# Floating Step-Down Converter with a Novel Lossless Snubber
## 2 Main Power Stage
### Operating Principle  
The following will describe the operating principle of the proposed circuit:  
State 1: [\(t_{0}\leq\) t \(<t_{1}\)]  
As shown in Figure 3, at \(t=t_{0}\), the switch \(Q_{1}\) conducts, and the diode \(D_{2}\) conducts, but \(D_{1}\) and \(D_{3}\) are cut off. At this time, the voltage of \(V_{in}-V_{o}\) across \(L_{o}\) increases, causing \(L_{o}\) to be magnetized. At the same time, there is a resonant loop from \(V_{in}\) to \(V_{o}\) to \(D_{2}\) to \(L_{r}\) to \(C_{r}\) to \(Q_{1}\) and then to \(V_{in}\). During this state, the input terminal will temporarily store energy in the snubber. At time \(t=t_{1}\), the resonant capacitor voltage \(v_{Cr}\) is charged to \(V_{in}\), and the diode \(D_{3}\) conducts, proceeding to the next state. Note that the snubber components \(L_{r}\) and \(C_{r}\) do not have any initial value, so the equivalent circuit is shown in Figure 4.  
Figure 3: Current flow in state 1. Figure 2: Illustrated waveforms for the proposed circuit. From the results of Figures 3 and 4, the resonant capacitor voltage \(v_{Cr}(t)\), the resonant inductor current \(i_{Lr}(t)\), the characteristic impedance \(Z_{a}\), and the resonant radian frequency \(\omega_{a}\) can be expressed as follows:  
\[v_{Cr}(t)=(V_{in}-V_{o})[1-\cos\omega_{a}(t-t_{0})] \tag{1}\]  
\[i_{Lr}(t)=\frac{V_{in}-V_{o}}{Z_{a}}\sin\omega_{a}(t-t_{0}) \tag{2}\]  
\[Z_{a}=\sqrt{\frac{L_{r}}{C_{r}}} \tag{3}\]  
\[\omega_{a}=\frac{1}{\sqrt{L_{r}C_{r}}} \tag{4}\]  
Finally, by substituting the boundary condition of the resonant capacitor voltage \(v_{Cr}(t)\), the expression of the experienced time interval \(\Delta t_{10}\) can be obtained as shown in (6):  
\[V_{Cr1}=(V_{in}-V_{o})[1-\cos\omega_{a}(t_{1}-t_{0})]=V_{in} \tag{5}\]  
\[\Delta t_{10}=t_{1}-t_{0}=\frac{1}{\omega_{a}}\cos^{-1}(\frac{-V_{o}}{V_{in}- V_{o}}) \tag{6}\]  
State 2: [\(t_{1}\leq t<t_{2}\)]  
As shown in Figure 5, the switch \(Q_{1}\) conducts, and the diodes \(D_{2}\) and \(D_{3}\) conduct, but \(D_{1}\) cuts off. At this time, the voltages across the output inductor \(L_{o}\) and the resonant inductor \(L_{r}\) are \(V_{in}-V_{o}\) and \(-V_{o}\), respectively, so the output inductor \(L_{o}\) and the resonant inductor \(L_{r}\) are in the magnetizing and demagnetizing states, respectively. At time \(t=t_{2}\), due to the existence of diode \(D_{2}\), the resonant inductor current \(i_{Lr}\) will not flow backwards after it drops to zero, proceeding to the next state. Figure 4: Equivalent circuit for state 1. Figure 3:0 Current flow in state 2. Figure 3:1  
Figure 3:2  
Figure 3:3  
It is worth mentioning that \(\Delta t_{20}=\Delta t_{21}+\Delta t_{10}\
eq\pi/\omega_{a}\) because state 2 is not a resonant state. State 3: [\(t_{2}\leq t<t_{3}\)]  
As shown in Figure 6, the switch \(Q_{1}\) conducts, and the diodes \(D_{1}\), \(D_{2}\), and \(D_{3}\) are cut off. At this time, the voltage across the output inductor \(L_{o}\) is \(V_{in}-V_{o}\), so the output inductor \(L_{o}\) is magnetized. At time \(t=t_{3}\), the switch \(Q_{1}\) is cut off, and due to the presence of resonant capacitor \(C_{r}\), the slope of the rise in the voltage \(v_{ds}\) will begin to slowly go up, so the switch \(Q_{1}\) can realize ZVS cutoff. Based on the result of state 1 and state 2, the expression of the experienced time interval \(\Delta t_{32}\) can be obtained as  
Figure 3:4  
State 4: [\(t_{4}\leq t\leq t_{5}\)]  
As shown in Figure 7, at the moment \(t=t_{3}\), the switch \(Q_{1}\) is cut off at approximately zero voltage, and the diode \(D_{3}\) conducts, but the diodes \(D_{1}\) and \(D_{2}\) are cut off. The energy stored in the snubber will be released to the output, and the following equation can be obtained to be  
Figure 3:5  
Figure 3:6 Current flow in state 3. Figure 3:7 Current flow in state 4.