*2.1. Operating Mode of the Proposed Converter*

The proposed converter has two different operating modes as follows:

• When the switch *M*<sup>1</sup> is ON: the diode *D* is reversed biased. Figure 3a shows the equivalent circuit and current directions of this mode. The energy is transferred and stored in the coupling capacitor *C*1. Meanwhile, both inductors *L*<sup>1</sup> and *L*<sup>2</sup> are energized. The current slope in the inductors is given according to the following equations:

$$\frac{dI\_{L\_1}}{dt} = \frac{V\_{in}}{L\_1} \tag{1}$$

$$\frac{dI\_{L\_2}}{dt} = \frac{V\_{\text{in}} + V\_{\text{C}\_1}}{L\_2} \tag{2}$$

From (1) and (2), the switch current is the sum of two inductors' current. This can be written as (3). Instantaneously, the voltage of capacitor *C*<sup>1</sup> is given by (4).

$$I\_{M1} = I\_{L1} + I\_{L2} \tag{3}$$

$$V\_{\mathbb{C}1} = V\_{\text{in}} - V\_{\text{L2}} \tag{4}$$

• When the switch *M*<sup>1</sup> is OFF: the diode is forward and conducts the current. Figure 3b shows this mode's equivalent circuit and current directions. All the energy stored in *C*1, *L*1, and *L*<sup>2</sup> is transferred to the load. The current slope in the inductors is given according to the following equations:

$$\frac{dI\_{L\_1}}{dt} = \frac{V\_o + V\_{C\_1}}{L\_1} \tag{5}$$

$$\frac{dI\_{L\_2}}{dt} = \frac{V\_{\text{in}} + V\_{\text{C}\_1}}{L\_2} \tag{6}$$

From (5) and (6), the diode current is the sum of two inductors' current. This can be written as on (7). Instantaneously, the voltage of capacitor *C*<sup>1</sup> is given by (8).

$$I\_D = I\_{L1} + I\_{L2} \tag{7}$$

$$V\_{\mathbb{C}1} = V\_{\mathbb{L}1} - V\_o \tag{8}$$

**Figure 3.** (**a**) Mode 1, when switch *M*<sup>1</sup> is on, (**b**) Mode 2, when switch *M*<sup>2</sup> is off.

In the steady state, the average inductors' voltages are zero. Based on (4) and (8) and the capacitance *C*<sup>1</sup> is large enough, then, the average *C*<sup>1</sup> voltage has to be equal to the following equation:

$$V\_{\mathbb{C}1} = V\_{\mathbb{L}1} - V\_{\mathbb{L}2} \tag{9}$$

According to (9), it can be seen that the capacitor voltage *C*<sup>1</sup> depends on the voltage difference between the two inductors. In the proposed converter, the coupling capacitor's average voltage is equal to zero. This means all the energy stored during the turn-on period is dissipated during the turn-off period. More details will be discussed in the Simulation Results Section.

#### *2.2. Duty Cycle and Voltage Gain of the Proposed Converter*

The duty cycle of the proposed converter can be calculated based on the same procedure used in other DC-DC converters. In a steady state, the average inductor voltages over one switching cycle (*T*s) must equal zero. Then, once the switch *M*<sup>1</sup> is turned on, the inductor *L*<sup>1</sup> is energized from the input DC voltage *Vin*. On the other hand, when the switch *M*<sup>1</sup> is turned off, the energy stored in *L*<sup>1</sup> is delivered to the load through the coupling capacitor *C*<sup>1</sup> and the diode *D*. Based on that, the inductor voltage function is given by:

$$V\_{L1}(t) = \begin{cases} \begin{array}{l} V\_{\text{in}\_{\prime}} \quad 0 < t < DT\_{\text{s}} \\ -V\_{o} \quad DT\_{\text{s}} < t < T\_{\text{s}} \end{array} \tag{10}$$

$$V\_{L2}(t) = \begin{cases} \begin{array}{c} V\_{in\_{\prime}} \\ -V\_{o\prime} \end{array} \begin{array}{c} 0$$

Calculating the average voltages of *L*<sup>1</sup> and *L*<sup>2</sup> results in (12). By solving (12), the voltage gain can be produced as per (13).

$$
\lambda \left< V\_{L1} \right> = DT\_s V\_{in} + (1 - D) T\_s V\_o = 0 \tag{12}
$$

$$V\_G = \frac{V\_o}{V\_{in}} = -\frac{D}{1 - D} \tag{13}$$

where *VG* is the voltage gain of the proposed converter.

Equation (12) shows that the voltage gain of the proposed converter is the same as Buck–Boost, Cuk, and SEPIC converters [32]. Nevertheless, the proposed converter has an inverted output voltage, which can be used to step up and down by selecting a proper duty cycle value (see (13)). Additionally, it can be seen that the critical value between the step-up and step-down is *D* = 50%. Figure 4 plots the voltage gain of the proposed converter and the converter duty cycle.

**Figure 4.** The voltage gain change based on the converter duty cycle.
