# Comparative Soft Switching Analysis of Star-Star and Star-Delta Connected Transformer Three-Phase Dual Active Bridge DC-DC Converter

Cevher Sami Birsin, Erdem Akboy

► To cite this version:

Cevher Sami Birsin, Erdem Akboy. Comparative Soft Switching Analysis of Star-Star and Star-Delta Connected Transformer Three-Phase Dual Active Bridge DC-DC Converter. The 26th European Conference on Power Electronics and Applications, GDR SEEDS France & EPE Association, Mar 2025, Paris, France. 10.34746/epe2025-0089 . hal-05088913

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Submitted on 28 May 2025

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# Comparative Soft Switching Analysis of Star-Star and Star-Delta Connected Transformer Three-Phase Dual Active Bridge DC-DC Converter

Cevher Sami Birsin, Erdem Akboy

Yildiz Technical University / Department of Electrical Engineering
Istanbul, Türkiye

Tel.: +90 / 507 686 88 22, +90 / 212 383 5843

E-Mail: cevhersami@gmail.com, eakboy@yildiz.edu.tr

URL: https://elm.yildiz.edu.tr

***Index Terms***—Three-phase dual active bridge (DAB) Converter, Single phase-shift control (SPS), Bidirectional DC-DC converter, Zero voltage switching (ZVS).

***Abstract***—This article discusses the soft switching performance of three-phase DAB converters with star-star and star-delta transformers. It explores how different transformer combinations can maintain soft switching under high voltage differences and light loads, supported by theoretical models and PSIM simulations

Despite the known advantages of DAB converters, there is a lack of comprehensive comparative studies on the impact of different transformer connections on the soft switching performance. This study aims to fill this gap by providing a detailed analysis of the star-star and star-delta connected transformers in three-phase DAB converters by examining the ZVS operation conditions and developing simulation models in PSIM software.

## I. INTRODUCTION

The ever-increasing demand for efficient and reliable power conversion systems has led to significant advancements in the field of power electronics. These developments are critical to addressing global energy challenges, including the integration of renewable energy sources, the electrification of transportation, and the efficient utilization of energy storage systems. Among these, the Dual Active Bridge (DAB) DC-DC converter has garnered considerable attention due to its bidirectional power flow capabilities, high efficiency, and ability to achieve soft switching. Soft switching techniques, such as Zero Voltage Switching (ZVS), are critical in reducing switching losses and improving overall efficiency, making them highly desirable in modern power conversion applications.

Three-phase DAB converters, in particular, offer enhanced power handling capabilities and improved efficiency compared to their single-phase counterparts. These converters are often utilized in applications such as electric vehicle (EV) charging stations[8],[9], renewable energy systems, and energy storage solutions[4].

This paper is structured as follows. Chapter II describes the mathematical modeling and operation principles of Three-Phase DAB Converters with star-star and star-delta connected transformers. The Three-Phase DAB Converter with star-delta connected transformers was analyzed for phase shift angles ranging from 0 to π/6 and from π/6 to π/2 to explore wider operation conditions. Chapter III presents the ZVS performance of Three-Phase DAB converters. Chapter IV shows the simulation results to verify the theoretical analysis and mathematical models of the proposed Three-Phase DAB converters. Simulation parameters were selected under the assumption that the converter would be used in 30 kW modules for EV charging stations. Finally, conclusions are presented in Chapter V.

By addressing these objectives, this research contributes to the field of power electronics by enhancing the understanding of transformer configuration impacts on DAB converter performance, thereby aiding in the design of more efficient and reliable power conversion systems
## II. OPERATING PRINCIPLES OF DAB CONVERTERS

### A. Star-Star Connected Transformer

<figure><img src="image_2.png" /><figcaption>Fig. 1. Schematic of Star-Star Connected Three Phase DAB</figcaption></figure>

Figure 1 illustrates the three-phase DAB converter with star-star connected transformer. Transformer conversion ratio is one, so output voltage equals to reflected output voltage.

$$n = 1, \quad nU_{out} = U_{out} \quad (1)$$

1) Star-Star Connected Transformer ($0 \le \phi \le \frac{\pi}{3}$): The difference of input and reflected output voltage gives us the inductance voltage. This voltage difference at the leg inductance creates a current curve and power flow. During the operation, inductance voltage changes and creates six intervals in a half period. Six different intervals, creates a six different current curve on series inductance. These curves obtained at Figure 2, and the formulas of the current curves are given at Table 1. To obtain current curve equation inductance formula 3 is used.

$$I_L = I_{L_0} + \frac{V}{L}t, \quad \theta = \omega t, \quad t = \frac{\theta}{\omega} \quad (2)$$

$$I_L = I_{L_0} + \frac{V}{\omega L} \theta \quad (3)$$

As we can see on the Figure 2's current waveform, current is symmetrical between first and second half of the period. Because of the symmetrical current waveform.Negative value of $I_{L_A}(0)$ is equal to $I_{L_A}(\pi)[1],[3],[6]$.

$$-I_{L_A}(0) = I_{L_A}(\pi) \quad (4)$$

This equation 4 will be our starting point for the whole analysis. $I_{L_A}(0)$ and $I_{L_A}(\pi)$ are the starting and ending points of the current curves in a half period.

<figure><img src="image_3.png" /><figcaption>Fig. 2. Operating Waveform of Star-Star Connected Three Phase DAB With ($0 \le \phi \le \frac{\pi}{3}$) Phase Shift</figcaption></figure>

<table><caption>TABLE I
CURRENT CURVE EQUATIONS OF Y-Y DAB ($0 \le \phi \le \frac{\pi}{3}$)</caption><tr><td>1</td><td>$I_{L_A}(0) + \frac{U_{in}+U_{out}}{3\omega L}\theta$</td><td>$0 \le \theta \le \varphi$</td></tr><tr><td>2</td><td>$I_{L_A}(\varphi) + \frac{U_{in}-U_{out}}{3\omega L}(\theta - \varphi)$</td><td>$\varphi \le \theta \le \frac{\pi}{3}$</td></tr><tr><td>3</td><td>$I_{L_A}(\frac{\pi}{3}) + \frac{2U_{in}-U_{out}}{3\omega L}(\theta - \frac{\pi}{3})$</td><td>$\frac{\pi}{3} \le \theta \le \varphi + \frac{\pi}{3}$</td></tr><tr><td>4</td><td>$I_{L_A}(\varphi + \frac{\pi}{3}) + \frac{2U_{in}-2U_{out}}{3\omega L}(\theta - \varphi - \frac{\pi}{3})$</td><td>$\varphi + \frac{\pi}{3} \le \theta \le \frac{2\pi}{3}$</td></tr><tr><td>5</td><td>$I_{L_A}(\frac{2\pi}{3}) + \frac{U_{in}-2U_{out}}{3\omega L}(\theta - \frac{2\pi}{3})$</td><td>$\frac{2\pi}{3} \le \theta \le \varphi + \frac{2\pi}{3}$</td></tr><tr><td>6</td><td>$I_{L_A}(\varphi + \frac{2\pi}{3}) + \frac{U_{in}-U_{out}}{3\omega L}(\theta - \varphi - \frac{2\pi}{3})$</td><td>$\varphi + \frac{2\pi}{3} \le \theta \le \pi$</td></tr></table>

So, $I_{L_A}(\pi)$ is the sum of six current curve equations. By summing, putting angle $\pi$ and simplifying these six curve equations, we can find $I_{L_A}(0)$.

$$I_{L_A}(0) = \frac{2\pi(U_{out} - U_{in}) - 3U_{out}\varphi}{9\omega L} \quad (5)$$
After we find $I_{L_A}(0)$, we can find the current equation of $I_{L_A}(\varphi)$. $I_{L_A}(\varphi)$ is the last point of curve 1 in the Figure 2. By putting $I_{L_A}(0)$ equation and angle value $\varphi$ at curve 1 equation, we can find $I_{L_A}(\varphi)$ equation.

Once we have the equation for $I_{L_A}(\varphi)$, the equation for $I_{L_A}(\pi/3)$ can be found using the same process, continuing sequentially until we obtain the equation for $I_{L_A}(\pi)$. The equation for $I_{L_A}(\pi)$ represents the end of the process and should be equal to $-I_{L_A}(0)$. If it is correct, the entire process is also verified.

All the equation of current curve change points are given at Table II. These points give an idea for the ZVS capability, maximum and minimum current values, circulation current on the leg and component selection for the converter.

<table><caption>TABLE II
INDUCTOR CURRENT EQUATIONS OF Y-Y DAB ($0 \le \phi \le \frac{\pi}{3}$)</caption><tr><td>$I_{L_A}(0)$</td><td>$\frac{2\pi(U_{out}-U_{in})-3U_{out}\varphi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\varphi)$</td><td>$\frac{2\pi(U_{out}-U_{in})+3U_{in}\varphi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\pi/3)$</td><td>$\frac{\pi(U_{out}-U_{in})+3U_{out}\varphi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\varphi + \pi/3)$</td><td>$\frac{\pi(U_{out}-U_{in})+6U_{in}\varphi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(2\pi/3)$</td><td>$\frac{-\pi(U_{out}-U_{in})+6U_{out}\varphi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\varphi + 2\pi/3)$</td><td>$\frac{-\pi(U_{out}-U_{in})+3U_{in}\varphi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\pi)$</td><td>$\frac{-2\pi(U_{out}-U_{in})+3U_{out}\varphi}{9\omega L}$</td></tr></table>

Transferred power via Phase A can be found by followed integral equation;

$$P_A = \frac{1}{\pi} \left[ \begin{array}{l} \frac{U_{in}}{3} \int_{0}^{\frac{\pi}{3}} I_{L_A}(\theta) d\theta \\ + \frac{2U_{in}}{3} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} I_{L_A}(\theta) d\theta \\ + \frac{U_{in}}{3} \int_{\frac{2\pi}{3}}^{\pi} I_{L_A}(\theta) d\theta \end{array} \right] \quad (6)$$

This equation divided by three, because of transformer primary voltage. During the half period, the transformer primary voltage changes at every $\frac{\pi}{3}$ phase angle. During the first and second intervals, the transformer's primary voltage is $\frac{U_{in}}{3}$. Third and fourth intervals, transformers primary voltage is $\frac{2U_{in}}{3}$. Lastly, during the fifth and sixth intervals, the transformer's primary voltage is again $\frac{U_{in}}{3}$. These primary coil voltages multiplied with related integral of current curves to find transferred power via one phase. Instead of solving integral equation we can find area of under curves by using Table II

equations. Because all curves are linear and area under curves are triangle and trapezoid.

At this point we need a one more important equation to find area of under curve one. In this operation ($0 \le \varphi \le \frac{\pi}{3}$), the inductance current value always crosses zero in curve one. To find positive and negative triangular areas under curve one, we need to find alpha point equation.

$$I_{L_A}(\alpha) = 0 = I_{L_A}(0) + \frac{U_{in} + U_{out}}{3\omega L}\alpha \quad (7)$$

$$\alpha = \frac{2\pi(U_{in} - U_{out}) + 3U_{out}\varphi}{3(U_{in} + U_{out})} \quad (8)$$

After finding alpha, circulation power can be found. Negative values of current on curve one is the circulation current which is circulating through transformer and reduces efficiency but also enables soft switching. Area under curves gives us inductance charge current and multiplying this charge value with transformer primary voltage, we can find transferred power via one phase.

$$P_A = \frac{(U_{in} - U_{out})}{\omega L} \varphi \left(\frac{2}{9} - \frac{\varphi}{6\pi}\right) \quad (9)$$

After finding transferred power with one phase, other phases transferred power will be equal. To find power equation, we can sum all phase power;

$$\begin{aligned} P_T &= P_A + P_B + P_C \\ &= \frac{(U_{in} - U_{out})}{\omega L} \varphi \left(\frac{2}{3} - \frac{\varphi}{2\pi}\right) \end{aligned} \quad (10)$$

## B. Star-Delta Connected Transformer

<figure><img src="image_4.png" /><figcaption>Fig. 3. Schematic of Star-Delta Connected Three Phase DAB</figcaption></figure>

1) Star-Delta Connected Transformer ($0 \le \phi \le \frac{\pi}{6}$): In this operation, the phase shifting angle is operated between 0 and $\frac{\pi}{6}$.

Due to the nature of the delta-connected transformer secondary winding, the transformer secondary coil has
<figure><img src="image_5.png" /><figcaption>Fig. 4. Operating Waveform of Star-Delta Connected Three Phase DAB With ($0 \le \phi \le \frac{\pi}{6}$) Phase Shift</figcaption></figure>

a $\frac{\pi}{6}$ radian natural phase shift. During the operation, inductance voltage changes and creates five different intervals in a half period. These intervals, crates a five different current curve on series inductance. These curves obtained at Figure 4, and the formulas of the curves are given at Table III.

<table><caption>TABLE III
CURRENT CURVE EQUATIONS OF Y-Δ DAB ($0 \le \phi \le \frac{\pi}{6}$)</caption><tr><td>1</td><td>$I_{L_A}(0) + \frac{U_{in}}{3\omega L}\theta$</td><td>$0 \le \theta \le \varphi + \frac{\pi}{6}$</td></tr><tr><td>2</td><td>$I_{L_A}(\varphi + \frac{\pi}{6}) + \frac{U_{in}-3U_{out}}{3\omega L}(\theta - \varphi - \frac{\pi}{6})$</td><td>$\varphi + \frac{\pi}{6} \le \theta \le \frac{\pi}{3}$</td></tr><tr><td>3</td><td>$I_{L_A}(\frac{\pi}{3}) + \frac{2U_{in}-3U_{out}}{3\omega L}(\theta - \frac{\pi}{3})$</td><td>$\frac{\pi}{3} \le \theta \le \frac{2\pi}{3}$</td></tr><tr><td>4</td><td>$I_{L_A}(\frac{2\pi}{3}) + \frac{U_{in}-3U_{out}}{3\omega L}(\theta - \frac{2\pi}{3})$</td><td>$\frac{2\pi}{3} \le \theta \le \varphi + \frac{5\pi}{6}$</td></tr><tr><td>5</td><td>$I_{L_A}(\varphi + \frac{5\pi}{6}) + \frac{U_{in}}{3\omega L}(\theta - \varphi - \frac{5\pi}{6})$</td><td>$\varphi + \frac{5\pi}{6} \le \theta \le \pi$</td></tr></table>

Same as with the Y-Y configuration waveforms, the current is also symmetrical between the first and second halves of the period in the Y-Δ configuration. With this symmetrical current waveform, we can also see that the negative value of $I_{L_A}(0)$ is equal to $I_{L_A}(\pi)$. Thus, equation 4 is also valid for the Y-Y configuration.

Same as with the Y-Y configuration, $I_{L_A}(\pi)$ is the sum of six current curve equations. By summing, substituting the angle $\pi$, and simplifying these six curve equations, we can find $I_{L_A}(0)$ for the Y-Δ configuration.

$$I_{L_A}(0) = \frac{\pi (3U_{out} - 2U_{in})}{9\omega L} \quad (11)$$

After we find $I_{L_A}(0)$, we can determine the current equation of $I_{L_A}(\phi + \frac{\pi}{6})$, then $I_{L_A}(\frac{\pi}{3})$, and the other two points one by one until $I_{L_A}(\pi)$. We can also verify the whole process if $I_{L_A}(\pi) = -I_{L_A}(0)$. All the equation of current curves edge points are given at Table IV.

<table><caption>TABLE IV
INDUCTOR CURRENT EQUATIONS OF Y-Δ DAB ($0 \le \phi \le \frac{\pi}{6}$)</caption><tr><td>$I_{L_A}(0)$</td><td>$\frac{\pi(3U_{out}-2U_{in})}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\phi + \frac{\pi}{6})$</td><td>$\frac{\pi(3U_{out}-\frac{3}{2}U_{in})+3U_{in}\phi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\frac{\pi}{3})$</td><td>$\frac{\pi(\frac{3}{2}U_{out}-U_{in})+9U_{out}\phi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\frac{2\pi}{3})$</td><td>$\frac{\pi(-\frac{3}{2}U_{out}+U_{in})+9U_{out}\phi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\phi + \frac{5\pi}{6})$</td><td>$\frac{\pi(-3U_{out}+\frac{3}{2}U_{in})+3U_{in}\phi}{9\omega L}$</td></tr><tr><td>$I_{L_A}(\pi)$</td><td>$\frac{\pi(-3U_{out}+2U_{in})}{9\omega L}$</td></tr></table>

Although, the transformer seconder coils connected delta, primary coils are still star connected. In this way, transferred power via Phase A can be found by same integral equation (6) with II.A.1.

This time, during the first and second intervals, the transformer's primary voltage is $\frac{U_{in}}{3}$. At the third interval, the transformer's primary voltage is $\frac{2U_{in}}{3}$. Lastly, during the fourth and fifth intervals, the transformer's primary voltage is again $\frac{U_{in}}{3}$. Phase A power equation could be found by multiplying the primary coil voltages with the related integral of current curves.

Same as with II.A.1, we can find triangle or trapezoid area of under curves by using Table IV equations. At this point alfa value can be found by following equation.

$$I_{L_A}(\alpha) = 0 = I_{L_A}(0) + \frac{U_{in}}{3\omega L}\alpha \quad (12)$$

$$\alpha = \frac{\pi (2U_{in} - 3U_{out})}{3U_{in}} \quad (13)$$
Finding alpha reveals that there is no phase-shifting variable ($\varphi$) in equation (13). It means circulation power doesn't affect by phase shifting angle and transferred power. It is affected only by the voltage gain.

We can find transferred power via one phase (14) by multiplying with transformer primary voltage and area under curves, that is, inductance charge current.

$$P_A = \frac{U_{in} \cdot U_{out}}{3\omega L} \varphi \quad (14)$$

After finding transferred power equation with Phase A, total output power will be three times this equation;

$$P_T = P_A + P_B + P_C = 3P_A = \frac{U_{in} \cdot U_{out}}{\omega L} \varphi \quad (15)$$

2) Star-Delta Connected Transformer ($\frac{\pi}{6} \le \phi \le \frac{\pi}{2}$): In this operation, all the calculation process will be same with Chapter 2.B.1. To avoid repetition, only equations will be shown.

$$P_T = 3P_A = \frac{U_{in} \cdot U_{out}}{\omega L} \left( \frac{2}{3} \left( \varphi - \frac{\varphi^2}{\pi} \right) - \frac{\pi}{24} \right) \quad (16)$$

### III. ANALYSIS OF THE ZVS CHARACTERISTICS OF CONVERTERS

In this chapter, ZVS capability will be analyzed for star-star and star-delta connected transformer for different operations.

Most of the switching losses in MOSFET's occurs during turn on because of the MOSFET's parasitic capacitance. MOSFET turn on action causes short circuit during the reverse recovery time of the parasitic body diode and energy of parasitic capacity discharges through this short circuit. It creates EMI problem and additional switching loss. In order to prevent all these problems, ZVS turn on must be achieved by discharging the parasitic capacity through the parasitic body diode.

#### A. ZVS Capability of Star-Star Connected Transformer

1) Phase Shifting Operation Between ($0 \le \phi \le \frac{\pi}{3}$): In DAB converters, input and output bridges have different conditions to achieve ZVS. For the Input Bridge at Phase A; rising transition starts at 0. Therefore $I_{L_A}(0)$ should be less than zero[6],[10].

$$I_{L_A}(0) = \frac{2\pi(U_{out} - U_{in}) - 3U_{out}\varphi}{9\omega L} < 0 \quad (17)$$

We can find ZVS condition with $\varphi$ related equation

$$\varphi > \frac{2\pi}{3} \left( 1 - \frac{U_{in}}{U_{out}} \right) \quad (18)$$

Rewriting power equation;

$$P > \frac{U_{in}(U_{out}^2 - U_{in}^2)}{9fLU_{out}} \quad (19)$$

For the output bridge at Phase A, the rising transition starts at $\varphi$. Therefore, $I_{L_A}(\varphi)$ should be greater than zero.

$$I_{L_A}(\varphi) = \frac{2\pi(U_{out} - U_{in}) + 3U_{in}\varphi}{9\omega L} > 0 \quad (20)$$

$\varphi$ related equation is;

$$\varphi > \frac{2\pi}{3} \left( 1 - \frac{U_{out}}{U_{in}} \right) \quad (21)$$

$$P > \frac{U_{out}(U_{in}^2 - U_{out}^2)}{9fLU_{in}} \quad (22)$$

As we can see from ZVS equations, input bridge's ZVS boundary starts when output voltage is bigger than input and output bridge's ZVS boundary starts when input voltage is bigger than output.

According to the derived ZVS equations, increasing the series inductance in the transformers enables soft switching to be achieved at lower power levels. This expansion of the soft-switching region is illustrated in Figure 5. However, increasing the series inductance also reduces the maximum transferable power and increases the physical dimensions of the system.

<figure><img src="image_6.png" /><figcaption>Fig. 5. ZVS Area of Star-Star Connected Three Phase DAB With ($0 \le \phi \le \frac{\pi}{3}$) Phase Shift</figcaption></figure>

Figure 5 prepared with Table V values. ZVS area on the Figure 5 can be optimized by fine-tuning the Switching frequency and Inductance value for desired operation.
## B. ZVS Capability of Star-Delta Connected Transformer

1) Phase Shifting Operation Between ($0 \le \phi \le \frac{\pi}{6}$): In this Operation transformer primary coils still is star connected so input bridge rising transition also starts at 0. Therefore $I_{L_A}(0)$ should be less than zero.

$$I_{L_A}(0) < 0 \quad I_{L_A}(0) = \frac{\pi (3U_{\text{out}} - 2U_{\text{in}})}{9\omega L} < 0 \quad (23)$$

$$\frac{U_{\text{in}}}{U_{\text{out}}} > \frac{3}{2} \quad (24)$$

Unlike previous approaches, transformer secondary coils are delta connected. Because of delta connection two phase current flows at same leg. For example, phase A and phase C current flows at the same time on the phase leg A. The output bridge rising transition starts at $\varphi + \frac{\pi}{6}$ with the contribution of Phase A and Phase C.

While Phase A current flows positively, Phase C current flows negatively at same leg. Therefore, difference of Phase A to Phase C current should be greater than zero;

$$I_{L_A} \left(\varphi + \frac{\pi}{6}\right) - I_{L_C} \left(\varphi + \frac{\pi}{6}\right) > 0 \quad (25)$$

Because of phase delay, phase C current equal to the value of Phase A after 240 degrees.

$$\begin{aligned} I_{L_C} \left(\varphi + \frac{\pi}{6}\right) &= I_{L_A} \left(\varphi + \frac{\pi}{6} - \frac{4\pi}{3}\right) \\ &= I_{L_A} \left(\varphi + \frac{5\pi}{6}\right) \end{aligned} \quad (26)$$

After converting Phase C current to current value from Table IV, new condition will be;

$$I_{L_A} \left(\varphi + \frac{\pi}{6}\right) - I_{L_A} \left(\varphi + \frac{5\pi}{6}\right) > 0 \quad (27)$$

$$\frac{\pi \left(3U_{\text{out}} - \frac{3}{2}U_{\text{in}}\right) + 3U_{\text{in}}\varphi}{9\omega L} - \frac{\pi \left(-3U_{\text{out}} + \frac{3}{2}U_{\text{in}}\right) + 3U_{\text{in}}\varphi}{9\omega L} > 0 \quad (28)$$

Output bridge ZVS condition will be obtained with simplifying equation (28).

$$\frac{U_{\text{in}}}{U_{\text{out}}} < 2 \quad (29)$$

2) Phase Shifting Operation Between ($\frac{\pi}{6} \le \phi \le \frac{\pi}{2}$): In this operation, all the calculation process will be same with Chapter 3.B.1. Input bridge ZVS condition;

$$\varphi > \pi \left( \frac{1}{2} - \frac{2U_{\text{in}}}{9U_{\text{out}}} \right) \quad (30)$$

Output bridge ZVS condition;

$$\varphi > \pi \left( \frac{1}{2} - \frac{2U_{\text{out}}}{3U_{\text{in}}} \right) \quad (31)$$

<figure><img src="image_7.png" /><figcaption>Fig. 6. ZVS Area of Star-Delta Connected Three Phase DAB With ($0 \le \phi \le \frac{\pi}{2}$) Phase Shift</figcaption></figure>

As evident from equations (24) and (29), the soft switching condition is not directly influenced by the series leakage inductance. However, beyond a thirty-degree phase shift, as derived from equations (30) and (31), a larger leakage inductance contributes to an expansion of the ZVS region.

# IV. SIMULATION

Simulation models of star-star and star-delta connected Three Phase DAB Converters built in PSIM Software to verify previous calculations.

## A. Control Block Diagram and Simulation Parameters

Table V shows that calculation parameters of Figure 5, Figure 6 and Figure 7. Values of the Table V also gives us the simulation parameters of the DAB Converter.
<table><caption>TABLE V
SIMULATION PARAMETERS</caption><thead><tr><td>Parameter</td><td>Value</td><td>Unit</td></tr></thead><tbody><tr><td>P<sub>nom</sub></td><td>30</td><td>kW</td></tr><tr><td>U<sub>in</sub></td><td>800</td><td>V</td></tr><tr><td>L<sub>p_leak</sub></td><td>6</td><td>μH</td></tr><tr><td>L<sub>m</sub></td><td>1</td><td>mH</td></tr><tr><td>f</td><td>100</td><td>kHz</td></tr><tr><td>n = N<sub>p</sub>/N<sub>s</sub></td><td>1</td><td></td></tr></tbody></table>

<figure><img src="image_8.png" /><figcaption>Fig. 7. ZVS Area of Star-Star and Star-Delta Connected Three Phase DAB</figcaption></figure>

According to Table V and Chapter III, Figure 7 is prepared for simulation. Four points on the Figure 7 are selected to verify ZVS area of the DAB converter.

Simulation control block is taken from [2], to make sure converter working at targeted output power and voltage.

<figure><img src="image_9.png" /><figcaption>Fig. 8. Feed Forward Control Block Diagram for Three Phase DAB</figcaption></figure>

FF(Feed forward) block on the Figure 8 contains following equations to find calculated phase angle.

Star-star connected transformer feed forward equation is rewriting of power equation (10);

$$ \varphi = \frac{2\pi}{3} \left( 1 - \sqrt{1 - \frac{9fLI_{\text{out}}}{U_{\text{in}}}} \right) \quad (32) $$

For the star-delta connected transformer feed forward equation can be obtained from equation (15);

$$ \varphi = \frac{2\pi f LI_{\text{out}}}{U_{\text{in}}} \quad (33) $$

## B. Simulation Results

Simulation result of the four points on the Figure 7 are analyzed on following chapter. Figure 10 and Figure 11 shows that primary phase leg current, primary coil voltages and secondary coil voltages of the Y-Y connected Three Phase DAB. Figure 12, and Figure ?? shows that both primary and secondary phase leg current and coil voltages of the Y-Δ connected Three Phase DAB. As mentioned on Chapter III, input side switches had to be turned on while primary phase leg current negative and output side switches had to be turned on while primary phase leg current positive to achieve ZVS. The impact of series leakage inductance on the soft switching performance has also been simulated for the first operating point.

<figure><img src="image_10.png" /><figcaption>a</figcaption></figure>

<figure><img src="image_11.png" /><figcaption>b</figcaption></figure>

<caption>Fig. 9. Drain-Source voltage and drain current waveforms of MOSFET switches for series leakage inductances of 6 μH (a) and 4 μH (b).</caption>

Figure 9 shows how series leakage inductance affects ZVS performance. The results clearly indicate that
reducing the leakage inductance makes it harder to achieve ZVS in the input bridges, as observed in the first operating point. This happens because lower leakage inductance provides less energy to discharge the parasitic capacitances of the MOSFETs, which is necessary for ZVS.

These findings match the patterns shown in Figure 5, where the role of leakage inductance in expanding or limiting the ZVS region is highlighted.

1. Point One: For the point one, the output voltage is higher than the input voltage so input bridge's switches had a ZVS limit.

<figure><img src="image_12.png" /><figcaption>Fig. 10. Waveforms of Coil Voltages and Inductance Current for First Operation Point</figcaption></figure>

Figure 10, Shows that the phase current almost zero when the input bridge's switches turn on. If output voltage little bit higher, Input bridges lost ZVS capability. On the other hand, the phase current much higher than the zero when output switches turn on. It means output switches turn on after discharge all of their parasitic capacitance energy.

2. Point Two: For the point two, the output voltage is lower than the input voltage so output bridge's switches had a ZVS limit.

<figure><img src="image_13.png" /><figcaption>Fig. 11. Waveforms of Coil Voltages and Inductance Current for Second Operation Point</figcaption></figure>

On the Figure 11, when the output switches turn on, phase current very close to zero and output switches

almost lost ZVS capability. ZVS capability of output bridge is will be lost if output voltage decreases further. Unlike output switches, input switches turn on at high negative phase current so ZVS easily guaranteed.

3. Point Three: Unlike previous points, converter transformer coils connected star-delta, so secondary phase current is not equal to primary phase current. Therefore, secondary phase current waveform added to Figure 12 to analysis output switch ZVS performance.

<figure><img src="image_14.png" /><figcaption>Fig. 12. Waveforms of Coil Voltages and Inductance Current for Third Operation Point</figcaption></figure>

Point 3 is very close to the input bridge hard switching border and Figure 12 shows that primary phase current almost exceeding to zero and almost lost ZVS when input switches turning on. However, output switches turn on at high positive phase current and there is enough negative current on the output switches to manage ZVS.

4. Point Four: Star-delta connected DAB operated at output switch hard switching area border in Point 4.

<figure><img src="image_15.png" /><figcaption>Fig. 13. Waveforms of Coil Voltages and Inductance Current for Fourth Operation Point</figcaption></figure>

Figure 13 shows that, while the input switches turn on at enough negative current to provide ZVS, the output switches turn on at almost zero reverse discharge current. Therefore, ZVS on the output switches almost lost.
## V. CONCLUSION

In this paper, the operating principles and mathematical models of three-phase DAB converters with star-star and star-delta connected transformers are analyzed. The study reveals that star-star transformers offer better ZVS performance with a lower voltage difference between input and output, while star-delta transformers provide enhanced ZVS performance when the output voltage is lower, due to their inherent gain. The equations and analysis results obtained in this study provide a robust foundation for implementing advanced control methods to expand the ZVS regions of the converter. For instance, modulation techniques such as dual, extended, and triple phase shift, similar to those used in single-phase DAB converters, could be employed to enhance performance. While these analyses may not fully apply to all advanced control strategies, they can still serve as a theoretical framework and a valuable starting point for future developments.

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