December 8, 2022
• December 8, 2022

# Minimizing Current Constraints for an Isolated Active Dual-Bridge DC-DC Converter

on October 10, 2022 0

### Proposed phase shift command

Figure 1a illustrates the circuit diagram of the DAB converter. In the proposed technique, a phase shift angle (D1Js) is between S1 and S4. A three-level voltage is synthesized on the primary side Vh1 of the converter, unlike the traditional two-level voltage adopted in the conventional SPS control technique. This three-level voltage helps reduce the reverse reverse power to V1. Another phase angle (D2Js) is done in the second bridge between S5 and S6; this controls the amount of power transferred into the converter by creating an essential phase shift between the squared voltages of the two bridges. The change of D2 the phase shift widens the transmission power regulation range, increasing regulation flexibility. Therefore, D1 is the phase shift ratio between the drive gate signals S1 and S4 in the primary bridge and (0the {D}_{1}the 1)while D2 is the ratio between the drive gate signals S5 and S6 in the secondary bridge and (0the {D}_{2}the 1).

### Modes of operation of the DAB converter under the proposed phase shift modulation technique

To simplify the analysis of the two-way DAB converter, the device was considered under steady-state conditions. The converter can be modeled as follows (Fig. 1b): the value of the voltage of the secondary bridge is related to that of the primary. And ({mathrm{V}}_{1}={mathrm{knV}}_{2}) and ({mathrm{V}}_{1}>{mathrm{nV}}_{2}), where k is the voltage ratio and n is the transformation ratio of the transformer. In order to simplify the performance analysis of the proposed converter, the following assumptions are made.

1. 1.

All power devices are ideal. On-resistance and parasitic capacitances of power switches are ignored and forward voltage drops of diodes are neglected.

2. 2.

The torque transformer leakage inductors are much smaller than the magnetizing inductors, and hence they are neglected.

As shown in Fig. 2, the converter switching cycle can be divided into 6 working modes as follows:

1. (1)

Mode 1 (t0 −t1)

As shown in Figure 3a, the inductor current ({mathrm{i}}_{mathrm{L}}) is in the negative direction. at t0, them1 and S2 are activated in the primary bridge, and S5 and Sseven are activated in the secondary. According to the direction of the current, the current flows through S2 and D1 in the primary bridge and by S5 and Dseven in secondary school. Vh1 and Vh2 are zero at this time; thus the voltage across L becomes zero and a constant current flows through the inductor at ({mathrm{i}}_{mathrm{L}}={mathrm{i}}_{mathrm{L}0}).

1. (2)

Mode 2 (t1 −t2)

Figure 3b displays the equivalent circuit of mode 2. The current is always in the negative direction. S1S4S5and Sseven are activated. According to the direction of the current, the current passes through D1 and D4 in the primary bridge and by S5 and Dseven in secondary school. Vh1 is fixed at V1 while Vh2 is always zero; hence the voltage across L is fixed at V1. In this mode, the current decreases linearly and can be expressed as follows:

$${i}_{L}left(tright)={i}_{L1}+frac{kn{V}_{2}}{L}left(t-{t}_{ 1}right).$$

(1)

1. (3)

Mode 3 (t2 −t3)

Figure 3c shows the equivalent circuit of mode 3. The current polarity changes from negative to positive. In this mode, S1 and S4 are always on and S5 and S6 are activated. According to the direction of the current, the current flows through S1 and S4 in the primary bridge and through D5 and D6 in secondary school. Vh1 is always at V1 while Vh2 is fixed at nV2. Therefore, the voltage across L is fixed at ({mathrm{V}}_{1}-{mathrm{nV}}_{2}). The current in this mode increases linearly and can be expressed as:

$${i}_{L}left(tright)={i}_{L2}+frac{{V}_{1}-n{V}_{2}}{L}left (t-{t}_{2}right).$$

(2)

1. (4)

Mode 4 (t3 −t4)

Figure 3d illustrates the equivalent circuit of Mode 4. As the waveforms in Figure 2 show, Mode 4 is similar to Mode 1; ({mathrm{i}}_{mathrm{L}}) is in the positive direction. at t3S3 and S4 are activated while S8 and S6 are activated. According to the direction of the current, the current flows through S4 and D3 in the primary bridge and by S8 and D6 in secondary school. Since Vh1 and Vh2 are zero, the voltage across L becomes zero and the current is fixed at ({mathrm{i}}_{mathrm{L}}={mathrm{i}}_{mathrm{L}3}).

1. (5)

Mode 5 (t4 −t5)

Figure 3e displays the equivalent circuit of mode 5. The current is always in the positive direction. S2 and S3 are activated while the switches S6 and S8 are activated. According to the direction of the current, the current passes through D2 and D3 in the primary bridge and by S8 and D6 in secondary school. Vh1 is fixed at −V1 while Vh2 is always zero; thus, the voltage across L is fixed at -V1. The current decreases linearly and can be expressed as follows:

$${i}_{L}left(tright)={i}_{L4}+frac{-kn{V}_{2}}{L}left(t-{t}_ {4}right).$$

(3)

1. (6)

Mode 6 (t5 −t6)

Figure 3f shows the equivalent mode 6 circuit. The current polarity changes from positive to negative. S2 and S3 are always on and the switches Sseven and S8 are activated. According to the direction of the current, the current flows through S2 and S3 in the primary bridge and through Dseven and D8 in secondary school. Vh1 is always at −V1 while Vh2 is fixed at − nV2. Therefore, the voltage across L is fixed at ({-mathrm{V}}_{1}+{mathrm{nV}}_{2}). The current increases linearly and can be expressed as follows:

$${i}_{L}left(tright)={i}_{L5}+frac{{-V}_{1}+n{V}_{2}}{L} left(t-{t}_{5}right).$$

(4)