Hitachi J300 Series Instruction Manual

Powering Inverters from DC
It is possible to power inverters from a DC Power source, or to connect the DC Bus of multiple inverters together to
achieve energy savings, since inverters in power driving mode can use power from those that are in regeneration mode.
[1] Connection method
There are several ways for DC bus connection of the inverters. (Examples of 3-phase 200V or 400V class inverter.)
ØAdvantage and disadvantages of each connection method.
Item Contents Advantage Disadvantage
Œ
Connecting to + & - terminal ŸNo concern for the
rectifier bridge diodes. ŸThere will be no inrush current limiting.
•
Connecting to AC inputs and - terminal ŸIntegrated inrush current
limiting circuit is used. ŸRectifier bridge diodes of the main inverter may
need to be up-sized.
INV #1
M 1
INV #2
M 2
INV #n
M n
DC power
supply
Case 1 : Connected in parallel to a common DC bus
INV #2
M 2
INV #n
M n
AC power
supply
Case 2 : Connected in parallel to an A-fed inverter
INV #1
M 1
+
-
INV #2
M 2
INV #n
M n
AC power
supply
Case 3 : AC & DC Connected together
INV #1
M 1
+
-
DC supply connection methods
INV
M
+
-
ŒConnecting to + and - terminal
INV
M
+
-
•Connecting to AC inputs and - terminal

[2] DC voltage to be supplied
Supplied DC voltage should be between the UV voltage and the OV voltage (or BRD ON level) of the inverter.
Model name UV level BRD ON level OV level (regen.) OV level (source)
J100 200V class 140Vac ~ 160Vac (V-SET) +137.5V 390Vdc ±15Vdc
400V class 280Vac ~ 320Vac (V-SET) +275V 780Vdc ±30Vdc
J300 200V class 140Vac ~ 160Vac (AVR set) +138V 369Vdc ~ 404Vdc
400V class 280Vac ~ 320Vac (AVR set) +276V 756Vdc ~ 827Vdc
L100 200V class 190Vdc ±10Vdc 395Vdc ±20Vdc Aprx. 365Vdc for 100s
400V class 395Vdc ±20Vdc -790Vdc ±40Vdc Aprx. 730Vdc for 100s
SJ100 200V class 370Vdc +4%,-3%
400V class Same as L100 740Vdc +4%,-3% Same as L100 Same as L100
L300P 200V class 200Vdc ±10Vdc 395Vdc ±10Vdc Aprx. 380Vdc for 60s
400V class 400Vdc ±20Vdc -790Vdc ±20Vdc Aprx. 760Vdc for 60s
SJ300 200V class
400V class Same as L300P Adjustable by
[b096] Same as L300P Same as L300P
ØIf it is higher, inverter may trip with OV or BRD error.
ØIf it is lower, inverter may trip with UV.
[3] Cautions
Case 1 : Connected parallel to a common DC net
ØTake preventive measures to limit inrush current at power ON,
since the integrated inrush current limiting circuit is not used.
àOtherwise there will be a unexpected UV at net side or
damage to the inverter caused by ∆V=di/dt.
ØUse DC chokes for each inverter to avoid interaction due
to surge and/or harmonics.
àOtherwise there may be an unexpected failure of the
inverter or other attached equipment.
ØTake preventive measures to ensure sufficient time between UV
level and dead voltage of the DC/DC converter (*1) at power
OFF. This is to allow ample time for EEPROM to store
the existing data at power OFF. (∆t ; see below)
àOtherwise there is a possibility of an EEPROM error at
the next power ON.
DC power
supply
Big inrush current at
power ON
INV #1
M 1
INV #2
M 2
INV #n
M n
UV level
Threshold voltage of DC-DC converter
(This value depends on the inverter model.
But in general it is around 100Vdc)
Main power (AC)
For Cases 2 & 3 OFF
DC-DC conv.
(internal 5V)
EEPROM store
period
DC bus voltage
For Cases 1~3
∆t
Around 0.5~1s or more (*1) The required time depends on the parameters which have changed before power OFF.

Case 2 : DC Bus connected n parallel to a single AC-fed inverter
ØPay attention to the selection of the main inverter
(#1) because all the input current flows through
the rectifier bridge of this inverter. (*2)
ØNeed sufficient time for EEPROM to store the data.
(Refer to Case 1)
ØUse 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;
ØTotal rated current of the inverters
and
ØPossible highest total motor current
[Example of 4 inverters in parallel]
ØINV#1~#4=SJ300-040HFx (8.6A rated)
ØiM1(max) = iM2(max) = iM3(max) = iM4(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 = i1+ i2+ i3+ i4= 8.6 * 4 = 32.2 Arms
Total motor current under possible worst case = iM1(max) + iM2(max) + iM3(max) + iM4(max) = 36Arms
àMain inverter must therefore be SJ300-185HFx (38A) or larger. SJ300-220HFx is suggested
to provide additional safety margin.
AC input
Input current ∑ ==
n
kksii 2
1
i2
in
INV #2
M 2
INV #n
M n
AC power
supply
INV #1
M 1
+
-
iM1
iM2
iMn