Hitachi J300 Series Bedienungsanleitung

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

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

ŒConnecting to + and - terminal

INV

•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)

(*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 ∑

kksii 2

INV #2

M 2

INV #n

M n

AC power

supply

INV #1

M 1

iM1

iM2

iMn

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