GE CS300 User‘s guide User Manual page 17

Half controlled power supply for inverter dc-link
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1)
=
Having as a unit of measure the "L" inductance in Henry, the "C" capacitor in Farad and the "R" resistance in
Ohm, according to the above mentioned data:
= 1331.21 rad/S
2)
=
= 357.14
from which:
3)
=
t
M
t
= 0.00117 s
from which:
M
( t
states the time needed by the current to reach its maximum value )
M
4)
the peak current can be calculated with the following formula:
=
I
P
from which :
It is obvious that considering a 70V discharge of the DC-LINK (3-mS mains dip) the current is too high for the
converter. As a consequence, it is necessary to consider a lower voltage reduction (corresponding to a shorter
mains dip). Therefore, with a voltage reduction of 35V (1.5-mS mains dip), the new value will be:
I
= 286.1 A
P
Such value meets the needing of both the converter (which for short periods is able to bear a current value two
times the rated one) and the inductance, whose saturation current is higher than 300A.
Table of S1.1-4 Delay for thyristor switching off during mains dip.
——————— Half controlled power supply for inverter DC-Link ————————
2
1
R
( -
)
L
*
C
2
*
L
R
2
*
L
2
*
V
(
* )
t *
e
M
*
L
I
= 572.3A
P
CS300
17

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