5.2.3. Control-To-Module Filter Inductor Current Transfer Function

The filter inductor current and the duty cycle relationship is found by performing the following: Substituting (75) in (76):

$$\begin{cases} \begin{array}{ll} \hat{\imath}\_{L1} = \left(s\mathbf{C} + \frac{1}{R}\right)\mathfrak{\flat}\_{out1} + \frac{1}{R}\mathfrak{\flat}\_{out2} + \frac{1}{R}\mathfrak{\flat}\_{out3} \\ \hat{\imath}\_{L2} = \frac{1}{R}\mathfrak{\flat}\_{out1} + \left(s\mathbf{C} + \frac{1}{R}\right)\mathfrak{\flat}\_{out2} + \frac{1}{R}\mathfrak{\flat}\_{out3} \\ \hat{\imath}\_{L3} = \frac{1}{R}\mathfrak{\flat}\_{out1} + \frac{1}{R}\mathfrak{\flat}\_{out2} + \left(s\mathbf{C} + \frac{1}{R}\right)\mathfrak{\flat}\_{out3} \end{array} \tag{80}$$

Rewriting (80) in terms of the output voltages:

$$\begin{cases} \begin{array}{c} \mathfrak{d}\_{\text{out1}} = \mathfrak{g}\_{1}\mathring{\mathfrak{i}}\_{L1} - \mathfrak{g}\_{2}\mathring{\mathfrak{i}}\_{L2} - \mathfrak{g}\_{2}\mathring{\mathfrak{i}}\_{L3} \\ \upcirc\_{\text{out2}} = \mathfrak{g}\_{2}\mathring{\mathfrak{i}}\_{L1} - \mathfrak{g}\_{1}\mathring{\mathfrak{i}}\_{L2} - \mathfrak{g}\_{2}\mathring{\mathfrak{i}}\_{L3} \\ \upcirc\_{\text{out3}} = \mathfrak{g}\_{2}\mathring{\mathfrak{i}}\_{L1} - \mathfrak{g}\_{2}\mathring{\mathfrak{i}}\_{L2} - \mathfrak{g}\_{1}\mathring{\mathfrak{i}}\_{L3} \end{array} \tag{81}$$

where

$$\text{g}\_1 = \frac{2 + \text{sRC}}{S^2 \text{C}^2 R + 3 \text{sC}} \quad \text{g}\_2 = \frac{1}{S^2 \text{C}^2 R + 3 \text{sC}} \tag{82}$$

Rewriting (81) in a matrix form:

$$
\begin{bmatrix}
\hat{v}\_{\text{out}1} \\
\hat{v}\_{\text{out}2} \\
\hat{v}\_{\text{out}3}
\end{bmatrix} = \begin{bmatrix}
\text{g}\_1 & -\text{g}\_2 & -\text{g}\_2 \\
\end{bmatrix} \begin{bmatrix}
\hat{i}\_{L1} \\
\hat{i}\_{L2} \\
\hat{i}\_{L3}
\end{bmatrix} \tag{83}
$$

Substituting (67), (69), and (81) in (71) and assuming *v*ˆ*in* = 0 would result in:

$$\begin{cases} \frac{V\_{\text{in}}}{K} \dot{d}\_{1} = (sL + R\_{d} + \operatorname{g}\_{1})\dot{l}\_{L1} - \operatorname{g}\_{2}\dot{l}\_{L2} - \operatorname{g}\_{2}\dot{l}\_{L3} \\\ \frac{V\_{\text{in}}}{K} \dot{d}\_{2} = -\operatorname{g}\_{2}\dot{l}\_{L1} + (sL + R\_{d} + \operatorname{g}\_{1})\dot{l}\_{L2} - \operatorname{g}\_{2}\dot{l}\_{L3} \\\ \frac{V\_{\text{in}}}{K} \dot{d}\_{3} = -\operatorname{g}\_{2}\dot{l}\_{L1} - \operatorname{g}\_{2}\dot{l}\_{L2} + (sL + R\_{d} + \operatorname{g}\_{1})\dot{l}\_{L3} \end{cases} \tag{84}$$

Rearranging (84), the control-to-module filter inductor current can be represented as:

$$
\begin{bmatrix}
\hat{d}\_1\\ \hat{d}\_2\\ \hat{d}\_3
\end{bmatrix} = \begin{bmatrix}
\.\&\-\,\,\text{g}\,4 & \.\text{g}\,4\\ 
\begin{bmatrix}
\.\&\-\,\text{g}\,4 & \.\text{g}\,4\\ 
\.\quad\text{g}\,4 & \.\text{g}\,4
\end{bmatrix}
\begin{bmatrix}
\hat{l}\_{L1}\\ 
\hat{l}\_{L2}\\ 
\hat{l}\_{L3}
\end{bmatrix}
\end{bmatrix} \tag{85}
$$

where

$$\text{Sg3} = \frac{V\_{\text{in}}(sL + R\_d + g\_1 - g\_2)}{\text{K}(\text{g2}\_1 - \text{g}\_1 \text{g2}\_2 + 2 \text{g}\_1 \text{g5}\_5 - 2 \text{g2}\_2 - \text{g2}\_2 \text{g5}\_5 + g\_5^2)}\tag{86}$$

$$\mathcal{g}\_4 = \frac{V\_{in}\mathcal{g}\_2}{\mathcal{K}\{\mathcal{g}\_1^2 - \mathcal{g}\_1\mathcal{g}\_2 + 2\mathcal{g}\_1\mathcal{g}\_5 - 2\mathcal{g}\_2^2 - \mathcal{g}\_2\mathcal{g}\_5 + \mathcal{g}\_5^2\}}\tag{87}$$

$$\mathbf{g}\_5 = \mathbf{s}\mathbf{L} + \mathbf{R}\_d\tag{88}$$

### 5.2.4. Converter Output Impedance

Similarly, the IPOS converter output impedance can be found by modifying (74) such that:

$$\sum\_{j=1}^{3} \hat{i}\_{Lj} + \hat{i}\_{out} = \mathfrak{d}\_{out} \big( \mathbf{sC} + \frac{\mathfrak{d}}{R} \big) \tag{89}$$

The output voltage and the output current relationship is found by considering the same assumptions as in Section 2.2.4, summing Equations in (71), and substituting (67), (69), (75) and (89).

$$-R\_d \left(\mathfrak{d}\_{\rm out} \left(\mathrm{sC} + \frac{\mathfrak{Z}}{R}\right) - \hat{l}\_{\rm out}\right) = \mathrm{sL} \left(\mathfrak{d}\_{\rm out} \left(\mathrm{sC} + \frac{\mathfrak{Z}}{R}\right) - \hat{l}\_{\rm out}\right) + \mathfrak{d}\_{\rm out} \tag{90}$$

Simplifying (90) would result in (91):

$$Z\_{out} = \frac{\vartheta\_{out}}{\hat{I}\_{out}} = \frac{3(R\_d + sL)}{s^2LC + s\left(\frac{3d}{R} + R\_dC\right) + 3\frac{R\_d}{R} + 1} \tag{91}$$
