2.2.3. Control-To-Module Input Voltage Transfer Function

The module input voltage and the duty cycle relationship is found as follows: Subtracting the 3rd Equation in (6) from the 1st Equation in (6):

$$\frac{D\_{eff}}{K}(\mathfrak{d}\_{\rm cdl1} - \mathfrak{d}\_{\rm cdl2}) + \frac{V\_{\rm in}}{3K} \{\hat{d}\_{l1} + \hat{d}\_{v1} + \hat{d}\_{l1} - \hat{d}\_{l2} - \hat{d}\_{v2} - \hat{d}\_{l2}\} = sL\{\hat{l}\_{l1} - \hat{l}\_{l2}\} + \mathfrak{d}\_{\rm out1} - \mathfrak{d}\_{\rm out2} \tag{16}$$

Substituting (2) and (4) in (16):

$$\frac{D\_{eff}}{K} \left( 1 + \frac{3R\_d}{R} \right) (\hat{v}\_{\alpha 1} - \hat{v}\_{\alpha 2}) + \frac{V\_{\text{in}}}{3K} (\hat{d}\_1 - \hat{d}\_2) = (\text{s}L + R\_d) \left( \hat{l}\_{l.1} - \hat{l}\_{l.2} \right) + \hat{v}\_{\text{out}1} - \hat{v}\_{\text{out}2} \tag{17}$$

Subtracting the 2nd Equation in (7) from the 1st Equation in (7), and substituting (5):

$$\frac{K}{D\_{eff}}(s\mathbb{C}\_d\mathfrak{d}\_{cd2} - s\mathbb{C}\_d\mathfrak{d}\_{cd1}) = \frac{V\_{in}}{KR}(\mathfrak{d}\_{i1} + \mathfrak{d}\_{v1} + \mathfrak{d}\_1 - \mathfrak{d}\_{i2} - \mathfrak{d}\_{v2} - \mathfrak{d}\_2) + \left(\mathfrak{i}\_{l1} - \mathfrak{i}\_{l2}\right) \tag{18}$$

Substituting (2) and (4) in (18) and rearranging the equation:

$$\left(\frac{3\mathcal{R}\_d}{R} - 1\right)\left(\mathbf{\hat{i}}\_{L1} - \mathbf{\hat{i}}\_{L2}\right) = \frac{V\_{in}}{KR}\left(\mathbf{\hat{d}}\_1 - \mathbf{\hat{d}}\_2\right) + \left(\frac{9R\_{\text{d}}D\_{eff}}{KR^2} + \frac{\text{sKC}\_d}{D\_{eff}}\right)\left(\mathbf{\hat{v}}\_{\text{cd1}} - \mathbf{\hat{v}}\_{\text{cd2}}\right) \tag{19}$$

Substituting *v*ˆ*out* by *v*ˆ*out*<sup>1</sup> + *v*ˆ*out*<sup>2</sup> + *v*ˆ*out*<sup>3</sup> in (8):

$$\begin{cases} \begin{array}{ll} \hat{\imath}\_{L1} = \left(s\mathbf{C} + \frac{1}{K}\right)\mathfrak{\flat}\_{out1} + \frac{1}{K}\mathfrak{\flat}\_{out2} + \frac{1}{K}\mathfrak{\flat}\_{out3} \\ \hat{\imath}\_{L2} = \frac{1}{K}\mathfrak{\flat}\_{out1} + \left(s\mathbf{C} + \frac{1}{K}\right)\mathfrak{\flat}\_{out2} + \frac{1}{K}\mathfrak{\flat}\_{out3} \\ \hat{\imath}\_{L3} = \frac{1}{K}\mathfrak{\flat}\_{out1} + \frac{1}{K}\mathfrak{\flat}\_{out2} + \left(s\mathbf{C} + \frac{1}{K}\right)\mathfrak{\flat}\_{out3} \end{array} \tag{20}$$

The three Equations in (20) can be represented as:

$$
\begin{bmatrix}
\hat{\imath}\_{L1} \\
\hat{\imath}\_{L2} \\
\hat{\imath}\_{L3}
\end{bmatrix} = \begin{bmatrix}
\text{sC} + \frac{1}{\mathbb{R}} & \frac{1}{\mathbb{R}} & \frac{1}{\mathbb{R}} \\
& \frac{1}{\mathbb{R}} & \text{sC} + \frac{1}{\mathbb{R}} & \frac{1}{\mathbb{R}} \\
& \frac{1}{\mathbb{R}} & \frac{1}{\mathbb{R}} & \text{sC} + \frac{1}{\mathbb{R}}
\end{bmatrix} \begin{bmatrix}
\boldsymbol{\vartheta}\_{out1} \\
\hat{\imath}\_{out2} \\
\hat{\imath}\_{out3}
\end{bmatrix} \tag{21}
$$

Solving for the output voltages in (21) would give:

$$\begin{cases} \begin{array}{l} \mathfrak{d}\_{out1} = \operatorname{g}\_{1}\mathring{\mathfrak{i}}\_{L1} - \operatorname{g}\_{2}\big{(}\mathring{\mathfrak{i}}\_{L2} + \mathring{\mathfrak{i}}\_{L3}\big{)}\\ \up{\mathfrak{d}}\_{out2} = \operatorname{g}\_{1}\mathring{\mathfrak{i}}\_{L2} - \operatorname{g}\_{2}\big{(}\mathring{\mathfrak{i}}\_{L1} + \mathring{\mathfrak{i}}\_{L3}\big{)}\\ \upleftrightarrow \mathring{\mathfrak{g}}\_{out3} = \operatorname{g}\_{1}\mathring{\mathfrak{i}}\_{L3} - \operatorname{g}\_{2}\big{(}\mathring{\mathfrak{i}}\_{L1} + \mathring{\mathfrak{i}}\_{L2}\big{)} \end{array} \tag{22}$$

where

$$\mathcal{g}\_1 = \frac{2 + \text{sRC}}{s^2 \text{RC}^2 + 3 \text{sC}} \text{ and } \mathcal{g}\_2 = \frac{1}{s^2 \text{RC}^2 + 3 \text{sC}}$$

$$\dots \quad \dots \quad \dots$$

Accordingly;

$$
\mathfrak{d}\_{out1} - \mathfrak{d}\_{out2} = (\mathfrak{g}\_1 + \mathfrak{g}\_2)(\hat{\mathfrak{i}}\_{L1} - \hat{\mathfrak{i}}\_{L2}) \tag{23}
$$

Substituting (23) in (17):

$$\frac{D\_{eff}}{K} \left( 1 + \frac{3R\_d}{R} \right) (\mathfrak{v}\_{cl1} - \mathfrak{v}\_{cl2}) + \frac{V\_{\text{in}}}{3K} \left( \hat{d}\_1 - \hat{d}\_2 \right) = (\text{sL} + R\_d + \mathfrak{g}\_1 + \mathfrak{g}\_2) \left( \hat{\mathfrak{i}}\_{l1} - \hat{\mathfrak{i}}\_{l2} \right) \tag{24}$$

Substituting (19) in (24):

$$\begin{array}{c} \frac{D\_{eff}}{K} \left( 1 + \frac{3R\_d}{R} \right) \left( \frac{3R\_d}{R} - 1 \right) (\mathfrak{d}\_{\rm cdl} - \mathfrak{d}\_{\rm cdl}) + \frac{V\_{\rm cp}}{3K} \left( \frac{3R\_d}{R} - 1 \right) \left( \hat{d}\_1 - \hat{d}\_2 \right) = \\ \frac{3(sL + R\_d + \mathfrak{g}\_1 + \mathfrak{g}\_2)}{KR} \left( \hat{d}\_1 - \hat{d}\_2 \right) + (sL + R\_d + \mathfrak{g}\_1 + \mathfrak{g}\_2) \left( \frac{\mathfrak{d}R\_d D\_{eff}}{KR^2} + \frac{sK \mathbb{C}\_d}{D\_{eff}} \right) (\mathfrak{d}\_{\rm cdl} - \mathfrak{d}\_{\rm cdl}) \end{array} \tag{25}$$

Simplifying (25) would result in (26):

$$\begin{pmatrix} \left(\hat{\upsilon}\_{\text{c2}} - \hat{\upsilon}\_{\text{c1}}\right) \\\\ \left(\hat{\upsilon}\_{\text{c2}} - \hat{\upsilon}\_{\text{c1}}\right) \\\\ \left(-\frac{D\_{\text{eff}}\left(R^{2}\frac{V\_{\text{in}}}{2} + 2RV\_{\text{in}}\left(\mathbf{g}\_{1} + \mathbf{g}\_{2}\right) + 2sLRV\_{\text{in}}\right)}{R^{2}D\_{\text{eff}}^{2} + 16R\_{d}D\_{\text{eff}}^{2}\left(\mathbf{g}\_{1} + \mathbf{g}\_{2}\right) + 16sLR\_{d}D\_{\text{eff}}^{2}}\right) \left(\hat{d}\_{2} - \hat{d}\_{1}\right) \\\\ \left(-k^{2}R^{2}C\_{d}\left(\mathbf{s}\left(\mathbf{R}\_{d} + \mathbf{g}\_{1} + \mathbf{g}\_{2}\right) + s^{2}L\right) \end{pmatrix} \tag{26}$$

Equation (26) can be written as:

$$(\mathfrak{d}\_{\text{cd2}} - \mathfrak{d}\_{\text{cd1}}) = A(\mathbf{s})(\hat{d}\_2 - \hat{d}\_1) \tag{27}$$

where

$$A(s) = -\left(\frac{D\_{eff}\left(R^2\frac{V\_{in}}{2} + 2RV\_{in}(\mathcal{g}\_1 + \mathcal{g}\_2) + 2sLRV\_{in}\right)}{R^2D\_{eff}^2 + 16R\_dD\_{eff}^2(\mathcal{g}\_1 + \mathcal{g}\_2) + 16sLR\_dD\_{eff}^2}\right)$$

Rearranging (27) would result in (28):

$$
\psi\_{\rm cd2} = \mathfrak{d}\_{\rm cd1} + A(\mathbf{s}) \Big(\hat{d}\_2 - \hat{d}\_1\Big) \tag{28}
$$

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Assuming *v*ˆ*in* = 0, hence *v*ˆ*cd*<sup>1</sup> + *v*ˆ*cd*<sup>2</sup> + *v*ˆ*cd*<sup>3</sup> = 0, Equation (28) can be generalized as follows:

$$
\psi\_{cdj} = \psi\_{cd1} + A(s)(\hat{d}\_j - \hat{d}\_1) \tag{29}
$$

Setting *v*ˆ*in* = 0, and substituting (29) in (11) would result in:

$$\sum\_{j=1}^{3} \hat{v}\_{cd1} + A(s) \left(\hat{d}\_j - \hat{d}\_1\right) = 0 \tag{30}$$

$$3\mathfrak{d}\_{cd1} - 3A(s)\hat{d}\_1 + \sum\_{j=1}^{3} A(s)\hat{d}\_j = 0 \tag{31}$$

Therefore,

$$\vartheta\_{\rm cd1} = A(s)\hat{d}\_1 + \frac{A(s)}{3} \sum\_{j=1}^{3} \hat{d}\_j \tag{32}$$

Substituting (32) in (29) would result in:

$$\psi\_{\rm cdj} = A(s)\hat{d}\_{\hat{j}} + \frac{A(s)}{3} \sum\_{j=1}^{3} \hat{d}\_{\hat{j}} \tag{33}$$

Presenting (33) in a matrix form would give:

$$
\begin{bmatrix}
\mathfrak{d}\_{cd1} \\
\mathfrak{d}\_{cd2} \\
\mathfrak{d}\_{cd3}
\end{bmatrix} = \begin{bmatrix}
\frac{2A(s)}{3} & \frac{-A(s)}{3} & \frac{-A(s)}{3} \\
\frac{-A(s)}{3} & \frac{2A(s)}{3} & \frac{-A(s)}{3} \\
\frac{-A(s)}{3} & \frac{-A(s)}{3} & \frac{2A(s)}{3}
\end{bmatrix} \begin{bmatrix}
\hat{d}\_1 \\
\hat{d}\_2 \\
\hat{d}\_3
\end{bmatrix} \tag{34}
$$

## 2.2.4. Converter Output Impedance

As studied in the SSM presented in [39,40], the ISOS converter output impedance can be found by modifying (8), such that:

$$\begin{cases} \begin{array}{l} \hat{\mathbf{i}}\_{L1} + \hat{\mathbf{i}}\_{out} = s\mathbf{C}\boldsymbol{\upbeta}\_{out1} + \frac{\boldsymbol{\upbeta}\_{out}}{R} \\ \hat{\mathbf{i}}\_{L2} + \hat{\mathbf{i}}\_{out} = s\mathbf{C}\boldsymbol{\upbeta}\_{out2} + \frac{\boldsymbol{\upbeta}\_{out}}{R} \\ \hat{\mathbf{i}}\_{L3} + \hat{\mathbf{i}}\_{out} = s\mathbf{C}\boldsymbol{\upbeta}\_{out3} + \frac{\boldsymbol{\upbeta}\_{out}}{R} \end{array} \tag{35}$$

Accordingly, Equation (9) is modified as follows:

$$\sum\_{j=1}^{3} \hat{\mathbf{i}}\_{Lj} = \hat{\mathbf{v}}\_{out} \left( \mathbf{s} \mathbf{C} + \frac{3}{R} \right) - 3 \hat{\mathbf{i}}\_{out} \tag{36}$$

The output voltage and the output current relationship is found by assuming *v*ˆ*in* = 0, and ˆ *dj* = 0, *j* = 1, 2, and 3, summing the KVL Equations in (6), and substituting (2), (4), (10), (11), and (36).

$$\left( (sL + R\_d) \left( sC + \frac{3}{R} \right) + 1 \right) \hat{v}\_{out} = 3 \left( R\_d + sL \right) \hat{i}\_{out} \tag{37}$$

Simplifying (37) would result in (38):

$$Z\_{out} = \frac{\mathfrak{d}\_{out}}{\mathfrak{d}\_{out}} = \frac{\mathfrak{d}(R\_d + sL)}{s^2LC + s\left(\frac{3L}{R} + R\_dC\right) + \frac{3R\_d}{R} + 1} \tag{38}$$
