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Case 2 : DC Bus connected n parallel to a single AC-fed inverter
AC power
supply
+
-
(*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
i
=
M1(max)
M2(max)
In this case, the total motor current under the possible worst case is higher than that of the inverters.
Total inverter rated current
Total motor current under possible worst case =
à Main inverter must therefore be SJ300-185HFx (38A) or larger. SJ300-220HFx is suggested
to provide additional safety margin.
INV #1
i
M1
M 1
INV #2
i
M2
M 2
INV #n
i
Mn
M n
i
i
=
=
= 9.0Arms
M3(max)
M4(max)
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>
Input current
AC input
i
i
i
i
=
+
+
+
= 8.6 * 4
1
2
3
4
i
i
i
+
+
M1( max)
M2(max)
M3(max)
n
i
i
s
1
k
k
2
i
2
i
n
= 32.2 Arms
i
+
= 36Arms
M4(max)
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