## 1 Introduction
### HG-SIQC--Mode II: Switch 'S' in OFF State  
During this switching interval, the switch 'S' is kept in the OFF state, as shown in Figure 6.  
The inductors connected at the input side discharge and transfer energy to capacitor \(C_{2}\). Meanwhile, capacitor \(C_{3}\) is discharged and supplies output capacitor \(C_{0}\). The voltage equations of the proposed converter during the OFF state of the switch can be obtained by applying the KVL to Figure 6.  
\[ \begin{cases}V_{in}-V_{L_{2}}+V_{C_{3}}-V_{C_{0}}=0\\ V_{in}-V_{L_{1}}+V_{C_{1}}-V_{C_{2}}=0\\ V_{in}-V_{L_{2}}-V_{L_{3}}-V_{C_{2}}=0\end{cases} . \tag{4}\]  
From Equation (4), the voltage across the inductors \(L_{1}\), \(L_{2}\), and \(L_{3}\) can be written as  
\[ \begin{cases}L_{1}\frac{dt_{1}}{dt}=V_{in}+V_{C_{1}}-V_{C_{2}}\\ L_{2}\frac{dt_{2}}{dt}=V_{in}+V_{C_{3}}-V_{0}\\ L_{3}\frac{dt_{3}}{dt}=-V_{C_{1}}\end{cases} , \tag{5}\]  
where \(V_{in}\) is the input voltage, \(V_{0}\) is the output voltage, and \(V_{C1}\) and \(V_{C3}\) are the voltages across the capacitors \(C_{1}\) and \(C_{3}\), respectively. Similarly, KCL is applied to the circuit shown in Figure 6 to yield the current equations. \[ \begin{cases}I_{C_{1}}=-\left[I_{C_{2}}-I_{L_{3}}\right]\\ I_{C_{2}}=I_{C_{1}}+I_{L_{3}}\\ I_{C_{3}}=I_{C_{0}}=I_{0}\end{cases} . \tag{6}\]  
Figure 6: HG-SIQBC—equivalent circuit for Mode II. (The conducting portion of the circuit is shown in Yellow color with the current flow in dotted lines). The total volt-seconds applied to the inductor \(L_{3}\) over one switching period are as follows [12]:  
\[\int\limits_{0}^{T}V_{L_{3}}(t)dt=(V_{\text{C}_{2}}-V_{\text{C}_{1}})DT+(-V_{ \text{C}_{1}})(1-D)T, \tag{7}\]  
where \(D\) is the duty cycle, and \(T\) is the switching period. The following equation can be obtained by equating Equation (7) to zero:  
\[(V_{\text{C}_{2}}D)-(V_{\text{C}_{1}}D)-V_{\text{C}_{1}}+(V_{\text{C}_{1}}D)=0. \tag{8}\]  
Therefore,  
\[V_{\text{C}_{2}}=\frac{V_{\text{C}_{1}}}{D}. \tag{9}\]  
The total volt-seconds applied to inductor \(L_{1}\) over one switching period are calculated as  
\[\int\limits_{0}^{T}V_{L_{1}}(t)dt=\bigg{(}\frac{V_{in}}{2}\bigg{)}DT+\big{(}V _{in}+V_{\text{C}_{1}}-V_{\text{C}_{2}}\big{)}(1-D)T. \tag{10}\]  
By equating Equation (10) to zero, the following equations can be obtained:  
Substituting \(V_{\text{C}_{2}}=\frac{V_{\text{C}_{1}}}{D}\),  
\[\bigg{(}\frac{V_{in}}{2}\bigg{)}D+\bigg{(}V_{in}+V_{\text{C}_{1}}-\frac{V_{ \text{C}_{1}}}{D}\bigg{)}(1-D)=0, \tag{11}\]  
\[\bigg{(}\frac{V_{in}D+2V_{in}-2V_{in}D}{2}\bigg{)}+\bigg{(}\frac{2V_{\text{C }_{1}}D-V_{\text{C}_{1}}D^{2}-V_{\text{C}_{1}}}{2}\bigg{)}, \tag{12}\]  
\[V_{\text{C}_{1}}\bigg{(}\frac{2D-D^{2}-1}{D}\bigg{)}=\frac{V_{in}(D-2)}{2}. \tag{13}\]  
Therefore,  
\[V_{\text{C}_{1}}=\frac{V_{in}D(D-2)}{2(1-D)^{2}}. \tag{14}\]  
Substituting the value of \(V_{\text{C}1}\) into Equation (9) yields  
\[V_{\text{C}_{2}}=\frac{V_{\text{C}_{1}}}{D}=\frac{V_{in}\left(D-2\right)}{2( 1-D)^{2}}. \tag{15}\]  
Similarly, \(V_{\text{C}3}\) is calculated using the following equation:  
\[\int\limits_{0}^{T}V_{L_{3}}(t)dt=(V_{\text{C}_{2}}-V_{\text{C}_{1}})DT+(-V_{ \text{C}_{1}})(1-D)T, \tag{7}\]0  
Subsequently, the voltage gain of the proposed converter can be obtained by applying the volt-sec principle to inductor \(L_{2}\). \[\int\limits_{0}^{T}V_{L_{3}}(t)dt=(V_{\text{C}_{2}}-V_{\text{C}_{1}})DT+(-V_{ \text{C}_{1}})(1-D)T, \tag{7}\]1  
By equating Equation (17) to zero, the following equations can be obtained:  
\[\int\limits_{0}^{T}V_{L_{3}}(t)dt=(V_{\text{C}_{2}}-V_{\text{C}_{1}})DT+(-V_{ \text{C}_{1}})(1-D)T, \tag{7}\]2\[\int\limits_{0}^{T}V_{L_{3}}(t)dt=(V_{\text{C}_{2}}-V_{\text{C}_{1}})DT+(-V_{ \text{C}_{1}})(1-D)T, \tag{7}\]3  
\[\int\limits_{0}^{T}V_{L_{3}}(t)dt=(V_{\text{C}_{2}}-V_{\text{C}_{1}})DT+(-V_{ \text{C}_{1}})(1-D)T, \tag{7}\]4  
where \(G\) is the voltage gain of the proposed HG-SIQBC converter, and \(D\) is the duty cycle of the converter switch.