**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 the Large 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)