# Active Clamp Boost Converter with Blanking Time Tuning Considered
## 2 Methodology
### Operation Principles
#### 2.2.1 Circuit Behavior
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]4  
According to Equation (20) and \(v_{ds2}(t_{6})=V_{c}\), the corresponding time elapsed \(T_{6}\) is  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]5  
Stage 7: (\(t_{6}\leq t\leq t_{7}\))  
As illustrated in Figures 3 and 10, \(S_{1}\) and \(S_{2}\) are still OFF, In the previous stage, \(v_{ds1}\) is equal to zero, whereas \(v_{ds2}\) is equal to \(V_{c}\). During this stage, \(i_{Lr}\) is still larger than \(I_{in}\), making \(D_{s1}\) conducted, hence \(v_{ds2}\) still clamped at \(V_{c}\). At the same time, the voltage across \(L_{r}\) is \(-\)\(V_{o}\), causing \(L_{r}\) to be linearly demagnetized. As soon as \(S_{1}\) is conducted, this stage comes to an end and the next stage begins. According to Figure 10b, one stage equation can be obtained as  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]6  
As the initial value of this stage is \(i_{Lr}(t_{6})=I_{Lr6}\), solving Equation (22) yields  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]7  
Because \(i_{Lr}(t_{7})=I_{Lr7}\), the corresponding time elapsed \(T_{7}\) is  
\[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]8  
Stage 8: (\(t_{7}\leq t\leq t_{8}\))  
As illustrated in Figures 3 and 11, \(S_{1}\) is conducted and \(S_{2}\) is still OFF. In the previous stage, \(D_{s1}\) is conducted so \(S_{1}\) is switched on with ZVS at \(t_{7}\). As \(i_{Lr}\) is still larger than \(I_{in}\), the voltage across \(L_{r}\) is still \(-\)\(V_{o}\), causing \(L_{r}\) to be still linearly demagnetized. As \(i_{Lr}=I_{in}\), this stage ends and the next stage starts. \[T_{2}=t_{2}-t_{1}=\frac{1}{\omega_{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{7}\]9 (**a**) Current path for stage 7; (**b**) equivalent of (**a**). According to Figure 11b, one stage equation can be obtained as  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]0  
As the initial value of this stage is \(i_{Lr}(t_{7})=I_{L7}\), solving Equation (25) yields  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]1  
Because \(i_{Lr}(t_{8})=I_{in}\), the corresponding time elapsed \(T_{8}\) is  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]2  
Stage 9: (\(t_{8}\leq t\leq t_{0}+T_{s}\))  
As illustrated in Figures 3 and 12, \(S_{1}\) is still ON but \(S_{2}\) is still OFF. Hence, \(v_{ds2}\) is clamped at \(V_{c}\). During this stage, \(i_{Lr}\) is smaller than \(I_{in}\), changing \(i_{ds1}\) to the positive direction. At the same time, the voltage across \(L_{r}\) is \(-V_{o}\), causing \(L_{r}\) to still be linearly demagnetized. From Figure 3, it can be seen that, as \(i_{Lr}=0\), the output diode \(D_{o}\) is cut off before the turn-off time point of \(v_{gs1}\), meaning that \(D_{o}\) has ZCS turn-off. Once \(i_{Lr}=0\), this stage ends with the next cycle repeated. According to Figure 12b, one stage equation can be obtained as  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]3  
As the initial value of this stage is \(i_{Lr}(t_{8})=I_{in}\), solving Equation (28) yields  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]4  
Because \(i_{Lr}(t_{0}+T_{s})=0\), the corresponding time elapsed \(T_{9}\) is  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]5  
\[L_{r}\frac{di_{Lr}(t)}{dt}=V_{c}-V_{o} \tag{8}\]6 (**a**) Current path for stage 8; (**b**) equivalent of (**a**). If \(T_{8}\) plus \(T_{9}\), which is about \(T_{9}\), is smaller than \(t_{on}\), i.e., \(DT_{s}\), the resonant inductor \(L_{r}\) works in DCM, meaning that there is no reverse recovery current during the turn-off period of \(D_{o}\). Because \(v_{gs1}\) and \(v_{gs2}\) have two blanking times between them per cycle, if \(T_{9}\) is larger than \(t_{on}\), then at the instant when \(S_{1}\) is cut off, but \(S_{2}\) is still OFF, \(i_{L}\), will flow through \(D_{s1}\). Once \(S_{2}\) is turned on, a large reverse recovery current will flow through \(S_{2}\) owing to a reverse voltage of \(V_{c}\) across \(D_{s1}\).