Three-phase Transformers Harmonics

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Three-phase Transformers Harmonics

The non-sinusoidal nature of the magnetising current in a three phase transformer produces some undesirable phenomena. The phase magnetising current should contain third and higher order harmonics which is necessary to produce a sinusoidal flux.

If the phase voltage across each phase is sinusoidal, then the phase magnetising currents are given as follows −

$$\mathrm{I_{RN} \: = \: I_{m} \: \sin \: \omega t \: + \: I_{3m} \sin(3\omega t \: + \: \phi_{3}) \: + \: I_{5m} \: \sin(5 \omega t \: + \: \phi_{5}) \: + \: \dotso (1)}$$

$$\mathrm{I_{YN} \: = \: I_{1m} \: \sin(\omega t \: - \: 120^{\circ}) \: + \: I_{3m} \: \sin[3(\omega t \: -\: 120^{\circ}) \: + \: \phi_{3}] \: + \: I_{5m} \: \sin[5(\omega t \: - \: 120^{\circ}) \: + \: \phi_{5}] \: + \: \dotso}$$

$$\mathrm{\Rightarrow \: I_{YN} \: = \: I_{1m} \: \sin(\omega t - 120^{\circ}) \: + \: I_{3m} \: \sin(3\omega t \: + \: \phi_{3}) \: + \: I_{5m} \: \sin(5\omega t \: + \: 120^{\circ} \: + \: \phi_{5}) \: + \: \dotso \:(2)}$$

And

$$\mathrm{I_{BN} \: = \: I_{1m} \: \sin(\omega t - 240^{\circ}) \: + \: I_{3m} \: \sin[3(\omega t - 240^{\circ}) \: + \: \phi_{3}] \: + \: I_{5m} \: \sin[5(\omega t - 240^{\circ}) \: + \: \phi_{5}] \: + \: \dotso}$$

$$\mathrm{\Rightarrow \: I_{BN} \: = \: I_{1m} \: \sin(\omega t - 240^{\circ}) \: + \: I_{3m} \: \sin(3 \omega t \: + \: \phi_{3}) \: + \: I_{5m} \: \sin(5\omega t \: + \: 240^{\circ} \: + \: \phi_{5}) \: + \: \dotso \: (3)}$$

Hence, from the eqns. (1), (2) and (3), it is clear that the third harmonics in the three phase currents are co-phasal i.e. they are in the same phase while the fifth harmonics have different phases.

Now, let us see the phenomena of third harmonics in delta connection and star connection of the transformer.

Case 1 - Delta Connection

If the currents IRN, IYN and IBN represents the phase magnetising currents in a delta connected transformer, then the line currents can be determined by subtracting two phase currents as follows −

$$\mathrm{I_{RYN} \: = \: I_{RN} \: - \: I_{YN}}$$

$$\mathrm{\Rightarrow \: I_{RYN} \: = \: \surd 3\:I_{1m}\:sin(\omega t \: + \: 30^{\circ})\: - \: \surd 3 \: I_{5m} \: \sin(5\omega t - 30^{\circ} \: + \: \phi_{5}) \: + \: \dotso (4)}$$

Therefore, from the equation (4) it can be seen that the third harmonic which is present in the phase magnetising current of a delta connected 3-phase transformer is not present in the line current. Since the third harmonics are co-phasal and hence cancelled out in the line currents.

Although the third harmonic currents flow around the closed loop of the delta connection. Hence, the delta connection allows a sinusoidal flux in the core of the transformer and the voltage with no third harmonic components in the supply line. Due to this reason, most of the 3-phase transformers have a delta-connected winding and in the cases where it is not possible to have either the primary or secondary connected in delta, a tertiary winding, which is connected in delta is provided. This tertiary winding provides the path for circulating the third harmonic currents which is required to produce the sinusoidal flux in the core of the transformer.

Case 2 - Star Connection

If the currents I_RN, I_YN and I_BN represents the phase magnetising currents in a star connected transformer, then, the neutral current is given by

$$\mathrm{I_{N} \: = \: I_{RN} \: + \: I_{YN} \: + \: I_{BN} \: + \: \dotso \: (5)}$$

Now, from the eqns. (1), (2) and (3), we have,

$$\mathrm{I_{RN} \: + \: I_{YN} \: + \: I_{BN} \: = \: 3I_{3m} \: \sin(3\omega t \: + \: \phi_{3}) \: \dotso \: (6)}$$

Thus, the eqn. (6) shows that under balanced conditions, the current in the neutral wire is a third harmonic current, which is having magnitude equal to the three times of the magnitude of each third harmonic phase current. These third harmonic currents produce interference with the neighbouring communication circuits.

Also, if the neutral wire is not present i.e. the star connection is of three wire, then the neutral current must be zero i.e.

$$\mathrm{3I_{3m} \: \sin(3\omega t \: + \: \phi_{3}) \: = \: 0}$$

$$\mathrm{\Rightarrow \: I_{3m} \: = \: 0}$$

Hence, it can be seen that the 3-wire star connected system supresses the flow of third harmonic currents. Whereas, a 4-wire star connected system, allows the third harmonic currents to flow through the neutral wire.