Modeling and Design of Split-Pi Converter

Abstract

High-power bidirectional dc–dc converters are being widely employed in renewable energy interfacing, energy storage, electric vehicle charging, military, aerospace, and marine applications. Among various bidirectional topologies documented in the literature for dc–dc power conversion, the split-pi converter invites special attention with regard to applications involving multi-phase systems requiring high-power density. This paper endeavors to present the small-signal modeling of the split-pi converter in its various operating modes. Subsequently, the dynamic characteristics of the converter are studied, and appropriate control design is presented for stable operation of the converter. Frequency response plots are illustrated, and a hardware prototype model of the converter is designed and implemented.

1. Introduction

Multiple applications involving aerospace, military, marine, and transport electrification and renewable energy integration necessitate high-efficiency dc power converters with stable bidirectional operation and high-power density. Quite a lot of topologies have been proposed by researchers for bidirectional dc–dc conversion in the past few years to suit various applications [1,2,3,4,5,6,7,8,9,10,11,12]. These topologies are either isolated or nonisolated allowing a bidirectional power flow. Further, these topologies operating in boost, buck, or buck–boost mode depending on the placement of passive elements. Amid various topologies proposed in the literature, the split-pi converter (SPC) is the one that draws exclusive interest owing to its remarkable features viz., high conversion efficiency, capability to operate in various switching modes, and bidirectional power flow. Additionally, its nonisolated structure and high-frequency operation along with the lower ratings of passive elements make it an apt choice for applications demanding high-power density [13,14]. Unlike the classic buck–boost topology, this topology overcomes the voltage gain limitations of the converter and is prone to minimal voltage and current fluctuations. Parallel connection is also possible with this converter topology and hence can be used in multiphase systems. A reduction in switching noise is achieved as the LC filters at either port of the converter favor small reactive components. Moreover, the control circuit of SPC topology involves simple duty cycle control of the PWM signal and is hence less laborious [15,16,17].

The aforementioned advantages of the SPC makes it ideally suited for electric vehicles and energy storage applications, among others. With the direction of current flow changing from instant to instant in these applications, a dynamic analysis of the converter is essential. In addition, an appropriate control design is essential in enhancing the overall performance of the system. Hence, this paper intends to present in detail the small-signal modeling of the SPC along with the control design for all the switching modes. Though the SPC topology was introduced in 2014 [18], very little literature is available related to the modeling and control design of the converter. A state–space averaging model of the SPC followed by a type-II compensator design is described in [19]. However, the state–space model is derived only for boost mode of operation of the converter, and the prototype design and experimental results are not presented. In [20], the authors have proposed a multilevel and symmetrical structure for the SPC and compared it with the interleaved topology. Few details pertaining to the control design with controlled current or voltage feedback and dc link voltage have been presented. Elaborate state–space models of the SPC with respect to energy storage interface of stiff and nonstiff dc microgrids have been discussed in [21]. A unique control strategy for controlling the flywheel system using the SPC in rural applications has been described in [22]. Thus, a generalized modeling and control design for different operation modes of the conventional SPC topology is not available in the literature according to the authors.

This paper attempts to explain in detail the small-signal modeling of the SPC topology followed by the control design for various operating modes. In addition, a stability analysis was carried out through frequency response plots, and a stable control design is presented for both the buck and boost modes of operation of the converter. An experimental prototype of the SPC was developed with the designed closed loop control, and the results are demonstrated to validate the design.

2. Small-Signal Modeling: Split-Pi Converter

The schematic of SPC topology is illustrated in Figure 1. The circuit comprises a boost topology followed by a buck topology. Thus, this converter is capable of operating in three different modes, namely, the boost, buck, and pass-through modes. Of the four switches available, only two conducts at any point of time in all the three switching modes provide the required output. While the buck and boost modes of operation are already familiar, the pass-through mode simply passes the input to the output. The mathematical analysis of the different modes of operation is presented below followed by the small-signal modeling.

Figure 1. Circuit topology of SPC.

2.1. Boost Mode

In the boost mode of operation, switch S4 is always ON, and S3 is always OFF. S1 and S2 are switched alternatively so that boost voltage is available at the output. The two modes of operation are described below

Mode 1:

The circuit schematic during mode 1 is depicted in Figure 2. Switches S1 and S3 are OFF, and S2 and S4 are turned ON. The differential equations governing the circuit operation are listed as follows:

Figure 2. Circuit schematic of SPC during mode 1 under boost operation.

Applying KVL and KCL to the circuit in Figure 2, we have the following set of equations for the voltage across the inductors and current through the capacitors:

$$L_1\frac{d{i}_1}{dt}=V_{in}-V_{c2}=V_{c1}-V_{c2}$$

(1)

$$L_2\frac{d{i}_2}{dt}=V_{c2}-V_{c3}$$

(2)

$$C_1\frac{d{v}_{c1}}{dt}=I_{in}-I_1$$

(3)

$$I_{in}-C_1\frac{d{v}_{c1}}{dt}=I_1$$

(4)

$$C_2\frac{d{v}_{c2}}{dt}=I_1-I_2$$

(5)

$$C_3\frac{d{v}_{c3}}{dt}=I_2-\frac{V_o}{R}$$

(6)

The state Equations (1)–(5) can be expressed as

$$\dot{X}=A_{1}X+B_{1}U$$

(7)

$$\left[\begin{array}{c}\dot{I_1}\\ \dot{I_2}\\ \dot{V_{c1}}\\ \dot{V_{c2}}\\ \dot{V_{c3}}\end{array}\right]=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\left[I_{in}\right]$$

$$\mathrm{where}{A}_1=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\mathrm{and}{B}_1=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]$$

The output equation for this mode is

$$V_{o }=V_{c3}$$

(8)

Equation (8) can be expressed as

$$Y={\complement }_{1}X+D_1$$

(9)

$$\mathrm{Therefore},V_{o }=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]\mathrm{where} {\complement }_1=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]$$

Mode 2:

The circuit schematic during mode 2 is depicted in Figure 3. Switches S2 and S3 are OFF, and S1 and S4 are turned ON. The differential equations governing the circuit operation are listed as follows:

Figure 3. Circuit schematic of SPC during mode 2 under boost operation.

Applying KVL and KCL to the circuit above, we have the following set of equations for voltage across the inductors and current through the capacitors:

$$L_1\frac{d{i}_1}{dt}=V_{c1}$$

(10)

$$L_2\frac{d{i}_2}{dt}=V_{c2}-V_{c3}$$

(11)

$$C_1\frac{d{v}_{c1}}{dt}=I_{in}-I_1$$

(12)

$$C_2\frac{d{v}_{c2}}{dt}=-I_2$$

(13)

$$C_3\frac{d{v}_{c3}}{dt}=I_2-\frac{V_o}{R}$$

(14)

Combining Equations (10)–(14) in matrix form,

$$\dot{X}=A_{2}X+B_{2}U$$

(15)

$$\left[\begin{array}{c}\dot{I_1}\\ \dot{I_2}\\ \dot{V_{c1}}\\ \dot{V_{c2}}\\ \dot{V_{c3}}\end{array}\right]=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ 0& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\left[I_{in}\right]$$

$$\mathrm{where}{A}_2=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ 0& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\mathrm{and}{B}_2=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]$$

The output equation is the same as Equation (8).

Applying state–space averaging, the state coefficient matrix A can be expressed as

$$\begin{array}{c}\overline{A }=A_{1}d+A_2\left(1-d\right)\\ =\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\times d\\ +\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ 0& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\times \left(1-d\right)\end{array}$$

$$\mathrm{Therefore},\overline{A}=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]$$

(16)

where d is the duty cycle of switch S1.

The averaged source coefficient matrix is given by

$$\overline{B }=B_{1}d+B_2\left(1-d\right)=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\times d+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\times \left(1-d\right)=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]$$

(17)

Similarly,

$$\overline{\complement }={\complement }_{1}d+{\complement }_2\left(1-d\right)=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]$$

(18)

Thus, the state–space averaged equations in matrix form are given by

$$\dot{X }=\overline{A}X+\overline{B }U=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\left[I_{in}\right]$$

(19)

The above matrix can be expanded to give the averaged state–space equations:

$$\frac{d{i}_1}{dt}=\frac{V_{c1}}{L_1}-\frac{d\times V_{c2}}{L_1}$$

(20)

$$\frac{d{i}_2}{dt}=\frac{V_{c2}}{L_2}-\frac{V_{c3}}{L_2}$$

(21)

$$\frac{d{V}_{c1}}{dt}=\frac{-I_1}{C_1}+\frac{I_{in}}{C_1}$$

(22)

$$\frac{d{V}_{c2}}{dt}=\frac{d\times I_1}{C_2}-\frac{I_2}{C_2}$$

(23)

$$\frac{d{V}_{c3}}{dt}=\frac{I_2}{C_3}-\frac{V_{c3}}{R\times C_3}$$

(24)

The small-signal model is used for deriving the compensators that can realize the control objective of keeping the output voltage constant while drawing constant power from the sources. In order to derive small-signal models, the averaged equations must contain small signal variables. This is done by replacing each variable with both a DC and an AC value. Assuming a stiff voltage across the capacitor C2 and applying small-signal perturbations to other electrical quantities, the Equations (20)–(22) and (24) are represented as

$$\frac{d\left(i_1+{\widehat{i}}_1\right)}{dt}=\frac{(V_{c1}+{\widehat{V}}_{c1})}{L_1}-\frac{(d+\widehat{d})\times V_{c2}}{L_1}=\frac{(V_{c1}-d\times V_{c2})}{L_1}+\frac{{\widehat{V}}_{c1}-\widehat{d}\times V_{c2}}{L_1}$$

(25)

$$\frac{d\left(i_2+{\widehat{i}}_2\right)}{dt}=\frac{V_{c2}}{L_2}-\frac{\left(V_{c3}+{\widehat{V}}_{c3}\right)}{L_2}=\frac{V_{c2}-V_{c3}}{L_2}-\frac{{\widehat{V}}_{c3}}{L_2}$$

(26)

$$\frac{d(V_{c1}+{\widehat{V}}_{c1})}{dt}=\frac{-(I_1+{\widehat{I}}_1)}{C_1}+\frac{(I_{in}+{\widehat{I}}_{in})}{C_1}=\frac{I_{in}-I_1}{C_1}+\frac{{\widehat{I}}_{in}-{\widehat{I}}_1}{C_1}$$

(27)

$$\frac{d(V_{c3}+{\widehat{V}}_{c3})}{dt}=\frac{(I_2+{\widehat{I}}_2)}{C_3}-\frac{(V_{c3}+{\widehat{V}}_{c3})}{R\times C_3}=\frac{I_2-V_{c3}/R}{C_3}+\frac{{\widehat{I}}_2-{\widehat{V}}_{c3}/R}{C_3}$$

(28)

The dc or steady-state solutions of the above set of equations are:

$$V_{c1}-d\times V_{c2}=0\to V_{c1}=d\times V_{c2}$$

(29)

$$V_{c2}-V_{c3}=0\to V_{c2}=V_{c3}$$

(30)

$$I_{in}-I_1=0\to I_1=I_{in}$$

(31)

$$I_2-\frac{V_{c3}}{R}=0\to I_2=\frac{V_{c3}}{R}$$

(32)

The dynamic or ac solutions are:

$$L_1\frac{d{\widehat{i}}_1}{dt}={\widehat{V}}_{c1}-\widehat{d}\times V_{c2}$$

(33)

$$L_2\frac{d{\widehat{i}}_2}{dt}=-{\widehat{V}}_{c3}$$

(34)

$$C_1\frac{d{\widehat{V}}_{c1}}{dt}={\widehat{I}}_{in}-{\widehat{I}}_1$$

(35)

$$C_3\frac{d{\widehat{V}}_{c3}}{dt}={\widehat{I}}_2-\frac{{\widehat{V}}_{c3}}{R}$$

(36)

Representing Equations (33)–(36) in matrix form,

$$\left[\begin{array}{c}\frac{d{\widehat{i}}_1}{dt}\\ \frac{d{\widehat{i}}_2}{dt}\\ \frac{d{\widehat{V}}_{c1}}{dt}\\ \frac{d{\widehat{V}}_{c3}}{dt}\end{array}\right]=\left[\begin{array}{cccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{\widehat{I}}_1\\ {\widehat{I}}_2\\ {\widehat{V}}_{c1}\\ {\widehat{V}}_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ {\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}_1\\ 0\end{array}\right]\left[{\widehat{I}}_{in}\right]+\left[\begin{array}{c}\raisebox{1ex}{$-V_{c2}$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.\\ 0\\ 0\\ 0\end{array}\right]\left[\widehat{d}\right]$$

The above matrix equation is of the form

$$\begin{array}{c}\dot{\hat{X}}=\tilde{A}\hat{X}+\tilde{B}\tilde{U}+E\widehat{d},\mathrm{where}\tilde{A}=\left[\begin{array}{cccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right],\tilde{B}=\left[\begin{array}{c}0\\ 0\\ {\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}_1\\ 0\end{array}\right]\mathrm{and}\\ E=\left[\begin{array}{c}\raisebox{1ex}{$-V_{c2}$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.\\ 0\\ 0\\ 0\end{array}\right]\end{array}$$

Laplace transform of the above equation yields

$$s\hat{X}\left(s\right)=\tilde{A}\hat{X}\left(s\right)+\tilde{B}\tilde{U}\left(s\right)+E\widehat{d}\left(s\right)$$

(37)

$$\hat{X}\left(s\right)={\left(sI-\tilde{A}\right)}^{-1}\tilde{B}\tilde{U}\left(s\right)+{\left(sI-\tilde{A}\right)}^{-1}E\widehat{d}\left(s\right)$$

(38)

where $I$ is the identity matrix. In Equation (38), $\widehat{d}$ is a function of x and u and termed as the control law. The control law is usually nonlinear, and its linearized model can be written as

$$\widehat{d}\left(s\right)=F^T\left(s\right)X\left(s\right)+Q^T\left(s\right)U\left(s\right)$$

(39)

where $F^T\left(s\right)=\left[\begin{array}{cccc}0& 0& \raisebox{1ex}{$-H\left(s\right)$}\!\left/ \!\raisebox{-1ex}{$V_p$}\right.& \raisebox{1ex}{$-H\left(s\right)$}\!\left/ \!\raisebox{-1ex}{$V_p$}\right.\end{array}\right]$ and $Q^T\left(s\right)=\left[\begin{array}{cccc}0& 0& 0& 0\end{array}\right]$ are the coefficient matrices for voltage mode PWM control, and H(s) denotes the transfer function of the error amplifier and its compensation network, and $V_p$ is the peak amplitude of the sawtooth signal [23].

Substituting Equation (39) into Equation (38), we have

$$\hat{X}\left(s\right)={\left[sI-\tilde{A}-E{F}^T\left(s\right)\right]}^{-1}\left[\tilde{B}+E{Q}^T\left(s\right)\right]\tilde{U}\left(s\right)$$

(40)

The source-to-state transfer functions for the SPC can be obtained by substituting for $F^T\left(s\right)$ and $Q^T\left(s\right)$ into Equation (40). Thus, the state variable matrix is

$$\left[\begin{array}{c}{\widehat{I}}_1\\ {\widehat{I}}_2\\ {\widehat{V}}_{c1}\\ {\widehat{V}}_{c3}\end{array}\right]={\left[\begin{array}{cccc}s& 0& \frac{1}{L_1}-\frac{H\left(s\right)}{V_p}\times \frac{V_{c2}}{L_1}& -\frac{H\left(s\right)}{V_p}\times \frac{V_{c2}}{L_1}\\ 0& s& 0& -\frac{1}{L_2}\\ \frac{-1}{C_1}& 0& s& 0\\ 0& \frac{1}{C_3}& 0& s-\frac{1}{R{C}_3}\end{array}\right]}^{-1}\left[\begin{array}{c}0\\ 0\\ {\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}_1\\ 0\end{array}\right]\left[{\widehat{I}}_{in}\right]$$

(41)

From Equation (41), the small-signal transfer functions can be obtained,

$$\frac{{\widehat{I}}_1}{{\widehat{I}}_{in}}=\frac{-\left(V_p-H\left(s\right)V_{c2}\right)}{C_1{L}_1{V}_p{s}^2+V_p-\left(H\left(s\right)V_{c2}\right)}$$

$$\frac{{\widehat{V}}_{c1}}{{\widehat{I}}_{in}}=\frac{L_1{V}_{p}s}{C_1{L}_1{V}_p{s}^2+V_p-\left(H\left(s\right)V_{c2}\right)}$$

The small-signal control transfer function can be obtained by substituting $\tilde{U}\left(s\right)=0$ in Equation (38).

Therefore,

$$\left[\begin{array}{c}{\widehat{I}}_1\\ {\widehat{I}}_2\\ {\widehat{V}}_{c1}\\ {\widehat{V}}_{c3}\end{array}\right]={\left[\begin{array}{cccc}s& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& s& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& s& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& s-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]}^{-1}\left[\begin{array}{c}\raisebox{1ex}{$-V_{c2}$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.\\ 0\\ 0\\ 0\end{array}\right]\left[\widehat{d}\right]$$

(42)

The open-loop transfer function of inductor current L₁ is thus given by

$$\frac{{\widehat{I}}_1}{\widehat{d}}=-\frac{s{C}_1{V}_{c2}}{1+s^2{C}_1{L}_1}$$

(43)

and the open-loop transfer function of input capacitor voltage C₁ is given by

$$\frac{{\widehat{V}}_{c1}}{\widehat{d}}=-\frac{V_{c2}}{1+s^2{C}_1{L}_1}$$

(44)

2.2. Buck Mode

In the buck mode of operation, switch S2 is always ON, and S1 is always OFF. S3 and S4 are switched alternatively so that the buck voltage is available at the output. The two operation modes are described below [24].

Mode 1:

The circuit schematic during mode 1 is depicted in Figure 4. Switches S1 and S4 are OFF, and S2 and S3 are turned ON. The differential equations governing the circuit operation are listed as follows:

$$L_1\frac{d{i}_1}{dt}=V_{in}-V_{c2}=V_{c1}-V_{c2}$$

(45)

$$L_2\frac{d{i}_2}{dt}=-V_{c3}$$

(46)

$$C_1\frac{d{v}_{c1}}{dt}=I_{in}-I_1$$

(47)

$$C_2\frac{d{v}_{c2}}{dt}=I_1$$

(48)

$$C_3\frac{d{v}_{c3}}{dt}=I_2-\frac{V_o}{R}$$

(49)

Figure 4. Circuit schematic of SPC during mode 1 under buck operation.

The state Equations (45)–(49) can be expressed as

$$\begin{gathered} \dot{X}=A_{3}X+B_{3}U \\ \left[\begin{array}{c}\dot{I_1}\\ \dot{I_2}\\ \dot{V_{c1}}\\ \dot{V_{c2}}\\ \dot{V_{c3}}\end{array}\right]=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\left[I_{in}\right] \end{gathered}$$

(50)

$$\mathrm{where}{A}_3=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right] \mathrm{and}{B}_3=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]$$

The output equation for this mode is

$$V_{o }=V_{c3}$$

(51)

Equation (51) can be written in matrix form as

$$Y={\complement }_{3}X+D_{3}U$$

$$V_{o }=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]\mathrm{where} {\complement }_3=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]$$

Mode 2:

Figure 5 represents the circuit schematic during mode 2. Switches S1 and S3 are OFF, and S2 and S4 are turned ON. The differential equations describing the circuit operation are given below [25]:

Figure 5. Circuit schematic of SPC during mode 2 under buck operation.

Applying KVL and KCL to the circuit above, we have the following set of equations for the voltage across the inductors and current through the capacitors:

$$L_1\frac{d{i}_1}{dt}=V_{in}-V_{c2}=V_{c1}-V_{c2}$$

(52)

$$L_2\frac{d{i}_2}{dt}=V_{c2}-V_{c3}$$

(53)

$$C_1\frac{d{v}_{c1}}{dt}=I_{in}-I_1$$

(54)

$$C_2\frac{d{v}_{c2}}{dt}=I_1-I_2$$

(55)

$$C_3\frac{d{v}_{c3}}{dt}=I_2-\frac{V_o}{R}$$

(56)

The state Equations (52)–(56) can be written in matrix form as

$$\dot{X}=A_{4}X+B_{4}U$$

(57)

$$\left[\begin{array}{c}\dot{I_1}\\ \dot{I_2}\\ \dot{V_{c1}}\\ \dot{V_{c2}}\\ \dot{V_{c3}}\end{array}\right]=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\left[I_{in}\right]$$

$$\mathrm{where}{A}_4=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\mathrm{and}{B}_4=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]$$

The output equation for this mode is

$$V_{o }=V_{c3}$$

(58)

Equation (58) can be expressed as

$$Y={\complement }_{4}X+D_{4}U$$

(59)

$$V_{o }=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]\mathrm{where}{\complement }_4=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]$$

Applying state–space averaging, the state coefficient matrix A can be written as

$$\begin{array}{c}\overline{A }=A_{3}d+A_4\left(1-d\right)\\ =\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\times d\\ +\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\times \left(1-d\right)\end{array}$$

$$\mathrm{Therefore},\overline{A}=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$\left(1-d\right)$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-\left(1-d\right)$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]$$

(60)

where d is the duty cycle of the converter.

The averaged source coefficient matrix is given by

$$\overline{B }=B_{3}d+B_4\left(1-d\right)=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\times d+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\times \left(1-d\right)=\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]$$

(61)

Similarly,

$$\overline{\complement }={\complement }_{3}d+{\complement }_4\left(1-d\right)=\left[\begin{array}{ccccc}0& 0& 0& 0& 1\end{array}\right]$$

(62)

Thus, the state–space averaged equations in matrix form are given by

$$\dot{X }=\overline{A}X+\overline{B }U=\left[\begin{array}{ccccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& \raisebox{1ex}{$\left(1-d\right)$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0& 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& \raisebox{1ex}{$-\left(1-d\right)$}\!\left/ \!\raisebox{-1ex}{$C_2$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ V_{c1}\\ V_{c2}\\ V_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.\\ 0\\ 0\end{array}\right]\left[I_{in}\right]$$

The above matrix can be expanded to give the averaged state–space equations:

$$\frac{d{i}_1}{dt}=\frac{V_{c1}}{L_1}-\frac{V_{c2}}{L_1}$$

(63)

$$\frac{d{i}_2}{dt}=\frac{\left(1-d\right)V_{c2}}{L_2}-\frac{V_{c3}}{L_2}$$

(64)

$$\frac{d{V}_{c1}}{dt}=\frac{-I_1}{C_1}+\frac{I_{in}}{C_1}$$

(65)

$$\frac{d{V}_{c2}}{dt}=\frac{d\times I_1}{C_2}-\frac{\left(1-d\right)I_2}{C_2}$$

(66)

$$\frac{d{V}_{c3}}{dt}=\frac{I_2}{C_3}-\frac{V_{c3}}{R\times C_3}$$

(67)

Assuming a stiff voltage across the capacitor C₂ and applying small-signal perturbations to other electrical quantities, the Equations (63)–(65) and (67) are represented as

$$\frac{d\left(i_1+{\widehat{i}}_1\right)}{dt}=\frac{(V_{c1}+{\widehat{V}}_{c1})}{L_1}-\frac{V_{c2}}{L_1}=\frac{(V_{c1}-V_{c2})}{L_1}+\frac{{\widehat{V}}_{c1}}{L_1}$$

(68)

$$\begin{gathered} \frac{d\left(i_2+{\widehat{i}}_2\right)}{dt}=\frac{(1-\left(d+\widehat{d)}\right)V_{c2}}{L_2}-\frac{\left(V_{c3}+{\widehat{V}}_{c3}\right)}{L_2} \\ =\frac{1-d\times V_{c2}-V_{c3}}{L_2}-\frac{\widehat{d}\times V_{c2}+{\widehat{V}}_{c3}}{L_2} \end{gathered}$$

(69)

$$\frac{d(V_{c1}+{\widehat{V}}_{c1})}{dt}=\frac{-(I_1+{\widehat{I}}_1)}{C_1}+\frac{(I_{in}+{\widehat{I}}_{in})}{C_1}=\frac{I_{in}-I_1}{C_1}+\frac{{\widehat{I}}_{in}-{\widehat{I}}_1}{C_1}$$

(70)

$$\frac{d(V_{c3}+{\widehat{V}}_{c3})}{dt}=\frac{(I_2+{\widehat{I}}_2)}{C_3}-\frac{(V_{c3}+{\widehat{V}}_{c3})}{R\times C_3}=\frac{I_2-V_{c3}/R}{C_3}+\frac{{\widehat{I}}_2-{\widehat{V}}_{c3}/R}{C_3}$$

(71)

The dc or steady-state solutions of the above set of equations are

$$V_{c1}-V_{c2}=0\to V_{c1}=V_{c2}$$

(72)

$$1-d\times V_{c2}-V_{c3}=0\to 1-d\times V_{c2}=V_{c3}$$

(73)

$$I_{in}-I_1=0\to I_1=I_{in}$$

(74)

$$I_2-\frac{V_{c3}}{R}=0\to I_2=\frac{V_{c3}}{R}$$

(75)

The dynamic or ac solutions are:

$$L_1\frac{d{\widehat{i}}_1}{dt}={\widehat{V}}_{c1}$$

(76)

$$L_2\frac{d{\widehat{i}}_2}{dt}=-\widehat{d}\times V_{c2}-{\widehat{V}}_{c3}$$

(77)

$$C_1\frac{d{\widehat{V}}_{c1}}{dt}={\widehat{I}}_{in}-{\widehat{I}}_1$$

(78)

$$C_3\frac{d{\widehat{V}}_{c3}}{dt}={\widehat{I}}_2-\frac{{\widehat{V}}_{c3}}{R}$$

(79)

Representing Equations (76)–(79) in matrix form,

$$\left[\begin{array}{c}\frac{d{\widehat{i}}_1}{dt}\\ \frac{d{\widehat{i}}_2}{dt}\\ \frac{d{\widehat{V}}_{c1}}{dt}\\ \frac{d{\widehat{V}}_{c3}}{dt}\end{array}\right]=\left[\begin{array}{cccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]\left[\begin{array}{c}{\widehat{I}}_1\\ {\widehat{I}}_2\\ {\widehat{V}}_{c1}\\ {\widehat{V}}_{c3}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ {\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}_1\\ 0\end{array}\right]\left[{\widehat{I}}_{in}\right]+\left[\begin{array}{c}0\\ -\raisebox{1ex}{$V_{c2}$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ 0\\ 0\end{array}\right]\left[\widehat{d}\right]$$

The above matrix equation is of the form

$$\begin{gathered} \dot{\hat{X}}=\tilde{A}\hat{X}+\tilde{B}\tilde{U}+E\widehat{d}\mathrm{where}\tilde{A}=\left[\begin{array}{cccc}0& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& 0& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& 0& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right],\tilde{B}=\left[\begin{array}{c}0\\ 0\\ {\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}_1\\ 0\end{array}\right] \\ \mathrm{and}E=\left[\begin{array}{c}0\\ -\raisebox{1ex}{$V_{c2}$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ 0\\ 0\end{array}\right] \end{gathered}$$

Referring to Equation (40), the state variable matrix is

$$\left[\begin{array}{c}{\widehat{I}}_1\\ {\widehat{I}}_2\\ {\widehat{V}}_{c1}\\ {\widehat{V}}_{c3}\end{array}\right]={\left[\begin{array}{cccc}s& 0& \frac{1}{L_1}-\frac{H\left(s\right)}{V_p}\times \frac{V_{c2}}{L_2}& -\frac{H\left(s\right)}{V_p}\times \frac{V_{c2}}{L_2}\\ 0& s& 0& -\frac{1}{L_2}\\ \frac{-1}{C_1}& 0& s& 0\\ 0& \frac{1}{C_3}& 0& s-\frac{1}{R{C}_3}\end{array}\right]}^{-1}\left[\begin{array}{c}0\\ 0\\ {\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}_1\\ 0\end{array}\right]\left[{\widehat{I}}_{in}\right]$$

(80)

From Equation (79), the small-signal transfer functions can be obtained,

$$\frac{{\widehat{I}}_1}{{\widehat{I}}_{in}}=\frac{-\left(L_2{V}_p-H\left(s\right)V_{c2}{L}_1\right)}{L_1{L}_2{V}_p{C}_1{s}^2+L_2{V}_p-L_1{V}_{c2}H\left(s\right)}$$

(81)

$$\frac{{\widehat{V}}_{c1}}{{\widehat{I}}_{in}}=\frac{L_1{L}_2{V}_{p}s}{L_1{L}_2{V}_p{C}_1{s}^2+L_2{V}_p-L_1{V}_{c2}H\left(s\right)}$$

(82)

The small-signal control transfer function can be obtained by substituting $\tilde{U}\left(s\right)=0$ in Equation (38).

Therefore,

$$\left[\begin{array}{c}{\widehat{I}}_1\\ {\widehat{I}}_2\\ {\widehat{V}}_{c1}\\ {\widehat{V}}_{c3}\end{array}\right]={\left[\begin{array}{cccc}s& 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_1$}\right.& 0\\ 0& s& 0& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_1$}\right.& 0& s& 0\\ 0& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$C_3$}\right.& 0& s-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R{C}_3$}\right.\end{array}\right]}^{-1}\left[\begin{array}{c}0\\ -\raisebox{1ex}{$V_{c2}$}\!\left/ \!\raisebox{-1ex}{$L_2$}\right.\\ 0\\ 0\end{array}\right]\left[\widehat{d}\right]$$

(83)

The inductor current L₂ and the capacitor voltage C₃ are thus given by

$$\frac{{\widehat{I}}_2}{\widehat{d}}=-\frac{V_{c2}\left(C_{3}Rs-1\right)}{s^2{C}_3{L}_{2}R-s{L}_2+R}$$

(84)

$$\frac{{\widehat{V}}_{c3}}{\widehat{d}}=-\frac{R{V}_{c2}}{s^2{C}_3{L}_{2}R-s{L}_2+R}$$

(85)

2.3. Pass-Through Mode

In pass-through mode, switches S1 and S3 are permanently switched OFF, and switches S2 and S4 are permanently switched ON. The input voltage is directly passed on to the output. Hence no controller is required for this mode.

3. Control Design of Split-Pi Converter

Following the mathematical analysis and small-signal modeling of the SPC, the open loop transfer functions of the input current and output voltage are derived in the previous section. Now, with the available transfer functions, a sample case study of the control design is carried out using MATLAB SISOTOOL for both the boost and buck modes of operation [26].

The converter ratings for the sample case study are furnished in Table 1.

Table 1. Ratings of the SPC.

Sl. No.	Parameter	Rating
1	Input Voltage (Vin)	35.7 to 48 V
2	Output Voltage (Vo)	48 V
3	Frequency (f)	25 kHz
4	Inductors (L1 = L2)	100 mH
5	Capacitors (C1 = C2)	100 µF
6	Capacitor (C3)	80 µF
7	Duty Cycle (d)	0.75

3.1. Boost Mode of Operation

With the converter ratings given in Table 1, the controller for boost mode operation can be designed. Substituting the converter ratings in Equation (44), we have

$$\frac{{\widehat{V}}_{c1}}{\widehat{d}}=-\frac{48}{1+1\times {10}^{-5}{s}^2}=G_{vd}$$

(86)

The system represented by the transfer function in (85) is marginally stable with open-loop poles on the imaginary axis. The root locus and Bode plots for $\frac{{\widehat{V}}_{c1}}{\widehat{d}}$ are presented in Figure 6 and Figure 7, respectively.

Figure 6. Root locus plot of G_vd represented in Equation (86).

Figure 7. Open-loop Bode plot of G_vd represented in Equation (86).

It is understood from the above plots that the open-loop system represented by the transfer function in Equation (86) is unstable. Therefore, a suitable compensator has to be inserted to make it stable.

The compensator design is performed through MATLAB SISOTOOL. The PID tuning method was adopted for the design. The compensator transfer function after suitable tuning is given by

$$C\left(s\right)=-\frac{1.186 \left(1+0.002s\right)\left(1+0.063s\right)}{s}$$

(87)

with a pole at origin (p = 0) and two zeros on the negative real axis (z = −499.848, −15.8774) The closed-loop transfer function of the system with the compensator is given by

$$G\left(s\right)=\frac{C\left(s\right)\ast G_{vd}\left(s\right)}{1+C\left(s\right)\ast G_{vd}\left(s\right)\ast H\left(s\right)}$$

(88)

$$G\left(s\right)=\frac{7.173\times {10}^{-8}{s}^5+3.7\times {10}^{-5}{s}^4+7.742\times {10}^{-3}{s}^3+3.7s^2+56.93s}{1\times {10}^{-10}{s}^6+7.173\times {10}^{-8}{s}^5+5.7\times {10}^{-5}{s}^4+7.742\times {10}^{-3}{s}^3+4.7s^2+56.93s}$$

(89)

The pole-zero map, root locus, and Bode plots of the compensated system are depicted in Figure 8, Figure 9 and Figure 10, respectively.

Figure 8. Pole-zero map of G_vd with the designed compensator.

Figure 9. Root locus plot of G_vd with the designed compensator.

Figure 10. Bode plot of G_vd with the designed compensator.

The compensated system is observed to have an infinite gain margin at a phase crossover frequency of 916 rad/s and a phase margin of 120°. The system is thus stable.

3.2. Buck Mode of Operation

The controller for the buck mode of operation can be designed in this section. Substituting the converter ratings given in Table 1 into Equation (84), we have

$$\frac{{\widehat{I}}_2}{\widehat{d}}=-\frac{0.0432s-54}{8\times {10}^{-5}{s}^2-100\times {10}^{-3}s+10}=G_{id}$$

(90)

The system represented by the transfer function in Equation (90) is unstable with two poles (p = 1140, 109.6) and a zero (z = 1250) on the positive real axis. The root locus and Bode plots for the system are shown in Figure 11 and Figure 12, respectively.

Figure 11. Root locus plot of G_id represented in Equation (92).

Figure 12. Bode plot of G_id represented in Equation (92).

It is understood from the above plots that the closed-loop system represented by the transfer function in Equation (92) is unstable. Therefore, a suitable compensator has to be inserted to make it stable.

The compensator design is performed through MATLAB SISOTOOL. The LQG synthesis method was adopted for the design. The compensator transfer function after suitable tuning is given by

$$C\left(s\right)=\frac{0.011801 \left(1-0.01s\right)\left(1-15s\right)}{s\left(1+1.7\times {10}^{-5}s+1.764\times {10}^{-9}{s}^2\right)}$$

(91)

with three poles (p = 0,$-4819\pm 23317i$) and two zeros (z = 99.933, 0.0667).

The closed-loop transfer function of the system with the compensator is given by

$$G\left(s\right)=\frac{C\left(s\right)\ast G_{id}\left(s\right)}{1+C\left(s\right)\ast G_{id}\left(s\right)\ast H\left(s\right)}$$

(92)

$$G\left(s\right)=\frac{-1.079\times {10}^{-17}{s}^8-7.59\times {10}^{-14}{s}^7-5.87\times {10}^{-9}{s}^6+1.57\times {10}^{-5}{s}^5-0.01182s^4+1.987s^3-95.65s^2+6.373s}{1.27\times {10}^{-17}{s}^8+8.128\times {10}^{-14}{s}^7+3.25\times {10}^{-11}{s}^6+9.33\times {10}^{-8}{s}^5-0.00029s^4-0.01s^3+4.35s^2+6.373s}$$

(93)

The pole-zero map, root locus, and Bode plots of the compensated system are depicted in Figure 13, Figure 14 and Figure 15.

Figure 13. Pole-zero map of G_id with the designed compensator.

Figure 14. Root locus plot of G_id with the designed compensator.

Figure 15. Bode plot of G_id with the designed compensator.

4. Experimental Prototype of Split-Pi Converter

An experimental hardware of 500 W SPC was constructed as shown in Figure 16 to verify its operation. The hardware ratings and component specifications are listed in Table 2 The STM32 controller was used to generate switching pulses to the converter. A generic PCB board was used. Two MOSFET switches were stacked in series to obtain the required voltage range. The design of the controller is vital to overcome the nonlinearity of the system and to obtain the desired output voltage with good dynamics. The direct pole placement scheme was used here. The control block diagram is given in Figure 17. Here, Hv(s) and Hi(s) denote the error amplifiers [27].

Figure 16. Experimental prototype of SPC.

Table 2. Ratings and component specifications of the converter hardware.

Rated Power	500 W
Voltage Range	24–48 V
Switching Frequency	25 kHz
Load	4.6 Ω
Inductances L1, L2	178 µH
Capacitances C1, C3	100 µF
Capacitance C2	80 µF

Figure 17. Control block diagram of SPC.

The converter was operated in both the forward and reverse modes to verify its bidirectional property. Switches S1 and S2 were alternatively triggered, while S3 and S4 were kept OFF and ON, respectively, to allow the converter operation in boost mode. The input and output voltage waveforms were recorded for both the forward and reverse modes and are depicted in Figure 18 and Figure 19, respectively. The input voltage was maintained at 12 V, and a 24 volt output was measured as seen in Figure 18 for forward mode of operation. A different input voltage of 18 volts was maintained as the input side for the reverse mode of operation, and a 36 volt output was measured as seen in Figure 19.

Figure 18. Input and output voltage waveforms under forward mode of operation.

Figure 19. Input and output voltage waveforms under reverse mode of operation.

The input current and voltage waveforms under both forward and reverse modes of operation of the converter were also recorded and are presented in Figure 20 and Figure 21, respectively. The prototype of the split-pi converter was tested for half and full load conditions also, and the results show good matching with the simulation studies performed. The current ripples were less than 5%, and the voltage ripples were less than 2% in all loading conditions. The efficiency at the indicated voltages and currents is nearly 88.7%, and the temperatures reached by the power components after full load were only 2 degrees more than the ambient.

Figure 20. Input current and voltage waveforms under forward mode of operation.

Figure 21. Input current and voltage waveforms under reverse mode of operation.

5. Conclusions

In this paper, the steady-state modeling for different modes of operation of the SPC was studied in detail. A state–space averaging scheme was adopted to formulate the small-signal open-loop transfer functions of the input current and output voltage. The small-signal transfer functions were derived for both the boost and buck modes of operation of the converter. Sample case studies of compensator design for stable operation of the converter under various operating modes were presented. Frequency response plots were illustrated for both the uncompensated and compensated systems. An experimental prototype of the SPC with closed-loop control proving its bidirectional capability was developed, and the recorded results were presented.