(23)
\({i}_{1}\ \)is resulted from \(L_{1}\)and \(L_{3}\)current ripple. Also\({i}_{2}\) is resulted from \(L_{2}\)and\(\ L_{4}\) current ripple. Therefore, (24)-(27) are concluded.
\({i}_{L1}=\frac{V_{\text{in}}}{L_{1}}\frac{(1-D)}{F_{s}}\)                                                                                     (24)
\({i}_{L3}=\frac{V_{\text{in}}}{L_{3}}\ \frac{D}{F_{s}}\)                                                                                            (25)
\({i}_{L2}=\frac{V_{\text{in}}}{L_{2}}\frac{(1-D)}{F_{s}}\)                                                                                      (26)
\({i}_{L4}=\frac{V_{\text{in}}}{L_{4}}\frac{D}{F_{s}}\)                                                                                              (27)
Therefore:
\({i}_{L1}={i}_{L2}=\frac{V_{\text{in}}}{F_{s}}(\frac{D}{L_{3}}-\frac{1-D}{L_{1}})\)                                                        (28)
In (28), it is considered that \(L_{1}=L_{2}\) and\(L_{3}=L_{4}\).
In (23), If\({\ i}_{\text{in}}=0\), then it is resulted that\({i}_{1}\ \)and\({\ i}_{2}\ \)is zero simultaneously or\({i}_{1}+{i}_{2}\ \) is zero. Firstly, it’s considered that
\(\ {i}_{1}={i}_{2}=0\), therefore:
\(L_{3}=L_{1}\frac{D}{1-D}\) (29)
In the operation point of\(\ D=D_{0}\):
\(L_{1}=L_{3}\frac{1-D_{0}}{D_{0}}\) (30)
By considering\(\ K=\frac{1-D_{0}}{D_{0}}\):
\(L_{1}=KL_{3}\) (31)
Therefore the current ripple of cell1 in the operation point of\(D_{0}\) is calculated as follows:
\({i}_{1}=\frac{V_{\text{in}}}{F_{s}L_{3}}(\frac{K+1}{K}D-\frac{1}{K})\)(32)
Secondly, for\(\ {i}_{1}+{i}_{2}=0\), (based on Fig. 7), in the time intervals of\(\ 0\leq t\leq(1-D)T_{s}\), the increased rate of\(i_{1}\) is:
\(\frac{V_{\text{in}}}{L_{1}}+\ \frac{\left(V_{\text{in}}-V_{c1}\right)}{L_{3}}=\frac{1}{L_{3}}(\frac{V_{\text{in}}\left(1+K\right)}{K}-V_{c1})\)(33)
And in the time intervals of\(\ 0\leq t\leq(1-D)T_{s}\), the decreased rate of\(\text{\ i}_{1}\ \)is:
\(\frac{V_{\text{in}}}{L_{4}}+\frac{V_{\text{in}}-V_{c2}}{L_{2}}=\frac{1}{L_{4}}(\frac{1+K}{K}V_{\text{in}}-\frac{1}{K}V_{c2})\)(34)
So based on (33) and (34), by considering (18), (19) and Fig. 7, the input current ripple is resulted from (35).
\({i}_{\text{in}}=\frac{V_{\text{in}}}{F_{s}L_{3}}(\frac{-2\left(K+1\right)D^{2}+\left(K+3\right)D-1}{\text{KD}})\)(35)
  1. Inductor Sizing Calculation
In this section, the inductors are calculated due to the operation of the proposed converter. The inductors are calculated in the boundary conduction. It is important to note that in , the average current of\(L_{1}\) is lower than \(L_{3}\) and by considering that theand are the same, therefore in this section, \(L_{1}\)is calculated. The voltage-current equation of inductor \(L_{1}\) is as (36).
\(V_{L1}=L_{1}\frac{{i}_{L1}}{t}\) (36)