gefran GPC - 4 Examples and Application Notes; Setting Parameters LS.A1;LS.A2;LS.A3 and HS.A1;HS.A2; Hs.a3; Full Input Voltage Range

81901A “MSW_GPC-40/600A”_03-2021_ENG_page 135
4. EXAMPLES AND APPLICATION NOTES
4.1. Setting parameters LS.A1/LS.A2/LS.A3 and HS.A1/HS.A2/HS.A3
The default values(LS.A1/LS.A2/LS.A3 = 0.0 and
HS.A1/HS.A2/HS.A3 = 100.0) can be changed to obtain
different assignments between the physical input value (V
or mA) and the engineering value (In.A1/In.A2/In.A3).
In automatic operation the engineering value (In.A1/In.A2/
In.A3) is allocated, for values between 0.0 and 100.0, to
power Ou.P.
4.1.1. Full input voltage range
Initial data
V
IN
= 0...10 V
tyP = 1
In this case the input voltage V
IN
covers the entire range that
the device can control. There is therefore no need to modify
LS.A and HS.A, which can be left at their default values.
4.1.2. Input voltage range limited down-
wards
Initial data
V
IN
= 2...10 V
tyP = 1
In this case, the V
IN
input voltage only covers the upper
80% of the range that the device can control. This range
should therefore be extended downwards so that 80%
of the modified range corresponds to the useful range
(0...100). 80% of 125 corresponds to 100.
Setting LS.A = -25.0 and HS.A = 100.0 gives the desired
result.
00%
2 4 6 8 10
100
50
100%
LS.A = 0.0
HS.A = 100.0
V
IN
Ou.P
00%
2 4 6 8 10
100
50
100%
LS.A = -25.0
HS.A = 100.0
V
IN
Ou.P
4.1.3. Input voltage range limited upwards
Initial data
V
IN
= 0...9 V
tyP = 1
In this case the V
IN
input voltage only covers the lower 90%
of the range that the device can control. This range should
therefore be extended upwards so that 90% of the modified
range corresponds to the useful range (0...100). 90% of
111.1 corresponds to 100.
Setting LS.A = 0.0 and HS.A = 111.1 gives the desired
result.
4.1.4. Formulas for calculating LS.Ax and
HS.Ax
Given the following definitions:
min
d
= lower limit of the range controllable by the input
device
max
d
= upper limit of the range controllable by the input
device
min
in
= real lower limit of the input signal
max
in
= real upper limit of the input signal
we have
LS.Ax = 100
min
d
- min
in
max
in
- min
in
HS.Ax = 100
max
d
- min
in
max
in
- min
in
00%
2 4 6 8 10
100
50
100%
LS.A = 0.0
HS.A = 111.1
V
IN
Ou.P