**4. Study of the Power Efficiency**

Now, the analysis and expressions for computing the power losses of each element for the proposed converter are provided. According to [23], the corresponding circuit including the parasitics of all elements of the configuration is exhibited in Figure 8. The resistances *RL*<sup>1</sup> , *RL*<sup>2</sup> , *RC<sup>p</sup>* , *RC*<sup>0</sup> , *RD*1, *RD*2, *RON*<sup>1</sup> , and *RON*<sup>2</sup> are the equivalent series resistance (ESR) of *L*1, *L*2, *Cp*, *C*0, *DS*<sup>1</sup> , *DS*<sup>2</sup> , *S*1, and *S*2, respectively. The voltages *VFDS*<sup>1</sup> and *VFDS*<sup>2</sup> are the forward voltage drops of *DS*<sup>1</sup> , and *DS*<sup>2</sup> . The gate voltage for *S*<sup>1</sup> and *S*<sup>2</sup> is *Vg*. In the interval *tON*, both active switches are conducting. The respective circuit for this interval is described in Figure 9. On the other hand, the respective circuit for the interval *tOFF* (both active switches are not conducting) is depicted in Figure 10.

Using the volt-second balance principle over *L*<sup>1</sup> and *L*2, two expressions for *VC<sup>p</sup>* are derived. From these two expressions, it is possible to find the relationship for computing the duty cycle (*Dloss*) and the losses of each component of the converter. To precisely quantify the losses associated with the parasitics of the converter, it is necessary to recalculate the duty cycle (*Dloss* > *D*). The resulting equation for *Dloss* has a fourth-order behavior:

$$D\_{loss}^4 + aD\_{loss}^3 + bD\_{loss}^2 + cD\_{loss} + d = 0\tag{14}$$

where:

$$\begin{aligned} a &= -\frac{4RV\_0 + RV\_{\text{F}\_{\text{DS1}}} + 4RV\_{\text{F}\_{\text{DS2}}} + R\_{D2} - R\_{ON2}V\_0}{RV\_0 + RV\_{\text{F}\_{\text{DS2}}}} \\\\ b &= \frac{6RV\_0 + 3RV\_{\text{F}\_{\text{DS1}}} + 6RV\_{\text{F}\_{\text{DS2}}} + 3R\_{D2}V\_0 + R\_{L\_2}V\_0 - 2R\_{ON2}V\_0 + ER}{RV\_0 + RV\_{\text{F}\_{\text{DS2}}}} \end{aligned}$$

$$\mathcal{L} = \frac{(R\_{ON1} + R\_{ON2} - 2R\_{L\_2} - 4R - R\_{D1} - 3R\_{D2})V\_0 + 2ER - 3RV\_{\text{FoS1}} - 4RV\_{\text{FoS2}}}{RV\_0 + RV\_{\text{FoS2}}}$$

$$d = \frac{RV\_0 + RV\_{F\_{DS1}} + R\_{D1}V0 + RV\_{F\_{DS2}} + R\_{D2}V\_0 + R\_{L\_1}V\_0 + R\_{L\_2}V\_0 - ER}{RV\_0 + RV\_{F\_{DS2}}}$$

to obtain the duty cycle *Dloss*, it is necessary to solve Equation (14) and choose the adequate root.

The expressions for the power losses of each element are shown in Table 2 (conduction and switching losses are included). The diode junction capacitances of *DS*<sup>1</sup> and *DS*<sup>2</sup> are *Cj*\_*DS*<sup>1</sup> and *Cj*\_*DS*<sup>2</sup> , respectively. The time intervals are *trr*<sup>1</sup> = *tr*<sup>1</sup> + *tON*<sup>1</sup> , *trr*<sup>2</sup> = *tr*<sup>2</sup> + *tON*<sup>2</sup> , *tf f* <sup>1</sup> = *t<sup>f</sup>* <sup>1</sup> + *tOFF*<sup>1</sup> , and *tf f* <sup>2</sup> = *t<sup>f</sup>* <sup>2</sup> + *tOFF*<sup>2</sup> , where *tr*1, *tON*<sup>1</sup> , *tr*2, *tON*<sup>2</sup> , *t<sup>f</sup>* <sup>1</sup> , *tOFF*<sup>1</sup> , *t<sup>f</sup>* <sup>2</sup> , and *tOFF*<sup>2</sup> are the rise time, turn ON delay time, fall time, and turn OFF delay time of *S*1, and *S*2, respectively. The input capacitances of *S*<sup>1</sup> and *S*<sup>2</sup> are *Ciss*<sup>1</sup> and *Ciss*<sup>2</sup> , respectively. The RMS current values can be obtained as:

$$\begin{array}{ll} I\_{L1\_{RMS}}^2 = \frac{V\_0^2}{R^2(1 - D\_{loss})^4}, \quad I\_{L2\_{RMS}}^2 = \frac{V\_0^2}{R^2(1 - D\_{loss})^2} \\\\ I\_{S1\_{RMS}}^2 = I\_{L1\_{RMS}}^2 D\_{loss}, \quad I\_{D1\_{RMS}}^2 = I\_{L1\_{RMS}}^2 (1 - D\_{loss}) \\\\ I\_{S2\_{RMS}}^2 = I\_{L2\_{RMS}}^2 D\_{loss}, \quad I\_{D2\_{RMS}}^2 = I\_{L2\_{RMS}}^2 (1 - D\_{loss}) \\\\ I\_{CP\_{RMS}}^2 = \frac{V\_0 D\_{loss} (D\_{loss}^2 - D\_{loss} + 1)}{R^2 (1 - D\_{loss})^6} \\\\ I\_{C0\_{RMS}}^2 = \frac{V\_0^2 D\_{loss} (2 - D\_{loss})^2}{R^2 (1 - D\_{loss})^3} \end{array}$$

**Figure 8.** Equivalent circuit with the parasitics of elements.

**Figure 9.** Equivalent circuit for the interval *tON*.

**Figure 10.** Equivalent circuit for the interval *tOFF*.


**Table 2.** Power losses calculation.

The total power loss is:

$$\begin{split} P\_{T\_{\text{loss}}} &= P\_{\text{loss\\_L1}} + P\_{\text{loss\\_L2}} + P\_{\text{loss\\_CP}} + P\_{\text{loss\\_C0}} + P\_{\text{loss\\_D1}} \\ &+ P\_{\text{loss\\_D2}} + P\_{\text{loss\\_S1}} + P\_{\text{loss\\_S2}} + P\_{\text{loss\\_d\^T}} + P\_{\text{loss\\_S2}}. \end{split} \tag{15}$$

The expression for computing the power efficiency is:

$$\eta = \frac{P\_0}{P\_0 + P\_{T\_{loss}}} \tag{16}$$

where *P*<sup>0</sup> is defined as *P*<sup>0</sup> = *V* 2 0 /*R*.
