Effect of Winding Configuration on the kVA Rating of Wye-connected Autotransformer Applied to 12- pulse Rectifier

Abstract

This paper presents the effect of winding configuration on the kVA rating of a wye-connected autotransformer applied to a 12-pulse rectifier. To describe the winding configuration of the wye-connected autotransformer, position and proportional parameters are defined and their quantitative relation is calculated. The voltages across and currents through the windings are measured under different winding connections. Consequently, a relation between the kVA rating and position parameter is established in accordance with the analysis, and the optimal winding configuration is obtained on the basis of this relation. A wye-connected autotransformer with the least equivalent kVA rating and simplest winding configuration is designed and applied to the 12-pulse rectifier. Simulations and experiments are conducted to validate the theoretical analysis.

I. INTRODUCTION

Multi-pulse rectifiers (MPRs) with low electromagnetic interference are widely used in high-power rectifications due to their superior performances in harmonic elimination at AC input and ripple reduction at DC output [1]-[3]. Generally, the MPR is composed of a phase-shifting transformer and two or more three-phase rectifiers [4]. The phase-shifting transformer is used to produce two or more sets of three-phase voltages and to feed the three-phase rectifiers. Moreover, the primary magnetic device in the MPR is the phase-shifting transformer, which determines the MPR volume. To improve the power density of the MPR, the volume of the phase-shifting transformer should be reduced [5]-[7]. In general, the volume of the phase-shifting transformer is determined by its kVA rating. Therefore, reducing the kVA rating of the phase-shifting transformer has become a popular topic in the design of MPRs.

The phase-shifting transformer is classified into two types, namely, isolated transformer and autotransformer [8], [9]. If the required output voltage is considerably higher or lower than the input voltage, then the isolated transformer is preferred in the viewpoint of protection. For example, in electroplating, the load voltage is only approximately 10 or more volts, which is considerably lower than the input voltages. In this application, the isolated transformer is preferred. The isolated transformer uses the magnetic coupling to transmit the load power; hence, the apparent power of the isolated transformer is equal to or more than the load power [10]. Therefore, if the isolated transformer is used as the phase-shifting transformer, then the MPR is bulky. If the required output voltage of the phase-shifting transformer is approximately equal to its input voltage, then the autotransformer is the best option as the phase-shifting transformer. As the windings of the autotransformer are interconnected, only a small fraction of the load power is transmitted by magnetic coupling [11]-[13]. Therefore, the apparent power of the autotransformer is less than the load power, thereby reducing the bulk of the phase-shifting transformer.

In MPRs, the autotransformer is classified into six, nine, 12, or more phases in accordance with the phase number. Among them, the six-phase autotransformer is frequently used because of its least kVA rating under the same load power. This autotransformer includes delta-, wye-, and polygon-connected autotransformers. Although the autotransformers are often used for high-power rectification, not all arbitrarily configurated autotransformers can reduce the MPR bulk. In [11], Meng et al. analyzed the winding configuration on the kVA rating of a delta-connected autotransformer; they reported that the bulk can only be reduced when the autotransformer is optimally designed. In this study, we analyze the winding configuration on the kVA rating of the wye-connected autotransformer and provide an interesting and useful confirmation that the well-used circuit is optimal in terms of kVA rating. Finally, we provide an additional insight into these devices and how they may be analyzed.

II. REQUIREMENT OF 12-PULSE RECTIFIER ON THE WYE-CONNECTED AUTOTRANSFORMER

Fig. 1 shows a 12-pulse rectifier with a wye-connected autotransformer. In Fig. 1, an inter-phase reactor (IPR) is used to absorb the instantaneous difference of the output voltages and ensure the independent operation of the two bridge rectifiers [14]. A zero-sequence blocking transformer is utilized to promote a 120° conduction for each rectifier diode due to high impedance to zero sequence currents [14]. The main function of the wye-connected autotransformer in Fig. 1 is to produce two sets of three-phase voltages with appropriate phase-shift angles. As discussed in [15], to eliminate the (12k±1)th (k is an odd number) harmonics, the phase-shift angle should meet

\(2 \alpha=\frac{60^{\circ}}{M},\)       (1)

where M denotes the number of three-phase diode bridge rectifiers.

Fig. 1. 12-pulse rectifier with wye-connected autotransformer.

Therefore, the phase-shift angle between the sets of three-phase voltages should be 30°. Assume the input three-phase voltage of the wye-connected autotransformer as

\(\left\{\begin{array}{l} u_{\mathrm{a}}=\sqrt{2} U_{\mathrm{m}} \sin \omega t \\ u_{\mathrm{b}}=\sqrt{2} U_{\mathrm{m}} \sin \left(\omega t-120^{\circ}\right) \\ u_{\mathrm{c}}=\sqrt{2} U_{\mathrm{m}} \sin \left(\omega t+120^{\circ}\right), \end{array}\right.\)       (2)

where U_m is the amplitude of the three-phase input voltage.

From Eqs. (1) and (2), the two sets of three-phase output voltages of the wye-connected autotransformer can be expressed as

\(\left\{\begin{array}{l} u_{\mathrm{a1}}=\sqrt{2} U_{\mathrm{n}} \sin \left(\omega t+15^{\circ}\right) \\ u_{\mathrm{b1}}=\sqrt{2} U_{\mathrm{n}} \sin \left(\omega t-120^{\circ}+15^{\circ}\right) \\ u_{\mathrm{c1}}=\sqrt{2} U_{\mathrm{n}} \sin \left(\omega t+120^{\circ}+15^{\circ}\right), \end{array}\right.\)       (3)

\(\left\{\begin{array}{l} u_{a 2}=\sqrt{2} U_{\mathrm{n}} \sin \left(\omega t-15^{\circ}\right) \\ u_{\mathrm{b} 2}=\sqrt{2} U_{\mathrm{n}} \sin \left(\omega t-120^{\circ}-15^{\circ}\right) \\ u_{c 2}=\sqrt{2} U_{\mathrm{n}} \sin \left(\omega t+120^{\circ}-15^{\circ}\right), \end{array}\right.\)       (4)

where U_n represents the amplitude of the three-phase output voltage.

III. WINDING CONFIGURATIONS OF THE WYE-CONNECTED AUTOTRANSFORMER

The windings of the wye-connected autotransformer can be divided into three types, namely, wye-connected, auxiliary, and extended windings, as shown in Fig. 1. On the basis of positional relations, four winding configurations of the wye-connected autotransformer are available, as shown in Fig. 2. Fig. 3 presents the corresponding phasor diagrams of these winding configurations.

Fig. 2. Winding configurations of the wye-connected autotransformer under different k1 and k₂ values. (a) k₁ ≥ 0 and k₂ ≥ 0. (b) k₁ ≤ 0 and k₂ ≥ 0. (c) k₁ ≥ 0 and k₂ ≤ 0. (d) k₁ ≤ 0 and k₂ ≤ 0.

Fig. 3. Phasor diagram of the wye-connected autotransformer under different k₁ and k₂ values. (a) k₁ ≥ 0 and k₂ ≥ 0. (b) k₁ ≤ 0 and k₂ ≥ 0. (c) k₁ ≥ 0 and k₂ ≤ 0. (d) k₁ ≤ 0 and k₂ ≤ 0.

In Fig. 3, with phase a as an example, \(\dot{U}_{\mathrm{a}}\) is the phasor of the input phase voltage u_a; \(\dot{U}_{\mathrm{a1}}\) and \(\dot{U}_{\mathrm{a2}}\) denote the phasors of output voltages u_a1 and u_a2, respectively. Given the wye connection of the winding configuration, the voltage across the wye-connected winding is equal to u_a, the voltage across the auxiliary winding is equal to u_a3a, and the voltage across the extended winding is equal to u_a1a3.

In Figs. 3(a) and 3(b), \(\dot{U}_{\mathrm{c}}\), \(\dot{U}_{\mathrm{a1a3}}\), \(\dot{U}_{\mathrm{a1a}}\), and \(\dot{U}_{\mathrm{a3a}}\) are the phasors of output voltages u_c, u_a1a3, u_a1a, and u_a3a, respectively. Figs. 3(a) and 3(b) show that \(\dot{U}_{\mathrm{a1a3}}\) is parallel to \(\dot{U}_{\mathrm{c}}\), and \(\dot{U}_{\mathrm{a3a}}\) is parallel to \(\dot{U}_{\mathrm{a}}\), respectively. The relation among \(\dot{U}_{\mathrm{a}}\), \(\dot{U}_{\mathrm{a1}}\), \(\dot{U}_{\mathrm{a3a}}\), \(\dot{U}_{\mathrm{a1a3}}\), and \(\dot{U}_{\mathrm{a1a}}\) meets

\(\left\{\begin{array}{l} \dot{U}_{\mathrm{a1}}=\dot{U}_{\mathrm{a}}+\dot{U}_{\mathrm{a1a}} \\ \dot{U}_{\mathrm{a1a}}=\dot{U}_{\mathrm{a} 3 \mathrm{a}}+\dot{U}_{\mathrm{a1a3}} \\ \dot{U}_{\mathrm{a1a}}=k_{1} \dot{U}_{\mathrm{a}}+k_{2} \dot{U}_{\mathrm{c}}. \end{array}\right.\)       (5)

Figs. 3(c) and 3(d) illustrate that \(\dot{U}_{\mathrm{a1a3}}\) is parallel to \(\dot{U}_{\mathrm{b}}\), and \(\dot{U}_{\mathrm{a3a}}\) is parallel to \(\dot{U}_{\mathrm{a}}\), respectively. The relation among \(\dot{U}_{\mathrm{a}}\), \(\dot{U}_{\mathrm{a1}}\), \(\dot{U}_{\mathrm{a3a}}\), \(\dot{U}_{\mathrm{a1a3}}\), and \(\dot{U}_{\mathrm{a1a}}\) meets

\(\left\{\begin{array}{l} \dot{U}_{\mathrm{a1}}=\dot{U}_{\mathrm{a}}+\dot{U}_{\mathrm{a1a}} \\ \dot{U}_{\mathrm{a1a}}=\dot{U}_{\mathrm{a} 3 \mathrm{a}}+\dot{U}_{\mathrm{a1a3}} \\ \dot{U}_{\mathrm{a1a}}=k_{1} \dot{U}_{\mathrm{a}}+k_{2} \dot{U}_{\mathrm{b}}, \end{array}\right.\)       (6)

where k₁ represents the position parameter of the auxiliary winding, which reflects the position relative to the wye-connected windings and turn numbers of the auxiliary winding, and k₂ refers to the proportional parameter of the extended winding, which reflects the polarity relative to the wye-connected windings and turn numbers of the extended winding.

In Eq. (6), k₁ and k₂ are scalars, which can be greater or smaller than zero. As shown in Fig. 2 and Fig. 3, to set phase a as an example, when k1 is greater than zero, \(\dot{U}_{\mathrm{a3a}}\) is in the direction of \(\dot{U}_{\mathrm{a}}\); the auxiliary winding a3a and wye-connected winding ao are different and connected to point a. When k₁ is less than zero, \(\dot{U}_{\mathrm{a3a}}\) is opposite to \(\dot{U}_{\mathrm{a}}\); the auxiliary winding a3a is part of wye-connected winding ao. When k₂ is greater than zero, \(\dot{U}_{\mathrm{a1a3}}\) and \(\dot{U}_{\mathrm{a2a3}}\) are parallel to \(\dot{U}_{\mathrm{c}}\) and \(\dot{U}_{\mathrm{b}}\), respectively; the extended winding and corresponding wye-connected winding have the same polarity. When k2 is lower than zero, \(\dot{U}_{\mathrm{a1a3}}\) and \(\dot{U}_{\mathrm{a2a3}}\) are opposite to \(\dot{U}_{\mathrm{b}}\) and \(\dot{U}_{\mathrm{c}}\), respectively; the extended winding and corresponding wye-connected winding have opposite polarity.

As shown in Figs. 2(a) and 3(a), when k₁ ≥ 0 and k₂ ≥ 0, the relation among N₁, N_Y, N_q, k₁, and k₂ meets

\(\frac{N_{1}}{N_{\mathrm{Y}}}=k_{1} \quad \frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}}=k_{2},\)       (7)

where N_Y is the turn number of the wye-connected winding, N₁ represents the turn number of the auxiliary winding, and N_q denotes the turn number of the extended winding.

As shown in Figs. 2(b) and 3(b), when k₁ ≤ 0 and k₂ ≥ 0, the relation among N₁, N_Y, N_q, k₁, and k₂ meets

\(\frac{N_{1}}{N_{\mathrm{Y}}}=-k_{1} \quad \frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}}=k_{2}.\)       (8)

As shown in Figs. 2(c) and 3(c), when k₁ ≥ 0 and k₂ ≤ 0, the relation among N₁, N_Y, N_q, k₁, and k₂ meets

\(\frac{N_{1}}{N_{\mathrm{Y}}}=k_{1} \quad \frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}}=-k_{2}.\)       (9)

As shown in Figs. 2(d) and 3(d), when k₁ ≤ 0 and k₂ ≤ 0, the relation among N₁, N_Y, N_q, k₁, and k₂ meets

\(\frac{N_{1}}{N_{\mathrm{Y}}}=-k_{1} \quad \frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}}=-k_{2}.\)       (10)

On the basis of Eqs. (5) and (6), the wye-connected, auxiliary, and extended windings are duly connected to produce output voltages with an appropriate phase-shift angle. Therefore, to achieve the phase-shift angle, k₁ and k₂ should meet some expressions

A. Constraint Condition Between Parameters

Assume that U_m is equal to 1. The turn ratio of the wye-connected autotransformer is defined as

\(k=\frac{U_{\mathrm{a1}}}{U_{\mathrm{a}}}=\frac{U_{\mathrm{n}}}{U_{\mathrm{m}}}=U_{\mathrm{n}}.\)       (11)

Fig. 3(a) shows the phasor diagram when k₁ ≥ 0 and k₂ ≥ 0. Triangle oa1a3 proves that

\(\frac{k_{2}}{\sin 15^{\circ}}=\frac{1+k_{1}}{\sin 105^{\circ}}=\frac{U_{\mathrm{a1}}}{\sin 60^{\circ}}=\frac{k}{\sin 60^{\circ}}.\)       (12)

Fig. 3(b) presents the phasor diagram when k₁ ≤ 0 and k₂ ≥ 0. Triangle oa1a3 shows that

\(\frac{k_{2}}{\sin 15^{\circ}}=\frac{1+k_{1}}{\sin 105^{\circ}}=\frac{U_{\mathrm{a1}}}{\sin 60^{\circ}}=\frac{k}{\sin 60^{\circ}}.\)       (13)

On the basis of Eqs. (12) and (13), when k₂ ≥ 0, k₂ and k can be expressed as

\(\left\{\begin{array}{l} k_{2}=(2-\sqrt{3})\left(1+k_{1}\right) \\ k=\frac{\sqrt{6}}{2}(\sqrt{3}-1)\left(1+k_{1}\right). \end{array}\right.\)       (14)

Fig. 3(c) displays the phasor diagram when k₁ ≥ 0 and k₂ ≤ 0. Triangle oa1a3 demonstrates that

\(\frac{-k_{2}}{\sin 15^{\circ}}=\frac{1+k_{1}}{\sin 45^{\circ}}=\frac{U_{\mathrm{a1}}}{\sin 120^{\circ}}=\frac{k}{\sin 120^{\circ}}.\)       (15)

Fig. 3(d) exhibits the phasor diagram when k₁ ≤ 0 and k₂ ≤ 0. Triangle oa1a3 establishes that

\(\frac{-k_{2}}{\sin 15^{\circ}}=\frac{1+k_{1}}{\sin 45^{\circ}}=\frac{U_{\mathrm{a1}}}{\sin 120^{\circ}}=\frac{k}{\sin 120^{\circ}}.\)       (16)

On the basis of Eqs. (15) and (16), when k₂ ≤ 0, k₂ and k can be expressed as

\(\left\{\begin{array}{l} k_{2}=-\frac{(\sqrt{3}-1)}{2}\left(1+k_{1}\right) \\ k=\frac{\sqrt{6}}{2}\left(1+k_{1}\right). \end{array}\right.\)       (17)

As shown in Fig. 3, k₁ should be greater than −1. Fig. 4(a) presents the relation between k₁ and k₂, whereas Fig. 4(b) visualizes the relation between k₁ and turn ratio k.

Fig. 4. Relation among k₁, k₂, and k. (a) Relation diagram between k₁ and k₂. (b) Relation between k₁ and k.

B. Currents Through Windings under Different Winding Configurations

The current through the windings under different winding configurations is calculated.

In Fig. 2(a), from the Ampere-turn equal law,

\(\left\{\begin{array}{l} N_{\mathrm{q}} i_{\mathrm{c} 2}+N_{\mathrm{q}} i_{\mathrm{b} 1}+N_{1} i_{11}=N_{\mathrm{Y}} i_{1} \\ N_{\mathrm{q}} i_{\mathrm{a} 2}+N_{\mathrm{q}} i_{\mathrm{c1}}+N_{1} i_{22}=N_{\mathrm{Y}} i_{2} \\ N_{\mathrm{q}} i_{\mathrm{b} 2}+N_{\mathrm{q}} i_{\mathrm{a1}}+N_{1} i_{33}=N_{\mathrm{Y}} i_{3}. \end{array}\right.\)       (18)

From Kirchhoff’s current law,

\(\left\{\begin{array}{l} i_{\mathrm{a}}=i_{11}+i_{1} \\ i_{\mathrm{b}}=i_{22}+i_{2} \\ i_{\mathrm{c}}=i_{33}+i_{3} \end{array} \quad\left\{\begin{array}{l} i_{11}=i_{\mathrm{a} 2}+i_{\mathrm{a1}} \\ i_{22}=i_{\mathrm{b} 2}+i_{\mathrm{b} 1} \\ i_{33}=i_{\mathrm{c} 2}+i_{\mathrm{c1}}. \end{array}\right.\right.\)       (19)

Therefore, to set the core limb (a) as an example, the currents through the auxiliary and extended windings and the input line current i_a are calculated as

\(\left\{\begin{array}{l} i_{1}=\frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}} i_{\mathrm{c} 2}+\frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}} i_{\mathrm{b1}}+\frac{N_{1}}{N_{\mathrm{Y}}}\left(i_{\mathrm{a} 2}+i_{\mathrm{a1}}\right) \\ i_{11}=i_{\mathrm{a} 2}+i_{\mathrm{a1}} \\ i_{\mathrm{a}}=i_{\mathrm{a} 2}+i_{\mathrm{a1}}+\frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}} i_{\mathrm{c} 2}+\frac{N_{\mathrm{q}}}{N_{\mathrm{Y}}} i_{\mathrm{b} 1}+\frac{N_{1}}{N_{\mathrm{Y}}}\left(i_{\mathrm{a} 2}+i_{\mathrm{a1}}\right). \end{array}\right.\)       (20)

Substituting Eq. (7) into Eq. (20) yields

\(\left\{\begin{array}{l} i_{1}=k_{2}\left(i_{\mathrm{c} 2}+i_{\mathrm{b} 1}\right)+k_{1}\left(i_{\mathrm{a} 2}+i_{\mathrm{a1}}\right) \\ i_{11}=i_{\mathrm{a} 2}+i_{\mathrm{a1}} \\ i_{\mathrm{a}}=\left(1+k_{1}\right)\left(i_{\mathrm{a} 2}+i_{\mathrm{a1}}\right)+k_{2}\left(i_{\mathrm{c} 2}+i_{\mathrm{b} 1}\right). \end{array}\right.\)       (21)

Under large inductive load, the load current can be assumed to be a constant value I_d, and the output currents of the bridge rectifiers can be expressed as

\(i_{\mathrm{d} 1}=i_{\mathrm{d} 2}=0.5 I_{\mathrm{d}}.\)       (22)

Output currents i_a1, i_b1, i_c1, i_a2, i_b2, and i_c2 of the autotransformer meet

\(\left\{\begin{array}{l} i_{\mathrm{a1}}=S_{\mathrm{a1}} i_{\mathrm{d1}} \\ i_{\mathrm{b1}}=S_{\mathrm{b} 1} i_{\mathrm{d1}} \\ i_{\mathrm{c1}}=S_{\mathrm{c1}} i_{\mathrm{d} 1} \end{array} \quad\left\{\begin{array}{l} i_{\mathrm{a} 2}=S_{\mathrm{a} 2} i_{\mathrm{d1}} \\ i_{\mathrm{b} 2}=S_{\mathrm{b} 2} i_{\mathrm{d1}} \\ i_{\mathrm{c} 2}=S_{\mathrm{c} 2} i_{\mathrm{d1}}, \end{array}\right.\right.\)       (23)

where S_a1, S_b1, S_c1, S_a2, S_b2, and S_c2 are the switching functions of phases a1, b1, c1, a2, b2, and c2, respectively.

Fig. 5 illustrates switching function S_a1 when the three-phase output voltages of the autotransformer meet Eqs. (3) and (4). The relation among the switching functions meets

\(\left\{\begin{array}{l} S_{\mathrm{b} 1}=S_{\mathrm{a1}} \angle-120^{\circ} \\ S_{\mathrm{c1}}=S_{\mathrm{a1}} \angle+120^{\circ} \end{array} \quad\left\{\begin{array}{l} S_{\mathrm{a} 2}=S_{\mathrm{a1}} \angle-30^{\circ} \\ S_{\mathrm{b} 2}=S_{\mathrm{b} 1} \angle-30^{\circ} \\ S_{\mathrm{c} 2}=S_{\mathrm{c1}} \angle-30^{\circ}. \end{array}\right.\right.\)       (24)

Fig. 5. Switching function of S_a1.

Substituting Eqs. (7) and (23) into Eq. (21) yields

\(\left\{\begin{array}{l} i_{1}=0.5 I_{\mathrm{d}}\left[k_{2}\left(S_{\mathrm{c} 2}+S_{\mathrm{b} 1}\right)+k_{1}\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)\right] \\ i_{11}=0.5 I_{\mathrm{d}}\left(S_{\mathrm{a1}}+S_{\mathrm{a} 2}\right) \\ i_{\mathrm{a}}=0.5 I_{\mathrm{d}}\left[\left(1+k_{1}\right)\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)+k_{2}\left(S_{\mathrm{c} 2}+S_{\mathrm{b} 1}\right)\right]. \end{array}\right.\)       (25)

Fig. 6 shows currents i₁ and i₁₁, the input line current, and the spectrum when k₁ = 1 and k₂ = \(2(2-\sqrt{3})\).

Fig. 6. Currents through the windings and input line current when k₁ = 1 and  k₂ = \(2(2-\sqrt{3})\). (a) Current through the wye-connected winding. (b) Current through the extended winding. (c) Input line current and (d) its spectrum.

Similarly, as shown in Figs. 2(b) and 3(b), when k₁ ≤ 0 and k₂ ≥ 0, the currents through the wye-connected and auxiliary windings can be calculated as

\(\left\{\begin{array}{l} i_{1}=0.5 I_{\mathrm{d}}\left[k_{2}\left(S_{\mathrm{c} 2}+S_{\mathrm{b} 1}\right)+k_{1}\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)\right] \\ i_{\mathrm{a}}=0.5 I_{\mathrm{d}}\left[k_{2}\left(S_{\mathrm{c} 2}+S_{\mathrm{b} 1}\right)+\left(1+k_{1}\right)\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)\right]. \end{array}\right.\)       (26)

Fig. 7 shows currents i₁ and i_a and the spectrum of the input line current when k₁ = −0.5 and k₂ = \(0.5 \times (2-\sqrt{3})\).

Fig. 7. Currents through the windings and input line current when k₁ = −0.5 and  k₂ = \(0.5 \times (2-\sqrt{3})\). (a) Current through the bottom winding of the wye-connected winding. (b) Current through the top winding of the wye-connected winding, which is also the input line current. (c) Spectrum of the input line current.

Similarly, as shown in Figs. 2(c) and 3(c), when k₁ ≥ 0 and k₂ ≤ 0, the currents through the auxiliary and extended windings and input line current i_a are calculated as

\(\left\{\begin{array}{l} i_{1}=0.5 I_{\mathrm{d}}\left[k_{1}\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)+k_{2}\left(S_{\mathrm{c1}}+S_{\mathrm{b} 2}\right)\right] \\ i_{11}=0.5 I_{\mathrm{d}}\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right) \\ i_{\mathrm{a}}=0.5 I_{\mathrm{d}}\left[\left(1+k_{1}\right)\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)+k_{2}\left(S_{\mathrm{c1}}+S_{\mathrm{b} 2}\right)\right]. \end{array}\right.\)       (27)

Fig. 8 shows currents i₁ and i₁₁, input line current, and the spectrum when k₁ = 1 and k₂ = \((1-\sqrt{3})\).

Fig. 8. Currents through the windings and input line current when k1 = 1 and  k₂ = \((1-\sqrt{3})\). (a) Current through the wye-connected winding. (b) Current through the extended winding. (c) Input line current and (d) its spectrum.

Similarly, as shown in Figs. 2(d) and 3(d), when k₁ ≤ 0 and k₂ ≤ 0, the currents through the wye-connected and auxiliary windings are calculated as

\(\left\{\begin{array}{l} i_{1}=0.5 I_{\mathrm{d}}\left[k_{1}\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)+k_{2}\left(S_{\mathrm{c1}}+S_{\mathrm{b} 2}\right)\right] \\ i_{\mathrm{a}}=0.5 I_{\mathrm{d}}\left[\left(1+k_{1}\right)\left(S_{\mathrm{a} 2}+S_{\mathrm{a1}}\right)+k_{2}\left(S_{\mathrm{c1}}+S_{\mathrm{b} 2}\right)\right]. \end{array}\right.\)       (28)

Fig. 9 shows currents i₁ and i_a and the spectrum of i_a when k₁ = −0.5 and k₂ = \(0.25 \times (1-\sqrt{3})\).

Fig. 9. Currents through the windings and input line current when k1 = −0.5 and  k₂ = \(0.25 \times (1-\sqrt{3})/4\). (a) Current through the bottom winding of the wye-connected winding. (b) Current through the top winding of the wye-connected winding, which is also the input line current. (c) Spectrum of the input line current.

Figs. 6(d), 7(c), 8(d), and 9(c) show that the THD of the input line current is approximately 15.21% under different winding configurations, and the winding configuration does not affect the THD.

IV. RELATION BETWEEN CONFIGURATION PARAMETER AND KVA RATING

The kilovolt-ampere rating of the autotransformer is determined by the voltage across and current through the windings. The kilovolt-ampere rating can be expressed as

\(S_{\mathrm{kVA}}=0.5 \sum U I,\)       (29)

where U and I are the voltage across and current through the windings of the transformer.

In accordance with modulation theory, the output voltages of the two bridge rectifiers can be expressed as

\(\left\{\begin{array}{l} u_{\mathrm{d1}}=u_{\mathrm{a1}} S_{\mathrm{a1}}+u_{\mathrm{b} 1} S_{\mathrm{b} 1}+u_{\mathrm{c1}} S_{\mathrm{c1}} \\ u_{\mathrm{d} 2}=u_{\mathrm{a} 2} S_{\mathrm{a} 2}+u_{\mathrm{b} 2} S_{\mathrm{b} 2}+u_{\mathrm{c} 2} S_{\mathrm{c} 2}. \end{array}\right.\)       (30)

On the basis of Fig. 1, the load voltage can be expressed as

\(u_{\mathrm{d}}=0.5\left(u_{\mathrm{d} 1}+u_{\mathrm{d} 2}\right).\)       (31)

From Eqs. (2)-(4), (11), and (14), when k₂ ≥ 0, the relation between U_n and U_m meets

\(U_{\mathrm{n}}=\frac{\sqrt{6}}{2}(\sqrt{3}-1)\left(1+k_{1}\right) U_{\mathrm{m}}.\)       (32)

Therefore, when k₂ ≥ 0, the two groups of three-phase output voltages can be expressed as

\(\begin{array}{l} \left\{\begin{array}{l} u_{\mathrm{a1}}=(3-\sqrt{3})\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t+15^{\circ}\right) \\ u_{\mathrm{b1}}=(3-\sqrt{3})\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t-105^{\circ}\right) \\ u_{\mathrm{c1}}=(3-\sqrt{3})\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t+135^{\circ}\right) \\ \end{array}\right. \\ \left\{\begin{array}{l} u_{\mathrm{a} 2}=(3-\sqrt{3})\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t-15^{\circ}\right) \\ u_{\mathrm{b} 2}=(3-\sqrt{3})\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t-135^{\circ}\right) \\ u_{\mathrm{c} 2}=(3-\sqrt{3})\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t+105^{\circ}\right). \end{array}\right. \end{array}\)       (33)

From Eqs. (24) and (30)–(33), when k₂ ≥ 0, load voltage u_d is calculated as

\(u_{\mathrm{d}}=\left\{\begin{array}{l} \frac{3}{\sqrt{2}}\left(1+k_{1}\right) U_{\mathrm{m}} \cos \left(\omega t-k 30^{\circ}\right) \\ \omega t \in\left[k 15^{\circ} \quad(k+1) 15^{\circ}\right], k=0,1,2,3 \ldots \\ \frac{3}{\sqrt{2}}\left(1+k_{1}\right) U_{\mathrm{m}} \cos \left(\omega t-k 30^{\circ}-30^{\circ}\right) \\ \omega t \in\left[(k+1) 15^{\circ} \quad(k+2) 15^{\circ}\right], k=0,1,2,3 \ldots. \end{array}\right.\)       (34)

From Eq. (34), when k₂ ≥ 0, the RMS value of load voltage is as follows:

\(U_{\mathrm{d}}=1.5 \sqrt{(\pi+3) / \pi}\left(1+k_{1}\right) U_{\mathrm{m}}.\)       (35)

From Eqs. (2)-(4), (11), and (17), when k₂ ≤ 0, the relation between U_n and U_m meets

\(U_{\mathrm{n}}=\frac{\sqrt{6}}{2}\left(1+k_{1}\right) U_{\mathrm{m}}.\)       (36)

Therefore, when k₂ ≤ 0, the two groups of three-phase output voltages can be expressed as

\(\begin{array}{l} \left\{\begin{array}{l} u_{\mathrm{a1}}=\sqrt{3}\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t+15^{\circ}\right) \\ u_{\mathrm{b1}}=\sqrt{3}\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t-105^{\circ}\right) \\ u_{\mathrm{c1}}=\sqrt{3}\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t+135^{\circ}\right) \\ \end{array}\right. \\ \left\{\begin{array}{l} u_{\mathrm{a} 2}=\sqrt{3}\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t-15^{\circ}\right) \\ u_{\mathrm{b} 2}=\sqrt{3}\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t-135^{\circ}\right) \\ u_{\mathrm{c} 2}=\sqrt{3}\left(1+k_{1}\right) U_{\mathrm{m}} \sin \left(\omega t+105^{\circ}\right). \end{array}\right. \end{array}\)       (37)

From Eqs. (24), (30), (31), and (37), when k₂ ≤ 0, the load voltage u_d is calculated as

\(u_{\mathrm{d}}=\left\{\begin{array}{l} \frac{3\sqrt{2}}{4}(\sqrt{3}+1)\left(1+k_{1}\right) U_{\mathrm{m}} \cos \left(\omega t-k 30^{\circ}\right) \\ \omega t \in\left[k 15^{\circ} \quad(k+1) 15^{\circ}\right], k=0,1,2,3 \ldots \\ \frac{3\sqrt{2}}{4}(\sqrt{3}+1)\left(1+k_{1}\right) U_{\mathrm{m}} \cos \left(\omega t-k 30^{\circ}-30^{\circ}\right) \\ \omega t \in\left[(k+1) 15^{\circ} \quad(k+2) 15^{\circ}\right], k=0,1,2,3 \ldots. \end{array}\right.\)       (38)

From Eq. (38), when k₂ ≤ 0, the RMS value of the load voltage is as follows:

\(U_{\mathrm{d}}=\frac{3 \sqrt{\pi+3}}{2(\sqrt{3}-1) \sqrt{\pi}}\left(1+k_{1}\right) U_{\mathrm{m}}.\)       (39)

The currents through the extended windings are equal to i_a1, i_b1, i_c1, i_a2, i_b2, and i_c2. Therefore, the RMS values of the currents through the extended windings are calculated as

\(I_{\mathrm{a1}}=I_{\mathrm{b1}}=I_{\mathrm{c1}}=I_{\mathrm{a} 2}=I_{\mathrm{b} 2}=I_{\mathrm{c} 2}=\frac{\sqrt{6}}{6} I_{\mathrm{d}}.\)       (40)

The detailed calculation processes of the kVA rating of the wye-connected autotransformer when k₁ ≥ 0 and k₂ ≥ 0 are shown as follows.

When k₁ ≥ 0 and k₂ ≥ 0, the voltages across the wye-connected windings are the three-phase input voltages, and their RMS value is equal to U_m. From Eq. (22), the RMS values of the currents through the wye-connected windings are as follows:

\(I_{1}=I_{\mathrm{d}} \sqrt{\frac{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}}{12}}.\)       (41)

On the basis of the definition of position parameter, the RMS value of the voltage across the auxiliary winding is equal to k₁U_m. Moreover, from Eq. (25), the RMS value of the current through the auxiliary winding is as follows:

\(I_{11}=\sqrt{\frac{7}{12}} I_{\mathrm{d}}.\)       (42)

On the basis of the definition of proportional parameter, the RMS value of the voltage across the extended winding is equal to k₂U_m, and the current through the extended winding meets Eq. (40).

Therefore, when k₁ ≥ 0 and k₂ ≥ 0, the kVA rating is calculated as

\(S_{\mathrm{kVA1}}=\frac{\sqrt{3}}{4}\left(\sqrt{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}}+\sqrt{7} k_{1}+2 \sqrt{2} k_{2}\right) U_{\mathrm{m}} I_{\mathrm{d}}.\)       (43)

On the basis of Eq. (35), S_kVA1 can also be expressed as

\(S_{\mathrm{kVA1}}=\frac{\sqrt{3 \pi} U_{\mathrm{d}} I_{\mathrm{d}}\left(\sqrt{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}}+\sqrt{7} k_{1}+2 \sqrt{2} k_{2}\right)}{6 \sqrt{\pi+3}\left(1+k_{1}\right)}.\)       (44)

The load power is defined as

\(P_{\mathrm{o}}=U_{\mathrm{d}} I_{\mathrm{d}}.\)       (45)

Substituting Eq. (30) into Eq. (44) yields

\(S_{\mathrm{kVA1}}=\frac{\sqrt{3 \pi}\left(\sqrt{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}}+\sqrt{7} k_{1}+2 \sqrt{2} k_{2}\right)}{6 \sqrt{\pi+3}\left(1+k_{1}\right)} P_{\mathrm{o}}.\)       (46)

The equivalent kVA rating S_eq,kVA is expressed as

\(S_{\mathrm{eq}, \mathrm{kVA}}=\frac{S_{\mathrm{kVA}}}{P_{\mathrm{o}}}.\)       (47)

S_eq,kVA describes the relation between the kVA rating and configuration parameter under the same load power.

From Eqs. (46) and (47), when k₁ ≥ 0 and k₂ ≥ 0, the equivalent kVA rating is as follows

\(S_{\mathrm{eq}, \mathrm{kVA1}}=\frac{\sqrt{3 \pi}\left(\sqrt{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}}+\sqrt{7} k_{1}+2 \sqrt{2} k_{2}\right)}{6 \sqrt{\pi+3}\left(1+k_{1}\right)}.\)       (48)

Similarly, when k₁ ≤ 0 and k₂ ≥ 0, the equivalent kVA rating is calculated as

\(\begin{aligned} S_{\mathrm{eq}, \mathrm{kV} \mathrm{A} 2}=& \frac{\sqrt{3 \pi}}{6 \sqrt{\pi+3}\left(1+k_{1}\right)}\left[\left(1+k_{1}\right) \sqrt{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}}\right.\\ &\left.-k_{1} \sqrt{7 k_{1}^{2}-4 k_{1} k_{2}+k_{2}^{2}+14 k_{1}-4 k_{2}+7}+2 \sqrt{2} k_{2}\right]. \end{aligned}\)       (49)

When k₁ ≥ 0 and k₂ ≤ 0, the equivalent kVA rating is calculated as

\(S_{\mathrm{eq,kV} \mathrm{A} 3}=\frac{\sqrt{\pi}\left(\sqrt{7 k_{1}^{2}-10 k_{1} k_{2}+4 k_{2}^{2}}+\sqrt{7} k_{1}-2 \sqrt{2} k_{2}\right)}{(3+\sqrt{3}) \sqrt{\pi+3}\left(1+k_{1}\right)}.\)       (50)

When k₁ ≤ 0 and k₂ ≤ 0, the equivalent kVA rating is calculated as

\(\begin{array}{c} S_{\mathrm{eq}, \mathrm{kVA} 4}=\frac{\sqrt{3 \pi}(\sqrt{3}-1)}{6 \sqrt{\pi+3}\left(1+k_{1}\right)}\left[\left(1+k_{1}\right) \sqrt{7 k_{1}^{2}-10 k_{1} k_{2}+4 k_{2}^{2}}\right. \\ \left.-k_{1} \sqrt{7 k_{1}^{2}-10 k_{1} k_{2}+4 k_{2}^{2}+14 k_{1}-10 k_{2}+7}-2 \sqrt{2} k_{2}\right]. \end{array}\)       (51)

From Eqs. (48), (49), (50), and (51), the equivalent kVA rating under different k₁ and k₂ values is obtained, as shown in Fig. 10.

Fig. 10. Equivalent kVA rating under different k₁ and k₂ values.

From Fig. 10, conclusions can be obtained as follows.

(1) When k₂ ≥ 0 and k₁ = 0, S_eq,kVA is minimal, and the minimum is 0.2118. Therefore, when k₂ ≥0 and k₁ = 0, the kilovolt-ampere rating of the wye-connected autotransformer is approximately 21% of the load power. Fig. 11 shows the winding configuration and phasor diagram of the autotransformer when k₂ ≥ 0 and k₁ = 0. In comparison with other winding configurations when k₂ ≥ 0 as shown in Figs. 2(a) and 2(b), the autotransformer has the simplest winding configuration when k₂ ≥ 0 and k₁ = 0. In addition, every core limb only has three windings. When k₂ ≥ 0 and k₁ = 0, the wye-connected autotransformer has the optimal configuration.

Fig. 11 Winding configuration and phasor diagram of the autotransformer when k₂ ≥ 0 and k₁ = 0. (a) Winding configuration. (b) Phasor diagram.

(2) When k₂ ≥ 0 and k₁ = 0, k = \(\sqrt{6}(\sqrt{3}-1)/2\), which is less than one. Therefore, the autotransformer operates under a step-down condition in this case. When k₂ ≥ 0 and k₁ = 0, k₂ = \((2-\sqrt{3})\). Therefore, the ratio of N_Y to N_q in Fig. 11 (a) is \(1/(2-\sqrt{3})\).

(3) When the configuration of the wye-connected autotransformer is optimal, the turn ratio k is constant, indicating that the output voltages of the wye-connected autotransformer are constant. When various output voltages are required, the desired turn ratio is different from the optimal turn ratio. Furthermore, the appropriate k₁ and k₂ should be selected from Eqs. (14) and (17), respectively. In this manner, the configuration of the wye-connected autotransformer is not optimal.

(4) In fact, when k₂ ≤ 0 and k₁ = 0, the autotransformer has the simplest winding configuration, and every core limb only has three windings. Fig. 12 shows the winding configuration and phasor diagram when k₂ ≤ 0 and k₁ = 0, wherein the equivalent kilovolt-ampere rating of the autotransformer is approximately 0.2671. This value is greater than that of the autotransformer when k₂ ≥ 0 and k₁ = 0. Therefore, in the viewpoint of simplest winding configuration and least equivalent kilovolt-ampere rating, the winding configuration shown in Fig. 11(a) is optimal.

Fig. 12. Winding configuration and phasor diagram of the autotransformer when k₂ ≤ 0 and k₁ = 0. (a) Winding configuration. (b) Phasor diagram.

V. EXPERIMENTAL VALIDATION

In this section, eight wye-connected autotransformers with different connections under various input voltages and loads are designed to verify the previous theoretical analysis. Fig. 13 shows pictures of the prototype.

Fig. 13. Pictures of the wye-connected autotransformers. (a) k₂ ≥ 0 and k₁ = −0.5. (b) k₂ ≥ 0 and k₁ = 0. (c) k₂ ≥ 0 and k₁ = 0.5. (d) k₂ ≥ 0 and k₁ = 1. (e) k₂ ≤ 0 and k₁ = −0.5. (f) k₂ ≤ 0 and k₁ = 0. (g) k₂ ≤ 0 and k₁ = 0.5. (h) k₂ ≤ 0 and k₁ = 1.

Experimental results are shown as follows:

(1) k₂ ≥ 0 and k₁ = −0.5: Fig. 14 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≥ 0 and k₁ = −0.5. Assume that the autotransformer is symmetrical. The apparent power of the autotransformer is equal to 738.6 VA, and the load power is 1014.3 W. Therefore, the equivalent kVA is 0.728.

Fig. 14. Voltage and current. (a) Through the wye-connected winding ao. (b) Through auxiliary winding aa3. (c) Extended winding b1b3. (d) The load voltage and current when k₂ ≥ 0 and k₁ = −0.5.

(2) k₂ ≥ 0 and k₁ = 0: Fig. 15 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≥ 0 and k₁ = 0. The apparent power of the autotransformer is 211.6 VA, and the load power is 1011.8 W. Therefore, the equivalent kVA is 0.2091.

Fig. 15. Voltage and current. (a) Through the wye-connected winding ao. (b) Extended winding b1b. (c) The load voltage and current when k₂ ≥ 0 and k₁ = 0.

(3) k₂ ≥ 0 and k₁ = 0.5: Fig. 16 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≥ 0 and k₁ = 0.5. The apparent power of the autotransformer is 502 VA, and the load power is 1029.6 W. Therefore, the equivalent kVA is 0.4876.

Fig. 16. Voltage and current. (a) Through the wye-connected winding ao. (b) Through auxiliary winding aa3, (c) Extended winding b1b3. (d) Load voltage and current when k₂ ≥ 0 and k₁ = 0.5.

(4) k₂ ≥ 0 and k₁ = 1: Fig. 17 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≥ 0 and k₁ = 1. The apparent power of the autotransformer is 723.9 VA, and the load power is 1085.9 W. Therefore, the equivalent kVA is 0.72.

Fig. 17. Voltage and current. (a) Through the wye-connected winding ao. (b) Through auxiliary winding a3a, (c) Extended winding b1b3. (d) The load voltage and current when k₂ ≥ 0 and k₁ = 1.

(5) k₂ ≤ 0 and k₁ = −0.5: Fig. 18 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≤ 0 and k₁ = −0.5. The apparent power of the autotransformer is 554.64 VA, and the load power is 959.76 W. Therefore, the equivalent kVA is 0.5779.

Fig. 18. Voltage and current. (a) Through the wye-connected winding ao, (b) Through auxiliary winding aa3. (c) Extended winding b3b2. (d) The load voltage and current when k₂ ≤ 0 and k₁ = −0.5.

(6) k₂ ≤ 0 and k₁ = 0: Fig. 19 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≤ 0 and k₁ = 0. The apparent power of the autotransformer is 388.8 VA, and the load power is 1458.4 W. Therefore, the equivalent kVA is 0.2666.

Fig. 19. Voltage and current. (a) Through the wye-connected winding ao. (b) Extended winding b3b2. (c) The load voltage and current when k₂ ≤ 0 and k₁ = 0.

(7) k₂ ≤ 0 and k₁ = 0.5: Fig. 20 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≤ 0 and k₁ = 0.5. The apparent power of the autotransformer is 763.91 VA, and the load power is 1447.4 W. Therefore, the equivalent kVA is 0.5278.

Fig. 20. Voltage and current. (a) Through the wye-connected winding ao. (b) Through auxiliary winding aa3. (c) Extended winding b2b3. (d) The load voltage and current when k₂ ≤ 0 and k₁ = 0.5.

(8) k₂ ≤ 0 and k₁ = 1: Fig. 21 shows the voltages across and current through the windings of core limb (a) and the load current and voltage when k₂ ≤ 0 and k₁ = 1. The apparent power of the autotransformer is 1019 VA, and the load power is 1487.1 W. Therefore, the equivalent kVA is 0.6852.

Fig. 21. Voltage and current. (a) Through the wye-connected winding ao, (b) Through auxiliary winding aa3. (c) Extended winding b3b2. (d) The load voltage and current when k₂ ≤ 0 and k₁ = 1.

Experimental results illustrate the relation between the position and equivalent kVA rating, as shown in Fig. 22. The equivalent kVA rating is minimal when k₂ ≥ 0 and k₁ = 0, which is identical to the theoretical analysis.

Fig. 22. Relation between the position and equivalent kVA rating under experiments.

Theoretical analysis and simulation results show that, when k₂ ≥ 0 and k₁ = 0, the wye-connected autotransformer has the least equivalent kVA rating and the simplest winding configuration. When k₂ ≥ 0 and k₁ = 0, the proportional parameter is \((2-\sqrt{3})\), and the turn ratio is equal to 0.8966.

VI. CONCLUSIONS

This study analyzes the effect of winding configuration on the kVA rating of a wye-connected autotransformer applied to a 12-pulse rectifier. Conclusions are obtained as follows:

(1) The position and proportional parameters are defined to describe the winding configuration of the wye-connected autotransformer. When the proportional parameter is greater than zero and the position parameter is equal to zero, the winding configuration of the wye-connected autotransformer is optimal. The optimal autotransformer has three windings on every core limb. When the autotransformer is applied to the 12-pulse rectifier, its equivalent kVA rating is approximately 21%.

(2) The winding configuration of the wye-connected autotransformer remarkably affects its equivalent kVA rating. Not all autotransformers of arbitrary winding configuration can reduce the equivalent kVA rating. The autotransformer has the least equivalent kVA rating and the simplest winding configuration only when it is optimally designed.

(3) The proposed method provides an interesting and useful confirmation that the well-designed autotransformer is optimal in terms of kVA rating. The method also offers an additional insight into these autotransformers and how they may be analyzed.