GE CS300 manual = 1331.21 rad/S

CS300

1)

ω		1		R	2
	=		- (		)
		L * C		2 * L

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)
	α	=		R
			2 * L
from which:	α	= 357.14
3)
	t M =			π
				2 * ω
from which:	tM = 0.00117 s

( tM states the time needed by the current to reach its maximum value )

4)the peak current can be calculated with the following formula:

		V	α * t M
I P = (	ω	* L	)* e
from which :	IP = 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:

IP= 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.

——————— Half controlled power supply for inverter DC-Link ———————— 17