## 3 Mathematical Equations for Parameters Design
### Buck-Boost Converter  
Based on the previous assumptions, \(i_{Lm}\) can be regarded as a constant value \(I_{Lm}\) when \(L_{m}\) is sufficiently large. During Mode 1 and Mode 2, \(i_{L11}\) rises and \(i_{L12}\) decreases, and it is known that the increase in \(i_{L11}\) is equal to the decrease in \(i_{L12}\) by (1). Mode 3 describes the process in which the parasitic capacitance of \(S_{2}\) is discharged from \(V_{B}+V_{o}\) to zero volts. Because \(C_{S2}\) is very small and the drop in \(i_{L12}\) is also very small, therefore, at the end of Mode 3, \(i_{L12}\) is negative but almost equal to zero. Thus, it is known that the peaks of \(i_{L11}\) and \(i_{L12}\) are very close to \(I_{Lm}\).  
\[i_{L11,peak}=i_{L12,peak}\approx I_{Lm}, \tag{31}\]Operation Mode 2 ends when \(i_{LI2}\) decreases to zero. Using (13) and (31), the time required for \(i_{LI2}\) to decrease from the peak value to zero is equal to  
\[t_{f}=t_{2}-t_{0}\approx\frac{I_{Lm}L_{l}(L_{l}+2L_{m})}{L_{m}V_{B}+(L_{m}+L_{l} )V_{o}}, \tag{32}\]  
During the period from Mode 1 to Mode 2, \(v_{T}\) is negative and \(i_{Lm}\) decreases. On the contrary, during the period from Mode 3 to Mode 5, \(v_{T}\) is positive and \(i_{Lm}\) increases. Applying the volt-second balance law of inductance, the average value of \(v_{T}\) is equal to zero. From (11) and (25), we can obtain  
\[\overline{v_{T}}=\frac{L_{m}(V_{B}-V_{o})}{L_{l}+2L_{m}}\cdot t_{f}+\frac{2L_ {m}V_{B}}{L_{l}+2L_{m}}\cdot t_{r}=0, \tag{33}\]  
where \(t_{r}\) represents the rising time of \(i_{Lm}\). Ignoring the short charging and discharging time of \(C_{S1}\) (\(C_{S2}\)), we can obtain  
\[t_{f}+t_{r}\approx 0.5T_{s}, \tag{34}\]  
Combining (33) and (34), it can be deduced that \(t_{f}\) and \(t_{r}\) are, respectively, expressed as  
\[t_{f}=\frac{V_{B}}{V_{B}+V_{o}}T_{s}, \tag{35}\]  
\[t_{r}=\bigg{(}0.5-\frac{V_{B}}{V_{B}+V_{o}}\bigg{)}T_{s}, \tag{36}\]  
Because both \(t_{r}\) and \(t_{f}\) must be positive values, it can be seen from (36) that the design constraint for realizing the converter proposed in this study is that  
\[V_{o}>V_{B}, \tag{37}\]  
Substituting (35) into (32) yields  
\[I_{Lm}=\frac{V_{B}(L_{m}V_{B}+(L_{m}+L_{l})V_{o})}{L_{l}(L_{l}+2L_{m})(V_{B}+ V_{o})}T_{s}, \tag{38}\]  
From the analysis of the operation modes, it can be seen that the output power comes from two paths. One is that \(L_{p1}\) (\(L_{p2}\)) directly provides energy to the output terminal through \(D_{3}\) (\(D_{4}\)). The other is that \(L_{p1}\) (\(L_{p2}\)) first charges \(C_{B1}\) (\(C_{B2}\)) through \(D_{3}\) (\(D_{4}\)), and then \(C_{B1}\) (\(C_{B2}\)) provides energy to the output terminal. The power supplied to the output terminal by each of these two paths is equal to  
\[P_{1}=\frac{P_{o}V_{o}}{(V_{B}+V_{o})}, \tag{39}\]  
\[P_{2}=\frac{P_{o}V_{B}}{(V_{B}+V_{o})}, \tag{40}\]  
For a buck-boost converter, when the active switch is turned off, the inductor current flows through the flywheel diode and releases energy to the output. Combining (35) and (38) yields that the average of the descending portions of \(i_{L1}\) and \(i_{L2}\) added together is equal to  
\[\overline{i_{L1,\,fall}+\overline{i_{LI2,\,fall}}}=2\times\frac{t_{f}I_{Lm}}{2 T_{s}}=\frac{V_{B}^{2}(L_{m}V_{B}+(L_{m}+L_{l})V_{o})}{L_{l}(L_{l}+2L_{m})(V_{B}+ V_{o})^{2}f_{s}}, \tag{41}\]Equation (41) represents the average current delivered to the output by the buck-boost converters. Combining (40) and (41) yields  
\[\frac{V_{B}(L_{m}V_{B}+(L_{m}+L_{l})V_{o})}{L_{l}(L_{l}+2L_{m})(V_{B}+V_{o})f_{s} }=\frac{P_{o}}{V_{o}}, \tag{42}\]  
The coupling coefficient of the coupled inductor is defined as  
\[k_{c}\triangleq\frac{L_{m}}{L_{l}+L_{m}}, \tag{43}\]  
Combining (42) and (43) yields  
\[L_{l}=\frac{V_{B}V_{o}(k_{c}V_{B}+V_{o})}{(1+k_{c})(V_{B}+V_{o})P_{o}f_{s}}, \tag{44}\]