Case 2 : DC Bus connected n parallel to a single
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+ | INV #1 |
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| INV #n |
ØPay attention to the selection of the main inverter (#1) because all the input current flows through
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| iM1 | the rectifier bridge of this inverter. (*2) | |||
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| M 1 | Ø Need sufficient time for EEPROM to store the data. |
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| iM2 | Ø Use DC choke. (Refer to Case 1) | |||
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| <Selection of the main inverter kW> | ||||
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| M 2 | |
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| Input current is1 = åik |
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| k =2 |
AC input
iMn
M n
i2
(*2) Capacity of the main inverter
Rated current of the main inverter should be higher than;
ØTotal rated current of the inverters
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Ø | Possible highest total motor current |
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[Example of 4 inverters in parallel]
Ø
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
