© 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_6

6. Analysis of Voltage-Mode Controlled DCM-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

6.1 Introduction

Discontinuous conduction mode (DCM) typically occurs when the DC/DC converter operates at a light load. For low-power applications, it is necessary for the converter to operate at DCM, even when fully loaded [1]. In addition the power-factor-correction (PFC) circuit is intentionally designed to operate at DCM [2–4] for better performances. Hence, modeling and analysis of the converter operating at DCM are very important.

At present, the method of analyzing PWM converters with DCM operation is mainly the state-space-averaging method [5]. In the case of discontinuous inductor current mode, the inductor current decreases to zero before the end of the switching cycle and remains at zero until the beginning of the next cycle. Hence, the inductor current is always zero at the beginning and the end of each cycle. The traditional state space averaging method [6] uses this as a constraint to get an order-reduced model that no longer contains the inductor current. For the dynamic characteristics of the system, the order-reduced model is not accurate [1]. As the inductor current in PFC is the ultimate control objective, the disappearance of the inductor current in the model is not what we expect.

Thus a new model for analyzing the full-order equivalent duty cycle model of DCM-operated converter is proposed in [5], which is finally experimentally proved to analyze the dynamic characteristics of AC small signal accurately. And in the reference [7], the average switching model applies the averaging technique to the nonlinear part of the converter circuit and directly obtains the DC and AC small signal analysis models of the PWM converter. However, for large signal characteristics analysis, it is necessary to resort to the more complicated circuit Model [8] (i.e., the nth harmonic three-port model).

In the late 1980s, power factor correction (PFC) technology in switching power supply has drawn many scholars’ attention all over the world [9–18]. The PFC circuit is a strong nonlinear periodic time-varying system. Although several engineering analysis and numerical modeling methods have appeared in recent years, the analytical solution of periodic time-varying system has not been found [14–18]. In Chap. 2, we establish a unified nonlinear large-signal model of PWM converters operating at CCM and DCM based on the switching function. In this chapter, the Equivalent-Small-Parameter method is used to analyze this model for DCM-operated converters, and the analytic solution to the steady-state of the closed-loop PWM switching converter system is obtained. It is believed that this method is easy to be applied to the analysis of the PFC system.

6.2 Time-Varying Equation for Closed-Loop DCM-Operated Converter System

Discontinuous inductor current waveform is shown in Fig. 6.1b. For convenience, in this chapter, d1 represents the duty cycle when the transistor switch is on (the inductor current rises in this period), d2 represents the duty cycle when the inductor current drops and d3 represents the duty cycle when the inductor current remains at zero. The corresponding switching functions are δ1(t), δ2(t) and δ3(t), or abbreviated as δ1, δ2 and δ3. Their meaning is a little different from that of Chap. 4.
../images/419194_1_En_6_Chapter/419194_1_En_6_Fig1_HTML.gif
Fig. 6.1

DCM boost regulator and the inductor current waveform

According to the modeling method in Chap. 2, the PWM converter, operating at discontinuous inductor current, can be described by time-varying differential Eq. (6.1):
$${\mathbf{G}}_{0} (p){\mathbf{x}} + {\mathbf{G}}_{1} (p) \cdot {\mathbf{f}}^{(1)} ({\mathbf{x}}) + {\mathbf{G}}_{3} (p) \cdot {\mathbf{f}}^{(2)} ({\mathbf{x}}) = {\mathbf{u}}$$
(6.1)
where the nonlinear vector functions are defined as follows.
$${\mathbf{f}}^{(1)} = \delta_{1} ({\mathbf{x}} + {\mathbf{e}}_{1} )$$
(6.2)
$${\mathbf{f}}^{(2)} = \delta_{3} ({\mathbf{x}} + {\mathbf{e}}_{2} )$$
(6.3)
where the meaning of the square matrices Gi(p) (i = 0, 1, 3) are similar to those in Chap. 5, and e1, e2 are constant vectors. Switching functions δ1, δ3 are defined as:
$$\delta_{1} (t) = \left\{ {\begin{array}{*{20}l} 1 & {0 < t \le d_{1} (t)T} \\ 0 & {d_{1} (t)T < t \le T} \\ \end{array} } \right.$$
(6.4)
$$\delta_{3} (t) = \left\{ {\begin{array}{*{20}l} 0 & {0 < t \le [1 - d_{3} (t)]T} \\ 1 & {[1 - d_{3} (t)]T < t \le T} \\ \end{array} } \right.$$
(6.5)
where δ1 represents the on-off state of the main switch, d1 represents the duty cycle when the main switch is on, which is determined by the feedback control algorithm. According to the discussion in Chap. 5, the duty cycle d1 can be expressed as a linear function of the system’s state variables, namely:
$$d_{1} (t) = K_{0} + {\mathbf{K}}_{1} {\mathbf{x}}(d_{1} T)$$
(6.6)
where the meaning of each coefficient is the same with that in Chap. 5.
The switching function δ3 = 1 indicates that the main switch and diodes are disconnected, and δ3 = 0 indicates that either the main switch conducts or the diode conduct. The symbol d3 represents the duty cycle when the inductor current remains zero (or when both switches are turned off). The reason why we choose δ3 to establish the time-varying differential equations of the switching converter system operating at discontinuous conduction mode is as follows: (1) When δ3 = 0 is constantly true during one switching cycle, Eq. (6.1) is the time-varying differential equations of the switching converter system under CCM, in the meantime, G0(p), G1(p), f(1)(x) remain unchanged; (2) As can be seen from the following analysis, δ3 can be transformed into series forms as δ1. According to Sect. 5.​2, using the Equivalent-Small-Parameter Method to solve Eq. (6.1), the following iterative equations can be obtained.
$$\begin{array}{*{20}c} {\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} (p){\mathbf{x}}_{0} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{0m}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{f}}_{0m}^{(2)} = {\mathbf{u}}} \hfill & {\text{ (a)}} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{1} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{1m}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{f}}_{1m}^{(2)} = - [{\mathbf{G}}_{1} (p){\mathbf{R}}_{1}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{R}}_{1}^{(2)} ]} \hfill & {\text{ (b)}} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{2} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{2m}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{f}}_{2m}^{(2)} = - [{\mathbf{G}}_{{\mathbf{1}}} (p){\mathbf{R}}_{2}^{(1)} + G_{3} (p){\mathbf{R}}_{2}^{(2)} ]} \hfill & {\text{ (c)}} \hfill \\ \end{array} } \right.} \\ \\ \ldots \\ \end{array}$$
(6.7)

6.3 Determination of Switching Function and Duty Cycle

The method used to determine δ1, d1 is the same as the CCM analysis method in Chap. 5. The following focuses on the discussion of the method of determining d3, δ3. We propose two different methods, known as the waveform determination method, and traditional determination method respectively.

6.3.1 The Waveform-Based Determination Method for d3

Assuming that the inductor current is piecewise linear (note that there is only the assumption of the linear waveform, without existence of small ripple). The slope of the ascending phase is m1, and the slope of the descent phase is m2 (shown in Fig. 6.1b). The average value of the inductor current in one cycle during steady the state is:
$$\bar{i}_{L} = \frac{1}{T}\int\limits_{0}^{{d_{1} T}} {(I_{0} + m_{1} t)dt} + \int\limits_{0}^{{d_{2} T}} {[(I_{0} + m_{1} d_{1} T - m_{2} t]dt}$$
(6.8)
where I0 is the initial value of the inductor current at the beginning of a switching cycle T. During the steady state, there exists m1d1T = m2d2T. From the above formula, the following relationship can be obtained.
$$\bar{i}_{L} = I_{0} (d_{1} + d_{2} ) + \frac{{m_{1} T}}{2}d_{1}^{2} + \frac{{m_{1} T}}{2}d_{1} d_{2}$$
(6.9)
Setting m = m1T/2, then, the following relationships can be obtained.
$$d_{2} = \frac{{\bar{i}_{L} - (md_{1}^{2} + I_{0} d_{1} )}}{{I_{0} + md_{1} }},\quad d_{3} = 1 - d_{1} - \frac{{\bar{i}_{L} - (md_{1}^{2} + I_{0} d_{1} )}}{{I_{0} + md_{1} }}$$
(6.10)
For the basic second-order PWM converter, such as Boost, Buck, Buck-boost converter, when the inductor current is discontinuous, generally, I0 = 0. Not taking into account the circuit’s parasitic parameters, m1 and m2 usually are functions of the average input power and the average output voltage (capacitor voltage). Thus, d3 can be expressed as a function of the power supply voltage, the average state variables $${\bar{\mathbf{x}}}$$ and d1, i.e.,
$$d_{3} = h_{1} (E,d_{1} ,{\bar{\mathbf{x}}})$$
(6.11)
According to the series expansion of $${\bar{\mathbf{x}}}$$ and d1, which are defined as follows.
$${\bar{\mathbf{x}}} = {\bar{\mathbf{x}}}_{0} + \varepsilon {\bar{\mathbf{x}}}_{1} + \varepsilon^{2} {\bar{\mathbf{x}}}_{2} + \cdots$$
(6.12)
$$d_{1} = d_{10} + \varepsilon d_{11} + \varepsilon^{2} d_{12} + \cdots$$
(6.13)
Assuming that d3 can be transformed into the series expansion that consists of the main term and small terms, the following relationship can be obtained.
$$d_{3} = d_{30} + \varepsilon d_{31} + \varepsilon^{2} d_{32} + \cdots$$
(6.14)
The function $$h_{1} (E,d_{1} ,{\bar{\mathbf{x}}})$$ is approximated by the linear terms of its Taylor series expanded at $$(d_{10} ,{\bar{\mathbf{x}}}_{0} )$$, then $$d_{3i}$$ (i = 1, 2, …) can be approximated by the following equations as
$$\begin{array}{*{20}l} {d_{30} = h_{1} (E,d_{10} ,{\bar{\mathbf{x}}}_{0} )} \hfill \\ {d_{31} = \left( {\left. {\frac{{\partial h_{1} }}{{\partial d_{1} }}} \right|_{\begin{subarray}{l} d_{1} = d_{10} \\ {\bar{\mathbf{x}} = \bar{\mathbf{x}}}_{0} \end{subarray} } } \right) \cdot d_{11} + \left( {\left. {\frac{{\partial h_{1} }}{{\partial \bar{x}}}} \right|_{\begin{subarray}{l} d_{1} = d_{10} \\ {\bar{\mathbf{x}} = \bar{\mathbf{x}}}_{0} \end{subarray} } } \right) \cdot {\bar{\mathbf{x}}}_{1} = H_{11}^{1} \cdot d_{11} + H_{12}^{1} \cdot {\bar{\mathbf{x}}}_{{\mathbf{1}}} } \hfill \\ {d_{32} = \left( {\left. {\frac{{\partial h_{1} }}{{\partial d_{1} }}} \right|_{\begin{subarray}{l} d_{1} = d_{10} + d_{11} \\ {\bar{\mathbf{x}}} = {\bar{\mathbf{x}}}_{0} + {\bar{\mathbf{x}}}_{{\mathbf{1}}} \end{subarray} } } \right) \cdot d_{12} + \left( {\left. {\frac{{\partial h_{1} }}{{\partial \bar{x}}}} \right|_{\begin{subarray}{l} d_{1} = d_{10} + d_{11} \\ {\bar{\mathbf{x}} = \bar{\mathbf{x}}}_{{\mathbf{0}}} +{\bar{\mathbf{x}}}_{{\mathbf{1}}} \end{subarray} } } \right) \cdot {\bar{\mathbf{x}}}_{2} = H_{21}^{1} \cdot d_{12} + H_{22}^{1} \cdot {\bar{\mathbf{x}}}_{2} } \hfill\\ \\ \cdots \hfill \\ \end{array}$$
(6.15)
where $$H_{11}^{1}$$ and $$H_{21}^{1}$$ are real numbers or vectors with $$1 \times n$$ dimensions (if x is $$1 \times n$$-dimensional). In this way, δ3 can be transformed into series expansion the same as δ1, then the following relationship can be obtained.
$$\delta_{3} = \delta_{30} + \varepsilon \delta_{31} + \varepsilon^{2} \delta_{32} + \cdots$$
(6.16)
In which each term in (6.16) is chosen as
$$\begin{array}{*{20}c} {\begin{array}{*{20}l} {\delta_{30} = d_{30} + q_{10} e^{j\tau } + c.c} \hfill \\ {\delta_{31} = d_{31} + q_{11} e^{j\tau } + q_{20} e^{j2\tau } + q_{30} e^{j3\tau } + c.c} \hfill \\ {\delta_{32} = d_{32} + q_{12} e^{j\tau } + q_{21} e^{j2\tau } + q_{31} e^{j3\tau } + q_{40} e^{j4\tau } + q_{50} e^{j5\tau } + c.c} \hfill \\ \end{array} } \\ \cdots \\ \end{array}$$
where the coefficients in Eq. (6.16), according to the improved algorithm proposed in Sect. 5.​5, can be deduced as:
$$\begin{aligned} q_{mi} & = \tfrac{1}{2}(\gamma_{mi} - j\eta_{mi} )\quad (m = 1,2, \ldots ,i = 0,1,2, \ldots ) \\ \gamma_{m0} & = \frac{{\sin 2m\pi d_{30} }}{m\pi },\quad \eta_{m0} = - \frac{{1 - \cos 2m\pi d_{30} }}{m\pi } \\ \gamma_{mi} & \approx 2d_{3i} \cos 2m\pi (d_{30} + \varepsilon d_{31} + \cdots + \varepsilon^{i - 1} d_{3(i - 1)} )\quad (i = 1,2, \ldots ) \\ \eta_{mi} & \approx - 2d_{3i} \sin 2m\pi (d_{30} + \varepsilon d_{31} + \cdots + \varepsilon^{i - 1} d_{3(i - 1)} )\quad (i = 1,2, \ldots ) \\ \end{aligned}$$
(6.17)

6.3.2 Traditional Determination Method for d3

According to the basic principle of DC/DC converters, when the converter is at steady state, d2 can be represented by d1, as shown in Table 6.1.
Table 6.1

Relationship between d1 and d2 in steady-state basic converter

 

Buck

Boost

Buck/boost

Cuk

$$k_{c}$$

$$k_{c} = 2Lf_{s} /R$$

$$k_{c} = 2Lf_{s} /R$$

$$k_{c} = 2Lf_{s} /R$$

$$k_{c} = 2L_{e} f_{s} /R$$

$$d_{2}$$

$$\frac{{2k_{c} }}{{d_{1} (1 + \sqrt {1 + 4k_{c} /d_{1}^{2} } )}}$$

$$\frac{{k_{c} (1 + \sqrt {1 + 4d_{1}^{2} /k_{c} } )}}{{2d_{1} }}$$

$$\sqrt {k_{c} }$$

$$\sqrt {k_{c} }$$

Where $$L_{e} = {{L_{1} L_{2} } \mathord{\left/ {\vphantom {{L_{1} L_{2} } {(L_{1} + L_{2} )}}} \right. \kern-0pt} {(L_{1} + L_{2} )}}$$, denoting the equivalent inductance of the Cuk converter.

As shown in Table 6.1, using the DC analysis method of converters, the duty cycle d2 can be described as the function of d1. Then, d3 can also be expressed as the function of d1 as follows.
$$d_{3} = 1 - d_{1} - d_{2} = h_{2} (d_{1} )$$
Similarly, $$d_{3}$$ can be expanded into the series form as formula (6.14), and the term $$d_{3i}$$ is determined as follows.
$$\begin{array}{*{20}l} {d_{30} = h_{2} (d_{10} ),d_{31} = \left( {\left. {\frac{{\partial h_{2} }}{{\partial d_{1} }}} \right|_{{d_{1} = d_{10} }} } \right) \cdot d_{11} = H_{11}^{2} \cdot d_{11} ,} \hfill \\ {d_{32} = \left( {\left. {\frac{{\partial h_{2} }}{{\partial d_{1} }}} \right|_{{d_{1} = d_{10} + d_{11} }} } \right) \cdot d_{12} = H_{21}^{2} \cdot d_{12} ,} \hfill \\ \cdots \hfill \\ \end{array}$$
(6.18)
where $$H_{11}^{2}$$ and $$H_{21}^{2}$$ are real numbers. According to (6.18), $$\delta_{3}$$ can also be expanded into series form as Eq. (6.16). Thus, both methods have the same solving process, except for applying different methods to get the main term and the small terms of the duty cycle $$d_{3}$$.

6.4 Solution of Time-Varying Equation for Closed-Loop DCM-Operated System

Utilizing the equivalent small-parameter method, the time-varying equation of the closed-loop system under DCM operation can be solved in the same way as that of the closed-loop system in CCM operation.

6.4.1 Solution of the Main Component

The main component of state variable is DC component, the following assumptions are made, i.e.,
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00}$$
(6.19)
Then the nonlinear vectors are determined by
$$\begin{aligned} {\mathbf{f}}_{0}^{(1)} & = \delta_{10} {\mathbf{x}}_{0} \\ {\mathbf{f}}_{0}^{(2)} & = \delta_{30} ({\mathbf{x}}_{0} + {\mathbf{e}}_{2} ) \\ \end{aligned}$$
(6.20)
Then the main term and the remainder belonging to $${\mathbf{f}}_{0}^{(1)}$$ can be obtained as
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0m}^{(1)} = d_{10} {\mathbf{a}}_{00} } \hfill \\ {{\mathbf{R}}_{1}^{(1)} = b_{10} {\mathbf{a}}_{00} e^{j\tau } + c.c} \hfill \\ \end{array} } \right.$$
(6.21)
And those of $${\mathbf{f}}_{0}^{(2)}$$ are shown as
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0m}^{(2)} = d_{30} ({\mathbf{a}}_{00} + {\mathbf{e}}_{2} )} \hfill \\ {{\mathbf{R}}_{1}^{(2)} = q_{10} ({\mathbf{a}}_{00} + {\mathbf{e}}_{2} )e^{j\tau } + c.c} \hfill \\ \end{array} } \right.$$
(6.22)
Substituting $${\mathbf{x}}_{0}$$, $${\mathbf{f}}_{0m}^{(1)}$$, and $${\mathbf{f}}_{0m}^{(2)}$$ into (6.7a) with the above formulas, the following matrix equation can be obtained, i.e.,
$${\mathbf{G}}_{{\mathbf{0}}} (0){\mathbf{a}}_{00} + {\mathbf{G}}_{1} (0)d_{10} {\mathbf{a}}_{00} + {\mathbf{G}}_{3} (0)d_{30} ({\mathbf{a}}_{00} + {\mathbf{e}}_{{\mathbf{2}}} ) = {\mathbf{u}}$$
(6.23)
In which, d30 determined by Eq. (6.15) or (6.18). And according to (5.​38a), the term d0 satisfies the following equation as
$$d_{0} = K_{0} + {\mathbf{K}}_{{\mathbf{1}}} {\mathbf{x}}_{0} = K_{0} + {\mathbf{K}}_{1} {\mathbf{a}}_{00}$$
(6.24)

As can be seen from the above discussion that the formulas for solving $${\mathbf{a}}_{00}$$ and $$d_{10} ,d_{30}$$ are all nonlinear, requiring the application of symbol method or numerical method.

6.4.2 Solution of First-Order Correction

According to Eqs. (6.21) and (6.22), the first-order correction can be assumed by the following formula as
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} e^{j\tau } + {\bar{\mathbf{a}}}_{11} e^{ - j\tau }$$
(6.25)
And the first-order corrections for nonlinear vectors can be deduced as
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{1}^{(1)} = \delta_{11} {\mathbf{x}}_{0} + \delta_{10} {\mathbf{x}}_{1} } \hfill \\ {{\mathbf{f}}_{1}^{(2)} = \delta_{31} ({\mathbf{x}}_{0} + {\mathbf{e}}_{2} ) + \delta_{30} {\mathbf{x}}_{1} } \hfill \\ \end{array} } \right.$$
(6.26)
Then the main terms and the remainders belonging to $${\mathbf{f}}_{1}^{(1)}$$ and $${\mathbf{f}}_{1}^{(2)}$$ can be obtained as
$$\begin{aligned} & \left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{1m}^{(1)} = (d_{10} {\mathbf{a}}_{11} + b_{11} {\mathbf{a}}_{00} ) \cdot e^{j\tau } + c.c} \hfill \\ {{\mathbf{R}}_{2}^{(1)} = (b_{10} {\bar{\mathbf{a}}}_{11} + \bar{b}_{10} {\mathbf{a}}_{11} + d_{11} {\mathbf{a}}_{00} ) + (b_{10} {\mathbf{a}}_{11} + b_{20} {\mathbf{a}}_{00} ) \cdot e^{j2\tau } + b_{30} {\mathbf{a}}_{{{\mathbf{00}}}} \cdot e^{j3\tau } + c.c} \hfill \\ \end{array} } \right. \\ & \left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{1m}^{(2)} = [d_{30} {\mathbf{a}}_{11} + q_{11} ({\mathbf{a}}_{00} + {\mathbf{e}}_{{\mathbf{2}}} )] \cdot e^{j\tau } + c.c} \\ {{\mathbf{R}}_{2}^{(2)} = [q_{10} {\bar{\mathbf{a}}}_{11} + \bar{q}_{10} {\mathbf{a}}_{11} + d_{31} ({\mathbf{a}}_{00} + e_{2} )] + [q_{10} {\mathbf{a}}_{11} + q_{20} ({\mathbf{a}}_{00} + {\mathbf{e}}_{2} )] \cdot e^{j2\tau } } \\ {\quad \quad \quad + q_{30} ({\mathbf{a}}_{00} + e_{2} ) \cdot e^{j3\tau } + c.c} \\ \end{array} } \right. \\ \end{aligned}$$
(6.27)
Substituting $${\mathbf{f}}_{1m}^{(1)}$$, $${\mathbf{f}}_{1m}^{(2)}$$ and $${\mathbf{R}}_{1}^{(1)}$$, $${\mathbf{R}}_{1}^{(2)}$$ into Eq. (6.7b) would give the following matrix equation as
$$\begin{aligned} & [{\mathbf{G}}{}_{0}(j\omega ) + {\mathbf{G}}_{1} (j\omega ) \cdot d_{10} + {\mathbf{G}}_{3} (j\omega ) \cdot d_{30} ] \cdot {\mathbf{a}}_{11} \\ & \quad = - {\mathbf{G}}_{1} (j\omega ) \cdot (b_{11} + b_{10} ) \cdot {\mathbf{a}}_{00} - {\mathbf{G}}_{3} (j\omega )[(q_{11} + q_{10} )({\mathbf{a}}_{00} + {\mathbf{e}}_{{\mathbf{2}}} {\mathbf{)}}] \\ \end{aligned}$$
(6.28a)
According to Eqs. (5.​32b) and (6.17), the following equations can be obtained.
$$b_{11} = d_{11} \cdot (\cos 2\pi d_{10} - j\sin 2\pi d_{10} ) = d_{11} \cdot e^{{ - j\tau_{s0} }} \quad (\tau_{s0} = 2\pi d_{10} )$$
(6.28b)
$$q_{11} = d_{31} \cdot (\cos 2\pi d_{30} + j\sin 2\pi d_{30} ) = d_{31} \cdot e^{{j\sigma_{s0} }} \quad (\sigma_{s0} = 2\pi d_{30} )$$
(6.28c)
Since $${\bar{\mathbf{x}}}_{1} = 0$$, then according to Eqs. (6.15) or (6.18), one can get the following formula as
$$d_{31} = H_{1} d_{11}$$
in which $$H_{1} = H_{11}^{1} {\text{ or }}H_{11}^{2}$$.
And the term d11 can be obtained according to the Eq. (5.​18b), that is
$$d_{11} = {\mathbf{K}}_{1} \cdot ({\mathbf{a}}_{11} e^{{j\tau_{s0} }} + {\bar{\mathbf{a}}}_{11} e^{{ - j\tau_{s0} }} )$$
(6.29a)
Then setting the following symbolic expressions as
$$\begin{aligned} {\mathbf{G}}_{1} & = {\mathbf{G}}_{0} (j\omega ) + {\mathbf{G}}_{1} (j\omega ) \cdot d_{10} + {\mathbf{G}}_{3} (j\omega ) \cdot d_{30} \\ {\mathbf{A}}_{12} & = - {\mathbf{G}}_{1}^{ - 1} \cdot {\mathbf{G}}_{1} (j\omega ),{\mathbf{A}}_{13} = - {\mathbf{G}}_{1}^{ - 1} \cdot {\mathbf{G}}_{3} (j\omega ) \\ {\mathbf{B}}_{10} & = {\mathbf{A}}_{12} \cdot b_{10} \cdot e^{{j\tau_{s0} }} ,{\mathbf{Q}}_{10} = {\mathbf{A}}_{13} \cdot q_{10} \cdot e^{{j\tau_{s0} }} \\ F & = H_{1} \cdot e^{{j(\tau_{s0} + \sigma_{s0} )}} \\ B_{1} & = K_{1} [({\mathbf{B}}_{10} + {\bar{\mathbf{B}}}_{10} )a_{00} + ({\mathbf{Q}}_{10} + {\bar{\mathbf{Q}}}_{10} )({\mathbf{a}}_{00} + {\mathbf{e}}_{{\mathbf{2}}} )] \\ A_{1} & = 1 - {\mathbf{K}}_{1} [({\mathbf{A}}_{12} + {\bar{\mathbf{A}}}_{12} ){\mathbf{a}}_{00} + ({\mathbf{A}}_{13} F + {\bar{\mathbf{A}}}_{13} \bar{F})({\mathbf{a}}_{00} + {\mathbf{e}}_{{\mathbf{2}}} {\mathbf{)}}] \\ \end{aligned}$$
(6.29b)
And considering Eqs. (6.28) and (6.29), the following equations for solving the first-order corrections can be obtained, i.e.,
$$\begin{aligned} d_{11} & = {{B_{1} } \mathord{\left/ {\vphantom {{B_{1} } {A_{1} }}} \right. \kern-0pt} {A_{1} }},\quad d_{31} = H_{1} \cdot d_{11} , \\ b_{11} & = d_{11} \cdot e^{{ - j\tau_{s0} }} ,\quad q_{11} = d_{31} \cdot e^{{j\sigma_{s0} }} , \\ a_{11} & = - {\mathbf{G}}^{ - 1} (j\omega )[{\mathbf{G}}_{{\mathbf{1}}} (j\omega )(b_{11} + b_{10} )a_{00} + {\mathbf{G}}_{3} (j\omega )(q_{11} + q_{10} )({\mathbf{a}}_{{{\mathbf{00}}}} + {\mathbf{e}}_{2} )] \\ \end{aligned}$$
(6.30)

6.4.3 Solution of Second-Order Correction

According to (6.27), the second-order correction can be assumed by the following formula as
$${\mathbf{x}}_{2} = {\mathbf{a}}_{02} + ({\mathbf{a}}_{22} e^{j2\tau } + {\mathbf{a}}_{32} e^{j3\tau } + c.c)$$
(6.31)
And the second-order corrections for nonlinear vectors can be deduced as
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{2}^{(1)} = \delta_{12} {\mathbf{x}}_{0} + \delta_{11} {\mathbf{x}}_{1} + \delta_{10} {\mathbf{x}}_{2} } \hfill \\ {{\mathbf{f}}_{2}^{(2)} = \delta_{32} ({\mathbf{x}}_{0} + {\mathbf{e}}_{{\mathbf{2}}} ) + \delta_{31} {\mathbf{x}}_{1} + \delta_{30} {\mathbf{x}}_{2} } \hfill \\ \end{array} } \right.$$
Then the main terms belonging to $${\mathbf{f}}_{2}^{(1)}$$ and $${\mathbf{f}}_{2}^{(2)}$$ can be obtained as
$$\left\{ {\begin{array}{*{20}l} \begin{aligned} {\mathbf{f}}_{2m}^{(1)} & = (d_{10} {\mathbf{a}}_{02} + b_{11} {\bar{\mathbf{a}}}_{11} + \bar{b}_{11} {\mathbf{a}}_{11} + d_{12} {\mathbf{a}}_{00} ) + (d_{10} {\mathbf{a}}_{22} + b_{11} {\mathbf{a}}_{11} + b_{30} {\bar{\mathbf{a}}}_{11} + b_{21} {\mathbf{a}}_{00} )e^{j2\tau } \\ & \quad + (d_{0} {\mathbf{a}}_{32} + b_{10} {\mathbf{a}}_{22} + b_{20} {\mathbf{a}}_{11} + b_{31} {\mathbf{a}}_{00} )e^{j3\tau } + c.c \\ \end{aligned} \hfill \\ \begin{aligned} {\mathbf{f}}_{2m}^{(2)} & = [d_{30} {\mathbf{a}}_{02} + q_{11} {\bar{\mathbf{a}}}_{11} + \bar{q}_{11} {\mathbf{a}}_{11} + d_{32} ({\mathbf{a}}_{{{\mathbf{00}}}} + {\mathbf{e}}_{{\mathbf{2}}} )] + [d_{30} {\mathbf{a}}_{{{\mathbf{22}}}} + q_{11} {\mathbf{a}}_{{{\mathbf{11}}}} + q_{30} {\bar{\mathbf{a}}}_{{{\mathbf{11}}}} + q_{21} ({\mathbf{a}}_{{{\mathbf{00}}}} + {\mathbf{e}}_{{\mathbf{2}}} {\mathbf{)}}]e^{j2\tau } \\ & \quad + [d_{30} {\mathbf{a}}_{{{\mathbf{32}}}} + q_{10} {\mathbf{a}}_{{{\mathbf{22}}}} + q_{20} {\mathbf{a}}_{{{\mathbf{11}}}} + q_{31} ({\mathbf{a}}_{{{\mathbf{00}}}} + {\mathbf{e}}{}_{{\mathbf{2}}})]e^{j3\tau } + c.c \\ \end{aligned} \hfill \\ \end{array} } \right.$$

It should be noted that the remainders of $${\mathbf{f}}_{2}^{(1)}$$ and $${\mathbf{f}}_{2}^{(2)}$$ are no longer listed here, as they are used to solve the third-order correction x3. However, since the ESPM is adopted, it usually only needs to be iterated three times, i.e., only the terms x0, x1 and x2 are solved, and the obtained steady-state periodic solutions of the state variables already have sufficient accuracy.

Then substituting $${\mathbf{f}}_{2m}^{(1)} ,{\mathbf{f}}_{2m}^{(2)} ,{\mathbf{R}}_{2}^{(1)} ,{\mathbf{R}}_{2}^{(2)}$$ in Eq. (6.7c) with the above formulas, we can get equations for solving the second-order correction, that is,
$$a_{22} = - {\mathbf{G}}_{2}^{ - 1} \cdot ({\mathbf{B}}_{22} + {\mathbf{Q}}_{22} )$$
(6.32)
where the symbols are defined as
$$\begin{aligned} {\mathbf{G}}_{2} & = {\mathbf{G}}_{0} (j2\omega ) + {\mathbf{G}}_{1} (j2\omega ) \cdot d_{10} + {\mathbf{G}}_{3} (j2\omega ) \cdot d_{30} \\ {\mathbf{B}}_{22} & = {\mathbf{G}}_{1} (j2\omega ) \cdot [(b_{10} + b_{11} ){\mathbf{a}}_{{{\mathbf{11}}}} + (b_{20} + b_{21} ){\mathbf{a}}_{00} + b_{30} {\bar{\mathbf{a}}}_{{{\mathbf{11}}}} ] \\ {\mathbf{Q}}_{22} & = {\mathbf{G}}_{3} (j2\omega ) \cdot [(q_{10} + q_{11} ){\mathbf{a}}_{11} + (q_{20} + q_{21} )({\mathbf{a}}_{{{\mathbf{00}}}} + {\mathbf{e}}_{{\mathbf{2}}} ) + q_{30} {\bar{\mathbf{a}}}_{{{\mathbf{11}}}} ] \\ \end{aligned}$$
And
$${\mathbf{a}}_{32} = - {\mathbf{G}}_{3}^{ - 1} \cdot ({\mathbf{B}}_{32} + {\mathbf{Q}}_{32} )$$
(6.33)
In which the symbols are defined as
$$\begin{aligned} {\mathbf{G}}_{{\mathbf{3}}} & = {\mathbf{G}}_{0} (j3\omega ) + {\mathbf{G}}_{1} (j3\omega ) \cdot d_{10} + {\mathbf{G}}_{3} (j3\omega ) \cdot d_{30} \\ {\mathbf{B}}_{32} & = {\mathbf{G}}_{1} (j3\omega ) \cdot [b_{10} {\mathbf{a}}_{22} + b_{20} {\mathbf{a}}_{{{\mathbf{11}}}} + (b_{30} + b_{31} ){\mathbf{a}}_{00} ] \\ {\mathbf{Q}}_{32} & = {\mathbf{G}}_{3} (j3\omega ) \cdot [q_{10} {\mathbf{a}}_{22} + q_{20} {\mathbf{a}}{}_{11} + (q_{30} + q_{31} )({\mathbf{a}}_{00} + {\mathbf{e}}_{{\mathbf{2}}} )] \\ \end{aligned}$$

As we can see, that the formulas for solving a22 and a32 are linear.

And the DC component a02 in the second-order correction can be solved by the following equation as
$${\mathbf{G}}_{0} \cdot {\mathbf{a}}_{02} + {\mathbf{G}}_{1} (0) \cdot {\mathbf{a}}_{00} \cdot d_{12} + {\mathbf{G}}_{3} (0) \cdot ({\mathbf{a}}_{{{\mathbf{00}}}} + {\mathbf{e}}_{2} ) \cdot d_{32} = - ({\mathbf{B}}_{02} + {\mathbf{Q}}_{02} )$$
(6.34)
In which the symbols are defined as
$$\begin{aligned} {\mathbf{G}}_{0} & = {\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0) \cdot d_{10} + {\mathbf{G}}_{3} (0) \cdot d_{30} \\ {\mathbf{B}}_{02} & = {\mathbf{G}}_{1} (0) \cdot [(b_{10} + b_{11} ){\bar{\mathbf{a}}}_{11} + (\bar{b}_{10} + \bar{b}_{11} ){\mathbf{a}}_{11} + d_{11} {\mathbf{a}}_{00} ] \\ {\mathbf{Q}}_{02} & = {\mathbf{G}}_{3} (0) \cdot [(q_{10} + q_{11} ){\bar{\mathbf{a}}}_{11} + (\bar{q}_{10} + \bar{q}_{11} ){\mathbf{a}}_{11} + d_{31} ({\mathbf{a}}_{00} + {\mathbf{e}}_{2} )] \\ \end{aligned}$$
And the term d12 in Eq. (6.34) can be determined by (5.​38c). We assume that:
$$\begin{aligned} {\mathbf{P}}_{11} & = {\mathbf{a}}_{11} e^{{j\tau_{s0} }} ,{\mathbf{P}}_{22} = {\mathbf{a}}_{22} e^{{j\tau_{s1} }} ,{\mathbf{P}}_{32} = {\mathbf{a}}_{32} e^{{j\tau_{s1} }} ,\quad (\tau_{s0} = 2\pi d_{10} ,\tau_{s1} = 2\pi (d_{10} + d_{11} )) \\ M & = - {\mathbf{K}}_{1} \cdot 4\pi d_{11} \text{Im} ({\mathbf{P}}_{11} ) + 2{\mathbf{K}}_{1} [\text{Re} ({\mathbf{P}}_{22} ) + \text{Re} ({\mathbf{P}}_{32} )] \\ \end{aligned}$$
Among them, Im(.) and Re(.) respectively represent the imaginary part and the real part of the complex number. Then according to (5.​38c), the term d12 can be solved by the equation as follows.
$$d_{12} = M + {\mathbf{K}}_{1} \cdot {\mathbf{a}}_{02}$$
(6.35)
And according to (6.15), the term d32 can be obtained from the following equation as
$$d_{32} = H_{2} \cdot d_{12} + {\mathbf{N}}_{2} \cdot {\mathbf{a}}_{02}$$
(6.36)

In which there would be $$H_{2} = H_{21}^{1}$$ and $$N_{2} = H_{22}^{1}$$ for the first method, or $$H_{2} = H_{21}^{2}$$ and $$N_{2} = 0$$ for the second method of determining d3.

Then considering equations from (6.34) to (6.36), one can get the following equation for solving the term a02, as shown in Eq. (6.37).
$${\mathbf{a}}_{02} = {\mathbf{B}}_{2} /{\mathbf{A}}_{2}$$
(6.37)
In which the coefficients are determined by
$$\begin{aligned} {\mathbf{A}}_{2} & = {\mathbf{G}}_{{\mathbf{0}}} + {\mathbf{G}}_{{\mathbf{1}}} (0) \cdot {\mathbf{a}}_{00} \cdot {\mathbf{K}}_{1} + {\mathbf{G}}_{3} (0) \cdot ({\mathbf{a}}_{00} + {\mathbf{e}}_{2} ) \cdot (H_{2} \cdot {\mathbf{K}}_{1} + N_{2} ) \\ {\mathbf{B}}_{2} & = {\mathbf{B}}_{02} + {\mathbf{Q}}_{02} - {\mathbf{G}}_{1} (0) \cdot a_{00} \cdot M - {\mathbf{G}}_{3} (0)({\mathbf{a}}_{00} + {\mathbf{e}}_{2} ) \cdot H_{2} \cdot M \\ \end{aligned}$$

6.5 Analysis Example

Figure 6.1 shows the basic circuit of a Boost regulator along with the typical waveform of inductor current when operating at DCM. Then this circuit will be used as an example to illustrate the process of solving the steady-state periodic solution by ESPM.

The parameters of the main circuit are chosen as E = 16 V, L = 208 uH, C = 222 uF, R = 12.5 Ω; and parameters of the control circuit are chosen as VR = 2.5 V, R1 = 18 kΩ, R2 = 2 kΩ, R3 = 0.5 kΩ, R4 = 500 kΩ, Cf = 2 uF; the amplitude and frequency of the sawtooth wave are set respectively as Vp = 7.5 V and $$f_{s} = 2.5\;{\text{kHz}},(T = 1/f_{s} )$$. And state differential equation in matrix form of the system is the same as (6.1), in which the coefficient matrices are determined according to the topologies and parameters of the converter system, i.e.,
$${\mathbf{G}}_{0} (p) = \left[ {\begin{array}{*{20}c} p & {\tfrac{1}{L}} & 0 \\ { - \tfrac{1}{C}} & {p + \tfrac{1}{RC}} & 0 \\ 0 & { - \tfrac{{g_{2} }}{{R_{3} C_{f} }}} & {p + \tfrac{1}{{R_{4} C_{f} }}} \\ \end{array} } \right],\quad {\mathbf{G}}_{1} (p) = {\mathbf{G}}_{3} (p) = \left[ {\begin{array}{*{20}c} 0 & { - \tfrac{1}{L}} & 0 \\ { - \tfrac{1}{C}} & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} } \right]$$
The state variable vector, nonlinear function vectors, the constant vector, and the input vector are determined by
$${\mathbf{x}} = [\begin{array}{*{20}c} {i_{L} } & {v_{c} } & v \\ \end{array}_{f} ]^{Tr} ,\quad {\mathbf{f}}^{(1)} ({\mathbf{x}}) = \delta_{1} {\mathbf{x}},\quad {\mathbf{f}}^{(2)} ({\mathbf{x}}) = \delta_{3} ({\mathbf{x}} + {\mathbf{e}}_{2} ),$$
$${\mathbf{e}}_{2} = [\begin{array}{*{20}c} 0 & { - E} & {0]^{Tr} } \\ \end{array} \quad {\text{and}}\quad {\mathbf{u}} = \left[ {\begin{array}{*{20}c} {\tfrac{E}{L}} & 0 & {\tfrac{{ - V_{r} g_{1} }}{{R_{3} C_{f} }}} \\ \end{array} } \right]^{Tr}$$
And the coefficients g, g1 and g2 are got from:
$$g = \frac{1}{{R_{1} }} + \frac{1}{{R_{2} }} + \frac{1}{{R_{3} }},\quad g_{1} = 1 - \frac{1}{{R_{3} \cdot g}},\quad g_{2} = \frac{1}{{R_{1} \cdot g}}$$
The duty cycle d1 is determined by the Eq. (6.6), in which
$$K_{0} = V_{r} /V_{p} \quad {\text{and}}\quad {\mathbf{K}}_{1} = [\begin{array}{*{20}c} 0 & 0 & { - 1/V_{p} ]} \\ \end{array}$$
When using the first method to determine the term d3, the following equation can be obtained.
$$d_{3} = 1 - d_{1} - ({\mathbf{K}}_{11} {\bar{\mathbf{x}}} - md_{1}^{2} )/(md_{1} )$$
(6.38)
In which the coefficients are got by
$$K_{11} = [\begin{array}{*{20}c} 1 & 0 & {0]} \\ \end{array} ,\quad m = (E \cdot T)/(2L)$$

And when using the second method, the term d3 can be obtained according to Table 6.1.

Then when using the first method of determining d3, the steady-state periodic solutions to the state variables in the main circuit can be deduced as shown in (6.39), in which the DC components can be got as Idc = 3.2793 A, and Vdc = 25.0182 V.
$$\begin{aligned} i_{L} & = 3.2793 - 2.2452\cos \tau + 3.4186\sin \tau - 0.6579\cos 2\tau - 0.2409\sin 2\tau \\ & \quad - 0.0078\cos 3\tau - 0.3045\sin 3\tau \\ v_{c} & = 25.0182 - 0.3962\cos \tau - 1.0736\sin \tau + 0.2067\cos 2\tau + 0.0931\sin 2\tau \\ & \quad - 0.0318\cos 3\tau + 0.0835\sin 3\tau \\ \end{aligned}$$
(6.39)
And the duty cycle d1 and d3 can be solved as d1 = 0.2819 and d3 = 0.2438 respectively. Comparison of the state variable ripples got from the ESPM (dashed line) and the numerical simulations (solid line) is shown in Fig. 6.2a, b, in which the waveform obtained from the ESPM is the calculation of the first three harmonics of state variables in the main circuit during one steady-state switching cycle.
../images/419194_1_En_6_Chapter/419194_1_En_6_Fig2_HTML.gif
Fig. 6.2

Ripple comparison between the ESPM (dashed line) and numerical simulation (solid line)

When the second method of determining d3 is adopted, the steady-state periodic solutions to the state variables can be deduced as shown in (6.40), in which the DC components can be got as Idc = 3.0266 A, and Vdc = 25.0196 V.
$$\begin{aligned} i_{L} & = 3.0266 - 2.1535\cos \tau + 3.4044\sin \tau - 0.6708\cos 2\tau - 0.2228\sin 2\tau \\ & \quad - 0.0337\cos 3\tau - 0.3177\sin 3\tau \\ v_{c} & = 25.0196 - 0.4144\cos \tau - 1.0576\sin \tau + 0.2091\cos 2\tau + 0.0799\sin 2\tau \\ & \quad - 0.0242\cos 3\tau + 0.0868\sin 3\tau \\ \end{aligned}$$
(6.40)

Under this situation, the duty cycle d1 and d3 can be solved as d1 = 0.2776 and d3 = 0.2475 respectively. And similarly, comparison of the state variable ripples got from the ESPM (dashed line) and the numerical simulations (solid line) is shown in Fig. 6.2c, d, in which the waveform obtained from the ESPM is the calculation of the first three harmonics of state variables in the main circuit during one steady-state switching cycle.

The DC components got from numerical simulation are Idc = 3.151 A and Vdc = 24.939 V, which are consistent with the results of the above two methods, i.e., Idc = 3.2793 A, and Vdc = 25.0182 V for method1, and Idc = 3.0266 A, and Vdc = 25.0196 V for method 2. The ripples and the DC components from both of the ESP methods are close to each other, and they are also in good agreement with the results from numerical simulation. But the first ESP method, deriving from the waveform of the state variable, is of more generality and easy for algorithm improvement, that is, when considering the rising or falling slope of the inductor current, the influence of the harmonic component can be taken into account at the cost of that duty ratio d2 or d3 cannot be explicitly expressed, which increases the computational complexity.

6.6 Summary

In this chapter, the ESP symbolic analysis method of the closed-loop PWM converter system with CCM operation proposed in Chap. 5 is extended to the steady-state analysis of the closed-loop converter system with DCM operation. And thus the steady-state periodic solution in analytic form of the state variables with the DCM-operated closed-loop converter system can be obtained. When operating in a closed-loop mode, the turn-on time of the switching transistor in the PWM converter is determined by the feedback control circuit. For converters operating at DCM, determining the duty cycle d3, the ratio of the duration of the period when inductor current remains at zero and the entire switching period, is the key to the Equivalent-Small-Parameter Analysis method. With the help of waveform of inductor current and the DC analysis at steady-state, this chapter presents two methods for determining the term d3. The simulation shows that the results of both methods are consistent and have high accuracy.