3. Analysis of the proposed converter
3.1. Voltage Gain Analysis
According to (5) to (14), by considering that the average voltage of
inductor in one duty cycle in steady state is zero (15), therefore (16)
and (17) are resulted:
\(\int_{t}^{t+T_{s}}{V_{L}d(t)}=0\) (15)
\(L_{1}:\left(V_{\text{in}}-V_{c1}\right)\left(1-D\right)T_{s}+V_{\text{in}}DT_{s}=0\ \ \) (16)
\(L_{3}:V_{\text{in}}\left(1-D\right)T_{s}+\left(V_{\text{in}}-V_{c2}\right)DT_{s}=0\) (17)
So (18) and (19) are resulted from (16) and (17), respectively.
\(V_{c1}=\frac{1}{1-D}V_{\text{in}}\ \) (18)
\(V_{c2}=\frac{1}{D}V_{\text{in}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\) (19)
In the other operation mode of the proposed converter, the capacitors \(C_{1}\) and \(C_{3}\) are parallel, therefore the average voltage of
them is as same as each other.
\(V_{c1}=V_{c3}\) (20)
Output voltage is resulted from sum of capacitors voltages of \(V_{c2}\)and\(\text{\ V}_{c3}\).
\(V_{O}=V_{c2}+V_{c3}\) (21)
From (18), (19), (20) and (21), the voltage gain of the proposed
converter is as (22).
\(M=\frac{V_{O}}{V_{\text{in}}}=\frac{1}{D(1-D)}\) (22)
3.2. Input Current Ripple Analysis
According to Fig. 1, and due to the Kirchhoff’s current law (KCL), the
input current is resulted by adding the input currents of cell1 and cell2 (\(i_{1}\)and\(\ i_{2}\)). Therefore the input current ripple is
resulted from (23).
\(\Delta_{i_{in}}=\Delta_i_1+\Delta_i_2\) (23)
\(\Delta_{{i}_{1}}\ \) is resulted from \(L_{1}\)and \(L_{3}\)current ripple. Also \(\Delta_{{i}_{2}}\) is resulted from \(L_{2}\)and\(\ L_{4}\) current ripple.
Therefore, (24)-(27) are concluded.
\(\Delta_{{i}_{L1}}=\frac{V_{\text{in}}}{L_{1}}\frac{(1-D)}{F_{s}}\) (24)
\(\Delta_{{i}_{L3}}=\frac{V_{\text{in}}}{L_{3}}\ \frac{D}{F_{s}}\) (25)
\(\Delta_{{i}_{L2}}=\frac{V_{\text{in}}}{L_{2}}\frac{(1-D)}{F_{s}}\) (26)
\(\Delta_{{i}_{L4}}=\frac{V_{\text{in}}}{L_{4}}\frac{D}{F_{s}}\) (27)
Therefore:
\(\Delta_{{i}_{L1}}=\Delta_{{i}_{L2}}=\frac{V_{\text{in}}}{F_{s}}(\frac{D}{L_{3}}-\frac{1-D}{L_{1}})\) (28)
In (28), it is considered that \(L_{1}=L_{2}\) and\(L_{3}=L_{4}\).
In (23), If \(\Delta_{i_{in}}=0\), then it is resulted that \(\Delta_{{i}_{1}}\ \)and \(\Delta_{{\ i}_{2}}\ \)is zero simultaneously or \(\Delta_{{i}_{1}}+\Delta_{{i}_{2}}\ \) is zero. Firstly, it’s considered that \(\Delta_{\ {i}_{1}}=\Delta_{{i}_{2}}=0\), 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:
\(\Delta_{{i}_{1}}=\frac{V_{\text{in}}}{F_{s}L_{3}}(\frac{K+1}{K}D-\frac{1}{K})\) (32)
Secondly, for \(\Delta_{{i}_{1}}+\Delta_{{i}_{2}}=0\), (based on Fig. 7), in the time
intervals of\(\ 0\leq t\leq(1-D)T_{s}\), the increasing 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
decreasing 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).
\(\Delta_{{i}_{\text{in}}}=\frac{V_{\text{in}}}{F_{s}L_{3}}(\frac{-2\left(K+1\right)D^{2}+\left(K+3\right)D-1}{\text{KD}})\) (35)
3.3. 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 cell1 , the average current of\(L_{1}\) is lower than \(L_{3}\) and by considering that the cell1 and cell2 are the
same, therefore in this section, \(L_{1}\)is calculated. The
voltage-current equation of inductor \(L_{1}\) is as (36).
\(V_{L1}=L_{1}\frac{\Delta_{{i}_{L1}}}{t}\) (36)