# Active Clamp Boost Converter with Blanking Time Tuning Considered
## 2 Methodology
### Operation Principles  
#### 2.2.1 Circuit Behavior  
Prior to this section, the symbols and assumptions associated with the circuit displayed in Figures 1 and 2 are given as follows: (i) \(V_{in}\), \(V_{o}\), and \(V_{c}\) are the dc input voltage, dc output voltage, and the dc voltage clamp voltage, respectively; (ii) \(l_{in}\) is the dc input current; (iii) \(i_{Lr}\) is the current in \(L_{r}\) and \(D_{o}\); (iv) \(v_{ds1}\) and \(i_{ds1}\) are the voltage on and current in \(S_{1}\)  
Figure 1: Proposed active clamp boost converter. Figure 2: Equivalent circuit of the circuit shown in Figure 1.  
respectively; (v) \(v_{ds1}\) and \(i_{ds2}\) are the voltage on and current in \(S_{2}\), respectively; and (vi) all the elements are ideal except that switches have individual body diodes and parasitic capacitors. There are nine stages in the converter operating shown in Figure 3, where \(t_{on}\) is the turn-on time of \(v_{gs1}\), which is equal to \(DT_{s}\) with a switching period of \(T_{s}\) and a duty cycle of \(D\), whereas \(t_{off}\) is the turn-off time of \(v_{gs1}\), which is equal to \((1-D)T_{s}\). Stage 1: (\(t_{0}\leq t\leq t_{1}\))  
As illustrated in Figures 3 and 4, \(S_{1}\) is ON but \(S_{2}\) is OFF. During this stage, \(L_{in}\) is magnetized by \(V_{in}\). At the same time, \(v_{ds2}\) is clamped at \(V_{c}\). Once \(S_{1}\) is cut off, the operation moves to stage 2. Stage 2: (\(t_{1}\leq t\leq t_{2}\))  
As illustrated in Figures 3 and 5, \(S_{1}\) is cut off and \(S_{2}\) is still OFF. During this state, \(C_{s1}\) is abruptly charged to \(V_{c}\). As \(V_{c}=v_{ds1}+v_{ds2}\), \(C_{s2}\) is abruptly discharged to zero. At the same time, \(L_{r}\) is magnetized and the energy at input terminal is transferred to the output terminal via \(D_{o}\). Once the energy stored in \(C_{s2}\) is entirely exhausted, the operation proceeds to stage 3. Figure 3: Key waveforms pertaining to the working converter. According to Figure 5b, three stage equations can be obtained as  
\[\left\{ \begin{array}{l}I_{in}=C_{\rm s1}\frac{dv_{d1}(t)}{dt}+i_{Lr}(t)-C_{\rm s2 }\frac{dv_{d2}(t)}{dt}\\ V_{c}=v_{d81}(t)+v_{d82}(t)\\ v_{d81}(t)=L_{r}\frac{dt_{Lr}(t)}{dt}+V_{o}\end{array} \right. \tag{1}\]  
Figure 4: (a) Current path for stage 1; (b) equivalent of (a). Figure 5: (a) Current path for stage 2; (b) equivalent of (a). By assuming that two switches are identical, \(C_{s1}=C_{s2}=C_{s}\). Moreover, the initial values of this stage are \(i_{Lr}(t_{1})=0\), \(v_{ds1}(t_{1})=0\) and \(v_{ds2}(t_{1})=V_{c}\). By taking the Laplace transform of Equation (1), the following equations can be attained to be  
\[\left\{ \begin{array}{l}I_{Lr}(s)=(\frac{1}{s}-\frac{s}{s^{2}+\frac{1}{2 LrCs}} )I_{in}-(\frac{\frac{1}{r}}{s^{2}+\frac{1}{2 LrCs}})V_{o}\\ V_{ds1}(s)=(\frac{1}{s}-\frac{s}{s^{2}+\frac{1}{2 LrCs}})V_{o}+(\frac{\frac{1}{ cs}}{s^{2}+\frac{1}{2 LrCs}})I_{in}\\ V_{ds2}(s)=\frac{V_{c}}{s}-(\frac{1}{s}-\frac{s}{s^{2}+\frac{1}{2 LrCs}})V_{o}-( \frac{\frac{1}{cs}}{s^{2}+\frac{1}{2 LrCs}})I_{in}\end{array} \right. \tag{2}\]  
By taking the inverse Laplace transform of Equation (2), the following equations can be attained to be  
\[\left\{ \begin{array}{l}i_{Lr}(t)=I_{in}[1-\cos\;\omega_{1}(t-t_{1})]-\frac{V _{c}}{Z_{1}}\sin\;\omega_{1}(t-t_{1})\\ v_{ds1}(t)=V_{o}[1-\cos\;\omega_{1}(t-t_{1})]+I_{in}Z_{1}\sin\;\omega_{1}(t-t _{1})\\ v_{ds2}(t)=V_{c}-V_{o}[1-\cos\;\omega_{1}(t-t_{1})]-I_{in}Z_{1}\sin\;\omega_{1} (t-t_{1})\end{array} \right. \tag{3}\]  
where  
\[\omega_{1}=\sqrt{\frac{1}{2L_{r}C_{s}}}\;\text{and}\;Z_{1}=\sqrt{\frac{L_{r}} {2C_{s}}} \tag{4}\]  
If \(\omega_{1}(t-t_{1})\cong 0\), then \(\cos\;\omega_{1}(t-t_{1})\) and \(\sin\;\omega_{1}(t-t_{1})\) can be close to the following equations:  
\[\left\{ \begin{array}{l}\cos\;\omega_{1}(t-t_{1})\cong 1\\ \sin\;\omega_{1}(t-t_{1})\cong\omega_{1}(t-t_{1})\end{array} \right.