© Springer Nature Singapore Pte Ltd. 2019
Yanfeng Chen and Bo ZhangEquivalent-Small-Parameter Analysis of DC/DC Switched-Mode ConverterCPSS Power Electronics Serieshttps://doi.org/10.1007/978-981-13-2574-8_4

4. Analysis of Open-Loop PWM DC/DC Converters Based on ESPM

Yanfeng Chen1   and Bo Zhang1  
(1)
School of Electric Power, South China University of Technology, Guangzhou, Guangdong, China
 
 
Yanfeng Chen (Corresponding author)
Email: eeyfchen@scut.edu.cn
 
Bo Zhang
Email: epbzhang@scut.edu.cn

4.1 Introduction

Harmonic balance method [1] is one of the commonly used methods for analyzing the periodic solution of nonlinear systems. It transforms the problem of solving nonlinear differential equations into solving nonlinear algebraic equations, which greatly simplifies the analysis. However, the harmonic balance method has the following disadvantages: (1) It is only suitable for obtaining a steady-state solution but cannot be used for transient analysis; (2) If a periodic solution is approximated by harmonics containing orders from 0 (DC component) to N, then there will be (2N + 1) nonlinear algebraic equations to be solved. Hence, the amount of calculation using this method will increase sharply as N increases.

The general averaging method [2] proposed in [3] generalizes the harmonic balance method to transient analysis, and successfully analyzes the resonant converter and PWM (Pulse Width Modulation) switching converter. For a PWM switching converter with a duty cycle D close to 0 (or 1), it is not enough to estimate the waveform with only one harmonic. Therefore, it is often necessary to solve the higher harmonics. The general average method is not easy to solve higher harmonics. The reason is similar to the harmonic balance method, that is, if a solution needs to be approximated by harmonics containing orders from 0 (DC component) to N, then there will be (2N + 1) nonlinear algebraic equations to be solved. Thus, it is quite difficult to have the solutions, which usually need to be obtained numerically. For an open-loop PWM switching converter, although the problem is shown in the form of the linear differential equations, the number of orders is higher bringing about large amount of calculation. It is obvious that this method fails to overcome the second shortcoming of the harmonic balance method.

The equivalent-small-parameter method [4] introduces the perturbation technique into the harmonic balance method, and approximately converts the periodic solution into an expanded triangular series according to the (equivalent) small parameter. According to the order of the small parameters and the type of harmonics, the corresponding algebraic equations can be obtained. Hence, there is no need to solve nonlinear algebraic equations with many variables. In fact, the equivalent-small-parameter method is only applied to solve the nonlinear equations of the main oscillation. The other higher harmonics and corresponding correction terms can be obtained by solving the linear equations. The main oscillation only contains one or two harmonics, and the corresponding algebraic equation has fewer variables. It can be seen that the calculation amount is greatly reduced.

In this chapter, the equivalent-small-parameter method is used to analyze the PWM switching converters operating in CCM (continuous current mode) and DCM (discontinuous current mode), and the analytical expression of the steady-state periodic solution is obtained.

Furthermore, applying the idea in [2, 3], the equivalent-small-parameter method is generalized to the transient analysis of the PWM converter to obtain the equivalent differential (rather than algebraic) equations. Moreover, for open-loop PWM switching converters, only linear differential equations need to be solved. In [5], the progressive method is applied to the switching converter for analyzing transient process and ripple. However, the asymptotic method combines the transient and steady-state solutions, which results in complicated solution. The equivalent-small-parameter method clearly distinguishes the transient and steady-state solutions. Hence, the solution can be easily obtained.

4.2 General Method for Analysis of PWM Switching Power Converter by ESPM

According to the analysis in Chap. 2, the PWM switching power converter can be uniformly represented by the following time-varying differential equations as
$${\mathbf{G}}_{0} (p){\mathbf{x}} + \sum\limits_{n = 1}^{N} {{\mathbf{G}}_{n} (p) \cdot {\mathbf{f}}^{(n)} ({\mathbf{x}})} = {\mathbf{u}}$$
(4.1)
In which p =  d/dt represents the differential operator, G0(p) and G1(p) are coefficient matrices related to the circuit parameters and the operator p, u is the input vector, and the nonlinear vector function can be expressed as
$${\mathbf{f}}^{(n)} ({\mathbf{x}}) = \delta^{(n)} ({\mathbf{x}} + {\mathbf{e}}^{(n)} )$$
(4.2)
Here e(n) is a constant vector, and the switching function is defined as
$$\delta^{(n)} = \left\{ {\begin{array}{*{20}l} 1 \hfill & {(t_{n - 1} ,t_{n} )} \hfill \\ 0 \hfill & {others} \hfill \\ \end{array} } \right.\quad (n = 1,\,2,\, \ldots \,N)$$
(4.3)
where $$t_{0} = kT,\,t_{N} = (k + 1)T$$ (k is an integer), which means the converter occupies (N + 1) sub-topologies during one switching cycle. The duty cycle of each sub-topology is defined as $$d^{(n)} = (t_{n} - t_{n - 1} )/T,\;(n = 1,\,2,\, \ldots \,N)$$, and $$\sum\nolimits_{n}^{N} {d^{(n)} } = 1$$. The state variable vector $${\mathbf{x}}$$ is still expressed as the sum of the following main components and the order corrections:
$${\mathbf{x}} = {\mathbf{x}}{}_{0} + \sum\limits_{i = 1}^{\infty } {\varepsilon^{i} {\mathbf{x}}_{i} }$$
(4.4)
Here $${\mathbf{x}}_{0}$$ and $${\mathbf{x}}_{i}$$ is the main component and the ith order correction term respectively. The parameter $$\varepsilon = 1$$ is introduced temporarily to indicate that different corrections would have different amplitudes, for example, the indicator $$\varepsilon^{k}$$ means the $${\mathbf{x}}_{k}$$ is a small amount higher than $${\mathbf{x}}_{k - 1}$$, that is $$\left| {{\mathbf{x}}_{k} } \right| < \left| {{\mathbf{x}}_{k - 1} } \right| \ll \left| {{\mathbf{x}}_{0} } \right|$$. Substituting (4.4) into (4.2) gives
$${\mathbf{f}}^{(n)} ({\mathbf{x}}) = {\mathbf{f}}_{0}^{(n)} + \sum\limits_{i = 1}^{\infty } {\varepsilon^{i} {\mathbf{f}}_{i}^{(n)} }$$
(4.5)
where
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0}^{(n)} = {\mathbf{f}}^{(n)} (x_{0} )} \hfill \\ {{\mathbf{f}}_{i}^{(n)} = {\mathbf{f}}^{(n)} ({\mathbf{x}}_{0} ,{\mathbf{x}}_{1} , \ldots {\mathbf{x}}_{i} ),\quad (i = 1,\,2\, \ldots )} \hfill \\ \end{array} } \right.$$
(4.6)
The switching function $$\delta^{(n)}$$ characterizes the periodic time-varying properties of the converter system. Correct processing of $$\delta^{(n)}$$ is the key to extend the equivalent small-parameter method to the analysis of switching converter systems. Since $$\delta^{(n)}$$ is a periodic function, it is usually expanded into a Fourier series, as follows.
$$\delta^{(n)} = b_{0}^{(n)} + \sum\limits_{m = 1}^{\infty } {\left[ {b_{m}^{(n)} \exp (jm\tau ) + \bar{b}_{m}^{(n)} \exp ( - jm\tau )} \right]}$$
(4.7)
where
$$\begin{aligned} b_{0}^{(n)} & = d^{(n)} ,\quad b_{m}^{(n)} = \tfrac{1}{2}(\alpha_{m}^{(n)} - j\beta_{m}^{(n)} ),\quad m = 1,\,2\, \ldots \\ \alpha_{m}^{(n)} & = \left[ {\sin 2(\sum\limits_{k = 0}^{n} {d^{(k)} )m\pi } - \sin 2(\sum\limits_{k = 0}^{n - 1} {d^{(k)} )m\pi } } \right]/m\pi \\ \beta_{m}^{(n)} & = \left[ {\cos 2(\sum\limits_{k = 0}^{n - 1} {d^{(k)} )m\pi } - \cos 2(\sum\limits_{k = 0}^{n} {d^{(k)} )m\pi } } \right]/m\pi \\ \end{aligned}$$
(4.8)
Similarly, the switch function is expanded to the sum of the main terms and the correction terms as
$$\delta^{(n)} = \delta_{.0}^{(n)} + \varepsilon \delta_{1}^{(n)} + \varepsilon^{2} \delta_{2}^{(n)} + \cdots$$
(4.9)
Since the magnitude of the coefficient $$b_{m}^{(n)}$$ decreases as m increases, it is generally assumed that $$\delta_{.0}^{(n)}$$ contains a DC component and a fundamental component, and $$\delta_{i}^{(n)}$$ contains two adjacent higher harmonic components, namely:
$$\begin{aligned} \delta_{0}^{(n)} & = d^{(n)} + b_{1}^{(n)} e^{j\tau } + c.c \\ \delta_{i}^{(n)} & = b_{2i}^{(n)} e^{j2i\tau } + b_{(2i + 1)}^{(n)} e^{j(2i + 1)\tau } + c.c \\ \end{aligned}$$
(4.10)
In which $$\tau = \omega t$$, the symbol c.c represents the conjugate complex term (the meaning is the same below). Substituting (4.4) and (4.9) into (4.2) gives
$$\begin{aligned} & {\mathbf{f}}_{0}^{(n)} = \delta_{0}^{(n)} ({\mathbf{x}}_{0} + {\mathbf{e}}^{(n)} ) \\ & {\mathbf{f}}_{1}^{(n)} = \delta_{0}^{(n)} {\mathbf{x}}_{1} + \delta_{1}^{(n)} ({\mathbf{x}}_{0} + {\mathbf{e}}^{(n)} ) \\ & {\mathbf{f}}_{2}^{(n)} = \delta_{0}^{(n)} {\mathbf{x}}_{2} + \delta_{1}^{(n)} {\mathbf{x}}_{1} + \delta_{2}^{(n)} ({\mathbf{x}}_{0} + {\mathbf{e}}^{(n)} ) \\ & \cdots \\ \end{aligned}$$
(4.11)

Using the same method as in Chap. 3, the main term $${\mathbf{f}}_{im}^{(n)}$$ and the remainder $${\mathbf{R}}_{i}^{(n)}$$ of each nonlinear function vector $${\mathbf{f}}_{i}^{(n)}$$ are got according to Eq. (4.11), and then the steady-state periodic solution of the system is obtained according to Eq. (3.​46).

4.3 Analysis of the Open-Loop Boost Converter Under CCM Operation

4.3.1 Modeling of the CCM-Boost Converter

The principle circuit of Boost converter is shown in Fig. 4.1a. It’s assumed that the converter operates in continuous-current-mode (CCM), that is, the converter has two equivalent topologies during one switching cycle T, which correspond to the switch ST is turned on and the SD is turned off (Fig. 4.1b), the switch ST is turned off, and the SD is turned on (Fig. 4.1c). Before modeling, the following assumptions are made:
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig1_HTML.gif
Fig. 4.1

CCM-Boost converter with its two equivalent topologies during one cycle

  1. (1)

    The input power supply is ideal, that is, its internal resistance is ignored;

     
  2. (2)

    Inductors and capacitors are considered as ideal components, that is, their parasitic parameters are not taken into account;

     
  3. (3)

    Both the controllable switch ST and the diode SD are ideal switches, their on-resistance is zero, and the resistance is infinite when disconnected.

     
Let ST have an on time of ton in one switching cycle T and d as a duty cycle, which is defined as d = ton/T. According to the analysis in Chap. 2, a switching function $$\delta (t)$$ (abbreviated as $$\delta$$) is defined to describe the on/off state of the switch ST, that is, $$\delta (t) = 1$$ when ST is on, and $$\delta (t) = 0$$ when ST is off. For the Boost converter operating in CCM, it is assumed that the controllable switch ST is turned on at the beginning of each switching cycle. After ton time, ST is turned off, and remains off state until the end of this cycle, then the definition of the switching function can be described by the following formula:
$$\delta (t) = \left\{ {\begin{array}{ll} 1 & {t \in [nT,\,(n + d)T]} \\ 0 & {t \in [(n + d)T,\,(n + 1)T]} \\ \end{array} } \right.$$
(4.12)
Here n = 0, 1, 2 …, is an integer.
As can be seen from Fig. 4.1b and c, when ST is on, SD is naturally turned off due to the reverse bias voltage. During this period, the current flowing through the switch ST is equal to the inductor current iL, and the reverse bias voltage that the SD is subjected to is vC. When the ST is turned off, the SD is turned on due to the positive bias voltage. And during this period, the current flowing through the switch ST is 0, and the voltage drop across the SD is also 0 under the assumption of the ideal switch. Thus, in one switching cycle, the current flowing through ST is δiL, and the reverse bias voltage that SD is subjected to is δvC. Using a controlled current source with a value of δiL to replace the switch ST, and a controlled voltage source with a value of δvC to replace the diode SD, then the nonlinear equivalent circuit of the Boost converter with CCM operation can be obtained, as shown in Fig. 4.2.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig2_HTML.gif
Fig. 4.2

Nonlinear equivalent circuit of CCM-Boost converter

Taking the inductor current $$i_{L}$$ and the capacitor voltage $$v_{C}$$ as the state variables, according to Fig. 4.2, the state differential equation describing the CCM-Boost converter can be obtained as follows:
$$\left\{ {\begin{array}{*{20}l} {\frac{{{\text{d}}i_{L} }}{{{\text{d}}t}} + \frac{1}{L}v_{C} + \frac{ - 1}{L}\delta v_{C} = \frac{1}{L}E} \hfill \\ {\frac{ - 1}{C}i_{L} + \frac{{{\text{d}}v_{C} }}{{{\text{d}}t}} + \frac{1}{RC}v_{C} + \frac{1}{C}\delta i_{L} = 0} \hfill \\ \end{array} } \right.$$
(4.13)
Let $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L} } & {v_{C} } \\ \end{array} } \right]^{Tr}$$ and p = d/dt, here the superscript “Tr” denote the transposition of a matrix, then the state differential equation of the above formula can be written in matrix form as follows
$$G_{0} \left( p \right){\mathbf{x}} + G_{1} \left( p \right){\mathbf{f}} = {\mathbf{u}}$$
(4.14)
where
$$G_{0} \left( p \right) = \left[ {\begin{array}{*{20}c} p & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & {p + \frac{1}{RC}} \\ \end{array} } \right],\quad G_{1} \left( p \right) = \left[ {\begin{array}{*{20}c} 0 & {\frac{ - 1}{L}} \\ {\frac{1}{C}} & 0 \\ \end{array} } \right],\quad {\mathbf{u}} = \left[ {\begin{array}{*{20}c} {E/L} \\ 0 \\ \end{array} } \right]$$
(4.15)
and the nonlinear vector is
$${\mathbf{f}} = \delta {\mathbf{x}}$$
(4.16)

4.3.2 The Equivalent Mathematical Model Based on ESPM

According to the method introduced in Chap. 3 and Sect. 4.1, we assume that the state vector x and the switching function δ(t) in Eq. (4.12) could be expanded into the following series:
$${\mathbf{x}} = {\mathbf{x}}{}_{0} + \varepsilon^{{}} {\mathbf{x}}_{1} + \varepsilon^{2} {\mathbf{x}}_{2} + \cdots$$
(4.17)
and
$$\delta (t) = \delta_{0}^{{}} + \varepsilon \delta_{1}^{{}} + \varepsilon^{2} \delta_{2}^{{}} + \cdots$$
(4.18)
where the zero order approximation x0 and the ith order approximation xi are called as the main wave and corrections, and represent the main components and harmonics of state vector x respectively. The parameter εi, which can also be called as “intrinsic small parameter”, is introduced temporarily to indicate that the terms xi have different order of magnitude. Similarly, δ(t) is expressed as the sum of main wave δ0 and the corrections δi too, as shown in Eq. (4.18). Substituting Eqs. (4.17) and (4.18) into f(x) = δ(t)x, and combining terms which have the same order of indicator parameter εi, we find that the nonlinear function f(x) can be expressed in the same way with state vector x:
$${\mathbf{f}}^{{}} ({\mathbf{x}}) = {\mathbf{f}}_{0}^{{}} + \varepsilon^{1} {\mathbf{f}}_{i}^{{}} + \varepsilon^{2} {\mathbf{f}}_{2}^{{}} + \cdots$$
(4.19)
in which:
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{0} } \hfill \\ {{\mathbf{f}}_{1}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{1} + \delta_{1}^{{}} {\mathbf{x}}_{0} } \hfill \\ {{\mathbf{f}}_{2}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{2} + \delta_{1}^{{}} {\mathbf{x}}_{1} + \delta_{2}^{{}} {\mathbf{x}}_{0} } \hfill \\ \cdots \hfill \\ \end{array} } \right.$$
(4.20)

Similarly, f0 and fi represent the main components and corrections of nonlinear vector function f(x) respectively.

Generally the terms $${\mathbf{x}}_{0}$$ and $${\mathbf{x}}_{i}$$ in Eq. (4.17) for state variables x can be expressed in the Fourier series as follows
$$\begin{aligned} {\mathbf{x}}_{0} & = \sum\limits_{{n \in E_{0} }} {{\mathbf{x}}_{00} = {\mathbf{a}}_{00} + \sum\limits_{{}} {({\mathbf{a}}_{n0} e^{jn\tau } + {\bar{\mathbf{a}}}_{n0} e^{ - jn\tau } )} } \\ {\mathbf{x}}_{i} & = \sum\limits_{{k \in E_{i} }} {{\mathbf{x}}_{ki} = {\mathbf{a}}_{0i} + \sum\limits_{{}} {({\mathbf{a}}_{ki} e^{jk\tau } + {\bar{\mathbf{a}}}_{ki} e^{ - jk\tau } )} } \\ \end{aligned}$$
(4.21)
in which n and k are integers, and τ = ωt (where ω = 2π/T) is the normalized time. The terms a00 and a0i are the DC components of x0 and xi respectively, and the terms an0 and aki are used to determine the amplitudes of the nth and the kth harmonics that belong to x0 and xi separately. The spectral content E0 of vector x0, which consists of a set of numbers representing relative frequencies of harmonics, is dictated by the physical sense of the study object. For example, as DC/DC converters have low-pass filtering properties, we usually assume that the main wave x0 includes DC components only, i.e. x0 =  a00 = [I00 V00]Tr, thus the set {E0} is {0}. Similarly, the set Ei contains the spectral contents for each xi, yet it is not known in advance and is determined in the process of iterated calculations, i.e. it is determined by x0, x1, …, and xi−1 during the iteration. The term ai0 and aik represent the DC components and the kth harmonics that belong to xi respectively.
The periodic switch function $$\delta$$ defined by (4.12) can be expanded into Fourier series as
$$\delta = b_{0}^{{}} + \sum\limits_{m = 1}^{\infty } {\left[ {b_{m} \exp (jm\tau ) + \bar{b}_{m} \exp ( - jm\tau )} \right]}$$
(4.22)
where
$$\begin{aligned} b_{0}^{{}} & = d \\ b_{m}^{{}} & = \frac{1}{2}(\alpha_{m}^{{}} - j\beta_{m}^{{}} ),\quad m = 1,\,2\, \ldots \\ \alpha_{m}^{{}} & = \frac{2}{T}\int_{0}^{T} {\delta (t)\cos (m} \omega t)dt = \frac{\sin 2dm\pi }{m\pi } \\ \beta_{m}^{{}} & = \frac{2}{T}\int_{0}^{T} {\delta (t)\sin (m} \omega t)dt = \frac{1 - \cos 2dm\pi }{m\pi } \\ \end{aligned}$$
(4.23)
Generally for DC/DC converters, we can choose that $$\delta_{0}^{{}}$$ contains only the DC component and the fundamental component, and $$\delta_{i}^{{}}$$ contains the (2i)th and (2i + 1)th harmonic components, here i = 1, 2, 3 …, is a positive integer, that is,
$$\begin{aligned} \delta_{0}^{{}} & = d^{{}} + b_{1}^{{}} e^{j\tau } + c.c \\ \delta_{i}^{{}} & = b_{2i}^{{}} e^{j2i\tau } + b_{(2i + 1)}^{{}} e^{j(2i + 1)\tau } + c.c \\ \end{aligned}$$
(4.24)
in which c.c denotes conjugate items.
The nonlinear function vector is further expressed as a two-level form of the sum of the main item series and the remainder series, as follows.
$${\mathbf{f}} = ({\mathbf{f}}_{0m} + \varepsilon {\mathbf{f}}_{1m} + \varepsilon^{2} {\mathbf{f}}_{2m} + \cdots ) + (\varepsilon {\mathbf{R}}_{1} + \varepsilon^{2} {\mathbf{R}}_{2} + \ldots )$$
(4.25)
The method for determining the main terms $${\mathbf{f}}_{{0{\mathbf{m}}}}$$ and $${\mathbf{f}}_{{{\mathbf{im}}}}$$, and the residual $$R_{{\mathbf{i}}}$$ is the same as described in Chap. 3. We can also refer to the specific solution process in the next section.
Substituting (4.17) and (4.25) into (4.14), one can obtain the equivalent mathematical model describing the system i.e. the iterative equations shown as follows.
$$\begin{aligned} & \left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} (p){\mathbf{x}}_{0} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{0m} = {\mathbf{u}}} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{1} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{1m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{1} } \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{2} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{2m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{2} } \hfill \\ \end{array} } \right. \\ & \quad \quad \quad \quad \quad \cdots \cdots \\ \end{aligned}$$
(4.26)

Each equation in (4.26) can be solved step by step by using the method of harmonic balance, where the first equation is used to obtain the main wave x0, and following equations are used to solve the corrections x1, x2, …, etc. If the harmonic amplitude in the kth correction xk is much smaller than those in the (k − 1)th correction xk−1, the calculation process is terminated. Hence, according to discussions above, the steady-state solution for the vector state variable Eq. (4.14) can be approximated by x ≈ x0 + x1 + x2 + …. Usually, as the low-pass filtering property of DC/DC converters, the magnitudes of harmonics with much higher frequencies are small, so they are neglected, and only the first three equations in Eq. (4.26) need to be solved, these solutions suffice for most technical applications.

4.3.3 The Steady-State Periodic Solution of the Boost Converter Based on ESPM

4.3.3.1 Solution of the Main Term

As it is known that DC/DC converters have low-pass properties, the main wave x0 to the steady-state value of state variable x could be chosen as
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = \left[ {\begin{array}{*{20}c} {I_{00} } & {V_{00} } \\ \end{array} } \right]^{Tr}$$
(4.27)
here I00 and V00 are dc values. Substituting it into (4.20) gives
$${\mathbf{f}}_{0} = \delta_{0} {\mathbf{x}}_{0} = b_{0} {\mathbf{a}}_{00} + b_{1} {\mathbf{a}}_{00} e^{j\tau } + c.c$$
(4.28)
According to assumptions in Chap. 3, f0m should include the same harmonics as x0, and the rest harmonics in f0 belong to R1, hence we can get
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0m} = b_{0} {\mathbf{a}}_{00} } \hfill \\ {{\mathbf{R}}_{1} = b_{1} {\mathbf{a}}_{00} e^{j\tau } + c.c} \hfill \\ \end{array} } \right.$$
(4.29)
Introducing x0 and f0m into the first equation of Eq. (4.26), it can be attained
$$\left[ {{\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0)b_{0} } \right]{\mathbf{a}}_{00} = {\mathbf{u}}$$
(4.30)
Since the derivative of the dc component is zero, $${\mathbf{G}}_{0} (0)$$ and $${\mathbf{G}}_{1} (0)$$ can be obtained by setting the differential operator p = 0 in Eq. (4.15). For the converter with open loop operation, the duty ratio d is a known amount, i.e. d = D, then Eq. (4.30) can be rewritten in the matrix form as
$$\left( {\left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ { - \frac{1}{C}} & {\frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{D}{L}} \\ {\frac{D}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {I_{00} } \\ {V_{00} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {E/L} \\ 0 \\ \end{array} } \right]$$
(4.31)
Thereby the term a00 could be got from (4.30) or (4.31).
$$V_{00} = \frac{E}{1 - D}\quad \quad I_{00} = \frac{{V_{00} }}{(1 - D)R}$$
(4.32)

4.3.3.2 Solution of the First Correction Term

According to the content of harmonics in the term R1 (there only exits the first-harmonic), we can deduce that the spectral content of vector x1 is {E1} = {1}, and then the first order correction x1 can be assumed to be:
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} e^{j\tau } + {\bar{\mathbf{a}}}_{11} e^{ - j\tau }$$
(4.33)
in which $${\bar{\mathbf{a}}}_{11}$$ means complex-conjugate, and a11 = [I11 V11]Tr. Substituting x0, x1, δ0 and δ1 into the equation $${\mathbf{f}}_{1}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{1} + \delta_{1}^{{}} {\mathbf{x}}_{0}$$, we can get
$${\mathbf{f}}_{1}^{{}} = (b_{0} {\mathbf{a}}_{11} e^{j\tau } + b_{0} {\bar{\mathbf{a}}}_{11} e^{ - j\tau } ) + (\bar{b}_{1} {\mathbf{a}}_{11} + b_{1} {\bar{\mathbf{a}}}_{11} ) + (b_{1} {\mathbf{a}}_{11} + b_{2} {\mathbf{a}}_{00} )e^{j2\tau } + b_{3} {\mathbf{a}}_{00} e^{j3\tau } + c.c$$
And the following expressions for f1m and R2 can be got as
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{1m} = b_{0} {\mathbf{a}}_{11} e^{j\tau } + c.c} \hfill \\ {{\mathbf{R}}_{2} = \left( {b_{1} {\bar{\mathbf{a}}}_{11} + \bar{b}_{1} {\mathbf{a}}_{11} } \right) + \left( {b_{1} {\mathbf{a}}_{11} + b_{2} {\mathbf{a}}_{00} } \right)e^{j2\tau } + b_{3} {\mathbf{a}}_{00} e^{j3\tau } + c.c} \hfill \\ \end{array} } \right.$$
(4.34)
where f1m should consist of all the terms in f1 with the same harmonics as x1, and R2 includes the rest. Introducing x1, R1 and f1m into the second equation in Eq. (4.26), and using the method of harmonic balance, one can get:
$$[{\mathbf{G}}_{0} (j\omega ) + {\mathbf{G}}_{1} (j\omega )b_{0} ]{\mathbf{a}}_{11} = - {\mathbf{G}}_{1} (j\omega ) \cdot b_{1} {\mathbf{a}}_{00}$$
(4.35)
where the coefficient b1 can be obtained according to (4.23). $${\mathbf{G}}_{0} (j\omega )$$ and $${\mathbf{G}}_{1} (j\omega )$$ can be obtained by setting the differential operator $$p = j\omega$$ in Eq. (4.15), as for the derivative of the exponential function, there is pejkωt = (jkω)ejkωt (k is an integer, here k = 1). For the Boost converter under study, Eq. (4.35) can be rewritten in the matrix form as:
$$\left( {\left[ {\begin{array}{*{20}c} {j\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {j\omega + \frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{D}{L}} \\ {\frac{D}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {I_{11} } \\ {V_{11} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {b_{1} I_{00} } \\ {b_{1} V_{00} } \\ \end{array} } \right]$$
(4.36)
Thus the vector a11 could be found from (4.36)
$$\left\{ {\begin{array}{*{20}l} {I_{11} = \frac{{b_{1} V_{00} - (1 - D)V_{11} }}{j\omega L}} \hfill \\ {V_{11} = \frac{{b_{1} [(1 - D)V_{00} - j\omega LI_{00} ]}}{{(1 - D)^{2} + (j\omega )^{2} LC + j\omega L/R}}} \hfill \\ \end{array} } \right.$$
(4.37)

4.3.3.3 Solution of the Second Correction Term

It can be seen from Eq. (4.34) that there exist DC components, second- and third-harmonics in the term R2, thus the spectral content of x2 is determined, that is {E2} = {0, 2, 3}. Following the same procedure, we can choose the second order correction x2 as:
$${\mathbf{x}}_{2} = {\mathbf{a}}_{20} + {\mathbf{a}}_{22} e^{j2\tau } + {\bar{\mathbf{a}}}_{22} e^{ - j2\tau } + {\mathbf{a}}_{23} e^{j3\tau } + {\bar{\mathbf{a}}}_{23} e^{ - j3\tau }$$
(4.38)

Here a20 = [I20 V20]Tr is the vector of dc values, it will give the corrections of dc components in a00, and a22 = [I22 V22]Tr and a23 = [I23 V23]Tr correspond to the amplitudes second- and third-harmonics of state variable x respectively.

Similarly, substituting x0, x1, x2, and δ0, δ1, δ2 into $${\mathbf{f}}_{2} = \delta_{0} {\mathbf{x}}_{2} + \delta_{1} {\mathbf{x}}_{1} + \delta_{2} {\mathbf{x}}_{0}$$, we can get
$$\begin{aligned} {\mathbf{f}}_{2} & = b_{0} {\mathbf{a}}_{20} + (b_{1} {\mathbf{a}}_{00} + \bar{b}_{1} {\mathbf{a}}_{22} + b_{2} {\bar{\mathbf{a}}}_{11} )e^{j\tau} + (b_{0} {\mathbf{a}}_{22} + \bar{b}_{1} {\mathbf{a}}_{23} + b_{3} {\bar{\mathbf{a}}}_{11} )e^{j2\tau} \\ & \quad + (b_{0} {\mathbf{a}}_{23} + b_{1} {\mathbf{a}}_{22} + b_{2} {\mathbf{a}}_{11} )e^{j3\tau} + (b_{1} {\mathbf{a}}_{23} + b_{3} {\mathbf{a}}_{11} + b_{4} {\mathbf{a}}_{00} )e^{j4\tau} + b_{5} {\mathbf{a}}_{00} e^{j5\tau} + c.c \\ \end{aligned}$$
and the expressions for f2m, it should include the same harmonics as x2.
$${\mathbf{f}}_{2m} = b_{0} {\mathbf{a}}_{20} + (b_{0} {\mathbf{a}}_{22} + \bar{b}_{1} {\mathbf{a}}_{23} + b_{3} {\bar{\mathbf{a}}}_{11} )e^{j2\tau} + (b_{0} {\mathbf{a}}_{23} + b_{1} {\mathbf{a}}_{22} + b_{2} {\mathbf{a}}_{11} )e^{j3\tau} + c.c$$
(4.39)
Next, substituting x2, R2 and f2m into the third equation of Eq. (4.26), and still using the method of harmonic balance, one can obtain the following three equations:
$$[G_{0} (0) + G_{1} (0)b_{0} ]{\mathbf{a}}_{20} = - G_{1} (0)(b_{1} {\bar{\mathbf{a}}}_{11} + \bar{b}_{1} {\mathbf{a}}_{11} )$$
(4.40a)
$$[G_{0} (j2\omega ) + G_{1} (j2\omega )b_{0} ]{\mathbf{a}}_{22} = - G_{1} (j2\omega ) \cdot (b_{2} {\mathbf{a}}_{00} + b_{1} {\mathbf{a}}_{11} + b_{3} {\bar{\mathbf{a}}}_{11} )$$
(4.40b)
and
$$[G_{0} (j3\omega ) + G_{1} (j3\omega )b_{0} ]{\mathbf{a}}_{23} = - G_{1} (j3\omega ) \cdot (b_{3} {\mathbf{a}}_{00} + b_{2} {\mathbf{a}}_{11} + b_{1} {\mathbf{a}}_{22} )$$
(4.40c)

Likewise, the coefficient matrix $${\mathbf{G}}_{0} (jk\omega )$$ and $${\mathbf{G}}_{1} (jk\omega )$$ can be obtained by setting the differential operator $$p = jk\omega$$ in Eq. (4.15), as for the derivative of the exponential function, there is pejkωt = (jkω)ejkωt (here k = 0, 2, 3).

It should be noticed that during the derivation process, the term $$\bar{b}_{1} {\mathbf{a}}_{23}$$ in Eq. (4.40b) can be omitted with respect to the term $$(b_{0} {\mathbf{a}}_{22} + b_{3} {\bar{\mathbf{a}}}_{11} )$$, as the magnitudes of harmonics with higher frequencies are smaller than those with lower frequencies.

So Eq. (4.40a) can be rewritten in matrix form as:
$$\left( {\left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ { - \frac{1}{C}} & {\frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{D}{L}} \\ {\frac{D}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {I_{20} } \\ {V_{20} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {\bar{b}_{1} I_{11} + b_{1} \bar{I}_{11} } \\ {\bar{b}_{1} V_{11} + b_{1} \bar{V}_{11} } \\ \end{array} } \right]$$
(4.41)
where b0 = D and b1 = [sin2Dπ − j(1 − cos2Dπ)]/2π, the vector a02 could be found from (4.41)
$$\left\{ {\begin{array}{*{20}l} {I_{20} = \frac{{R(\bar{b}_{1} I_{11} + b_{1} \bar{I}_{11} ) + V_{20} }}{R(1 - D)}} \hfill \\ {V_{20} = \frac{{\bar{b}_{1} V_{11} + b_{1} \bar{V}_{11} }}{1 - D}} \hfill \\ \end{array} } \right.$$
(4.42)
Then Eq. (4.40b) is rewritten in matrix form as:
$$\left( {\left[ {\begin{array}{*{20}c} {j2\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {j2\omega + \frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{D}{L}} \\ {\frac{D}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {I_{22} } \\ {V_{22} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {b_{1} I_{11} + b_{2} I_{00} + b_{3} \bar{I}_{11} } \\ {b_{1} V_{11} + b_{2} V_{00} + b_{3} \bar{V}_{11} } \\ \end{array} } \right]$$
(4.43)
And Eq. (4.40c) can be rewritten in matrix form as:
$$\left( {\left[ {\begin{array}{*{20}c} {j3\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {j3\omega + \frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{D}{L}} \\ {\frac{D}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {I_{23} } \\ {V_{23} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {b_{1} I_{22} + b_{2} I_{11} + b_{3} I_{00} } \\ {b_{1} V_{22} + b_{2} V_{11} + b_{3} V_{00} } \\ \end{array} } \right]$$
(4.44)
where bk = [sin2kDπ − j(1 − cos2kDπ)]/2kπ (k = 1, 2, 3). Then the vectors a22 and a23 could be found from (4.43) and (4.44), as shown in (4.45) and (4.46) respectively.
$$\left\{ {\begin{array}{*{20}l} {I_{22} = \frac{{b_{1} V_{11} + b_{2} V_{00} + b_{3} \bar{V}_{11} - (1 - D)V_{22} }}{j2\omega L}} \hfill \\ {V_{22} = \frac{{(1 - D)(b_{1} V_{11} + b_{2} V_{00} + b_{3} \bar{V}_{11} ) - j2\omega L(b_{1} I_{11} + b_{2} I_{00} + b_{3} \bar{I}_{11} )}}{{(1 - D)^{2} + (j2\omega )^{2} LC + j2\omega L/R}}} \hfill \\ \end{array} } \right.$$
(4.45)
$$\left\{ {\begin{array}{*{20}l} {I_{23} = \frac{{b_{1} V_{22} + b_{2} V_{11} + b_{3} V_{00} - (1 - D)V_{23} }}{j3\omega L}} \hfill \\ {V_{23} = \frac{{(1 - D)(b_{1} V_{22} + b_{2} V_{11} + b_{3} V_{00} ) - j3\omega L(b_{1} I_{22} + b_{2} I_{11} + b_{3} I_{00} )}}{{(1 - D)^{2} + (j3\omega )^{2} LC + j3\omega L/R}}} \hfill \\ \end{array} } \right.$$
(4.46)
Following this procedure, the term aik could be solved step by step, and detailed information is listed in Table 4.1.
Table 4.1

Values of aik for CCM-Boost converter

i

k

aik

Vik

Iik

0

0

$$\frac{E}{(1 - D)}$$

$$\frac{E}{{R(1 - D)^{2} }}$$

1

1

$$\frac{{(1 - D)b_{1} V_{00} - j\omega Lb_{1} I_{00} }}{\Delta (j\omega )}$$

$$\frac{{b_{1} V_{00} - (1 - D)V_{11} }}{(j\omega )L}$$

2

0

$$\frac{{\bar{b}_{1} V_{11} + b_{1} \bar{V}_{11} }}{{\left( {1 - D} \right)}}$$

$$\frac{{R(\bar{b}_{1} I_{11} + b_{1} \bar{I}_{11} ) + V_{20} }}{{R\left( {1 - D} \right)}}$$

2

$$\frac{{(1 - D)(b_{1} V_{11} + b_{2} V_{00} + b_{3} \bar{V}_{11} ) - j2\omega L(b_{1} I_{11} + b_{2} I_{00} + b_{3} \bar{I}_{11} )}}{\Delta (j2\omega )}$$

$$\frac{{(b_{1} V_{11} + b_{2} V_{00} + b_{3} \bar{V}_{11} ) - (1 - D)V_{22} }}{j2\omega L}$$

3

$$\frac{{(1 - D)(b_{1} V_{22} + b_{2} V_{11} + b_{3} V_{00} ) - (j3\omega L(b_{1} I_{22} + b_{2} I_{11} + b_{3} I_{00} )}}{\Delta (j3\omega )}$$

$$\frac{{(b_{1} V_{22} + b_{2} V_{11} + b_{3} V_{00} ) - (1 - D)V_{23} }}{j3\omega L}$$

In Table 4.1, the ∆(jkω) is defined as:
$$\Delta (jk\omega ) = (1 - D)^{2} + (jk\omega )^{2} LC + jk\omega L/R$$
(4.47)
And the coefficient bi in Eq. (4.23) is rewritten as follows
$$\left\{ {\begin{array}{*{20}l} {b_{0}^{{}} = d} \hfill \\ {b_{m}^{{}} = \frac{\sin 2dm\pi - j(1 - \cos 2dm\pi )}{2m\pi },\;m = 1,\,2\, \ldots } \hfill \\ \end{array} } \right.$$
(4.48)
$$\bar{b}_{m}$$ is the conjugate complex of $$b_{m}$$.
Hence, according to the discussions above, the approximate periodic solutions for CCM-Boost converter could be expressed as
$$\begin{aligned} {\mathbf{x}} & = \left[ {\begin{array}{*{20}c} {i_{L} } & {v_{C} } \\ \end{array} } \right]^{Tr} \\ & = \left( {{\mathbf{a}}_{00} + {\mathbf{a}}_{02} } \right) + \left( {{\mathbf{a}}_{11} e^{j\tau } + {\mathbf{a}}_{22} e^{j2\tau } + {\mathbf{a}}_{32} e^{j3\tau } + c.c} \right) \\ \end{aligned}$$
(4.49)
The components of this vector are
$$\begin{aligned} i_{L} & \approx \left( {I_{00} + I_{02} } \right) + 2(\text{Re} \,I_{11} \cos \omega t - \text{Im} \,I_{11} { \sin }\,\omega t \\ & \quad + \text{Re} \,I_{22} \cos 2\omega t - \text{Im} \,I_{22} { \sin }2\omega t + \text{Re} \,I_{32} \cos 3\omega t - \text{Im} \,I_{32} { \sin }\,3\omega t )\\ \end{aligned}$$
(4.50a)
in amperes, and
$$\begin{aligned} v_{C} & \approx \left( {V_{00} + V_{02} } \right) + 2(\text{Re} \,V_{11} \cos \omega t - \text{Im} \,V_{11} { \sin }\,\omega t \\ & \quad + \text{Re} \,V_{22} \cos 2\omega t - \text{Im} \,V_{22} { \sin }2\omega t + \text{Re} \,V_{32} \cos 3\omega t - \text{Im} \,V_{32} { \sin }\,3\omega t{ )} \\ \end{aligned}$$
(4.50b)
in volts. In Eqs. (4.50a) and (4.50b) Re xik and Im xik represent the real- and imaginary-part of the complex term xik separately. Obviously, the DC values and harmonic components of state variables can be obtained from Eqs. (4.49) or (4.50a).

4.3.4 Simulations

The circuit parameters of the open-loop Boost converter are list in Table 4.2.
Table 4.2

Circuit parameters of the open-loop Boost converter

Parameters

Values

Input voltage E

37.5 V

Switching frequency f

1 kHz

Inductance L

6 mH

Capacitance C

45 μF

Load resistance R

30 Ω

Duty ratio D

0.25

According to (4.48), when the duty ratio is known, the coefficients bm can be obtained. Then according to Table 4.1, the approximate solution of the state variable $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L} } & {v_{C} } \\ \end{array} } \right]^{Tr}$$ can be obtained as follows, here $$i_{L}$$ is the inductor current and $$v_{C}$$ is the voltage over the output capacitor.
$$\begin{aligned} i_{L} & = 2.11 - 0.487\cos \tau + 0.392\sin \tau - 0.205\cos 2\tau - 0.002\sin 2\tau \\ & \quad - 0.0476\cos 3\tau - 0.0408\sin 3\tau \quad \quad \quad \quad ( {\text{A)}} \\ v_{C} & = 48.08 + 0.942\cos \tau - 3.945\sin \tau + 1.24\cos 2\tau - 0.026\sin 2\tau \\ & \quad + 0.1878\cos 3\tau + 0.3523\sin 3\tau \quad \quad \quad \quad {\text{ (V)}} \\ \end{aligned}$$
(4.51)
For the analysis of the steady state solution of the state variable, the ESP analysis method (dashed line) and the simulation results (solid line) are compared in Figs. 4.3 and 4.4. Figure 4.3 shows the comparison of the inductor current waveform and the capacitor voltage ripple waveform with $$f_{s} = 1\,{\text{kHz}}$$, and Fig. 4.4 shows the comparison waveforms with $$f_{s} = 10\,{\text{kHz}}$$. It can be seen that the results from the ESPM are in good coincidence with those from simulations, indicating that the ESPM can analyze the steady-state solution of the open-loop converter well. Even when the operating frequency of the converter is small, such as $$f_{s} = 1\,{\text{kHz}}$$, the state variable ripple is relatively large, the ESPM can still effectively obtain the analytical expression of the periodic solution of the converter, and only three iterations are needed to obtain enough precision. It can be seen that the equivalent small-parameter method is effective for analyzing the strong nonlinear system of the switching converter.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig3_HTML.gif
Fig. 4.3

Steady-state ripple waveforms of state variables for CCM-Boost converter with fs = 1 kHz

../images/419194_1_En_4_Chapter/419194_1_En_4_Fig4_HTML.gif
Fig. 4.4

Steady-state ripple waveforms of state variables for CCM-Boost converter with fs = 10 kHz

4.4 Analysis of the Open-Loop Buck Converter Under CCM Operation

4.4.1 Modeling of the CCM-Buck Converter

The principle circuit of Buck converter is shown in Fig. 4.5a. It’s assumed that the converter operates in continuous-current-mode (CCM), that is, the converter has two equivalent topologies during one switching cycle T, which correspond to the switch ST is turned on and the SD is turned off (Fig. 4.5b), the switch ST is turned off, and the SD is turned on (Fig. 4.5c).
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig5_HTML.gif
Fig. 4.5

CCM-Buck converter with its two equivalent topologies during one cycle

Introduce a switching function to describe the on/off state of the switch ST, which is defined as (4.12). Then according to the method in Chap. 2, the Buck converter with CCM operation can be replaced by a nonlinear equivalent circuit, as shown in Fig. 4.6.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig6_HTML.gif
Fig. 4.6

Nonlinear equivalent circuit of CCM-Buck converter

Taking the inductor current $$i_{L}$$ and the capacitor voltage $$v_{C}$$ as the state variables, according to Fig. 4.6, the state differential equation describing the CCM-Buck converter can be obtained as follows:
$$\left\{ \begin{aligned} \frac{{{\text{d}}i_{L} }}{{{\text{d}}t}} + \frac{1}{L}v_{C} + \frac{ - 1}{L}\delta E = 0 \hfill \\ \frac{ - 1}{C}i_{L} + \frac{{{\text{d}}v_{C} }}{{{\text{d}}t}} + \frac{1}{RC}v_{C} = 0 \hfill \\ \end{aligned} \right.$$
(4.52)
Let $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L} } & {v_{C} } \\ \end{array} } \right]^{Tr}$$ and p = d/dt, then the state differential equation of the above formula can be written in matrix form as follows
$$G_{0} \left( p \right){\mathbf{x}} + G_{1} \left( p \right){\mathbf{f}} = 0$$
(4.53)
Here the nonlinear vector is
$${\mathbf{f}} = \delta {\mathbf{e}}$$
(4.54)
The constant vector e and the coefficient matrices are given as follows
$$G_{0} \left( p \right) = \left[ {\begin{array}{*{20}c} p & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & {p + \frac{1}{RC}} \\ \end{array} } \right],\quad G_{1} \left( p \right) = \left[ {\begin{array}{*{20}c} 0 & {\frac{ - 1}{L}} \\ 0 & 0 \\ \end{array} } \right],\quad {\mathbf{e}} = \left[ {\begin{array}{*{20}c} 0 \\ E \\ \end{array} } \right]$$
(4.55)

4.4.2 The Equivalent Mathematical Model Based on ESPM

Using the same method introduced in Sect. 4.2, we expand the state vector x, the switching function δ(t), and the nonlinear vector function f into the series as shown in Eqs. (4.17), (4.18) and (4.19), respectively. It should be noted that for the Buck converter, the main term and the correction amount of the nonlinear vector function are determined according to Eq. (4.54), it should be
$${\mathbf{f}}_{i}^{{}} = \delta_{i}^{{}} {\mathbf{e}}$$
(4.56)
Using a similar processing method to solve Eq. (4.53), one can obtain the equivalent mathematical model describing the Buck converter, the iterative equations shown as follows, i.e.
$$\begin{aligned} & \left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} (p){\mathbf{x}}_{0} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{0m} = 0} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{1} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{1m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{1} } \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{2} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{2m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{2} } \hfill \\ \end{array} } \right. \\ & \quad \quad \quad \quad \quad \cdots \cdots \\ \end{aligned}$$
(4.57)

4.4.3 The Steady-State Periodic Solution of the Buck Converter Based on ESPM

Similar to Sect. 4.3.3, we can solve Eq. (4.57) step by step. First, the main wave x0 to the steady-state value of state variable x is chosen as
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = \left[ {\begin{array}{*{20}c} {I_{00} } & {V_{00} } \\ \end{array} } \right]^{Tr}$$
(4.58)
The main wave of nonlinear vector is $${\mathbf{f}}_{0} = \delta_{0} {\mathbf{e}} = b_{0} {\mathbf{e}} + b_{1} {\mathbf{e}} \cdot e^{j\tau } + c.c$$, and hence we can get
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0m} = b_{0} {\mathbf{e}}} \hfill \\ {{\mathbf{R}}_{1} = b_{1} {\mathbf{e}} \cdot e^{j\tau } + c.c} \hfill \\ \end{array} } \right.$$
(4.59)
Introducing x0 and f0m into the first equation of Eq. (4.57), it can be attained
$${\mathbf{G}}_{0} (0){\text{a}}_{00} = - {\mathbf{G}}_{1} (0){\text{b}}_{0} {\mathbf{e}}$$
(4.60)
And Eq. (4.60) can be rewritten in the matrix form as
$$\left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ { - \frac{1}{C}} & {\frac{1}{RC}} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {I_{00} } \\ {V_{00} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{{b_{0} }}{L}} \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 0 \\ E \\ \end{array} } \right]$$
(4.61)
Thereby the term a00 could be got from (4.61).
$$V_{00} = b_{0} E\quad \quad I_{00} = \frac{{V_{00} }}{R}$$
Then according to the content of harmonics in the term R1, the first order correction x1 can be assumed to be:
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} e^{j\tau } + {\bar{\mathbf{a}}}_{11} e^{ - j\tau }$$
(4.62)
The first-order correction term of the nonlinear vector is $${\mathbf{f}}_{1}^{{}} = \delta_{1} {\mathbf{e}} = b_{2} {\mathbf{e}} \cdot e^{j2\tau } + b_{3} {\mathbf{e}} \cdot e^{j3\tau } + c.c$$, and the following expressions for f1m and R2 can be got as
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{1m} = 0} \hfill \\ {{\mathbf{R}}_{2} = b_{2} {\mathbf{e}} \cdot e^{j2\tau } + b_{3} {\mathbf{e}} \cdot e^{j3\tau } + c.c} \hfill \\ \end{array} } \right.$$
(4.63)
Introducing x1, R1 and f1m into the second equation in Eq. (4.57), and using the method of harmonic balance, one can get
$${\mathbf{G}}_{0} (j\omega ){\mathbf{a}}_{{{\mathbf{11}}}} = - {\mathbf{G}}_{1} (j\omega )b_{1} {\mathbf{e}}\quad$$
(4.64)
And Eq. (4.64) can be rewritten in the matrix form as:
$$\left[ {\begin{array}{*{20}c} {j\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {j\omega + \frac{1}{RC}} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {I_{11} } \\ {V_{11} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{{b_{1} }}{L}} \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 0 \\ E \\ \end{array} } \right]$$
(4.65)
Thus the vector a11 could be found from (4.65).
Next according to the harmonics in the term R2, we can choose the second order correction x2 as:
$${\mathbf{x}}_{2} = {\mathbf{a}}_{22} e^{j2\tau } + {\bar{\mathbf{a}}}_{22} e^{ - j2\tau } + {\mathbf{a}}_{23} e^{j3\tau } + {\bar{\mathbf{a}}}_{23} e^{ - j3\tau }$$
(4.66)
Similarly, from $${\mathbf{f}}_{2} = \delta_{2} {\mathbf{e}}$$, there is $${\mathbf{f}}_{2m} = 0$$, thus one can obtain the following two equations
$${\mathbf{G}}_{0} ({\text{j}}2\omega ){\mathbf{a}}_{22} = - {\mathbf{G}}_{1} ({\text{j}}2\omega )b_{2} {\mathbf{e}}\quad$$
(4.67a)
$$G_{0} (j3\omega ){\mathbf{a}}_{23} = - G_{1} (j3\omega )b_{3} {\mathbf{e}}$$
(4.67b)
And they can be rewritten in matrix form as:
$$\left[ {\begin{array}{*{20}c} {j2\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {j2\omega + \frac{1}{RC}} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {I_{22} } \\ {V_{22} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{{b_{2} }}{L}} \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 0 \\ E \\ \end{array} } \right]$$
(4.68a)
And
$$\left[ {\begin{array}{*{20}c} {j3\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {j3\omega + \frac{1}{RC}} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {I_{23} } \\ {V_{23} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{{b_{3} }}{L}} \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} 0 \\ E \\ \end{array} } \right]$$
(4.68b)
Then the vector a22 and a23 be found.
Following this process, the term aik could be solved step by step, and detailed information is listed in Table 4.3.
Table 4.3

Values of aik for CCM-Buck converter

i

k

aik

Vik

Iik

0

0

$$V_{00} = b_{0} E \,$$

$$I_{00} = \frac{{V_{00} }}{R}$$

1

1

$$V_{11} = \frac{{b_{1} E }}{{1 + (j\omega )^{2} LC + j\omega L/R}}$$

$$I_{11} = \frac{{b_{1} E - V_{11} }}{j\omega L}$$

 

2

$$V_{22} = \frac{{b_{2} E }}{{1 + (j2\omega )^{2} LC + j2\omega L/R}}$$

$$I_{22} = \frac{{b_{2} E - V_{11} }}{j2\omega L}$$

3

$$V_{23} = \frac{{b_{3} E }}{{1 + (j3\omega )^{2} LC + j3\omega L/R}}$$

$$I_{23} = \frac{{b_{3} E - V_{11} }}{j3\omega L}$$

And the coefficient bi are still determined by Eq. (4.48).

4.4.4 Simulations

The circuit parameters of the open-loop Buck converter is list in Table 4.5.

According to (4.48), when the duty ratio is known, the coefficients bm can be obtained. Then according to Table 4.4, the approximate solution of the state variable $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L} } & {v_{C} } \\ \end{array} } \right]^{Tr}$$ can be obtained as follows, here $$i_{L}$$ is the inductor current and $$v_{C}$$ is the voltage over the output capacitor.
Table 4.4

Circuit parameters of the open-loop Buck converter

Parameters

Values

Input voltage E

15 V

Switching frequency f

50 kHz

Inductance L

150 μH

Capacitance C

4.7 μF

Load resistance R

10 Ω

Duty ratio D

0.35

$$\begin{aligned} i_{L} & = 0.5267 - 0.1631\cos \tau + 0.0833\sin \tau - 0.0333\cos 2\tau - 0.0242\sin 2\tau \, ( {\text{A)}} \\ v_{C} & = 5.25 - 0.0636\cos \tau - 0.1062\sin \tau + 0.0078\cos 2\tau - 0.0308\sin 2\tau \quad ( {\text{V)}} \\ \end{aligned}$$
(4.69)

It should be noted that, as the amplitude of the 3rd harmonic is quite small, which is ignored in (4.69).

The ESP analysis method (solid line) and the simulation results (dashed line) are compared in Fig. 4.7, which shows the comparison of the inductor current ripple and the capacitor voltage ripple waveform during one cycle. It can be seen that the results from ESPM are in good coincidence with those from simulations, indicating that the ESPM can analyze the steady-state solution of the open-loop converter well (Fig. 4.7).
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig7_HTML.gif
Fig. 4.7

Steady-state ripple waveforms of state variables for CCM-Buck converter

4.5 Analysis of the Open-Loop Cuk Converter Under CCM Operation

4.5.1 Modeling of the CCM-Cuk Converter

The principle circuit of Cuk converter is shown in Fig. 4.8a. It’s assumed that the converter operates in continuous-current-mode (CCM), that is, the converter has two equivalent topologies during one switching cycle T, which correspond to the switch ST is turned on and the SD is turned off (Fig. 4.8b), the switch ST is turned off and the SD is turned on (Fig. 4.8c).
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig8_HTML.gif
Fig. 4.8

CCM-Cuk converter with its two equivalent topologies during one cycle

Introduce a switching function to describe the on/off state of the switch ST, which is defined as (4.12 ). Then according to the method in Chap. 2, the Cuk converter with CCM operation can be replaced by a nonlinear equivalent circuit, as shown in Fig. 4.9.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig9_HTML.gif
Fig. 4.9

Nonlinear equivalent circuit of CCM-Cuk converter

Taking the inductor currents $$i_{1}$$, $$i_{2}$$ and the capacitor voltages $$v_{1}$$,$$v_{2}$$ as the state variables, according to Fig. 4.9, the state differential equation describing the CCM-Cuk converter can be obtained as follows:
$$\left\{ {\begin{array}{*{20}l} {\frac{{{\text{d}}i_{1} }}{{{\text{d}}t}} + \frac{1}{{L_{1} }}v_{1} + \frac{ - 1}{{L_{1} }}\delta v_{1} = \frac{1}{{L_{1} }}E} \hfill \\ {\frac{{{\text{d}}i_{2} }}{{{\text{d}}t}} + \frac{1}{{L_{2} }}v_{2} + \frac{ - 1}{{L_{2} }}\delta v_{1} = 0} \hfill \\ {\frac{ - 1}{{C_{1} }}i_{1} + \frac{{{\text{d}}v_{1} }}{{{\text{d}}t}} + \frac{1}{{C_{1} }}\delta (i_{1} + i_{2} ) = 0} \hfill \\ {\frac{ - 1}{C}i_{2} + \frac{{{\text{d}}v_{2} }}{{{\text{d}}t}} + \frac{1}{RC}v_{2} = 0} \hfill \\ \end{array} } \right.$$
(4.70)
Let $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{1} } & {i_{2} } & {v_{1} } & {v_{2} } \\ \end{array} } \right]^{Tr}$$ and p = d/dt, then the state differential equation of the above formula can be written in matrix form as follows
$$G_{0} \left( p \right){\mathbf{x}} + G_{1} \left( p \right){\mathbf{f}} = {\mathbf{u}}$$
(4.71)
Here the nonlinear vector is
$${\mathbf{f}} = \delta {\mathbf{x}}$$
(4.72)
The constant vector u and the coefficient matrices are given as follows
$$G_{0} \left( p \right) = \left[ {\begin{array}{*{20}l} p \hfill & 0 \hfill & {\frac{1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & p \hfill & 0 \hfill & {\frac{1}{{L_{2} }}} \hfill \\ {\frac{ - 1}{{C_{1} }}} \hfill & 0 \hfill & p \hfill & 0 \hfill \\ 0 \hfill & {\frac{ - 1}{C}} \hfill & 0 \hfill & {p + \frac{1}{RC}} \hfill \\ \end{array} } \right],\quad G_{1} \left( p \right) = \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\frac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\frac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ {\frac{1}{{C_{1} }}} \hfill & {\frac{1}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right],\quad {\mathbf{u}} = \left[ {\begin{array}{*{20}c} {E/L_{1}} \\ \begin{aligned} 0 \hfill \\ 0 \hfill \\ 0 \hfill \\ \end{aligned} \\ \end{array} } \right]$$
(4.73)

Equations (4.71) and (4.72) are identical in form to Eqs. (4.14) and (4.16) respectively. And the equivalent mathematical model of the CCM-operated Cuk converter is also the same as Eq. (4.26), which can be solved in exactly the same way as in Sect. 4.3.

4.5.2 The Steady-State Periodic Solution of the Cuk Converter Based on ESPM

According to the basic principle and the low-pass properties of DC/DC converter, the main wave x0 to the steady-state value of state variable x could be chosen as
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = \left[ {\begin{array}{*{20}c} {I_{100} } & {I_{200} } & {V_{100} } & {V_{200} } \\ \end{array} } \right]^{Tr}$$
And a00 can be got by solving the following matrix equation:
$$\left[ {{\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0)b_{0} } \right]{\mathbf{a}}_{00} = {\mathbf{u}}$$
(4.74)
Here $${\mathbf{G}}_{0} (0)$$ and $${\mathbf{G}}_{1} (0)$$ can be obtained by setting the differential operator p = 0 in Eq. (4.73). For the converter with open loop operation, the duty ratio d is a known amount, i.e. d = D, then Eq. (4.74) can be rewritten in the matrix form as
$$\left( {\left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & {\tfrac{1}{{L_{2} }}} \hfill \\ {\tfrac{ - 1}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & {\tfrac{ - 1}{C}} \hfill & 0 \hfill & {\tfrac{1}{RC}} \hfill \\ \end{array} } \right] + \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{D}{{C_{1} }}} \hfill & {\tfrac{D}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]} \right) \cdot \left[ {\begin{array}{*{20}c} {I_{100} } \\ {I_{200} } \\ {V_{100} } \\ {V_{200} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\tfrac{1}{{L_{1} }}E} \\ \begin{aligned} 0 \hfill \\ 0 \hfill \\ 0 \hfill \\ \end{aligned} \\ \end{array} } \right]$$
(4.75)
From (4.75), the solutions of DC components of the state variables can be obtained as
$$V_{100} = \frac{E}{1 - D},\quad V_{200} = \frac{{L_{2} }}{{L_{1} }}\frac{DE}{(1 - D)},\quad I_{200} = \frac{{V_{200} }}{R},\quad I_{100} = \frac{{ DI_{200} }}{1 - D}$$
The first order correction x1 can be assumed to be:
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} e^{j\tau } + {\bar{\mathbf{a}}}_{11} e^{ - j\tau }$$
in which $${\bar{\mathbf{a}}}_{11}$$ means complex-conjugate, and a11 = [I111 I211 V111 V211]Tr. And a11 can be got from the following equation as
$$[{\mathbf{G}}_{0} (j\omega ) + {\mathbf{G}}_{1} (j\omega )b_{0} ]{\mathbf{a}}_{11} = - {\mathbf{G}}_{1} (j\omega ) \cdot b_{1} {\mathbf{a}}_{00}$$
(4.76)
where the coefficient b1 can be obtained according to (4.23). $${\mathbf{G}}_{0} ({\text{j}}\omega )$$ and $${\mathbf{G}}_{1} ({\text{j}}\omega )$$ can be obtained by setting the differential operator $${\text{p}} = {\text{j}}\omega$$ in Eq. (4.73). For the Cuk converter under study, Eq. (4.76) can be rewritten in the matrix form as:
$$\left( {\left[ {\begin{array}{*{20}l} j\omega \hfill & 0 \hfill & {\tfrac{1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & j\omega \hfill & 0 \hfill & {\tfrac{1}{{L_{2} }}} \hfill \\ {\tfrac{ - 1}{{C_{1} }}} \hfill & 0 \hfill & j\omega \hfill & 0 \hfill \\ 0 \hfill & {\tfrac{ - 1}{C}} \hfill & 0 \hfill & {j\omega+\tfrac{1}{RC}} \hfill \\ \end{array} } \right] + \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{D}{{C_{1} }}} \hfill & {\tfrac{D}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]} \right) \cdot \left[ {\begin{array}{*{20}c} {I_{111} } \\ {I_{211} } \\ {V_{111} } \\ {V_{211} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{1}{{C_{1} }}} \hfill & {\tfrac{1}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {b_{1} I_{100} } \\ \begin{aligned} b_{1} I_{200} \hfill \\ b_{1} V_{100} \hfill \\ b_{1} V_{200} \hfill \\ \end{aligned} \\ \end{array} } \right]$$
(4.77)
Following the same procedure, we can choose the second order correction x2 as:
$${\mathbf{x}}_{2} = {\mathbf{a}}_{20} + {\mathbf{a}}_{22} e^{j2\tau } + {\bar{\mathbf{a}}}_{22} e^{ - j2\tau } + {\mathbf{a}}_{23} e^{j3\tau } + {\bar{\mathbf{a}}}_{23} e^{ - j3\tau }$$

Here a20 = [I120 I220 V120 V220]Tr is the vector of DC values, it will give the corrections of DC components in a00, and a22 = [I122 I222 V122 V222]Tr and a23 = [I123 I223 V123 V223]Tr correspond to the amplitudes of second- and third-harmonics of state variable x respectively.

Similarly, a20, a22, a23 can be obtained from the following equations.
$$[G_{0} (0) + G_{1} (0)b_{0} ]{\mathbf{a}}_{20} = - G_{1} (0)(b_{1} {\bar{\mathbf{a}}}_{11} + \bar{b}_{1} {\mathbf{a}}_{11} )$$
(4.78a)
$$[G_{0} (j2\omega ) + G_{1} (j2\omega )b_{0} ]{\mathbf{a}}_{22} = - G_{1} (j2\omega ) \cdot (b_{2} {\mathbf{a}}_{00} + b_{1} {\mathbf{a}}_{11} + b_{3} {\bar{\mathbf{a}}}_{11} )$$
(4.78b)
$$[G_{0} (j3\omega ) + G_{1} (j3\omega )b_{0} ]{\mathbf{a}}_{23} = - G_{1} (j3\omega ) \cdot (b_{3} {\mathbf{a}}_{00} + b_{2} {\mathbf{a}}_{11} + b_{1} {\mathbf{a}}_{22} )$$
(4.78c)
The coefficient matrices $${\mathbf{G}}_{0} ({\text{jk}}\,\omega )$$ and $${\mathbf{G}}_{1} ({\text{jk}}\,\omega )$$ can be obtained by setting the differential operator $${\text{p}} = {\text{jk}}\,\omega$$ in Eq. (4.73) (here k = 0, 2, 3). Equation (4.78) can be rewritten in matrix form as:
$$\left( {\left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & {\tfrac{1}{{L_{2} }}} \hfill \\ {\tfrac{ - 1}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & {\tfrac{ - 1}{C}} \hfill & 0 \hfill & {\tfrac{1}{RC}} \hfill \\ \end{array} } \right] + \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{D}{{C_{1} }}} \hfill & {\tfrac{D}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {I_{120} } \\ {I_{220} } \\ {V_{120} } \\ {V_{220} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{1}{{C_{1} }}} \hfill & {\tfrac{1}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right] \cdot \left[ {\begin{array}{*{20}l} {\bar{b}_{1} I_{111} + b_{1} \bar{I}_{111} } \hfill \\ {\bar{b}_{1} I_{211} + b_{1} \bar{I}_{211} } \hfill \\ {\bar{b}_{1} V_{111} + b_{1} \bar{V}_{111} } \hfill \\ {\bar{b}_{1} V_{211} + b_{1} \bar{V}_{211} } \hfill \\ \end{array} } \right]$$
(4.79a)
$$\begin{aligned} & \left( {\left[ {\begin{array}{*{20}l} {j2\omega } \hfill & 0 \hfill & {\tfrac{1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & {j2\omega } \hfill & 0 \hfill & {\tfrac{1}{{L_{2} }}} \hfill \\ {\tfrac{ - 1}{{C_{1} }}} \hfill & 0 \hfill & {j2\omega } \hfill & 0 \hfill \\ 0 \hfill & {\tfrac{ - 1}{C}} \hfill & 0 \hfill & {j2\omega + \tfrac{1}{RC}} \hfill \\ \end{array} } \right] + \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{D}{{C_{1} }}} \hfill & {\tfrac{D}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]} \right) \cdot \left[ {\begin{array}{*{20}c} {I_{122} } \\ {I_{222} } \\ {V_{122} } \\ {V_{222} } \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{1}{{C_{1} }}} \hfill & {\tfrac{1}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right] \cdot \left[ {\begin{array}{*{20}l} {b_{1} I_{111} + b_{2} I_{100} + b_{3} \bar{I}_{111} } \hfill \\ {b_{1} I_{211} + b_{2} I_{200} + b_{3} \bar{I}_{211} } \hfill \\ {b_{1} V_{111} + b_{2} V_{100} + b_{3} \bar{V}_{111} } \hfill \\ {b_{1} V_{211} + b_{2} V_{200} + b_{3} \bar{V}_{211} } \hfill \\ \end{array} } \right] \\ \end{aligned}$$
(4.79b)
$$\begin{aligned} & \left( {\left[ {\begin{array}{*{20}l} {j3\omega } \hfill & 0 \hfill & {\tfrac{1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & {j3\omega } \hfill & 0 \hfill & {\tfrac{1}{{L_{2} }}} \hfill \\ {\tfrac{ - 1}{{C_{1} }}} \hfill & 0 \hfill & {j3\omega } \hfill & 0 \hfill \\ 0 \hfill & {\tfrac{ - 1}{{C_{{}} }}} \hfill & 0 \hfill & {j3\omega + \tfrac{1}{RC}} \hfill \\ \end{array} } \right] + \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - D}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{D}{{C_{1} }}} \hfill & {\tfrac{D}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]} \right) \cdot \left[ {\begin{array}{*{20}c} {I_{123} } \\ {I_{223} } \\ {V_{123} } \\ {V_{223} } \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}l} 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & {\tfrac{ - 1}{{L_{1} }}} \hfill & 0 \hfill \\ {\tfrac{1}{{C_{1} }}} \hfill & {\tfrac{D}{{C_{1} }}} \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} } \right]\left[ {\begin{array}{*{20}l} {b_{1} I_{122} + b_{2} I_{111} + b_{3} I_{100} } \hfill \\ {b_{1} I_{222} + b_{2} I_{211} + b_{3} I_{200} } \hfill \\ {b_{1} V_{122} + b_{2} V_{111} + b_{3} V_{100} } \hfill \\ {b_{1} V_{222} + b_{2} V_{211} + b_{3} V_{200} } \hfill \\ \end{array} } \right] \\ \end{aligned}$$
(4.79c)

According to Eqs. (4.75), (4.77) and (4.79), the analytic form solutions of the state variables can be obtained by using some symbolic analysis software tools.

4.5.3 Simulations

The circuit parameters of the open-loop Cuk converter is list in Table 4.5.
Table 4.5

Circuit parameters of the open-loop Cuk converter

Parameters

Values

Input voltage E

25 V

Switching frequency f

50 kHz

Inductance L1, L2

1.9, 0.96 mH

Capacitance C1, C

850, 47 μF

Load resistance R

30 Ω

Duty ratio D

0.55

According to (4.48), when the duty ratio is known, the coefficients bm can be obtained. Then according to Eqs. (4.75), (4.77) and (4.79), when the circuit parameters are introduced, the approximate solution of the state variable $${\mathbf{x}}$$ can be obtained as follows, where $$i_{1}$$ and $$i_{2}$$ correspond the inductor currents of L1 and L2, $$v_{1}$$ and $$v_{2}$$ are the voltages across C1 and C (output filter capacitor), respectively.
$$\begin{aligned} i_{1} & = 1.306 + 0.578\cos \tau + 0.0092\sin \tau + 0.0014\cos 2\tau - 0.0044\sin 2\tau \\ & \quad + 0. 0 0 5 2\cos 3\tau + 0.0027\sin 3\tau \quad \quad \quad \quad ( {\text{A)}} \\ i_{2} & = 1.0108 + 0.1144\cos \tau + 0.0181\sin \tau + 0.0028\cos 2\tau - 0.0086\sin 2\tau \\ & \quad + 0. 0 1 0 3\cos 3\tau + 0.0053\sin 3\tau \quad \quad \quad \quad ( {\text{A)}} \\ v_{1} & = 55.55 - 0.0054\cos \tau - 0.0006\sin \tau - 0.0003\cos 2\tau + 0.0004\sin 2\tau \\ & \quad - 0.0005\cos 3\tau - 0.0002\sin 3\tau \quad \quad \quad \quad ( {\text{V)}} \\ v_{2} & = 30.55 - 0.0012\cos \tau + 0.0078\sin \tau + 0.0003\cos 2\tau + 0.0001\sin 2\tau \\ & \quad - 0.0001\cos 3\tau + 0.0002\sin 3\tau \quad \quad \quad \quad ( {\text{V)}} \\ \end{aligned}$$
(4.80)
The DC components obtained from the ESPM and the numerical simulation are listed in Table 4.6, from which it can be seen analytical results are very close to those from simulations.
Table 4.6

DC values of CCM-Cuk converter from the ESPM and simulations

 

DC components of state variables

ESPM

I1 = 1.306 A, I2 = 1.0108 A, V1 = 55.55 V, V2 = 30.55 V

Numerical simulation

I1 = 1.250 A, I2 = 1.030 A, V1 = 55.56 V, V2 = 30.56 V

For the AC components (ripples) of the state variables, the ESP analysis results (dashed line) and the simulation results (solid line) are compared in Fig. 4.10. Figure 4.10a, b show the inductor current ripples, and Fig. 4.10c, d the capacitor voltage ripples. It can be seen that the results from the ESPM are in good coincidence with those from simulations. Thus, for the Cuk converter, although the orders of the circuits are increased to four, the complexity of the analysis is not increased due to the matrix analysis calculation, indicating that even for higher orders converters, the ESPM can analyze the steady-state solutions with sufficient accuracy.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig10_HTML.gif
Fig. 4.10

Comparison of ripples of state variables of Cuk in CCM

The analysis method for the steady-state periodic solution of the converter operating in discontinuous-conduction-mode (DCM) is similar and will not be repeated in this chapter. We can also refer to Chap. 6, which provides a detailed description of the steady-state periodic solution of a closed-loop converter operating in DCM.

4.6 Transient Analysis of the Open-Loop PWM Converter by ESPM

4.6.1 The Solution Procedure

Similar to the analysis of the steady-state solution of the switching converter by the equivalent small-parameter method, the high-order strong nonlinear system to be solved is described by the state differential equation with the following matrix form as:
$$G_{0} \left( p \right){\mathbf{x}} + G_{1} \left( p \right){\mathbf{f}} = {\mathbf{u}}$$
(4.81)
Here the l*1-ordered vector $${\mathbf{x}} = [x_{1} ,x_{2} , \cdots x_{l} ]^{Tr}$$ (the superscript “Tr” indicates matrix transposition). The l*l-ordered coefficient matrices $${\mathbf{G}}_{0} (p)$$ and $${\mathbf{G}}_{1} (p)$$ are all related to the differential operator p and the circuit parameters, where the highest number of p is 1. The term $${\mathbf{u}}$$ a input vector with l*1 order, and the l*l-ordered nonlinear vector function $${\mathbf{f}}({\mathbf{x}})$$ is related to the state variable vector $${\mathbf{x}}$$.
According to the method introduced in Chap. 3 and Sect. 4.2, we still assume that the state vector x could be expanded into the following series:
$${\mathbf{x}} = {\mathbf{x}}{}_{0} + \varepsilon^{{}} {\mathbf{x}}_{1} + \varepsilon^{2} {\mathbf{x}}_{2} + \cdots$$
(4.82)
Substituting x into the nonlinear function $${\mathbf{f}}({\mathbf{x}})$$ gives:
$${\mathbf{f}}^{{}} ({\mathbf{x}}) = {\mathbf{f}}_{0}^{{}} + \varepsilon^{1} {\mathbf{f}}_{i}^{{}} + \varepsilon^{2} {\mathbf{f}}_{2}^{{}} + \cdots$$
(4.83)
in which:
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{0} \;\quad } \hfill \\ {{\mathbf{f}}_{1}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{1} + \delta_{1}^{{}} {\mathbf{x}}_{0} } \hfill \\ {{\mathbf{f}}_{2}^{{}} = \delta_{0}^{{}} {\mathbf{x}}_{2} + \delta_{1}^{{}} {\mathbf{x}}_{1} + \delta_{2}^{{}} {\mathbf{x}}_{0} \;} \hfill \\ {\; \cdots } \hfill \\ \end{array} } \right.$$
(4.84)
In Eqs. (4.82) and (4.83), $$i$$ is an approximate order; $$\varepsilon$$ is a small parameter marker.
Generally the terms $${\mathbf{x}}_{0}$$ and $${\mathbf{x}}_{i}$$ in Eq. (4.82) for state variables x can be expressed in the Fourier series as follows
$$\begin{aligned} {\mathbf{x}}_{0} & = \sum\limits_{{n \in E_{0} }} {{\mathbf{x}}_{n0} = \sum\limits_{{n \in E_{0} }} {\left( {{\mathbf{a}}_{n0} (\tau )e^{jn\tau } + {\bar{\mathbf{a}}}_{n0} (\tau )e^{ - jn\tau } } \right)} } \\ {\mathbf{x}}_{i} & = \sum\limits_{{k \in E_{i} }} {{\mathbf{x}}_{ki} = \sum\limits_{{k \in E_{i} }} {\left( {{\mathbf{a}}_{ki} (\tau )e^{jk\tau } + {\bar{\mathbf{a}}}_{ki} (\tau )e^{ - jk\tau } } \right)} } \\ \end{aligned}$$
(4.85)
where the normalized time $$\tau = \omega t = 2\pi f_{s} t$$, the terms $${\bar{\mathbf{a}}}_{n0} (\tau )$$ and $${\bar{\mathbf{a}}}_{ki} (\tau )$$ are the conjugate plurals of $${\mathbf{a}}_{n0} (\tau )$$ and $${\mathbf{a}}_{ki} (\tau )$$ respectively. It should be noted that, unlike the steady-state periodic solution, the harmonic coefficients $${\mathbf{a}}_{n0} (\tau )$$ and $${\mathbf{a}}_{ki} (\tau )$$ are variables related to the normalized time $$\tau$$. In the case of a steady-state periodic solution, the coefficients $${\mathbf{a}}_{n0}$$ and $${\mathbf{a}}_{ki}$$ are unknown constants to be solved, they are independent of time. Similar to the solution of the steady-state periodic solution, in Eq. (4.85), the spectral set E0 of the principal component is determined by the physical knowledge associated with the system under study. For example, when the switching power converter system is stable, the main component of its output voltage and inductor current is DC, so $$E_{0} = \left\{ 0 \right\}$$ can be selected; and for the weak nonlinear system we can choose $$E_{0} = \left\{ 1 \right\}$$, which means that only the fundamental wave are contained in the main component. The frequency set $$E_{i}$$ of the correction term $${\mathbf{x}}_{i}$$ is determined step by step during the iteration.
Substituting $${\mathbf{x}}_{\text{i}}$$ into $${\mathbf{f}}_{\text{i}}$$ and decomposing $${\mathbf{f}}_{\text{i}} \,(i = 0,\,1,\,2\, \ldots )$$ into the sum of the main term and the remainder, the following formula can be obtained.
$${\mathbf{f}}_{i} = {\mathbf{f}}\left( {{\mathbf{x}}_{0} ,{\mathbf{x}}_{1} , \cdots ,{\mathbf{x}}_{i} } \right) = {\mathbf{f}}_{im} + \varepsilon {\mathbf{R}}_{i + 1}$$
(4.86)
In which the main term $${\mathbf{f}}_{im}$$ has the same spectral set as the term $${\mathbf{x}}_{{\mathbf{i}}}$$, that is, all items in $${\mathbf{f}}_{\text{i}}$$ with the same harmonic components as $${\mathbf{x}}_{{\mathbf{i}}}$$ belong to $${\mathbf{f}}_{im}$$, while other harmonic components belong to the remainder $${\mathbf{R}}_{i + 1}$$. Therefore, $${\mathbf{R}}_{i + 1}$$ can be regarded as a small amount with higher order than the term fim, and in front of which the small amount indicator $$\varepsilon$$ is introduced again. The terms $${\mathbf{f}}_{im}$$ and $${\mathbf{R}}_{i + 1}$$ can be expressed into Fourier series as follows:
$${\mathbf{f}}_{0m} = \sum\limits_{{k \in E_{0} }} {\left[ {{\mathbf{h}}_{n0} (\tau )e^{jn\tau } + {\bar{\mathbf{h}}}_{n0} (\tau )e^{ - jn\tau } } \right] \, }$$
(4.87a)
$${\mathbf{f}}_{im} = \sum\limits_{{k \in E_{i} }} {\left[ {{\mathbf{h}}_{ki} (\tau )e^{jk\tau } + {\bar{\mathbf{h}}}_{ki} (\tau )e^{ - jk\tau } } \right] \, }$$
(4.87b)
$${\mathbf{R}}_{i + 1} = \sum\limits_{{k \in E_{i + 1} }} {\left[ {{\mathbf{h}}_{k(i + 1)}^{{\prime }} (\tau )^{jk\tau } + {\bar{\mathbf{h}}}_{k(i + 1)}^{{\prime }} (\tau )e^{ - jk\tau } } \right]}$$
(4.87c)

It still should be noted that, the amplitude coefficients of each harmonic in the above equations are time-dependent variables, which are quite different with those in the steady-state periodic solution.

Substituting (4.86) into (4.83) gives
$${\mathbf{f}} = ({\mathbf{f}}_{0m} + \varepsilon {\mathbf{f}}_{1m} + \varepsilon^{2} {\mathbf{f}}_{2m} + \cdots ) + (\varepsilon {\mathbf{R}}_{1} + \varepsilon^{2} {\mathbf{R}}_{2} + \cdots )$$
(4.88)
Substituting (4.82) and (4.88) into (4.81) and making the coefficients of the same mark $$\varepsilon^{i}$$ on both sides of the equation equal, you can get the following iterative equations as
$$\begin{aligned} & \left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} (p){\mathbf{x}}_{0} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{0m} = {\mathbf{u}}} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{1} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{1m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{1} } \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{2} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{2m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{2} } \hfill \\ \end{array} } \right. \\ & \quad \quad \quad \quad \quad \cdots \cdots \\ \end{aligned}$$
(4.89)
The above iterative Eq. (4.89) is exactly what is needed to solve the transient solution. The main term x0 and the correction term xi can be obtained by solving each equation in (4.89) using the harmonic balance method step by step. It can be seen that Eq. (4.89) is identical in form to the iterative Eq. (3.​43) for finding the steady-state periodic solution of the nonlinear system. However, it should be noted that Eq. (3.​43) is used to solve the steady-state periodic solution as an algebraic equation, and Eq. (4.89) is used to find the transient solution as a differential equation.
$$p({\mathbf{a}}_{ki} e^{jk\omega t} ) = \frac{{d({\mathbf{a}}_{ki} e^{jk\omega t} )}}{dt} = jk\omega \,({\mathbf{a}}_{ki} e^{jk\omega t} )$$
(4.90)
Therefore, when finding the coefficients $${\mathbf{a}}_{ki}$$ of the harmonic components with the steady-state solution, the differential operator $$p = jk\omega$$ in Eq. (3.​43), where k = 0, 1 …, represents the order of harmonics.
While in the transient solution, since the coefficients $${\mathbf{a}}_{ki} (\tau )$$ of the harmonic components are time-dependent variables, there exists the following equation about the differential operation.
$$\begin{aligned} p[{\mathbf{a}}_{ki} (\tau )e^{jk\omega t} ] & = \frac{{d[{\mathbf{a}}_{ki} (\tau )e^{jk\omega t} ]}}{dt} = \frac{{d[{\mathbf{a}}_{ki} (\tau )]}}{dt} \cdot e^{jk\omega t} + jk\omega \cdot {\mathbf{a}}_{ki} (\tau )\,e^{jk\omega t} \\ & \,{ = }\, (p{ + }jk\omega ){\mathbf{a}}_{ki} (\tau )\,e^{jk\omega t} \\ \end{aligned}$$
(4.91)

Therefore, when finding the coefficients $${\mathbf{a}}_{ki} (\tau )$$ of the harmonic components with the transient solution, we use (p + jkω) instead of the differential operator p in Eq. (4.89).

For open-loop switching converters, the differential equations in Eq. (4.89) are usually linear and can also be expressed as follows.
$$\quad \frac{{d{\mathbf{x}}}}{dt} = {\mathbf{Ax}} + {\mathbf{Bg}}(t)$$
(4.92)
The solution of the equation above is
$$\quad {\mathbf{x}}(t) = e^{{{\mathbf{A}}t}} {\mathbf{x}}(0) + e^{{{\mathbf{A}}t}} \int_{0}^{t} {e^{{ - {\mathbf{A}}\tau }} {\mathbf{Bg}}(\tau } )d\tau$$
(4.93)
in which the term $${\mathbf{x}}\left( 0 \right)$$ represents the initial values of the state variable vector.

4.6.2 Initial Value Determination

In order to find the transient solution of the nonlinear system, the initial values need to be determined. In the case of accurate solution, the initial values should satisfy the following conditions as
$${\mathbf{x}}\left( 0 \right) = {\mathbf{x}}_{0} \left( 0 \right) + \varepsilon {\mathbf{x}}_{1} \left( 0 \right) + \varepsilon^{2} {\mathbf{x}}_{2} \left( 0 \right) + \cdots = {\mathbf{x}}_{0} \left( 0 \right) + {\mathbf{x}}_{1} \left( 0 \right) + {\mathbf{x}}_{2} \left( 0 \right) + \cdots$$
(4.94)
In which the small amount indicator $$\varepsilon$$ = 1.

In the case of simple estimation, $${\mathbf{x}}_{0} \left( 0 \right) = {\mathbf{x}}\left( 0 \right)$$ can be used to obtain the transient solution $${\mathbf{x}}_{0} \left( \tau \right)$$ of the first equation in Eq. (4.89), and then the forced solutions $${\mathbf{x}}_{1} \left( \tau \right)$$ and $${\mathbf{x}}_{2} \left( \tau \right)$$ … of the second and the third equations in Eq. (4.89) …, etc. However, the transient solution obtained in this way usually does not satisfy the initial conditions (4.94).

Usually, similar to the literature [3], we determine the initial values as follows:
  1. (1)

    Let $${\mathbf{x}}_{0} \left( 0 \right) = {\mathbf{x}}\left( 0 \right)$$, and find $${\mathbf{x}}_{0} \left( \tau \right)$$ by the first equation of Eq. (4.89);

     
  2. (2)

    Solve the second equation of Eq. (4.89) and get the special solution $${\mathbf{x}}_{1} \left( \tau \right)$$, then let $${\mathbf{x}}_{0} \left( 0 \right) = {\mathbf{x}}\left( 0 \right) - {\mathbf{x}}_{1} \left( 0 \right)$$;

     
  3. (3)

    Similarly, after solving Eq. (4.89) to obtain the special solutions $${\mathbf{x}}_{2} \left( \tau \right)$$ and $${\mathbf{x}}_{3} \left( \tau \right)$$, …, etc., let $${\mathbf{x}}_{0} \left( 0 \right) = {\mathbf{x}}\left( 0 \right) - {\mathbf{x}}_{1} \left( 0 \right) - {\mathbf{x}}_{2} \left( 0 \right) - {\mathbf{x}}_{3} \left( 0 \right) \cdots$$. Then use the modified $${\mathbf{x}}_{0} \left( 0 \right)$$ as the initial value to find $${\mathbf{x}}_{0} \left( \tau \right)$$, and according to the correction equation, the special solutions $${\mathbf{x}}_{1} \left( \tau \right)$$ and $${\mathbf{x}}_{2} \left( \tau \right)$$, …, etc. are obtained one by one, and finally the transient solution of Eq. (4.89) is obtained.

     

It must be noted that when using the equivalent small-parameter method, as long as the appropriate main component is selected, the initial $${\mathbf{x}}_{0} \left( 0 \right)$$ has little effect on the special solution $${\mathbf{x}}_{i} \left( \tau \right)$$. Thus, if $${\mathbf{x}}\left( 0 \right) = 0$$ is selected, then $${\mathbf{x}}_{i} \left( 0 \right) = 0\,\left( {i = 1,\,2,\, \ldots } \right)$$ can be made to simplify the analysis process of the transient solution.

4.6.3 Transient Analysis of Open-Loop PWM Boost Converter

We still take the Boost converter shown in Fig. 4.1 as an example to illustrate the analysis of the transient solution of a switching converter using the ESPM. The circuit parameters are chosen as: L = 6 mH, C = 45 μF, R = 30 Ω, E = 37.5 V, fs = 1 kHz, and the duty ratio is set to be D = 0.25.

According to the analysis in Sect. 4.3, the CCM-operated Boost converter can be described by a nonlinear vector differential equation as shown in (4.14), which is rewritten as follows
$$G_{0} \left( p \right){\mathbf{x}} + G_{1} \left( p \right){\mathbf{f}} = {\mathbf{u}}$$
(4.95)
where the coefficient matrices and the input vector are shown as follows.
$$G_{0} \left( p \right) = \left[ {\begin{array}{*{20}c} p & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & {p + \frac{1}{RC}} \\ \end{array} } \right],\quad G_{1} \left( p \right) = \left[ {\begin{array}{*{20}c} 0 & {\frac{ - 1}{L}} \\ {\frac{1}{C}} & 0 \\ \end{array} } \right],\quad {\mathbf{u}} = \left[ {\begin{array}{*{20}c} {E/L} \\ 0 \\ \end{array} } \right]$$
(4.96a)
the nonlinear vector is
$${\mathbf{f}} = \delta {\mathbf{x}}$$
(4.96b)
And the switching function $$\delta (t)$$ is defined as shown in Eq. (4.12).
Considering that the initial condition is zero, there is $${\text{x}}_{\text{i}} \left( 0 \right) = 0$$ , so the zero-order approximate solution can be assumed to be
$${\mathbf{x}}_{0} = a_{00} (t) = \left[ {i_{00} ,v_{00} } \right]^{Tr}$$
Substituting x 0 into the first equation in Eq. (4.89) gives
$$\left[ {{\mathbf{G}}_{0} (p) + {\mathbf{G}}_{1} (p)b_{0} } \right]{\mathbf{a}}_{00} (\tau ) = {\mathbf{u}}$$
(4.97)
Equation (4.97) can be written in matrix form as
$$\left( {\left[ {\begin{array}{*{20}c} p & {\frac{1}{L}} \\ { - \frac{1}{C}} & {p + \frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{{b_{0} }}{L}} \\ {\frac{{b_{0} }}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {i_{00} } \\ {v_{00} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {E/L} \\ 0 \\ \end{array} } \right]$$
(4.98)
in which b0 = D. Equation (4.98) can be expressed as shown in Eq. (4.99), since it is a linear differential equation.
$$\frac{{d{\mathbf{x}}_{0} }}{dt} = {\mathbf{A}}_{0} {\mathbf{x}}_{0} + {\mathbf{u}}$$
(4.99)
In which, the coefficient matrix and the input vector are shown as follows.
$${\mathbf{A}}_{0} = \left[ {\begin{array}{*{20}c} 0 & { - \frac{1 - D}{L}} \\ {\frac{1 - D}{C}} & {\frac{ - 1}{RC}} \\ \end{array} } \right],\quad {\mathbf{u}} = \left[ {\begin{array}{*{20}c} {E/L} \\ 0 \\ \end{array} } \right]$$
And then the following zero-order solutions, as shown in Eq. (4.100), can be obtained according to Eq. (4.93).
$$\begin{aligned} i_{00} & = 2.222 - {\text{e}}^{ - 0.059\tau } [2.222\,{ \cos }\,0.222\tau + 3.89\,{ \sin }\,0.222\tau ] \\ v_{00} & = 50 - {\text{e}}^{ - 0.059\tau } [50\,{ \cos }\,0.222\tau - 13.276\,{ \sin }\,0.222\tau ] \\ \end{aligned}$$
(4.100)
in which $$\tau = 2\pi f_{s} t$$ . The results from Eq. (4.100) (red dashed line) and the simulations (green solid line) are compared in Fig. 4.11, from which it can be seen that the zero-order approximation roughly grasps the characteristics of the transient process.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig11_HTML.gif
Fig. 4.11

Zero-order approximation of the transient solution of CCM-Boost converter

Similar to Eq. (4.33), the first-order correction term can be chosen as
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} (t)e^{j\tau } + {\bar{\mathbf{a}}}_{11} (t)e^{ - j\tau }$$
Here $${\mathbf{a}}_{11} (t) = \left[ {i_{11} ,v_{11} } \right]^{Tr}$$. By replacing (p + jω) with the (jω) in Eq. (4.35), the first-order correction equation corresponding to the second equation in Eq. (4.89) can be obtained as
$$[{\mathbf{G}}_{0} (p + {\text{j}}\omega ) + {\mathbf{G}}_{1} (p + {\text{j}}\omega ){\text{b}}_{0} ]{\mathbf{a}}_{11} = - {\mathbf{G}}_{1} (p + {\text{j}}\omega ) \cdot b_{1} {\mathbf{a}}_{00}$$
(4.101)
which can be rewritten in matrix from as
$$\left( {\left[ {\begin{array}{*{20}c} {p + j\omega } & {\frac{1}{L}} \\ { - \frac{1}{C}} & {p + j\omega + \frac{1}{RC}} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 & { - \frac{D}{L}} \\ {\frac{D}{C}} & 0 \\ \end{array} } \right]} \right)\left[ {\begin{array}{*{20}c} {i_{11} } \\ {v_{11} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 & {\frac{1}{L}} \\ {\frac{ - 1}{C}} & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {b_{1} i_{00} } \\ {b_{1} v_{00} } \\ \end{array} } \right]$$
(4.102)
Equation (4.102) is also a linear differential equation, and can be expressed as a standard form as follows:
$$\frac{{d{\mathbf{a}}_{11}(t) }}{dt} = {\mathbf{A}}_{1} {\mathbf{a}}_{11}(t) + {\mathbf{B}}_{1} {\mathbf{x}}_{0}$$
(4.103)
In which the coefficient matrices are shown as
$${\mathbf{A}}_{1} = \left[ {\begin{array}{*{20}c} { - j\omega } & { - \frac{1 - D}{L}} \\ {\frac{1 - D}{C}} & { - j\omega - \frac{1}{RC}} \\ \end{array} } \right],\quad {\mathbf{B}}_{1} = \left[ {\begin{array}{*{20}c} 0 & {\frac{{b_{1} }}{L}} \\ {\frac{{ - b_{1} }}{C}} & 0 \\ \end{array} } \right]$$
The initial value x1(0) = 0, that is, i11(0) = 0 and v11(0) = 0. Then the special solution x1(τ) can be obtained according to the calculation of Eq. (4.93), as shown in Eq. (4.104).
$$\begin{aligned} i_{11} & = {\text{e}}^{ - 0.059\tau } [0.2209\,{\text{cos(}}0.222\tau ) + 0.2547\,{\text{sin(}}0.222\tau ) + 0.7298\,\cos (1.222\tau ) \\ & \quad - 0.1333\,\sin (1.222\tau )] - 0.5007\,\cos \tau + 0.4031\,\sin \tau \\ v_{11} & = {\text{e}}^{ - 0.059\tau } [ - 2.188\,{\text{cos(}}0.222\tau ) + 3.22\,{\text{sin(}}0.222\tau ) + 1.22\,\cos ( 1.222\tau ) \\ & \quad + 3.365\,\sin ( 1.222\tau )] + 0.968\,\cos \tau - 3.945\,\sin \tau \\ \end{aligned}$$
(4.104)
Thus the first-order approximation of Eq. (4.95) can be obtained as
$${\mathbf{x}} \approx {\mathbf{x}}_{0} + {\mathbf{x}}_{1} = \left[ {\begin{array}{*{20}c} {i_{00} + i_{11} } \\ {v_{00} + v_{11} } \\ \end{array} } \right]$$
(4.105)
The results acquired from Eq. (4.105) (red dash line) are shown together with those from simulations (green solid line) in Fig. 4.12. It can be seen from Fig. 4.12 that the first-order approximation is close to the transient process, but the pulsating component of the state variable is still not well estimated.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig12_HTML.gif
Fig. 4.12

First-order approximation of the transient solution of CCM-Boost converter

A similar method can be used to further obtain the second-order correction of the transient solution. Thus the second-order approximate transient solution of Eq. (4.95) is obtained as
$${\mathbf{x}} = [\begin{array}{*{20}c} {i_{L} } & {v_{C} } \\ \end{array} ]^{Tr} \approx {\mathbf{x}}_{0} + {\mathbf{x}}_{1} + {\mathbf{x}}_{2}$$
(4.106)
The specific approximate solutions are shown in (4.107a) and (4.107b) respectively.
$$\begin{aligned} i_{L} & \approx 2.1734 + {\text{e}}^{ - 0.059\tau } [ - 2.1335\,{\text{cos(0}} . 2 2 2\tau ) + 4.1547\,{\text{sin(0}} . 2 2 2\tau ) + 0.021\tau \,{\text{cos(0}} . 2 2 2\tau ) \\ & \quad + 0.0018\tau \,\sin ( 0. 2 2 2\tau ) + 0.1810\,{ \cos }(0.778\tau ) - 0.0238\,\sin (0.778\tau ) \\ & \quad + 0.2733\,\cos (1.222\tau ) - 0.1333\,\sin (1.222\tau )] - 0.5007\,\cos \tau + 0.4031\,\sin \tau \\ & \quad - 0.2051\,\cos 2\tau - 0.002\,\sin 2\tau - 0.0476\,{ \cos }3\tau - 0.0408\,\sin 3\tau \\ \end{aligned}$$
(4.107a)
$$\begin{aligned} v_{C} & \approx 49.368 + {\text{e}}^{ - 0.059\tau } [ - 51.636\,{\text{cos(0}} . 2 2 2\tau ) - 12.352\,{\text{sin(0}} . 2 2 2\tau ) \\ & \quad + 0.0438\tau \,{\text{cos(0}} . 2 2 2\tau ) + 0.2431\tau \,\sin ( 0. 2 2 2\tau ) + 0.0768\,{ \cos }(0.778\tau ) \\ & \quad + 0.2672\,\sin (0.778\tau ) + 1.22\,\cos (1.222\tau ) + 3.3648\,\sin (1.222\tau )] + 0.0968\,\cos \tau \\ & \quad - 3.945\,\sin \tau + 1. 2 3 9 7\,\cos 2\tau - 0.026\,\sin 2\tau + 0.1878\,{ \cos }3\tau + 0.3523\,\sin 3\tau \\ \end{aligned}$$
(4.107b)
The results acquired from Eq. (4.107a) (red dash line) are shown together with those from simulations (blue solid line) in Fig. 4.13, from which we can see that the second-order approximation in very good coincidence with the transient process, especially it can estimate the ripple at steady state very well.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig13_HTML.gif
Fig. 4.13

Second-order approximation of the transient solution of CCM-Boost converter

4.6.4 Simplified Calculation

Since the equations for the first-order and second-order harmonic components are complex, the calculation process involves solving the second-order non-homogeneous linear differential equations, so the solution process is too complicated. Generally, the transient solution of the system can be regarded as the sum of two parts, one is the difference obtained by subtracting the DC component from the principal component of the transient solution, and the other is the steady-state periodic solution of the system. The steady-state solutions of the Boost converter, which can be easily solved by the ESPM, as introduced in Sect. 4.3, are shown again in (4.108).
$$\begin{aligned} i_{steady} & = 2.11 - 0.487\,{ \cos }\tau + 0.392\,{ \sin }\tau - 0.205\,{ \cos }2\tau \\ & \quad - 0.002\,{ \sin }2\tau - 0.0476\,{ \cos }3\tau - 0.0408\,{ \sin }3\tau \\ v_{steady} & = 48.08 + 0.942\,{ \cos }\tau - 3.945\,{ \sin }\tau + 1.24\,{ \cos }2\tau \\ & \quad - 0.026\,{ \sin }2\tau + 0.1878\,{ \cos }3\tau + 0.3523\,{ \sin }3\tau \\ \end{aligned}$$
(4.108)
Then, the main component of the transient solution of the system as shown in Eq. (4.100), is obtained by solving Eq. (4.89). The steady-state component (DC component) in Eq. (4.100) is removed, and the remaining result is added to the steady-state solution of Eq. (4.109) as the transient solution of the converter, as shown in (4.109).
$$\begin{aligned} i & \approx {\text{e}}^{ - 0.059\tau } [ - 2.222\,{ \cos }\,0.222\tau + 3.89\,{ \sin }\,0.222\tau ] + i_{steady} \\ v & \approx {\text{e}}^{ - 0.059\tau } [ - 50\,{ \cos }\,0.222\tau - 13.276\,{ \sin }\,0.222\tau ] + v_{steady} \\ \end{aligned}$$
(4.109)
The results acquired from Eq. (4.109) (red dash line) are shown together with those from simulations (green solid line) in Fig. 4.14, from which we can see that they agree well with each other. Therefore, the simplified algorithm can solve transient solutions more accurately and can greatly reduce the amount of calculation.
../images/419194_1_En_4_Chapter/419194_1_En_4_Fig14_HTML.gif
Fig. 4.14

Approximated transient solution of  CCM-Boost converter from simplified method

4.7 Summary

In this chapter, the equivalent-small-parameter method is used to analyze the PWM converters operating in CCM (continuous current mode) and DCM (discontinuous current mode) for obtaining the analytical expressions of steady-state periodic solutions and transient solutions. The equivalent-small-parameter method overcomes the shortcoming of dealing with large amount of calculation, which exists in the general average method. It is only necessary to solve the linear algebra or differential equations with lower order. It shows that even if the switching frequency is low and the ripple is large, the ESPM can still be applicable with high precision.