# Active Clamp Boost Converter with Blanking Time Tuning Considered
## 2 Methodology
### Operation Principles
#### 2.2.1 Circuit Behavior
\tag{5}\]  
Substituting Equation (5) into Equation (3) yields  
\[v_{ds2}(t)\cong V_{c}-I_{in}Z_{1}\omega_{1}(t-t_{1}) \tag{6}\]  
According to Equation (6) and \(v_{ds}(t_{2})=0\), the corresponding time elapsed \(T_{2}\) is  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]  
Stage 3: (\(t_{2}\leq t\leq t_{3}\))  
As illustrated in Figures 3 and 6, \(S_{1}\) and \(S_{2}\) are both still OFF. In the previous stage, \(v_{ds1}(t_{2})=V_{c}\) and \(v_{ds2}(t_{2})=0\). During this stage, \(i_{Lr}\) is smaller than \(I_{in}\). Hence, \(D_{S2}\) is conducted, making \(v_{ds1}\) still clamped at \(V_{c}\). At the same time, the voltage across \(L_{r}\) is \(V_{c}-V_{o}\), causing \(i_{Lr}\) to increase linearly. As soon as \(S_{2}\) is conducted, the operation goes to stage 4. According to Figure 6b, one stage equation can be obtained as  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]  
As the initial value of this stage is \(i_{Lr}(t_{2})=I_{Lr2}\), solving Equation (8) yields  
\[i_{Lr}(t)=\frac{(V_{c}-V_{o})}{L_{r}}(t-t_{2})+I_{Lr2} \tag{9}\]  
Since \(i_{Lr}(t_{3})=I_{Lr3}\), the corresponding time elapsed \(T_{3}\) is  
\[T_{3}=t_{3}-t_{2}=\frac{(I_{Lr3}-I_{Lr2})}{V_{c}-V_{o}}L_{r} \tag{10}\]  
Stage 4: (\(t_{3}\leq t\leq t_{4}\))  
As illustrated in Figures 3 and 7, \(S_{1}\) is still OFF, but \(S_{2}\) is conducted. In the previous stage, \(v_{ds2}(t_{3})=0\). Therefore, \(S_{2}\) is conducted at \(t_{3}\) with ZVS. As \(I_{in}\) is still smaller than \(i_{Lr}\), the voltage across \(L_{r}\) is still \(V_{c}-V_{o}\), causing \(L_{r}\) to still be linearly magnetized. Once \(i_{Lr}=I_{in}\), the operation proceeds to stage 5. One stage equation can be obtained as shown in Figure 7b. \[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{11}\]  
As the initial value of this stage is \(i_{Lr}(t_{3})=I_{Lr3}\), solving Equation (11) yields  
\[i_{Lr}(t)=\frac{(V_{c}-V_{o})}{L_{r}}(t-t_{3})+I_{Lr3} \tag{12}\]  
Because \(i_{Lr}(t_{4})=I_{in}\), the corresponding time elapsed \(T_{4}\) is  
\[T_{4}=t_{4}-t_{3}=\frac{(I_{in}-I_{Lr3})}{V_{c}-V_{o}}L_{r} \tag{13}\]  
Figure 6: (**a**) Current path for stage 3; (**b**) equivalent of (**a**). Stage 5: (\(t_{4}\leq t\leq t_{5}\))  
As illustrated in Figures 3 and 8, \(S_{1}\) is still OFF, but \(S_{2}\) is still ON. Hence, the voltage across \(S_{1}\) is clamped at \(V_{c}\). During this stage, \(i_{Lr}\) is larger than \(I_{in}\), changing \(i_{ds,2}\) to the positive direction. At the same time, the voltage across \(L_{r}\) is still \(V_{c}-V_{o}\), causing \(L_{r}\) to still be linearly magnetized. As soon as \(S_{2}\) is cut off, the operation goes to stage 6. One stage equation can be obtained as shown in Figure 8b. \[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{14}\]  
As the initial value of this stage is \(i_{Lr}(t_{4})=I_{in}\), solving Equation (14) yields  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]0  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]1 (**a**) Current path for stage 5; (**b**) equivalent of (**a**). \[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]2 (**a**) Current path for stage 4; (**b**) equivalent of (**a**). Because \(i_{Lr}(t_{5})=I_{Lr5}\), the corresponding time elapsed \(T_{5}\) is  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]3  
Stage 6: (\(t_{5}\leq t\leq t_{6}\))  
As illustrated in Figures 3 and 9, \(S_{1}\) is still OFF and \(S_{2}\) is cut off. During this stage, \(i_{Lr}\) is still large than \(I_{in}\), hence \(C_{s2}\) is abruptly charged to \(V_{c}\) and \(C_{s1}\) is abruptly discharged to zero. At the same time, \(L_{r}\) begins to be demagnetized. The moment \(C_{s1}\) is discharged to zero, this stage comes to an end and the next stage begins. Three stage equations can be obtained as shown in Figure 9b.