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Case 2 : DC Bus connected n parallel to a single AC-fed inverter
Ø Pay attention to the selection of the main inverter
(#1) because all the input current flows through
the rectifier bridge of this inverter. (*2)
Ø Need sufficient time for EEPROM to store the data.
(Refer to Case 1)
Ø Use DC choke. (Refer to Case 1)
<Selection of the main inverter kW>
(*2) Capacity of the main inverter
Rated current of the main inverter should be higher than;
Ø Total rated current of the inverters
and
Ø Possible highest total motor current
[Example of 4 inverters in parallel]
Ø INV#1~#4=SJ300-040HFx (8.6A rated)
Ø i
M1(max)
= i
M2(max)
= i
M3(max)
= i
M4(max)
= 9.0Arms
In this case, the total motor current under the possible worst case is higher than that of the inverters.
Total inverter rated current = i
1
+ i
2
+ i
3
+ i
4
= 8.6 * 4 = 32.2 Arms
Total motor current under possible worst case = i
M1(max)
+ i
M2(max)
+ i
M3(max)
+ i
M4(max)
= 36Arms
à Main inverter must therefore be SJ300-185HFx (38A) or larger. SJ300-220HFx is suggested
to provide additional safety margin.
AC input
Input current
∑
=
=
n
k
ks
ii
2
1
i
2
i
n
INV #2
M 2
INV #n
M n
AC power
supply
INV #1
M 1
+
-
i
M1
i
M2
i
Mn