Dimensioning of mains filter / mains choke for a second power supply

Question:

How can the current split and the voltage drop across a(n) mains filter / choke be calculated for a second supply?

Answer:

1. Calculation of the required DC input power and the resulting total current required

• see documentation in chapter 'Network of several drives'
• or U_mains = 400 V according to rough calculation I_mains = 2 x sum P_motor (Example: 2 x 26 kW => 52 A)

2. Selection of mains filter / choke L1 from operating instructions

• see documentation in chapter 'mains filter' (Example: EZN3A0088H024 => 3 x 24 A; 3 x 0.88 mH)

3. Calculation of voltage drop via L1

• XL1 = 2 PI x f x L (Example: 2 x 3.141 x 50 Hz x 0.88 mH = 0.276 Ohm)
• U = I x R => UL = I x XL1 (Example: 24 A x 0.276 Ohm = 6.6 V)

4. Calculation of the required inductance L2, estimated that the voltage via L1 and L2 is the same

• IL2 = Itotal - IL1 (Example: 52 A - 24 A = 28 A)
• XL2 = UL / IL2 (Example: 6.6 V / 28 A = 0.236 Ohm)
• L2 = XL2 / (2 PI x f) (Example: 0.236 Ohm / (2 x 3.141 x 50 Hz) = 0.75 mH)

L2 should be a mains filter or a mains choke with maximum 0.75 mH and a current capacity of >= 28 A  (e. g. ELN3-0075H045 (mains choke: 0.75 mH, 45 A) or EZN3B0055H060 (mains filter B: 0.55 mH, 60 A)).

Example (functional interconnection):

Note: The above calculation does not take into account tolerances of the inductances. This is typically +-10 %.