Suppression of harmonic currents by phase shift transformer

In some steel rolling lines that use two DC motors with a rigid shaft connection, a power transformer is often used to supply the motor control power circuit. In which, the power transformer has two secondary windings, one star-connected and one delta-connected. The use of such a power transformer allows to suppress some of the higher-order harmonics on the primary side.

The contents presented below are based on [1].

Phase shift of harmonic currents

Figure 1 is a diagram of the $\Delta/Y$ transformer supplying a load. Assume a voltage ratio $v_{AB}/v_{ab} = 1$ with a turn ratio of $N_{1}/N_{2} = \sqrt{2}$. The transformer has a phase angle of $\delta = \angle \overline{v}_{ab} – \angle \overline{v}_{AB} = -30^{\circ}$.

Figure 1. Investigation of harmonic currents in primary and secondary windings

For a balanced 3-phase system, the load line current is expressed as:

$$\begin{array} {rl} i_{a} &= \sum_{n=1,5,7,11,…}^{\infty} I_{n} sin (n\omega t) \nonumber \ \i_{b} &= \sum_{n=1,5,7,11,…}^{\infty} I_{n} sin (n(\omega t – 120^{\circ})) \label{ EQ:DIGPIDEQ} \\ i_{a} &= \sum_{n=1,5,7,11,…}^{\infty} I_{n} sin (n(\omega t – 240^{\circ} )) \end{array} $$

where $I_{n}$ is the peak value of the order current $n$. When $i_{a}$ and $i_{b}$ are converted to the primary side, the conversion currents $i_{ap}^{'}$ and $i_{bp}^{'}$ are described. described by

$$\begin{array} {rl} i_{ap}^{'} &= i_{a} \frac{N_{2}}{N_{1}} = \frac{1}{\sqrt{3} } (I_{1} sin (\omega t) + I_{5} sin (5\omega t) + I_{7} sin (7\omega t) + I_{11} sin (11\omega t) + \ cdots ) \nonumber \\ i_{bp}^{'} &= i_{b} \frac{N_{2}}{N_{1}} = \frac{1}{\sqrt{3}} (I_{ 1} sin (\omega t – 120^{\circ}) + I_{5} sin (5\omega t – 240^{\circ}) + I_{7} sin (7\omega t – 120^{\ circ}) \nonumber \\ & + I_{11} sin (11\omega t – 240^{\circ}) + \cdots \nonumber \end{array}$$

Then the primary line current is calculated as follows:

$$\begin{array} {rl} i_{a}^{'} &= i_{ap}^{'} – i_{bp}^{'} = I_{1} sin (\omega t + 30^ {\circ}) + I_{5} sin (5\omega t – 30^{\circ}) + I_{7} sin (7\omega t + 30^{\circ}) \nonumber \\ & + I_ {11} sin (11\omega t – 30^{\circ}) + \cdots \nonumber \\ &= \sum_{n=1,7,13,…}^{\infty} I_{n} sin ( n\omega t – \delta) + \sum_{n=5,11,17,…}^{\infty} I_{n} sin (n\omega t + \delta) \nonumber \end{array}$$

where $n = 1, 7, 13, 19, \cdots$ are the forward components, and $n = 5, 11, 17, 23, \cdots$ are the reverse order components. Thus, between the secondary line current (load current) and the primary conversion current, the phase difference angle is

$$\begin{array} {rl} \angle i_{an}^{'} &= \angle i_{an} – \delta \end{array} \qquad (1) $$

with $n = 1, 7, 13, 19, \cdots$ (forward waves)

$$\begin{array} {rl} \angle i_{an}^{'} &= \angle i_{an} + \delta \end{array} \qquad (2) $$

with $n = 5, 11, 17, 23, \cdots$ (reverse waves)

Eliminate harmonics

To illustrate how harmonic currents are suppressed by a phase shift transformer, consider an example in Figure 2. The angle of phase difference between the star and delta secondary windings is $0^{\circ}$ and $30^ {\circ}$. Let's say the voltage ratio is $v_{AB}/v_{ab} = v_{AB}/v_{\hat{a}\hat{b}} = 2$.

Figure 2. Phase shift transformer

The secondary line currents are shown as follows

$$\begin{array} {rl} i_{a} &= \sum_{n=1,5,7,11,…}^{\infty} I_{n} sin (n\omega t) \\ i_ {\hat{a}} &= \sum_{n=1,5,7,11,…}^{\infty} I_{n} sin (n(\omega t + \delta)) \end{array} $$

Because the first secondary coil is $Y/Y$, when converted to the primary, their phase angle is unchanged, so

$$\begin{array} {rl} i_{a} &= \frac{1}{2} = (I_{1} sin (\omega t) + I_{5} sin (5\omega t) + I_ {7} sin (7\omega t) + I_{11} sin (11\omega t) + \cdots ) \nonumber \end{array}$$

To convert $i_{\hat{a}}$ to the primary side, we note the phase difference angle of the forward and reverse order components according to formulas (1) and (2). Then

$$\begin{array} {rl} i_{a}^{'} &= \frac{1}{2}\left(\sum_{n=1,7,13,…}^{\infty} I_ {n} sin (n(\omega t – \delta) – \delta) + \sum_{n=5,11,17,…}^{\infty} I_{n} sin (n(\omega t + \) delta) + \delta) \right) \nonumber \\ &= \frac{1}{2}\left( I_{1} sin (\omega t) – I_{5} sin (5\omega t) – I_ {7} sin (7\omega t) + I_{11} sin (11\omega t) + \cdots \right) \nonumber \end{array}$$

with $\delta = 30^{\circ}$. The primary line current will be calculated as follows:

$$\begin{array} {rl} i_{A} &= i_{a}^{'} + i_{\hat{a}}^{'} = I_{1} sin (\omega t) + I_ {11} sin (11\omega t) + I_{13} sin (13\omega t) + I_{23} sin (23\omega t) + \cdots \nonumber \end{array}$$

[1] B. WuHigh-Power Converters and AC Drives. Wiley-IEEE Press, 2006.

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