PDU–4xx–W
©
Dinel, s.r.o.
38
Example 5: The square root characteristic
Let the input mode = 4-20 mA, and parameters “Lo C” and “Hi C” equal to -300 and 1200 respectively.
The calculations will be done for there different input currents from example 2.
a). I
in
=10 mA a I
n
= 0,375
Accordingly to expression on page 39 for square root characteristic:
sqrt(0,375) × [1200 -(- 300)] ≈ 919
and next, the “Lo C” value is added to the result , so the displayed value:
W ≈ 919 + (-300) = 619
b). I
in
= 2,5 mA a I
n
= -0,0938, normalized result is negative , so the displayed value is equal
to “Lo C” parameter: W ≈
“Lo C”
= -300.
c). I
in
= 20,5 mA a I
n
= 1,0313
W ≈ 1223
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 a I
n
= 0,375
Accordingly to expression on page 38 for square characteristic:
(0,375)2 × [1200 -(-300)] ≈ 211. and next, the “Lo C” value is added to the result , so the
displayed value:
W ≈ 211 + (-300) = -89
b). I
in
= 2,5 mA a I
n
= -0,0938.
W ≈ -287
c). I
in
= 20,5 mA a I
n
= 1,0313
W ≈ 1295
