GE CS300 - Manual (Page 17)

CS300
——————— Half controlled power supply for inverter DC-Link ————————
17
1)
)
L * 2
R
( -
C * L
1
=
2
ω
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:
ω
= 1331.21 rad/S
2)
L * 2
R
= α
from which:
α
= 357.14
3)
ω
π
* 2
= t
M
from which: t
M
= 0.00117 s
( t
M
states the time needed by the current to reach its maximum value )
4) the peak current can be calculated with the following formula:
e * )
L *
V
( = I t *
P
M
α
ω
from which : I
P
= 572.3A
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
P
= 286.1 A
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.