Manual
Regenerative power supply modules
Rated data (devices for 400/500V mains)
Current-time diagrams
7
276
EDS94SPP101-7.1
Application example
ƒ Drive task
– A current I
ol
= 4.0 A is required for a time t
ol
=3s.
– During the recovery phase there is a current I
re
=1.0A.
– During the overload phase a switching frequency of 4 kHz is accepted.
– The switching frequency during the recovery phase is to be 8 kHz.
– The mains voltage is 400 V.
– The times with output frequencies under 5 Hz are insignificant.
ƒ Problem
– How long does the recovery phase t
re
have to be when the E94AxxE0034 controller
is used?
ƒ The currents required are based on the rated device current:
–I
ol
= 4.0 A/2.5 A = 160 % - selected characteristic: 175 %
–I
re
=1.0A/2.5A=40%-selectedcharacteristic:50%
– Because of the short overcurrent time t
ol
= 3 s, the cutout diagram of the
E94AxxE0034 device is taken into consideration.
– The switching frequency f
chopp
= 8 kHz, var does not have to be considered
separately.
ƒ Procedure within the diagram
– c Starting from the time axis at t
ol
= 3 s, a perpendicular line is drawn to the
selected characteristic at 175 % (corresponds to the minimum required current I
ol
).
– d Draw a straight line from the point of intersection horizontally to the right to
the selected characteristic, the recovery phase, at 50 %.
– e Draw a perpendicular line from the point of intersection back to the time axis.
– f Read the minimum required duration of recovery on the time axis.
60 %
60 %
75 %
75 %
50 %
50 %
25 %
25 %
0 %
0%
350
%
350 %
400
%
400 %
300
%
300 %
275
%
275 %
250
%
250 %
225 %
225 %
200 %
200 %
175 %
175 %
150 %
150 %
876
5
4
3
21
0
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t [s]
ol
t [s]
re
SZD_E94A024_044 BSP
ƒ Result in the example
– The duration of recovery has to be at least t
re
=5s!