
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 I
p
= 0,9 for these values the displayed value W
@
795.
Example 7: Current output value calculation
Lets assume that we have active current output and its parameters are:
“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 35:
I
out
= (17.5-10.0) / (20.0-10.0) × 16 mA + 4 mA = 0.75 ·16 + 4 = 16 mA
48
