# Active Clamp Boost Converter with Blanking Time Tuning Considered
## 3 Design Considerations
### Design of Active Clamp Capacitor C\({}_{c}\)  
Regarding the design of the active clamp capacitor \(C_{c}\), it can be designed based on the required voltage ripple \(\Delta v_{c}\). From stages 2, 3, 4, 5, and 6 and Figure 3, it can be seen that \(C_{c}\) has the behavior of slight charge and discharge. Therefore, the product of \(C_{c}\) and \(\Delta v_{c}\) can be expressed as  
\[C_{c}\Delta v_{c}=\Delta Q_{c}=\frac{1}{2}(i_{d82,max})(T_{5}+T_{6}) \tag{46}\]  
where \(T_{5}\) and \(T_{6}\) are the times elapsed for stages 5 and 6, respectively. Because individual switch parasitic capacitors in stages 2 and 6 have charge and discharge, the corresponding times elapsed are so short. Therefore, if two switches are identical, then  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]  
Accordingly, based on Equation (16), \(T_{5}\) plus \(T_{6}\) can be represented by  
\[T_{5}+T_{6}=\frac{(I_{L5}-I_{in})}{V_{c}-V_{o}}L_{r}+0.02\ \mu \tag{48}\]  
In order to find the value of \(C_{c}\), the currents \(i_{d82,max}\) and \(I_{L75}\) should be figured out first. From stage 5 and Figure 8, the value of \(i_{d81,max}\) can be obtained to be  
\[i_{d82}(t_{5})=i_{d82,max}=i_{L7}(t_{5})-I_{in}=I_{L75}-I_{in} \tag{49}\]  
From Equation (49), the value of \(I_{L75}\) should be figured out first, via stages 2 and 6 in Section 2.2.1. Based on Equations (7), (21) and (47), the following equation can be obtained to be  
\[\frac{1}{\omega_{1}}[\frac{V_{c}}{(I_{L75}-I_{in})Z_{1}}]\cong\frac{1}{\omega _{1}}(\frac{V_{c}}{I_{in}Z_{1}}) \tag{50}\]  
Rearranging Equation (50) yields  
\[I_{L75}\cong 2I_{in} \tag{51}\]  
Finally, substituting Equation (51) into Equation (49) yields  
\[i_{d82,max}\cong 2I_{in}-I_{in}=I_{in} \tag{52}\]  
As \(I_{in}=I_{o,rated}/(1\text{-}D)\), \(I_{in}\) can be obtained to be 4.165A from Table 2, if the voltage ripple \(\Delta v_{c}\) has 5% of \(V_{c}\), then, substituting Equations (48), (51), and (52) into Equation (46), the value of \(C_{c}\) can be expressed to be  
\[C_{c}=\frac{86.94\ \mu+0.04\ \mu V_{c}}{0.05\ V_{c}(V_{c}-42)} \tag{53}\]  
To solve the value of \(C_{c}\), the value of \(V_{c}\) should be figured out first. From Figure 3, the following equation can be obtained:  
\[\sum_{n=2}^{7}T_{n}=(1-D)T_{s} \tag{54}\]Sequentially, from stage 2 in Section 2.2.1, the following equation can be obtained:  
\[I_{in}=i_{ds1}(t_{2})+i_{Lr}(t_{2})-i_{ds2}(t_{2}) \tag{55}\]  
From Section 2.2.1, it can be seen that \(i_{ds1}(t_{2})=0\) can be found and \(i_{Lr}(t_{2})\cong 0.22\) A can be found based on Equation (3) and \(T_{2}\cong 0.02\) us. Substituting \(i_{ds1}(t_{2})=0\) and \(i_{Lr}(t_{2})\cong 0.22\) A into Equation (55) yields  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]0  
In addition, from stages 3 to 5, the voltage across \(L_{r}\) is \(V_{c}-V_{o}\), thereby causing \(L_{r}\) to be linearly magnetized with \(i_{Lr}(t)=I_{in}+i_{ds2}(t)\), hence \(i_{ds2}(t)\) has the same slope from stages 3 to 5. Accordingly, based on the ampere-second balance, the following equation can be obtained:  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]1  
From Equations (33) and (47), the following equation can be obtained:  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]2  
Based on Equations (48), (54), (57), and (58), the following equation can be obtained:  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]3  
Based on Table 2 and the obtained values, Equation (59) can be rewritten to be  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]4  
Solving Equation (60) yields  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]5  
If \(\alpha=0.568\), this means that the auxiliary switch has only turn-on time of \(0.02\ \mu\)s. By considering the rising and falling times, \(\alpha\) is chosen to be \(0.19\). Substituting \(\alpha\) into Equation (32) yields  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]6  
Substituting Equation (62) into Equation (53) yields  
\[T_{6}\cong T_{2}=0.02\ \mu \tag{47}\]7  
Solving Equation (63) yields \(C_{s}\cong 1.33\ \mu\)F. Eventually, the value of \(C_{s}\) is selected as \(2.2\ \mu\)F.