# A High Gain Modified Quadratic Boost DC-DC Converter with Voltage Stress Half of Output Voltage
## 2 Evolution of the Proposed Modified Quadratic Boost Converter
### Proposed Modified Quadratic Boost Converter with Integrated Dual Voltage Lift Cell
#### 2.4.1 Operation Principle and Ideal Voltage Gain Calculation, Continuous Current Mode (CCM)  
Mode 1 [0, DT]: Switch ON, D1, D2, D3 ON  
When the switch was turned on by the PWM signal, the drain terminal was connected to the ground internally; as a result, the energy transfer process occurred from the source to the passive component and from the passive component to other passive components (i.e., capacitor to inductor). The corresponding circuit configuration is shown in Figure 6.  
During this mode of operation, the diodes D1, D2, and D3 will be forward biased and will carry the necessary current to transfer the energy. The diodes D4, D5, and D6 will be reverse biased during the first mode of operation. The inductor \(L_{1}\) and capacitor C1 are charged by the source voltage (i.e., \(V_{in}\)) through diode D2, where the charging current of the capacitor C1 is supplied by the diode D1. In the second integrated voltage lift cell, the capacitor C3 transfers the energy directly into the inductor L2, and the capacitor C2 is  
Figure 4: Circuit configuration of dual voltage lift proposed at [37].  
Figure 5: Circuit diagram of the Proposed converter.  
charged by the summation of the voltage source supplied by the capacitors C\({}_{3}\) and C\({}_{4}\). On the other hand, the output capacitor C\({}_{0}\) transfer the necessary load current to the load.  
The characteristics equations during the first mode of operation are as follows.  
\[V_{L1}=V_{in}=L_{1}\frac{di_{L1}}{dt} \tag{1}\]  
\[V_{\text{C1}}=V_{in} \tag{2}\]  
\[V_{L2}=V_{\text{C3}}=L_{2}\frac{di_{L2}}{dt} \tag{3}\]  
\[V_{\text{C2}}=V_{\text{C3}}+V_{\text{C4}} \tag{4}\]  
\[V_{\text{C0}}=-V_{0} \tag{5}\]  
Mode 2 [DT, T]: Switch OFF, D\({}_{4}\), D\({}_{5}\), D\({}_{6}\) ON  
During this interval period, the switch was turned off, ideally with the withdrawal of the PWM signal. The corresponding circuit configuration during mode 2 is shown in Figure 7. In this mode of operation, the energy is directly transferred from the source to the load through the passive component, and it transfers the energy to the other passive component (i.e., C\({}_{3}\), C\({}_{4}\)). The diodes D\({}_{1}\), D\({}_{2}\), and D\({}_{3}\) will be in reverse bias; on the other hand, the D\({}_{5}\) and D\({}_{6}\) will carry the necessary current for the load. The charging current for capacitor C\({}_{4}\) is provided by the inductor \(L_{2}\) through diode D\({}_{4}\).  
The characteristics equation during this mode is derived in the following section.  
\[V_{L1}=V_{in}+V_{\text{C1}}-V_{\text{C3}} \tag{6}\]  
Figure 6: Circuit configuration during the first mode of operation.  
Figure 7: Circuit configurations for the second mode of operation.  
\[V_{L1}=2V_{in}-V_{C3} \tag{7}\]  
\[V_{C4}=-V_{L2} \tag{8}\]  
\[V_{L2}=V_{C3}+V_{C2}-V_{0} \tag{9}\]  
Putting the Equation from (8) and (9) into the Equation (4)  
\[V_{C2}=V_{C3}+(-V_{C3}-V_{C2}+V_{0}) \tag{10}\]  
\[2V_{C2}=V_{0} \tag{11}\]  
\[V_{C2}=\frac{V_{0}}{2} \tag{12}\]  
To obtain the voltage gain, the volt-second method is applied on both Inductor \(L_{1}\) and \(L_{2}\). The equations are as follows.  
\[\int_{0}^{T}V_{L1}(t)\cdot dt=0 \tag{13}\]  
\[V_{in}\times dT+(2V_{in}-V_{C3})\times(1-d)T=0 \tag{14}\]  
\[V_{C3}=\frac{(2-d)}{(1-d)} \tag{15}\]  
\[\int_{0}^{T}V_{L2}(t)\cdot dt=0 \tag{16}\]  
\[V_{C3}\times dT+(V_{C3}+V_{C2}-V_{0})\times(1-d)T=0 \tag{17}\]  
Putting the value of \(V_{C2}\) from Equations (12)-(17) the below equation is obtained.  
\[V_{C3}\times dT+\left(V_{C3}-\frac{V_{0}}{2}\right)\times(1-d)T=0 \tag{18}\]  
\[V_{0}=\frac{2}{(1-d)}\times V_{C3} \tag{19}\]  
Putting the value of \(V_{C3}\) from Equations (15)-(20) the ideal voltage gain of the proposed converter is calculated as follows.  
\[V_{0}=\frac{2(2-d)}{(1-d)^{2}}\times V_{in} \tag{20}\]  
\[Volta\ gain\ (M)=\frac{V_{0}}{V_{in}}=\frac{2(2-d)}{(1-d)^{2}} \tag{21}\]