Lenze 9400 Manual - Manual (Page 276)

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
= 3 s.
– During the recovery phase there is a current I
re
= 1.0 A.
– 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/2.5 A = 40 % - selected characteristic: 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 %
8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10
t [s]
ol
t [s]
re
‚ ‚
ƒ ƒ
„ „
 
SZD_E94A024_044 BSP
ƒ Result in the example
– The duration of recovery has to be at least t
re
= 5 s!