Hitachi AN091802-1 User Manual

Application Note: 
Powering Inverters from a 
DC Supply

Hitachi America, Ltd.

AN091802-1

Please refer also to the

Inverter Instruction Manual

© 2002 Hitachi America, Ltd.

Powering Inverters from DC

INV #2

M 2

INV #n

M n

INV #1

M 1

INV

M

INV

M

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.)

Case 1 : Connected in parallel to a common DC bus

DC power
supply

INV #1

M 1

INV #2

M 2

INV #n

M n

Case 3 : AC & DC Connected together

AC power
supply

Case 2 : Connected in parallel to an A-fed inverter

AC power
supply

INV #1

+

M 1

­INV #2

M 2

INV #n

M n

DC supply connection methods

+

-

+

-

ΠConnecting to + and - terminal

+

-

• Connecting to AC inputs and - terminal

Ø Advantage and disadvantages of each connection method.

Item Contents Advantage Disadvantage

Connecting to + & - terminal Ÿ No concern for the

Œ

Connecting to AC inputs and - terminal Ÿ Integrated inrush current

•

rectifier bridge diodes.

limiting circuit is used.

Ÿ There will be no inrush

current limiting.

Ÿ Rectifier bridge diodes

of the main inverter may
need to be up-sized.

[2] DC voltage to be supplied

Big inrush current at
INV #1

M 1

INV #2

M 2

INV #n

M n

UV level

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

400V class 280Vac ~ 320Vac (V-SET) +275V

J300 200V class 140Vac ~ 160Vac (AVR set) +138V 369Vdc ~ 404Vdc

400V class 280Vac ~ 320Vac (AVR set) +276V 756Vdc ~ 827Vdc

L100 200V class

400V class

SJ100 200V class 370Vdc +4%,-3%

400V class

L300P 200V class

400V class

SJ300 200V class

400V class

190Vdc ± 10Vdc 395Vdc ± 20Vdc
395Vdc ± 20Vdc

Same as L100

200Vdc ± 10Vdc 395Vdc ± 10Vdc
400Vdc ± 20Vdc

Same as L300P

740Vdc +4%,-3%

-

-

Adjustable by

[b096]

390Vdc ± 15Vdc
780Vdc ± 30Vdc

Aprx. 365Vdc for 100s

790Vdc ± 40Vdc

Same as L100 Same as L100

790Vdc ± 20Vdc

Same as L300P Same as L300P

Aprx. 730Vdc for 100s

Aprx. 380Vdc for 60s
Aprx. 760Vdc for 60s

Ø If it is higher, inverter may trip with OV or BRD error.
Ø If it is lower, inverter may trip with UV.

[3] Cautions

Case 1 : Connected parallel to a common DC net

power ON

DC power
supply

Main power (AC)
For Cases 2 & 3

OFF

Ø Take preventive measures to limit inrush current at power ON,

since the integrated inrush current limiting circuit is not used.
à Otherwise there will be a unexpected UV at net side or
damage to the inverter caused by ∆V=di/dt.

Ø Use DC chokes for each inverter to avoid interaction due

to surge and/or harmonics.
à Otherwise there may be an unexpected failure of the
inverter or other attached equipment.

Ø Take 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)
à Otherwise there is a possibility of an EEPROM error at
the next power ON.

DC bus voltage
For Cases 1~3

DC-DC conv.
(internal 5V)

EEPROM store
period

Threshold voltage of DC-DC converter
(This value depends on the inverter model.

∆t

Around 0.5~1s or more

But in general it is around 100Vdc)

(*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

INV #2

M 2

INV #n

M n

INV #1

M 1

AC power
supply

+

i

M1

-

i

M2

i

Mn

(*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

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

=

ii

∑

ks

1

=nk

2

i

2

i

n

[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 = i1 + i2 + i3 + i
Total motor current under possible worst case = i

M1( max)

+ i

= 8.6 * 4 = 32.2 Arms

4

M2(max)

+ i

M3(max)

+ i

M4(max)

à Main inverter must therefore be SJ300-185HFx (38A) or larger. SJ300-220HFx is suggested

to provide additional safety margin.

= 36Arms