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 Minimizing Current Constraints for an Isolated Active DualBridge DCDC Converter
Minimizing Current Constraints for an Isolated Active DualBridge DCDC Converter
Proposed phase shift command
Figure 1a illustrates the circuit diagram of the DAB converter. In the proposed technique, a phase shift angle (D_{1}J_{s}) is between S_{1} and S_{4}. A threelevel voltage is synthesized on the primary side V_{h1} of the converter, unlike the traditional twolevel voltage adopted in the conventional SPS control technique. This threelevel voltage helps reduce the reverse reverse power to V_{1}. Another phase angle (D_{2}J_{s}) is done in the second bridge between S_{5} and S_{6}; 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 D_{2} the phase shift widens the transmission power regulation range, increasing regulation flexibility. Therefore, D_{1} is the phase shift ratio between the drive gate signals S_{1} and S_{4} in the primary bridge and (0the {D}_{1}the 1)while D_{2} is the ratio between the drive gate signals S_{5} and S_{6} 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 twoway DAB converter, the device was considered under steadystate 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.
All power devices are ideal. Onresistance and parasitic capacitances of power switches are ignored and forward voltage drops of diodes are neglected.

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)
Mode 1 (t_{0} −t_{1})
As shown in Figure 3a, the inductor current ({mathrm{i}}_{mathrm{L}}) is in the negative direction. at t_{0}, them_{1} and S_{2} are activated in the primary bridge, and S_{5} and S_{seven} are activated in the secondary. According to the direction of the current, the current flows through S_{2} and D_{1} in the primary bridge and by S_{5} and D_{seven} in secondary school. V_{h1} and V_{h2} 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}).

(2)
Mode 2 (t_{1} −t_{2})
Figure 3b displays the equivalent circuit of mode 2. The current is always in the negative direction. S_{1}S_{4}S_{5}and S_{seven} are activated. According to the direction of the current, the current passes through D_{1} and D_{4} in the primary bridge and by S_{5} and D_{seven} in secondary school. V_{h1} is fixed at V_{1} while V_{h2} is always zero; hence the voltage across L is fixed at V_{1}. 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)

(3)
Mode 3 (t_{2} −t_{3})
Figure 3c shows the equivalent circuit of mode 3. The current polarity changes from negative to positive. In this mode, S_{1} and S_{4} are always on and S_{5} and S_{6} are activated. According to the direction of the current, the current flows through S_{1} and S_{4} in the primary bridge and through D_{5} and D_{6} in secondary school. V_{h1} is always at V_{1} while V_{h2} is fixed at nV_{2}. 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)

(4)
Mode 4 (t_{3} −t_{4})
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 t_{3}S_{3} and S_{4} are activated while S_{8} and S_{6} are activated. According to the direction of the current, the current flows through S_{4} and D_{3} in the primary bridge and by S_{8} and D_{6} in secondary school. Since V_{h1} and V_{h2} are zero, the voltage across L becomes zero and the current is fixed at ({mathrm{i}}_{mathrm{L}}={mathrm{i}}_{mathrm{L}3}).

(5)
Mode 5 (t_{4} −t_{5})
Figure 3e displays the equivalent circuit of mode 5. The current is always in the positive direction. S_{2} and S_{3} are activated while the switches S_{6} and S_{8} are activated. According to the direction of the current, the current passes through D_{2} and D_{3} in the primary bridge and by S_{8} and D_{6} in secondary school. V_{h1} is fixed at −V_{1} while V_{h2} is always zero; thus, the voltage across L is fixed at V_{1}. 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)

(6)
Mode 6 (t_{5} −t_{6})
Figure 3f shows the equivalent mode 6 circuit. The current polarity changes from positive to negative. S_{2} and S_{3} are always on and the switches S_{seven} and S_{8} are activated. According to the direction of the current, the current flows through S_{2} and S_{3} in the primary bridge and through D_{seven} and D_{8} in secondary school. V_{h1} is always at −V_{1} while V_{h2} is fixed at − nV_{2}. 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)