# A Voltage Multiplier Circuit Based Quadratic Boost Converter for Energy Storage Application
## Proposed Topology
### Working  
**Mode 1**: In this mode, both the switches are ON, and diode D\({}_{3}\) conducts. This interval exists for a time interval of DT where D is the duty ratio, and T is the time period. The equivalent circuit for the first mode of operation is shown in Figure 2. During the first mode of operation, the inductor L\({}_{1}\) is energized by the capacitor C\({}_{1}\), and the charge of the capacitor C\({}_{1}\) decreases during the first mode of operation, as shown in Figure 3. While the DC supply energizes inductor L\({}_{2}\), capacitor C\({}_{2}\) discharges into C\({}_{3}\), and the energy of capacitor C\({}_{4}\) is transferred to the load.  
Figure 1: (**a**) Conventional quadratic boost converter [18]. (**b**) Twice the Quadratic boost converter proposed in [30] (**c**). Proposed converter.  
Figure 2: The first mode of operation.  
The related equations during the first mode of operation are as follows:  
\[V_{L1}=V_{in}+V_{\rm C1} \tag{1}\]  
\[V_{L2}=V_{in} \tag{2}\]  
\[V_{\rm C3}=V_{in}+V_{\rm C2} \tag{3}\]  
**Mode 2:** When both the switches are turned off during the second mode of operation, the diode D\({}_{1}\), D\({}_{2}\), and D\({}_{4}\) are conducting while diode D\({}_{3}\) is reversed biased. During the second mode of operation, the capacitor \(C_{1}\) is charged by the DC power supply, the energy of the inductors is transferred to the capacitor \(C_{2}\) while the capacitor \(C_{3}\) charges \(C_{4}\). The equivalent circuit for the second mode of operation is shown in Figure 4. The related equations are as follows:  
\[V_{L1}=V_{\rm C1}+V_{\rm C3}-V_{O} \tag{4}\]  
\[V_{L2}=V_{in}-V_{\rm C1} \tag{5}\]  
\[V_{\rm C3}=V_{O}-V_{\rm C2} \tag{6}\]  
Figure 4: The second mode of operation.  
Figure 3: Related waveforms of the proposed converter.  
Now applying volt-sec balance in inductor \(L_{2}\)  
\[\int_{0}^{T}V_{L2}(t).\,dt=0 \tag{7}\]  
\[V_{in}\times DT+(V_{in}-V_{\text{C1}})\times(1-D)T=0 \tag{8}\]  
\[V_{\text{C1}}=\frac{V_{in}}{1-D} \tag{9}\]  
Now applying volt-sec balance in inductor \(L_{1}\)  
\[\int_{0}^{T}V_{L1}(t).\,dt=0 \tag{10}\]  
\[(V_{in}+V_{\text{C1}})\times DT+(V_{\text{C1}}+V_{\text{C3}}-V_{O})\times(1-D)T=0 \tag{11}\]  
After combining the above equations, the voltage gain (\(M\)) of the converter can be written as:  
\[M=\frac{V_{O}}{V_{in}}=\frac{\left(3-D^{2}\right)}{\left(1-D\right)^{2}} \tag{12}\]  
The voltage across the capacitors are as follows:  
\[V_{\text{C1}}=\frac{V_{in}}{1-D}=\frac{V_{O}(1-D)}{(3-D^{2})} \tag{13}\]  
\[V_{\text{C2}}=\frac{V_{in}\Big{(}1+D-2D^{2}\Big{)}}{\left(1-D\right)^{2}}= \frac{V_{O}\Big{(}1+D-2D^{2}\Big{)}}{\left(3-D^{2}\right)} \tag{14}\]  
\[V_{\text{C3}}=\frac{V_{in}\left(2-D\right)}{\left(1-D\right)^{2}}=\frac{V_{O }(2-D)}{\left(3-D^{2}\right)} \tag{15}\]  
For the continuous mode of operation, the value of the inductors and capacitors should be selected as follows  
\[L_{1}\geq\frac{R(1-D)^{2}(2-D)D}{4f_{5}(3-D^{2})} \tag{16}\]  
\[L_{2}\geq\frac{R(1-D)^{4}D}{4f_{5}(3-D^{2})} \tag{17}\]  
\[\begin{array}{l}C_{1}=\frac{2V_{O}D}{R(1-D)f_{5}AV_{\text{C1}}}=\frac{2V_{in }\left(3-D^{2}\right)D}{R(1-D)^{3}f_{5}AV_{\text{C1}}}\\ C_{2}=\frac{V_{O}}{Rf_{5}AV_{\text{C2}}}=\frac{V_{in}\left(3-D^{2}\right)}{R(1 -D)^{2}f_{5}AV_{\text{C2}}}\\ C_{3}=\frac{V_{O}}{Rf_{5}AV_{\text{C3}}}=\frac{V_{in}\left(3-D^{2}\right)}{R(1 -D)^{2}f_{5}AV_{\text{C3}}}\\ C_{4}=\frac{V_{O}D}{Rf_{5}AV_{\text{C4}}}=\frac{V_{in}\left(3-D^{2}\right)D}{R( 1-D)^{2}f_{5}AV_{\text{C4}}}\end{array} \tag{18}\]