2.2.3. Output Impedance

The generalized converter output impedance for the hybrid ISIP-OSOP multimodule DC-DC converter can be found by considering two groups of equations. The primary group is the *L* number of KCL equations presented in (12). However, the secondary group is the *M* number of KCL equations presented in (12).

To find the generalized converter output impedance, the KCL equation in (12) can be rewritten as follows: ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ˆˆˆˆ

$$\begin{bmatrix} i\_{L111} + i\_{L121} + \cdots + i\_{L1\ell\_{11}1} + i\_{outL} = g\_L \hat{v}\_{outL1} + \frac{v\_{outL}}{R} \\ i\_{L112} + i\_{L122} + \cdots + i\_{L1\ell\_{11}2} + i\_{outL} = g\_L \hat{v}\_{outL2} + \frac{v\_{outL}}{R} \\ \vdots \\ i\_{L1\ell\_{11}} + i\_{L12\ell\_{11}} + \cdots + i\_{L1\ell\_{11}\ell\_{11}} + i\_{outL} = g\_L \hat{v}\_{outL1} + \frac{v\_{outL}}{R} \\ i\_{L1\ell\_{11}} + i\_{L1\ell\_{11}2} + \cdots + i\_{L\ell\_{11}\ell\_{11}} + i\_{outM} = g\_M \hat{v}\_{outM1} + \frac{v\_{outM}}{R} \\ \vdots \\ i\_{L1\ell\_{11}} + i\_{L1\ell\_{11}2} + \cdots + i\_{L\ell\_{11}\ell\_{11}} + i\_{outM} = g\_M \hat{v}\_{outM2} + \frac{v\_{outM}}{R} \\ \vdots \\ i\_{L1\ell\_{11}} + i\_{L1\ell\_{11}2} + \cdots + i\_{L\ell\_{11}\ell\_{11}} + i\_{outM} = g\_M \hat{v}\_{outM1} + \frac{i\_{outM}}{R} \end{bmatrix} \tag{35}$$

where; *gL* = *sCL sRcLCL*+1 and ݃ெ ൌ ௦ಾ ௦ோಾಾାଵ . Accordingly, the KCL equation in (14) can be modified as follows:

$$\sum\_{i=1}^{a\_{L1}} \sum\_{j=1}^{b\_{L1}} \hat{\imath}\_{LLij} + \hat{\imath}\_{outL} = \hat{\upsilon}\_{outL} \left( \frac{sRC\_L + s\nu\_{L1}R\_{cL}C\_L + b\_{L1}}{R(1 + sR\_{cL}C\_L)} \right) \tag{36}$$

The relationship between the output voltage and the output current for the *L* modules is obtained by adding the *L* equations in (11), assuming *<sup>v</sup>*<sup>ˆ</sup>*inL* = 0, and ˆ *dLj* = 0, *j* = 1, 2, ... , *L*, and substituting (2), (4), (17), (18) and (36).

*Deff* 1 *K*1 γ*L<sup>v</sup>* ˆ *inL* + β*L*2 *Vin* β*L*1 *K*1 ⎛⎜⎜⎜⎜⎜⎜⎜⎝ −β*L*1*K*1*RdL* β*L*2 *Vin v*<sup>ˆ</sup>*outLsRCL*+*sbL*1*RcLCL*+*bL*<sup>1</sup> *<sup>R</sup>*(<sup>1</sup>+*sRcLCL*) − ˆ*ioutL*+ *aL*2*bL*1β*L*1*RdLDeff* 1 *aL*1*bL*2β*L*2 *RVin* γ*Lv*<sup>ˆ</sup>*inL* + ˆ*dL*1 ⎞⎟⎟⎟⎟⎟⎟⎟⎠ = *sLLsRCL*+*sbL*1*RcLCL*+*bL*<sup>1</sup> *<sup>R</sup>*(<sup>1</sup>+*sRcLCL*) *v*<sup>ˆ</sup>*outL* − ˆ*ioutL* + *cLv*<sup>ˆ</sup>*outL* (37)

$$\begin{split} -R\_{\rm dL} \Big( \mathfrak{H}\_{\rm outL} \Big( \frac{sR \, \mathrm{C}\_{\rm L} + s \, \mathrm{i} \, \mathrm{I}\_{\rm L} \mathrm{R}\_{\rm L} \mathrm{C}\_{\rm L} + \mathrm{i} \, \mathrm{i} \, \mathrm{I}\_{\rm L}}{R \, (1 + s \, \mathrm{R}\_{\rm L} \mathrm{C}\_{\rm L})} \Big) - \hat{\mathfrak{I}}\_{\rm outL} \Big) \\ = sL\_{\rm L} \Big( \Big( \frac{sR \, \mathrm{C}\_{\rm L} + s \, \mathrm{i} \, \mathrm{I}\_{\rm L} \, \mathrm{R}\_{\rm L} \mathrm{C}\_{\rm L} + \mathrm{i} \, \mathrm{I}\_{\rm L}}{R \, (1 + s \, \mathrm{R}\_{\rm L} \mathrm{C}\_{\rm L})} \Big) \mathfrak{H}\_{\rm outL} - \hat{\mathfrak{I}}\_{\rm outL} \Big) + c\_{\rm L} \mathfrak{H}\_{\rm outL} \end{split} \tag{38}$$

Simplifying (38) would result in (39).

$$\begin{split} Z\_{\text{outL}} &= \frac{\theta\_{\text{outL}}}{s\_{\text{wall}}}\\ &= \frac{s\_{\text{L}1} \left(R\_{\text{dL}} + sL\_{L}\right) \left(1 + sR\_{\text{L}}C\_{L}\right)}{s^{2}L\_{L}C\_{L}\left(1 + \frac{b\_{\text{L}1}R\_{\text{L}}}{K}\right) + s\left(\frac{b\_{\text{L}1}L\_{L}}{K} + R\_{\text{dL}}C\_{L}\left(1 + \frac{b\_{\text{L}1}R\_{\text{L}}}{K}\right) + c\_{\text{L}}R\_{\text{dL}}C\_{L}\right) + \frac{b\_{\text{L}1}R\_{\text{L}}}{K} + c\_{\text{L}}}\end{split} \tag{39}$$

Similarly, the KCL equation in (16) can be modified as follows:

$$\sum\_{l=1}^{641} \sum\_{f=1}^{641} \mathbb{I}\_{LMf} + \mathbb{I}\_{outM} = \mathbb{\partial}\_{outM} \left( \frac{\mathbf{s}RC\_M + \mathbf{s}\mathbf{b}\_{M1}R\_{cM}C\_M + \mathbf{b}\_{M1}}{R\left(1 + sR\_{cM}C\_M\right)} \right) \tag{40}$$

The relationship between the output voltage and the output current for the *M* modules is derived by summing the *M* equations in (11), assuming ˆ *vinM* = 0, and ˆ *dMj* = 0, *j* = 1, 2, ... , *M*, and substituting (7), (9), (19), (20) and (40).

$$\begin{aligned} \frac{B\_{off2}}{K\_2} \mathbf{y}\_M \boldsymbol{\hat{\upmu}}\_{\text{inM}} + \frac{\boldsymbol{\hat{\upmu}}\_{M1} \boldsymbol{V}\_{\text{in}}}{\mathcal{B}\_{M1} K\_2} \\ &= \left( -\frac{\boldsymbol{\mathcal{B}}\_{M1} \boldsymbol{K}\_2 \boldsymbol{R}\_{\text{dM}}}{\mathcal{B}\_{M2} \boldsymbol{V}\_{\text{in}}} \left( \boldsymbol{\hat{\upmu}}\_{\text{outM}} \left( \frac{s \boldsymbol{R} \boldsymbol{C}\_{M} + s \boldsymbol{\hat{\upmu}}\_{M1} \boldsymbol{R}\_{\text{cM}} \boldsymbol{C}\_{M} + \boldsymbol{\hat{\upmu}}\_{M1}}{\mathcal{R} (1 + s \boldsymbol{R}\_{\text{cM}} \boldsymbol{C}\_{M})} \right) - \boldsymbol{\hat{\upmu}}\_{\text{outM}} \right) + \right) \\ &\qquad \frac{a\_{M2} \boldsymbol{b}\_{M1} \boldsymbol{R}\_{\text{cM}} \boldsymbol{R}\_{\text{cM}} \boldsymbol{R}\_{\text{cM}} \boldsymbol{R}\_{\text{cM}}}{a\_{M1} \boldsymbol{b}\_{M2} \boldsymbol{R}\_{\text{cM}} \boldsymbol{R}\_{\text{cM}}} \mathbf{y}\_M \boldsymbol{\hat{\upmu}}\_{\text{inM}} + \boldsymbol{\hat{\upmu}}\_{M1} \end{aligned} \tag{41}$$

$$\begin{split} s\boldsymbol{L}\_{M} \Big( \Big( \frac{\mathrm{sRC}\_{M} + \mathrm{s\boldsymbol{b}}\_{M1}\mathrm{R}\_{cM}\mathrm{C}\_{M} + \mathrm{b}\_{M1}}{R\left(1 + \mathrm{s\boldsymbol{R}}\_{cM}\mathrm{C}\_{M}\right)} \Big) \boldsymbol{\vartheta}\_{outM} - \boldsymbol{\mathfrak{t}}\_{outM} \Big) + \mathbf{c}\_{\mathrm{H}} \,\boldsymbol{\mathfrak{\beta}}\_{outM} \\ & \qquad - \boldsymbol{R}\_{dM} \Big( \boldsymbol{\mathfrak{\beta}}\_{outM} \Big( \frac{\mathrm{s\boldsymbol{R}}\mathrm{C}\_{M} + \mathrm{s\boldsymbol{b}}\_{M1}\mathrm{R}\_{cM}\mathrm{C}\_{M} + \mathrm{b}\_{M1}}{R\left(1 + \mathrm{s\boldsymbol{R}}\_{cM}\mathrm{C}\_{M}\right)} \Big) - \boldsymbol{\mathfrak{t}}\_{outM} \Big) \\ & \qquad = \mathrm{s\boldsymbol{L}\_{M}} \Big( \Big( \frac{\mathrm{s\boldsymbol{R}}\mathrm{C}\_{M} + \mathrm{s\boldsymbol{b}}\_{M1}\mathrm{R}\_{cM}\mathrm{C}\_{M} + \mathrm{b}\_{M1}}{R\left(1 + \mathrm{s\boldsymbol{R}}\_{cM}\mathrm{C}\_{M}\right)} \Big) \boldsymbol{\upbeta}\_{outM} - \boldsymbol{\mathfrak{t}}\_{outM} \Big) + \boldsymbol{\mathfrak{c}}\_{\mathrm{M}} \boldsymbol{\mathfrak{\beta}}\_{outM} \end{split} \tag{42}$$

Simplifying (42) would result in (43):

$$\begin{split} Z\_{outM} &= \frac{\mathfrak{d}\_{outM}}{\mathfrak{f}\_{outM}}\\ &= \frac{b\_{M1}(R\_{dM} + sL\_{M})(1 + sR\_{CM}C\_{M})}{s^{2}L\_{M}C\_{M}\left(1 + \frac{b\_{M1}R\_{CM}}{R}\right) + s\left(\frac{b\_{M1}L\_{M}}{R} + R\_{dM}C\_{M}\left(1 + \frac{b\_{M1}R\_{CM}}{R}\right) + c\_{M}R\_{CM}C\_{M}\right) + \frac{b\_{M1}R\_{dM}}{R} + c\_{M}}\end{split} \tag{43}$$

By adding *ZoutL* and *ZoutM*, the output impedance transfer function can be found.
