*3.1. Converter Voltage Gain*

Applying the volt-second balance principle to *L*<sup>1</sup> and *L*<sup>2</sup> results in:

$$\begin{aligned} \left(EDT\_\text{s} + (E + V\_{\mathbb{C}\_p} - V\_0)(1 - D)T\_\text{s} = 0\\ \left(-V\_{\mathbb{C}\_p} + V\_0\right)DT\_\text{s} - V\_{\mathbb{C}\_p}(1 - D)T\_\text{s} = 0 \end{aligned} \tag{1}$$

By using the last equations, the expression for *VC<sup>p</sup>* can be derived:

$$V\_{\mathbb{C}\_p} = V\_0 - \frac{E}{(1 - D)} = DV\_0 \tag{2}$$

From the above expression, the voltage gain *M* is:

$$M = \frac{V\_0}{E} = \frac{1}{(1 - D)^2} \tag{3}$$

According to Equation (3), the voltage gain *M* has a quadratic conversion ratio.
