Creating a 4 lines 3 phases net in a 3 lines 3 phases net using a Zn-transformer? About operation mode of a Zn - transformer and its design. Typical application schemas  of a Zn-transformer is shown in the Fig.1. The Zn-transformer has 6 equal windings which are connected in zig - zag. If the 3 phase load is symetrical  ( linear load complex impedances ZA = ZB = ZC ) then the there is no current in the neutlal line ( In = 0 ) and the Zn-transformer runs in the no-load operation mode. If the 3 phase load is symetrical  ( l ZA = ZB = ZC) but non-linear (typical for some rectifier) then the currents IA, IB and Ic can have the following form: iA = I1m sin(ωt)           +I3m sin(3ωt)+I5m sin(5ωt)          +I7m sin(7ωt)             +I9m sin(9ωt)+... iB = I1m sin(ωt+120°) +I3m sin(3ωt) +I5m sin(5ωt+120°)+I7m sin(7ωt+120°)   +I9m sin(9ωt)+... iC = I1m sin(ωt+2400°) +I3m sin(3ωt)+I5m sin(5ωt+240°)+I7m sin(7ωt+2400°)+I9m sin(9ωt)+... and the current in the neutral line is: in = iA + iB + iC = 3 *I3m sin(3ωt) + 3*I9m sin(9ωt) + ... where I3m and I9m are max. values of the 3. and 9. current harmonics. For this operation mode the curents through the Zn transformer are: Ia = ib = ic = in/3 = I3m sin(3ωt) + I9m sin(0ωt) + ... Note that the Zn-transformer works in this operation mode as a very powerful filter of th 4., 9., ... current harmonicd.  Assune the rms values of the 3. and 9. harmoc in all 3 phases of a 3x400V, 50Hz,  three phase net are: I3rms=33A I9rms = 10A then: UA= UB = UC = 230 and the zig & zag voltage is 230/2/cos(30°) = 133V Run theLlarge Transformer èrogram and set the input as follows in the Fig.2: Rermsrks: Due to the fact that the transformer is driven by 50Hz you have to set a small  50Hz current. In order to set the 3. and 9. current harmonics you have to set Circuit = 41 Do not forget to encrease the eddy current factpr RacRdc Note that the turns of the zig and zag windings have to be equal. Normally the worst case for the Zn-transformer is unsymetrical operation mode when only one phase is on load: iA = I1m sin(ωt)  + I3m sin(3ωt)+I5m sin(5ωt)  + I7m sin(7ωt)             +I9m sin(9ωt)+... iB = 0 iC = 0 and the current in the neutral line is: in = iA  = I1m sin(ωt)  + I3m sin(3ωt) + I9m sin(9ωt) + ... For this operation mode the curents through the Zn transformer are: Ia = ib = ic = in/3 = (I1m/3)  sin(ωt)  + (I3m/3)  sin(3ωt)+(I5m/3)  sin(5ωt)  + (I7m/3)  sin(7ωt) ...  Assune the rms values of the current harmonics in the phase A in a 3x400V, 50Hz net are: Î1rms=90A I3rms=33A I5rms = 21A I7rms=15A then: UA= UB = UC = 230 and the zig & zag voltage is 230/2/cos(30°) = 133V Run theLlarge Transformer èrogram and set the input as follows in the Fig.4:   Rermsrks: In order to set the 3.  current harmonics you have to set Circuit = 41 Do not forget to encrease the eddy current factpr RacRdc Note that the turns of the zig and zag windings have to be equal. Finally note that you can use the Small Transformers Program if there are no current harmonics: iA = I1m sin(ωt)

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