*5.2. IPOS Small-Signal Analysis*

The SSM for the IPOS converter shown in Figure 6 is derived using the SSM presented in [37].

**Figure 6.** IPOS DC-DC converter SSM.

Since the input current per module is *Iin* <sup>3</sup> , and the output voltage per module is *Vo* <sup>3</sup> , accordingly, the load resistance per module is <sup>R</sup> <sup>3</sup> . Therefore, <sup>ˆ</sup> *dij*, ˆ *dv*<sup>j</sup> and *Ieq* presented in Figure 6 can be expressed as follows, where the subscript *j* = 1, 2, *and* 3:

$$d\_{ij} = -\frac{4}{K V\_{\rm in}} \mathbf{\hat{r}}\_{Lj} \tag{66}$$

Rewriting (66) in terms of *Rd* would result in:

$$\hat{d}\_{ij} = -\frac{\mathcal{K}\mathcal{R}\_d}{V\_{in}}\hat{\mathbf{i}}\_{Lj} \tag{67}$$

$$\hat{d}\_{v\circ} = \frac{12 \, L\_{\text{lk}} f\_{\text{s}} D\_{eff}}{k^2 R V\_{\text{in}}} \mathfrak{d}\_{\text{in}} \tag{68}$$

Similarly, rewriting (68) in terms of *Rd* would result in:

$$
\hat{d}\_{vj} = \frac{3\,\mathrm{R}\_d D\_{eff}}{\mathrm{RV}\_{\mathrm{in}}} \mathfrak{d}\_{\mathrm{in}} \tag{69}
$$

$$I\_{eq} = \frac{3\,\,V\_{in}}{\,\,K\,\, }\tag{70}$$

The following equations are obtained from Figure 6:

⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩

⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩

$$\begin{cases} \frac{D\_{eff}}{K}\mathfrak{v}\_{in} + \frac{V\_{in}}{K}(\hat{d}\_{l1} + \hat{d}\_{v1} + \hat{d}\_{1}) = sL\hat{I}\_{L1} + \mathfrak{v}\_{out1} \\ \frac{D\_{eff}}{K}\mathfrak{v}\_{in} + \frac{V\_{in}}{K}(\hat{d}\_{l2} + \hat{d}\_{v2} + \hat{d}\_{2}) = sL\hat{I}\_{L2} + \mathfrak{v}\_{out2} \\ \frac{D\_{eff}}{K}\mathfrak{v}\_{in} + \frac{V\_{in}}{K}(\hat{d}\_{l3} + \hat{d}\_{l3} + \hat{d}\_{3}) = sL\hat{I}\_{L3} + \mathfrak{v}\_{out3} \end{cases} \tag{71}$$

$$\begin{array}{c} \frac{K}{3\mathcal{D}\_{eff}} \left(\hat{\mathbf{i}}\_{\rm in} - s\mathbf{C}\_{d}\boldsymbol{\vartheta}\_{\rm in}\right) = I\_{\rm cq} \Big(\hat{d}\_{i1} + \hat{d}\_{\rm v1} + \hat{d}\_{1}\big) + \hat{\mathbf{i}}\_{L1} \\\ \frac{K}{3\mathcal{D}\_{eff}} \left(\hat{\mathbf{i}}\_{\rm in} - s\mathbf{C}\_{d}\boldsymbol{\vartheta}\_{\rm in}\right) = I\_{\rm cq} \Big(\hat{d}\_{i2} + \hat{d}\_{\rm v2} + \hat{d}\_{2}\big) + \hat{\mathbf{i}}\_{L2} \\\ \frac{K}{3\mathcal{D}\_{eff}} \left(\hat{\mathbf{i}}\_{\rm in} - s\mathbf{C}\_{d}\boldsymbol{\vartheta}\_{\rm in}\right) = I\_{\rm cq} \Big(\hat{d}\_{i3} + \hat{d}\_{\rm v3} + \hat{d}\_{3}\big) + \hat{\mathbf{i}}\_{L3} \end{array} \tag{72}$$

$$\begin{cases} \begin{array}{l} \hat{\imath}\_{L1} = s \text{C} \pounds\_{out1} + \frac{\hat{\imath}\_{nt}}{\mathcal{R}}\\ \hat{\imath}\_{L2} = s \text{C} \pounds\_{out2} + \frac{\hat{\imath}\_{nt}}{\mathcal{R}}\\ \hat{\imath}\_{L3} = s \text{C} \pounds\_{out3} + \frac{\hat{\imath}\_{nt}}{\mathcal{R}} \end{array} \tag{73}$$

Summing Equations in (73) would result in (74):

$$\sum\_{j=1}^{3} \hat{\imath}\_{Lj} = \imath\_{out} \left( s\mathbb{C} + \frac{3}{R} \right) \tag{74}$$

where

$$
\mathfrak{d}\_{out1} + \mathfrak{d}\_{out2} + \mathfrak{d}\_{out3} = \mathfrak{d}\_{out} \tag{75}
$$

### 5.2.1. Control-To-Output Voltage Transfer Function

The output voltage and the duty cycle relationship is found by summing up Equations in (71), considering the same assumptions as in Section 2.2.1, and substituting (67), (69), and (74).

$$\frac{3D\_{\varepsilon f f}}{K}\boldsymbol{\eth}\_{\rm in} + \frac{V\_{\rm in}}{K} \left( -\frac{K\mathcal{R}\_d}{V\_{\rm in}} \left( \boldsymbol{\eth}\_{\rm out} \left( \mathbf{sC} + \frac{3}{R} \right) \right) + \sum\_{j=1}^3 \frac{3 \, R\_d D\_{\varepsilon f f}}{R V\_{\rm in}} \boldsymbol{\eth}\_{\rm in} + \boldsymbol{\eth}\_1 \right) = \boldsymbol{sL} \left( \boldsymbol{\eth}\_{\rm out} \left( \mathbf{sC} + \frac{3}{R} \right) \right) + \boldsymbol{\eth}\_{\rm out} \tag{76}$$

Simplifying (76) would result in (77):

$$G\_{\rm rd} = \frac{\vartheta\_{\rm out}}{\hat{d}\_{\rm j}} = \frac{\frac{V\_{\rm jr}}{R}}{s^2LC + s\left(\frac{3L}{R} + R\_dC\right) + \frac{3R\_d}{R} + 1} \tag{77}$$

### 5.2.2. Control-To-Filter Inductor Current Transfer Function

The filter inductor current and the duty cycle relationship is found by substituting *v*ˆ*out* in terms of ˆ*iLj* using (74) in (76) and considering the same assumptions as in Section 2.2.1.

$$\frac{V\_{\rm in}}{K}\hat{d}\_{1} - R\_{d}\sum\_{j=1}^{3}\hat{\imath}\_{Lj} = s\mathcal{L}\sum\_{j=1}^{3}\hat{\imath}\_{Lj} + \frac{R}{s\mathcal{R}\mathcal{C} + 3}\sum\_{j=1}^{3}\hat{\imath}\_{Lj} \tag{78}$$

Simplifying (78) would result in (79):

$$G\_{\rm id} = \frac{\hat{t}\_L}{\hat{d}\_j} = \frac{\frac{V\_{\rm in}}{\mathcal{K}}(3 + sRC)}{R\left(s^2LC + s\left(\frac{3L}{\mathcal{K}} + R\_dC\right) + \frac{3R\_d}{\mathcal{K}} + 1\right)}\tag{79}$$
