© 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_5

5. Analysis of Voltage-Mode Controlled CCM-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

5.1 Introduction

State space averaging method [1] is a commonly used method for analyzing switching power converters. But there are obvious limitations [2–7] with the average model as it is independent on the switching frequency. In particular, it is not easy to analyze ripples, or explain the DC offset phenomenon when the switching power converter operates with quite low switching frequency [3]. And the proposed improvement of the average method or the general average method in [2–7] is too cumbersome and difficult to be understood. Equivalent small parameter method (ESPM) [8–11] combines the advantages of the perturbation method and the harmonic balance method, without the need for artificial introduction of small parameters, is a symbolic analysis method with higher precision and relatively simple analysis procedure, and is suitable for analysis of strong nonlinear high-order system. In [12–18], The ESPM is used for analysis of open-loop power converters with continuous current mode (CCM), discontinuous-current-mode (DCM), multi-topology mode operation (including multiple switch converters), as well as in the steady state analysis of Class E amplifiers. And the analytical expression of the converters output ripple can be obtained.

The basic principles of the ESPM together with its applications in the steady-state and transient solutions of open-loop PWM switching power converters are systematically illustrated in Chaps. 3 and 4 respectively. And this chapter begins to apply the equivalent small parameter method (ESP) to the analysis of closed-loop systems. In Chaps. 5 and 6, this symbol analysis method is extended to the closed-loop Voltage-Mode-Controlled (VMC) PWM switching power converter system with CCM and DCM operation. The analytic solutions of the state variable ripples and the duty cycle of the closed-loop system can be obtained. Based on these analytical solutions, the effect of ripple on the duty cycle and the relationship between the DC offset in the converter and the operating frequency can be directly explained. At the same time, the examples show that increasing the integral capacitance in the feedback control loop can properly compensate for the DC drift existing in the system. In view of the fact that the Current-Mode-Control (CMC) has more advantages than the duty cycle programmed control (i.e., voltage-mode-control), we will establish the equivalent small-parameter symbol analysis method of the closed-loop PWM converter system with current-mode-control in Chap. 7. In Chap. 8, we further extend the ESPM to the analysis of frequency-modulated (FM) quasi-resonant converters to show the applicability of the method.

In Sects. 5.2 and 5.3, we first discuss in detail the establishment of the mathematical model of the closed-loop PWM converter system, and then propose the basic algorithm of the closed-loop system based on the equivalent small parameter method. Section 5.4 gives an example analysis and comparison of ESPM and numerical simulations. Section 5.5 proposes a double iterative method to improve the basic algorithm and improve its convergence and accuracy. The experimental results for further verification of the analysis for open-loop and closed-loop converters based on the ESPM are introduced in Sect. 5.6, and the summary of this chapter is given in Sect. 5.7 finally.

5.2 Modeling the Closed-Loop VMC-PWM Converter with CCM Operation

5.2.1 Mathematical Description of the Closed-Loop System

According to Chap. 2, the main circuit of the PWM DC/DC converter operating in CCM mode can be uniformly described as a time-varying state differential equation as
$${\mathbf{g}}_{1} (p){\mathbf{x}}_{1} + {\mathbf{g}}_{2} (p){\mathbf{f}}_{1} = {\mathbf{u}}_{1}$$
(5.1)
The feedback control circuit can also be expressed as a state differential equation:
$${\mathbf{g}}_{3} (p){\mathbf{x}}_{2} + {\mathbf{F}}{\mathbf{x}}_{1} = {\mathbf{u}}_{2}$$
(5.2)
Here differential operator p = d/dt, and the coefficient matrices g1(p) and g2(p) are associated with a particular circuit parameters and the operator p. The terms x1 and x2 are the state variable vectors of the main circuit and the control circuit respectively. F is the feedback coefficient vector, ant the terms u1 and u2 represent respectively the input power vectors of the main circuit and control circuit. The strong nonlinear vector function $${\mathbf{f}}_{1} = \delta (t)({\mathbf{x}}_{1} + {\mathbf{e}}_{1} )$$, in which e1 is a constant vector, and $${\mathbf{e}}_{1} = 0$$ when $${\mathbf{u}}_{1} \ne 0$$. According to (5.1) and (5.2), one can get the following matrix equation as
$$\left[ {\begin{array}{*{20}c} {{\mathbf{g}}_{1} (p)} & 0 \\ {\mathbf{F}} & {{\mathbf{g}}_{3} (p)} \\ \end{array} } \right]{\kern 1pt} \left[ {\begin{array}{*{20}c} {{\mathbf{x}}_{1} } \\ {{\mathbf{x}}_{2} } \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} {{\mathbf{g}}_{2} (p)} & 0 \\ 0 & 0 \\ \end{array} } \right]\delta (t)\left( {\left[ {\begin{array}{*{20}c} {{\mathbf{e}}_{1} } \\ 0 \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} {{\mathbf{x}}_{1} } \\ {{\mathbf{x}}_{2} } \\ \end{array} } \right]} \right) = \left[ {\begin{array}{*{20}c} {{\mathbf{u}}_{1} } \\ {{\mathbf{u}}_{2} } \\ \end{array} } \right]$$
(5.3)
Equation (5.3) can also be rewritten as
$${\mathbf{G}}_{0} (p){\mathbf{x}} + {\mathbf{G}}_{1} ({\mathbf{p}}){\mathbf{f}} = {\mathbf{u}}$$
(5.4)
In Eq. (5.4) the nonlinear vector $${\mathbf{f}} = \delta (t)({\mathbf{x}} + {\mathbf{e}})$$. The meaning of each item in Eq. (5.4) can be obtained by referring to Eq. (5.3), where x represents the state variables of the whole closed-loop system (including the power stage main circuit and the feedback control circuit). It can be seen that the mathematical model of the closed-loop system is exactly the same in form as that of the open-loop system. The switch function $$\delta (t) = 1(0)$$ (hereinafter, $$\delta (t)$$ is sometimes abbreviated as $$\delta$$) indicates the on (off) state of the main switch in the circuit, respectively. For CCM-operated converter, the switching function is defined as
$$\delta (t) = \left\{ {\begin{array}{*{20}l} 1 \hfill & {0\,<\,t \le d(t)T} \hfill \\ 0 \hfill & {d(t)T\,<\,t\le T} \hfill \\ \end{array} } \right.$$
(5.5)

Here T denotes the switching cycle, and d(t) (abbreviated as d) is duty cycle. In the open loop, d(t) is a constant value of D, while in the case of closed-loop, d(t) is determined by the state variable and the feedback control law.

This method is suitable for higher order converters, and Eqs. (5.1) and (5.2) can be generalized to more complex circuits. For the sake of simplicity, this chapter discusses the case where the main circuit is second order, that is, the main circuit has only one inductor L and one capacitor C, and the inductor current along with the capacitor voltage are taken as the main circuit state variable vector, i.e. x1 = [iLvc]Tr, where the superscript “Tr” indicates that the matrix is transposed.

5.2.2 Expression of the Duty Cycle d

5.2.2.1 Expression of d with Proportional Feedback Control

In this case, the state variables of the closed-loop system are the same as those in the open loop, i.e. x = [iLvc]Tr. When the parasitic parameters of the main circuit filter capacitor are not taken into account, the output voltage is v0 = vC. The feedback circuit is shown in Fig. 5.1a, where Vramp is the sawtooth wave voltage, the control voltage Vk of the comparator in Fig. 5.1b is linear with the state variable vector, and the relationship between the duty cycle d(t) and the state vector x can be deduced according to Fig. 5.1c. According to Fig. 5.1, we have
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig1_HTML.gif
Fig. 5.1

a Proportional feedback control, b proportional-integral feedback control, c diagram of duty cycle

$$V_{k} = V_{r} - g_{1} i_{L} - g_{2} v_{c}$$
(5.6)
where the terms g1 and g2 are the sampling coefficients of the output voltage and the inductor current respectively, and Vr is the reference voltage. Suppose the comparator has a state flip at t = ts, that is, ts is defined as
$$t_{s} = t|_{{v_{k} = v_{ramp} }} = nT + t_{on}$$
(5.7)
In which ton denotes the time that the switch is turned on, here the integer n = 0, 1,2,…, means the number of switching cycles. For simplicity, this chapter takes one steady-state cycle for analysis, i.e., n = 0. Thus, the duty cycle d can be expressed as
$$d = \frac{{t_{on} }}{T} = \frac{{t_{s} }}{T} = \frac{{V_{k} - V_{L} }}{{V_{u} - V_{L} }} = \frac{{V_{r} - g_{1} i_{L} - g_{2} v_{c} - V_{L} }}{{V_{u} - V_{L} }} = K_{0} + {\mathbf{K}}_{1} {\mathbf{x}}(t_{s} )$$
(5.8)

Here the coefficient $$K_{0} = \frac{{V_{r} - V_{L} }}{{V_{u} - V_{L} }}\quad$$, and the vector $${\mathbf{K}}_{1} = \left[ {\begin{array}{*{20}c} {\frac{{ - g_{1} }}{{V_{u} - V_{L} }}} & {\frac{{ - g_{2} }}{{V_{u} - V_{L} }}} \\ \end{array} } \right]$$, in which Vu and VL are the maximum and minimum values of the sawtooth wave, respectively.

5.2.2.2 Expression of d with Proportional-Integral Feedback Control

For the sake of simplicity, the PI adjustment circuit shown in Fig. 5.1b is used as an example. In fact, the analysis method is the same for a more complex feedback circuit. Let vf denote the voltage over the feedback capacitor, then the relationship between vf and the reference voltage Vr, along with the output voltage vC, can be obtained according to the PI feedback network, as shown in Eq. (5.9), it is a polynomial of the differential operator p. It can be seen from Fig. 5.1b that the control voltage signal Vk = Vr − vf, and the duty cycle d can still be deduced from Fig. 5.1c, as shown in Eq. (5.10)
$$\frac{{g_{1} }}{{R_{1} C_{f} }}i_{L} + \frac{{g_{2} }}{{R_{1} C_{f} }}v_{c} + \left( {p + \frac{1}{{R_{2} C_{f} }}} \right)v_{f} = - \frac{1}{{R_{1} C_{f} }}V_{r}$$
(5.9)
$$d = \frac{{t_{on} }}{T} = \frac{{V_{k} - V_{L} }}{{V_{u} - V_{L} }} = \frac{{V_{r} - V_{L} - v_{f} }}{{V_{u} - V_{L} }} = K_{0} + {\mathbf{K}}_{1} {\mathbf{x}}(t_{s} )$$
(5.10)

Here the state variables of the closed-loop system with PI control is chosen as $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L\;} } & {v_{C\;} } & {v_{f} } \\ \end{array} } \right]\;^{{{\text{T}}r}}$$, the coefficient $$K_{0} = \frac{{V_{r} - V_{L} }}{{V_{u} - V_{L} }}$$, and the coefficient row vector is. $${\mathbf{K}}_{1} = \quad \left[ {\begin{array}{*{20}c} 0 & 0 & {\frac{ - 1}{{V_{u} - V_{L} }}} \\ \end{array} } \right]$$.

As mentioned previously, the switching converter system has a low pass or bandpass filtering characteristic, so that the steady-state periodic solution $${\mathbf{x}}$$ can be developed into a series as:
$${\mathbf{x}} = {\mathbf{x}}_{0} + \varepsilon {\mathbf{x}}_{1} + \varepsilon^{2} {\mathbf{x}}_{2} + \cdots + \varepsilon^{i} {\mathbf{x}}_{i} + \cdots$$
(5.11)
where $${\mathbf{x}}_{0}$$ contains the main oscillation component of $${\mathbf{x}}$$. The mark $$\varepsilon^{i}$$ is used to indicate that $$\left| {x_{i} } \right| \ll x_{0}$$, that is $${\mathbf{x}}_{i}$$ is an i-ordered small amount of $${\mathbf{x}}_{0}$$. When a specific value is needed for calculation, let $$\varepsilon = 1$$. Similarly, expand the duty cycle $$d$$ into the sum of the main item and the correction quantity as follows.
$$d = d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots$$
(5.12)
Substituting (5.12) into (5.8) or (5.10) can give
$$d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots = K_{0} + K_{1} \mathbf{x}_{0} + \varepsilon K_{1} \mathbf{x}_{1} (t_{s} ) + \varepsilon^{2} K_{1} \mathbf{x}_{2} (t_{s} ) + \cdots$$
(5.13)
Then according to Fig. 5.1c, the switching instant $$t_{s}$$ of the switch ST can be obtained as
$$t_{s} = dT = (d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots )T$$
(5.14)
We can define the main term and the correction terms of the time $$t_{s}$$ as
$$\begin{array}{*{20}l} {t_{s0} = d_{0} T} \hfill \\ {t_{s1} = (d_{0} + \varepsilon d_{1} )T} \hfill \\ {t_{s2} = (d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} )T} \hfill \\ \cdots \hfill \\ \end{array}$$
(5.15)

It is easy to know from (5.15) that ts0 represents the moment when the comparator is flipped with only the principal component x0 being contained into the steady-state periodic solution x. And ts1 corresponds to the moment that the comparator is flipped when x contains the main component x0 and the first-order correction term x1. Similarly, ts2 corresponds to the flip moment when x contains the terms x0, x1 and x2, and so on.

Let $$\tau = \omega \,t$$ and $$\tau_{si} = \omega \,t_{si} \;(i = 0,1,2, \ldots )$$, here the angular frequency $$\omega = 2\pi /T$$, then according to (5.15), we can get that
$$\begin{array}{*{20}l} {\tau_{s0} = 2\pi d_{0} } \hfill \\ {\tau_{s1} = 2\pi (d_{0} + \varepsilon d_{1} )} \hfill \\ {\tau_{s2} = 2\pi (d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} )} \hfill \\ \cdots \hfill \\ \end{array}$$
(5.16)
For switching power converters, the duty cycle is determined primarily by $$t_{s0}$$. Considering that the general ripple has little effect on the duty cycle, which can be explained from the following example, to simplify the calculation, we use the first-order approximation of the Taylor series to find the term di. Meanwhile, according to iterative solution, when seeking to find $$d_{i}$$, the terms $$d_{0} ,d_{1} , \ldots d_{i - 1}$$ are already known. Therefore the following equations can be obtained.
$${\mathbf{x}}_{1} (t_{s} ) = {\mathbf{x}}_{1} (t_{s0} ) + \left. {\frac{{\partial {\mathbf{x}}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} \varepsilon d_{1} T + \left. {\frac{{\partial {\mathbf{x}}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} \varepsilon^{2} d_{2} T + \cdots$$
(5.17a)
$${\mathbf{x}}_{2} (t_{s} ) = {\mathbf{x}}_{2} (t_{s1} ) + \left. {\frac{{\partial {\mathbf{x}}_{2} }}{\partial t}} \right|_{{t = t_{s1} }} \varepsilon^{2} d_{2} T + \left. {\frac{{\partial {\mathbf{x}}_{2} }}{\partial t}} \right|_{{t = t_{s1} }} \varepsilon^{3} d_{3} T + \cdots$$
(5.17b)
$${\mathbf{x}}_{i} (t_{s} ) = {\mathbf{x}}_{i} (t_{s(i - 1)} ) + \left. {\frac{{\partial {\mathbf{x}}_{i} }}{\partial t}} \right|_{{t = t_{s(i - 1)} }} \varepsilon^{i} d_{i} T + \left. {\frac{{\partial {\mathbf{x}}_{i} }}{\partial t}} \right|_{{t = t_{s(i - 1)} }} \varepsilon^{i + 1} d_{i + 1} T + \cdots$$
(5.17c)
Substituting (5.17a) into (5.13), and equalizing the coefficients with the same power of $$\varepsilon$$ on both sides of the equation gives:
$$d{}_{0} = K_{0} + {\mathbf{K}}_{1} \cdot {\mathbf{x}}_{0}$$
(5.18a)
$$d_{1} = {\mathbf{K}}_{1} \cdot {\mathbf{x}}_{1} (t_{s0} )$$
(5.18b)
$$d_{2} = {\mathbf{K}}_{1} \cdot \left. {\frac{{\partial {\mathbf{x}}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} d_{1} T + {\mathbf{K}}_{1} \cdot {\mathbf{x}}_{2} (t_{s1} )$$
(5.18c)
$$d_{3} = {\mathbf{K}}_{1} \left( {\left. {\frac{{\partial {\mathbf{x}}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} + \left. {\frac{{\partial {\mathbf{x}}_{2} }}{\partial t}} \right|_{{t = t_{s1} }} } \right) \cdot d_{2} T + {\mathbf{K}}_{1} \cdot {\mathbf{x}}_{3} (t_{s2} )$$
(5.18d)

It can be seen from the above analysis that, for the closed-loop PWM switching converter system, no matter how complex the feedback control circuit is, take the power stage main circuit and feedback control circuit as a whole during analysis, then the state equation for a closed-loop system can be written in a uniform form similar to that of an open-loop system. And the duty cycle can be expressed as a linear combination of state variables (such as (5.8) or (5.10)).

5.2.3 Series Expansion of Switching Function δ(t) for Closed-Loop Systems

Similar to the open-loop system, the switch function $$\delta (t)$$ defined by (5.5) can still be expanded into a Fourier series as
$$\delta = b_{0} + \sum\limits_{m = 1}^{\infty } {[b_{m} \exp (jm\tau ) + \bar{b}_{m} \exp ( - jm\tau )]}$$
(5.19)
where the coefficient $$\bar{b}_{m}$$ represents the conjugate pair of the term $$b_{m}$$, which can be found by the following equation as
$$\begin{aligned} b_{0} & = \tfrac{1}{T}\int\limits_{0}^{T} {\delta (t)dt} = d(t),\quad b_{m} = \tfrac{1}{2}(\alpha_{m} - j\beta_{m} ),\quad m = 1,2 \ldots \\ \alpha_{m} & = \tfrac{2}{T}\int\limits_{0}^{T} {\delta (t) \cdot \cos (m\omega t)dt,} \quad \beta_{m} = \tfrac{2}{T}\int\limits_{0}^{T} {\delta (t) \cdot \sin (m\omega t)dt} \\ \end{aligned}$$
(5.20)
It should be noted that for closed-loop systems, the duty cycle d is not a known constant, but an unknown number to be determined, which can be approximated by the following series
$$b_{0} = d = d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots$$
(5.21)
According to the conduction interval of the switch ST during one switching cycle, the effective integration interval of Eq. (5.20) can be divided. Moreover, since d0 < 1, there is di ≪ 1. And because of the characteristics of the equivalent small-parameter method, the larger the value of m is, the smaller the corresponding di would be. Thus it can be assumed that
$$\begin{array}{*{20}l} {\sin (\varepsilon m\pi d_{1} ) \approx \varepsilon m\pi d_{1} \;(m = 1)} \hfill \\ {\sin (\varepsilon^{2} m\pi d_{2} ) \approx \varepsilon^{2} m\pi d_{2} \;(m \le 3)} \hfill \\ \cdots \hfill \\ \end{array}$$
(5.22)
Therefore the coefficient $$\alpha_{m}$$ in Eq. (5.20) can be expressed as
$$\begin{aligned} \alpha_{m} &amp; = \frac{1}{\pi }\int\limits_{0}^{{2\pi d_{0} }} {\cos m\tau d\tau + \frac{1}{\pi }} \int\limits_{{2\pi d_{0} }}^{{2\pi (d_{0} + \varepsilon d_{1} )}} {\cos m\tau d\tau + \frac{1}{\pi }} \int\limits_{{2\pi (d_{0} + \varepsilon d_{1} )}}^{{2\pi (d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} )}} {\cos m\tau d\tau + \cdots } \\ &amp; = \alpha_{m0} + \varepsilon \alpha_{m1} + \varepsilon^{2} \alpha_{m2} + \cdots \\ \end{aligned}$$
(5.23)
And from the equation above one can get
$$\begin{aligned} \alpha_{m0} &amp; = \frac{{\sin 2m\pi d_{0} }}{m\pi } \\ \alpha_{mi} &amp; \approx 2d_{i} \cos 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon^{i - 1} d_{i - 1} )\quad (i = 1,2, \ldots ) \\ \end{aligned}$$
(5.24)
When deriving (5.24), the following approximation is utilized, i.e.,
$$\tfrac{1}{2}\varepsilon^{i} d_{i} \ll \left( {d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots \varepsilon^{i - 1} d_{i - 1} } \right)$$
(5.25)
Similarly, the coefficient $$\beta_{m}$$ in Eq. (5.20) can be expressed as
$$\beta_{m} = \beta_{m0} + \varepsilon \beta_{m1} + \varepsilon^{2} \beta_{m2} + \cdots$$
(5.26)
where
$$\begin{aligned} \beta_{m0} &amp; = \frac{{1 - \cos 2m\pi d_{0} }}{m\pi } \\ \beta_{mi} &amp; \approx 2d_{i} \sin 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon^{i - 1} d_{i - 1} ),\quad (i = 1,2, \ldots ) \\ \end{aligned}$$
(5.27)
Thus, the coefficient $$b_{m}$$ in Eq. (5.20) can also be expressed as
$$\begin{aligned} b_{m} &amp; = \frac{1}{2}(\alpha_{m0} - j\beta {}_{m0}) + \varepsilon \cdot \frac{1}{2}(\alpha_{m1} - j\beta {}_{m1}) + \varepsilon^{2} \frac{1}{2}(\alpha_{m2} - j\beta {}_{m2}) + \cdots \\ &amp; = b_{m0} + \varepsilon^{1} b_{m1} + \varepsilon^{2} b_{m2} + \cdots \\ \end{aligned}$$
(5.28)
where $$b_{mi} = \tfrac{1}{2}(\alpha_{mi} - j\beta_{mi} )\quad (m = 1,2, \ldots ,\;i = 0,1,2, \ldots )$$
Substituting (5.21), (5.28) into (5.19) gives
$$\begin{aligned} \delta &amp; = \left[ {d_{0} + \sum\limits_{m = 1}^{\infty } {(b_{m0} e^{jm\tau } + c.c)} } \right] + \varepsilon \left[ {d_{1} + \sum\limits_{m = 1}^{\infty } {(b_{m1} e^{jm\tau } + c.c)} } \right] \\ &amp; \quad + \varepsilon^{2} \left[ {d_{2} + \sum\limits_{m = 1}^{\infty } {(b_{m2} e^{jm\tau } + c.c)} } \right] + \cdots \\ \end{aligned}$$
(5.29)
Each of the items in square brackets in (5.29) is expanded into the sum of the main item and the small quantity, that is:
$$\begin{aligned} &amp; d_{i} + \sum\limits_{m = 1}^{\infty } {(b_{mi} e^{jm\tau } + c.c)} = (d_{i} + b_{1i} e^{j\tau } + c.c) \\ &amp; \quad + \varepsilon (b_{2i} e^{j2\tau } + b_{3i} e^{j3\tau } + c.c) + \varepsilon^{2} (b_{4i} e^{j4\tau } + b_{5i} e^{j5\tau } + c.c) + \cdots \\ \end{aligned}$$
(5.30)
The reasons for selection of the main items and small quantities in Eq. (5.30) can be referred to Chap. 4. Substituting (5.30) into (5.29) gives
$$\delta = \delta_{0} + \varepsilon \delta_{1} + \varepsilon^{2} \delta_{2} + \cdots$$
(5.31)
In which the term $$\delta_{i}$$ is defined as
$$\begin{array}{*{20}c} {\left\{ {\begin{array}{*{20}l} {\delta_{0} = d_{0} + b_{10} e^{j\tau } + c.c} \hfill \\ {\delta_{1} = d_{1} + b_{11} e^{j\tau } + b_{20} e^{j2\tau } + b_{30} e^{j3\tau } + c.c} \hfill \\ {\delta_{2} = d_{2} + b_{12} e^{j\tau } + b_{21} e^{j2\tau } + b_{31} e^{j3\tau } + b_{40} e^{j4\tau } + b_{50} e^{j5\tau } + c.c \, } \hfill \\ \end{array} } \right.} \\ \cdots \\ \end{array}$$
(5.32a)
From (5.20) to (5.30), and the coefficients in (5.32a) are summarized as follows:
$$\begin{aligned} b_{mi} &amp; = \tfrac{1}{2}\,(\alpha_{mi} - j\beta_{mi} )\;(m = 1,2, \ldots ,i = 0,1,2, \ldots ) \\ \alpha_{m0} &amp; = \frac{{\sin 2m\pi d_{0} }}{m\pi },\quad \beta_{m0} = \frac{{1 - \cos 2m\pi d_{0} }}{m\pi } \\ \alpha_{mi} &amp; \approx 2d_{i} \cos 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon^{i - 1} d_{i - 1} )\; (i = 1,2, \ldots ) \\ \beta_{mi} &amp; \approx 2d_{i} \sin 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon^{i - 1} d_{i - 1} )\;(i = 1,2, \ldots ) \\ \end{aligned}$$
(5.32b)

So far, the discrete time-varying differential equations represented by (5.4) can be transformed into a state differential equation with continuous time by the transformation of (5.31). Therefore, the equivalent small parameter can be adopted to obtain the analytic solution of the closed-loop system.

5.3 Solution of the Time-Varying Closed-Loop System with CCM Operation

The closed-loop time-varying differential equation (5.4) is solved by using equivalent small parametric method now. Since (5.4) has the same form as the open-loop state equation (5.1), the basic principle of solving is the same as the open-loop system. The difference is that the duty cycle $$d(t)$$ is not a constant, but a variable associated with the feedback circuit and the state variable (see (5.8) or (5.10)). The switching function defined by (5.5) is expanded in series shown in Eq. (5.31). For the sake of simplicity, we only consider the case of $${\mathbf{e}} = 0$$ in Eq. (5.4).

The steady-state periodic solution $${\mathbf{x}}$$ is expressed as the sum of the main wave x0 and the corrections xi, as shown in (5.11). Similarly the nonlinear vector function f is still expanded into the sum of the main term f0 and the correction terms fi, and they are determined by the following expressions as
$$\begin{array}{*{20}l} {{\mathbf{f}}_{0} = \delta_{0} {\mathbf{x}}_{0} } \hfill \\ {{\mathbf{f}}_{1} = \delta_{0} {\mathbf{x}}_{1} + \delta_{1} {\mathbf{x}}_{0} } \hfill \\ {{\mathbf{f}}_{2} = \delta_{0} {\mathbf{x}}_{2} + \delta_{1} {\mathbf{x}}_{1} + \delta_{2} {\mathbf{x}}_{0} } \hfill \\ \cdots \hfill \\ \end{array}$$
(5.33a)
Using the same method as the analysis of open-loop converter system to solve the closed-loop converter system Eq. (5.4), the same iterative equations as in Eq. (4.​26) can be obtained, which is rewritten as follows:
$$\begin{array}{*{20}c} {\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} (p){\mathbf{x}}_{0} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{0m} = {\mathbf{u}}} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{1} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{1m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{1} } \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{2} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{2m} = - {\mathbf{G}}_{1} (p){\mathbf{R}}_{2} } \hfill \\ \end{array} } \right.} \\ \cdots \\ \end{array}$$
(5.33b)

5.3.1 Solution of Main Component

If the system can operate stably under the selected parameters, the main component of the state variable is the DC component, so it can be set:
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00}$$
(5.34)
Substituting (5.32a) and (5.34) into (5.33a) gives
$${\mathbf{f}}_{0} = \delta_{0} {\mathbf{x}}_{0} = d_{0} {\mathbf{a}}_{00} + b_{10} {\mathbf{a}}_{00} e^{j\tau } + c.c$$
(5.35)
The main term and the remaining item are as follows.
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{f}}_{0m} = d_{0} {\mathbf{a}}_{00} } \hfill \\ {{\mathbf{R}}_{1} = b_{10} {\mathbf{a}}_{00} e^{j\tau } + c.c} \hfill \\ \end{array} } \right.$$
(5.36)
Substituting $${\mathbf{x}}_{{\mathbf{0}}}$$ and $${\mathbf{f}}_{0m}$$ into (5.33b) gives
$$\left[ {{\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0)d_{0} } \right]{\mathbf{a}}_{00} = {\mathbf{u}}$$
(5.37)
And according to (5.18a), the main term of the duty cycle is determined by
$$d_{0} = K_{0} + K_{1} \mathbf{x}_{0} = K_{0} + K_{1} \mathbf{a}_{00}$$
(5.38)

The solutions of a00 and d0 can be obtained by combining Eqs. (5.38) and (5.37). Generally, for closed-loop systems, the solutions of a00 and d0 are non-linear and depend on the specific main circuit and feedback control law. The specific solution process can be analyzed with reference to the following examples.

5.3.2 Solution of the First-Order Correction

Since $$R_{1}$$ in (5.36) contains the fundamental wave only, the first-order correction x1 can be assumed that
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} e^{j\tau } + {\bar{\mathbf{a}}}_{11} e^{ - j\tau }$$
(5.39)
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
$$\begin{aligned} {\mathbf{f}}_{1m} &amp; = (d_{0} {\mathbf{a}}_{11} + b_{11} {\mathbf{a}}_{00} ) \cdot e^{j\tau } + c.c \\ {\mathbf{R}}_{2} &amp; = (b_{10} {\bar{\mathbf{a}}}_{11} + \bar{b}_{10} {\mathbf{a}}_{11} + d_{1} {\mathbf{a}}_{00} ) + (b_{10} {\mathbf{a}}_{11} + b_{20} {\mathbf{a}}_{00} ) \cdot e^{j2\tau } + b_{30} {\mathbf{a}}_{00} \cdot e^{j3\tau } + c.c \\ \end{aligned}$$
(5.40)
Substituting $${\mathbf{f}}_{1m}$$ and $${\mathbf{R}}_{1}$$ into the first order correction in Eq. (5.33b) gives
$$[{\mathbf{G}}{}_{0}(j\omega ) + {\mathbf{G}}_{1} (j\omega ) \cdot d_{0} ] \cdot {\mathbf{a}}_{11} = - {\mathbf{G}}_{1} (j\omega ) \cdot (b_{11} + b_{10} ) \cdot {\mathbf{a}}_{00}$$
(5.41a)
The coefficients b10 and b11 are determined according to (5.32b), i.e.,
$$\begin{aligned} b_{10} &amp; = [\sin 2\pi d_{0} - j(1 - \cos 2\pi d_{0} )]/2\pi \\ b_{11} &amp; = d_{1} \cdot (\cos 2\pi d_{0} - j\sin 2\pi d_{0} ) = d_{1} \cdot e^{{ - j\tau_{s0} }} \;(\tau_{s0} = 2\pi d_{0} ) \\ \end{aligned}$$
(5.41b)
And according to (5.18b), the term d1 is expressed as
$$d_{1} = {\mathbf{K}}_{1} \cdot ({\mathbf{a}}_{11} e^{{j\tau_{s0} }} + {\bar{\mathbf{a}}}_{11} e^{{ - j\tau_{s0} }} )$$
(5.41c)
The first-order corrections d1 and a11 can be obtained by combining the Eqs. (5.41a), (5.41b) and (5.41c), i.e.,
$$\begin{aligned} d_{1} &amp; = B_{1} /A_{1} \\ {\mathbf{a}}_{11} &amp; = {\mathbf{G}}^{ - 1} (j\omega ) \cdot [ - {\mathbf{G}}_{1} (j\omega ) \cdot (d_{1} e^{{ - j\tau_{s0} }} + b_{10} ) \cdot {\mathbf{a}}_{00} ] \\ \end{aligned}$$
(5.42)
where the superscript “−1” indicates the inversion, and the symbols are defined as
$${\mathbf{G}}(j\omega ) = {\mathbf{G}}_{0} (j\omega ) + {\mathbf{G}}_{1} (j\omega )d_{0} ,\quad {\mathbf{C}}_{1} = {\mathbf{G}}^{ - 1} (j\omega ){\mathbf{G}}_{1} (j\omega )$$
$$A_{1} = 1 + {\mathbf{K}}_{1} ({\mathbf{C}}_{1} + {\bar{\mathbf{C}}}_{1} ){\mathbf{a}}_{00} ,\quad {\mathbf{B}}_{1} = - {\mathbf{K}}_{1} ({\mathbf{C}}_{1} b_{10} e^{{j\tau_{s0} }} + {\bar{\mathbf{C}}}_{1} \bar{b}_{10} e^{{ - j\tau_{s0} }} ){\mathbf{a}}_{00}$$

As can be seen from (5.42), in the approximate calculation, the first order correction can be obtained by solving the linear equations.

5.3.3 Solution of the Second-Order Correction

According to the remaining term $${\mathbf{R}}_{2}$$ in (5.40), one can assume the second-order correction term x2 can be expressed as
$${\mathbf{x}}_{2} = {\mathbf{a}}_{02} + ({\mathbf{a}}_{22} e^{j2\tau } + {\mathbf{a}}_{32} e^{j3\tau } + c.c)$$
(5.43)
Introducing $${\mathbf{x}}_{{\mathbf{0}}} ,{\mathbf{x}}_{{\mathbf{1}}} ,{\mathbf{x}}_{{\mathbf{2}}}$$ and $$\delta_{0} ,\delta_{1} ,\delta_{2}$$ into $${\mathbf{f}}_{2} = \delta_{0} {\mathbf{x}}_{2} + \delta_{1} {\mathbf{x}}_{1} + \delta_{2} {\mathbf{x}}_{0}$$, one can get $${\mathbf{f}}_{{2{\text{m}}}}$$ as
$$\begin{aligned} {\mathbf{f}}_{2m} &amp; = (d_{0} {\mathbf{a}}_{02} + b_{11} {\bar{\mathbf{a}}}_{11} + \bar{b}_{11} {\mathbf{a}}_{11} + d_{2} {\mathbf{a}}_{00} ) + (d_{0} {\mathbf{a}}_{22} + b_{11} {\mathbf{a}}_{11} + b_{30} {\bar{\mathbf{a}}}_{11} + b_{21} {\mathbf{a}}_{00} )e^{j2\tau } \\ &amp; \quad + (d_{0} {\mathbf{a}}_{32} + b_{10} {\mathbf{a}}_{22} + b_{20} {\mathbf{a}}_{11} + b_{31} {\mathbf{a}}_{00} )e^{j3\tau } + c.c \\ \end{aligned}$$
(5.44)
In order to simplify the calculation, the higher-order small quantity $$\bar{b}_{10} a_{32}$$ of the second harmonic is omitted in the second bracket of the above Eq. (5.44). Then introducing (5.44) and (5.40) into (5.33b) gives the three equations of the second order correction term as follows.
$$[{\mathbf{G}}_{0} (j2\omega ) + {\mathbf{G}}_{1} (j2\omega )d_{0} ]{\mathbf{a}}_{22} = - {\mathbf{G}}_{1} (j2\omega )[(b_{21} + b_{20} ){\mathbf{a}}_{00} + (b_{11} + b_{10} ){\mathbf{a}}_{11} + b_{30} {\bar{\mathbf{a}}}_{11} ]$$
(5.45a)
$$[{\mathbf{G}}_{0} (j3\omega ) + {\mathbf{G}}_{1} (j3\omega )d_{0} ]{\mathbf{a}}_{32} = - {\mathbf{G}}_{1} (j3\omega ) \cdot [(b_{31} + b_{30} ){\mathbf{a}}_{00} + b_{10} {\mathbf{a}}_{22} + b_{20} {\mathbf{a}}_{11} ]$$
(5.45b)
$$[{\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0)d_{0} ]{\mathbf{a}}_{02} + {\mathbf{G}}_{1} (0){\mathbf{a}}_{00} d_{2} = - {\mathbf{G}}_{1} (0)[(b_{11} + b_{10} ){\bar{\mathbf{a}}}_{11} + (\bar{b}_{11} + \bar{b}_{10} ){\mathbf{a}}_{11} + d_{1} {\mathbf{a}}_{00} ]$$
(5.45c)
The solutions of the terms $${\mathbf{a}}_{22}$$ and $${\mathbf{a}}_{32}$$ can be easily obtained as Eqs. (5.45a) and (5.45b) are linear. While Eq. (5.45c) is nonlinear, which should be solved by combining Eq. (5.18c) for d2, that is
$$\begin{aligned} {\mathbf{a}}_{02} &amp; = - [{\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0)d_{0} + {\mathbf{G}}_{1} (0){\mathbf{a}}_{00} \cdot {\mathbf{K}}_{1} ]^{ - 1} \cdot {\mathbf{G}}_{{\mathbf{1}}} (0) \cdot [B_{1} + {\mathbf{a}}_{00} {\mathbf{B}}_{2} ] \\ d_{2} &amp; = B_{2} + {\mathbf{K}}_{1} {\mathbf{a}}_{02} \\ \end{aligned}$$
(5.46)
where the coefficients B1 and B2 are determined by
$$\begin{aligned} B_{1} &amp; = (b_{11} + b_{10} ){\bar{\mathbf{a}}}_{11} + (\bar{b}_{11} + \bar{b}_{10} ){\mathbf{a}}_{11} + d_{1} {\mathbf{a}}_{00} \\ B_{2} &amp; = - {\mathbf{K}}_{1} [4\pi d_{1} \cdot \text{Im} (a_{11} e^{{j\tau_{s0} }} ) - 2\text{Re} ({\mathbf{a}}_{22} e^{{j\tau_{s1} }} ) - 2\text{Re} ({\mathbf{a}}_{32} e^{{j\tau_{s1} }} )] \\ \end{aligned}$$
Here the symbols $$\text{Im} ( \cdot ),\text{Re} ( \cdot )$$ denote the imaginary and real parts of the complex. Generally the ripple has little effect on the duty cycle, especially the second harmonic and third harmonic components, whose amplitudes are usually very small. Therefore the term d2 is mainly determined by the DC component a02 in the second-order correction term x2. Equation (5.46) can be simplified as
$$\begin{aligned} {\mathbf{a}}_{02} &amp; = - [{\mathbf{G}}_{0} (0) + {\mathbf{G}}_{1} (0)d_{0} + {\mathbf{G}}_{1} (0){\mathbf{a}}_{00} \cdot {\mathbf{K}}_{1} ]^{ - 1} \cdot {\mathbf{G}}_{1} (0) \cdot {\mathbf{B}}_{1} \\ d_{2} &amp; = {\mathbf{K}}_{1} {\mathbf{a}}_{02} \\ \end{aligned}$$
(5.47)
Then according to equations from (5.34) to (5.47), the final expression of the steady-state periodic solution is shown as
$${\mathbf{x}} = {\mathbf{a}}_{00} + {\mathbf{a}}_{02} + ({\mathbf{a}}_{11} e^{j\tau } + {\mathbf{a}}_{22} e^{j2\tau } + {\mathbf{a}}_{32} e^{j3\tau } + c.c)$$
(5.48a)
Which can also expressed in form of trigonometric function as:
$${\mathbf{x}} = a_{0} + a_{1} \cos \tau + b_{1} \sin \tau + a_{2} \cos 2\tau + b_{2} \sin 2\tau + a_{3} \cos 3\tau + b_{3} \sin 3\tau + \cdots$$
(5.48b)
Among which, the steady-state DC solution is
$${\mathbf{x}}_{dc} = {\mathbf{a}}_{0} = {\mathbf{a}}_{00} + {\mathbf{a}}_{02}$$
(5.48c)

5.4 Examples

5.4.1 Boost Regulator with Proportional Control

The circuit diagram is shown in Fig. 5.2, the main circuit parameters are chosen as E = 5 V, L = 50 μH, C = 4.4 μF and R = 28 Ω; and the control circuit parameters are set as Vr = 0.13 V, g1 = 0.174, g2 = −0.0435, Vl= 0 and Vu = 1 V. The state differential equation and the duty cycle equation are described as (5.4) and (5.8), respectively, where the state variable vector is $${\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L} } &amp; {v_{C} } \\ \end{array} } \right]^{\text{Tr}}$$, the input vector $${\mathbf{u}} = \left[ {\begin{array}{*{20}c} {{E \mathord{\left/ {\vphantom {E L}} \right. \kern-0pt} L}} &amp; 0 \\ \end{array} } \right]^{Tr}$$, the nonlinear vector function $${\mathbf{f}} = \delta {\mathbf{x}}$$; and the feedback proportional coefficient $$K_{0} = V_{r}$$, and the coefficient matrix $${\mathbf{K}}_{1} = [ - g_{1} , - g_{2} ]$$. The matrices G0(p) and G1(p) are as
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig2_HTML.gif
Fig. 5.2

Boost regulator with proportional modulation

$$G_{0} (p) = \left[ {\begin{array}{*{20}c} p &amp; {\tfrac{1}{L}} \\ {\tfrac{ - 1}{C}} &amp; {p + \tfrac{1}{RC}} \\ \end{array} } \right],\quad G_{1} (p) = \left[ {\begin{array}{*{20}c} 0 &amp; {\tfrac{ - 1}{L}} \\ {\tfrac{1}{C}} &amp; 0 \\ \end{array} } \right]$$
(5.49)

5.4.1.1 Find the Main Term of the Steady State Solution

Similarly to the method introduced for analysis of open-loop converters, the main term of the steady state solution is DC component, that is:
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = [I_{00} ,V_{00} ]^{Tr}$$
(5.50)
According to (5.37) and (5.38), we can get:
$$V_{00} = \frac{E}{{1 - d_{0} }},\quad I_{00} = \frac{{V_{00} }}{{R(1 - d_{0} )}},\quad d_{0} = V_{r} - \frac{{g_{1} E}}{{(1 - d_{0} )^{2} R}} - \frac{{g_{2} E}}{{(1 - d_{0} )}}$$
(5.51)

From Eq. (5.51) we can see that for the closed-loop Boost converter with the inductor current being contained in the feedback amounts, the solution of d0 needs to solve a cubic equation, which can be solved by the numerical method or the commonly used formula of the cubic equation. There may be multiple real-value solutions that meet the condition of d0. In this case, they need to be determined by the stability analysis method of the equilibrium point (see Chap. 9).

However, if only the output voltage feedback is available, for the Boost circuit with the linear PWM feedback control, the solution of d0 only needs to solve the quadratic equation. Thus under this condition it is easy to determine whether the solution meets the requirements of the converter, and it is also indicates that potential instability factors would exist in the closed-loop converter system with current feedback control.

In this case, we obtain a real solution  d0 = 0.4091, and the main terms are
$$V_{00} = 8.4617\,(V),\quad I_{00} = 0.5114\,(A)$$
(5.52)

Here $$V{}_{00},\;I_{00}$$ are independent of the switching frequency of the circuit, which is the same as the steady-state DC solution obtained by the state space method.

5.4.1.2 Find the First-Order Correction

According to the analysis in the previous Sect. 5.3, we can see that only the fundamental component is included in the first-order remainder term R1, so the first-order correction x1 can be assumed that
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} e^{j\tau } + {\bar{\mathbf{a}}}_{11} e^{ - j\tau } ,\quad {\mathbf{a}}_{11} = [I_{11} ,V_{11} ]^{Tr}$$
(5.53)
According to (5.42), one can get
$$\begin{aligned} V_{11} &amp; = \frac{{(b_{11} + b_{10} ) \cdot [j\omega L \cdot I_{00} - (1 - d_{0} ) \cdot V_{00} ]}}{{\omega^{2} LC - (1 - d_{0} )^{2} - j\omega L/R}} \approx \frac{{(b_{11} + b_{10} ) \cdot [j\omega L \cdot I_{00} - (1 - d_{0} ) \cdot V_{00} ]}}{{\omega^{2} LC - (1 - d_{0} )^{2} }} \\ I_{11} &amp; = \frac{{(b_{11} + b_{10} ) \cdot V_{00} - (1 - d_{0} ) \cdot V_{11} }}{j\omega L} \\ \end{aligned}$$
(5.54)
$$d_{1} = \frac{{(1 - d_{0} ) \cdot (K_{0} - d_{0} )\sin \tau_{s0} - (1 - \cos \tau_{s0} )(g_{1} \omega C \cdot V_{00} - g_{2} \omega L \cdot I_{00} )}}{{\omega^{2} LC - (1 - d_{0} )^{2} - 2(1 - d{}_{0})(K_{0} - d_{0} )}}$$
(5.55)
With the three switching frequencies, the first-order correction can be obtained by the algorithm in this chapter, as shown in Table 5.1.
Table 5.1

First-order ripple component of state variables and correction of duty ratio

Frequency

$$I_{11}$$

$$V_{11}$$

$$d_{1}$$

fs = 50 kHz

−0.1372 − 0.0692j

0.0589 + 0.1093j

−0.0656

fs = 100 kHz

−0.0742 − 0.0292j

0.0437 + 0.0370j

−0.0325

fs = 1 MHz

−0.0078 − 0.0024j

0.0053 + 0.0018j

−0.0032

As can be seen from the above table, the impact of ripples on the duty cycle is relatively small and inversely proportional to the switching frequency. In fact, Eq. (5.55) can be approximated as
$$d_{1} \approx \frac{{ - (1 - \cos \tau_{s0} )(g_{1} C \cdot V_{00} - g_{2} L \cdot I_{00} )}}{\omega LC}$$
(5.56)

Obviously $$d_{1}$$ is inversely proportional to the switching frequency.

5.4.1.3 Find the Second-Order Correction

The amplitudes of the harmonics contained in the second-order correction can be obtained according to the previous mentioned Eq. (5.45a), i.e.,
$$\begin{array}{*{20}l} {V_{22} \approx \frac{{j2\omega L \cdot \Delta_{I22} - (1 - d_{0} ) \cdot \Delta_{V22} }}{{4\omega^{2} LC - (1 - d_{0} )^{2} }}} \hfill &amp; {I_{22} = \frac{{\Delta_{V22} - (1 - d_{0} ) \cdot V_{22} }}{j2\omega L}} \hfill \\ {V_{32} \approx \frac{{j3\omega L \cdot \Delta_{I32} - (1 - d_{0} ) \cdot \Delta_{V32} }}{{9\omega^{2} LC - (1 - d_{0} )^{2} }}} \hfill &amp; {I_{32} = \frac{{\Delta_{V32} - (1 - d_{0} ) \cdot V_{32} }}{j3\omega L}} \hfill \\ {V_{02} \approx \frac{{(g_{1} I_{00} + d^{\prime}_{0} ) \cdot \Delta_{V02} - g_{1} V_{00} \cdot \Delta {}_{I02}}}{{(g_{1} I_{00} + d^{\prime}_{0} )(g_{2} V_{00} + d^{\prime}_{0} ) + M_{02} }}} \hfill &amp; {I_{02} \approx \frac{{\Delta_{V02} - (g_{2} V_{00} + 1 - d_{0} )V_{02} }}{{g_{1} V_{00} }}} \hfill \\ \end{array}$$
(5.57)
where
$$\begin{aligned} \Delta_{I22} &amp; = (b_{21} + b_{20} )I_{00} + (b_{11} + b_{10} )I_{11} + b_{30} \bar{I}_{11} \\ \Delta_{V22} &amp; = (b_{21} + b_{20} )V_{00} + (b_{11} + b_{10} )V_{11} + b_{30} \bar{V}_{11} \\ \Delta {}_{I32} &amp; = (b_{31} + b_{30} )I_{00} + b_{20} I_{11} + b_{10} I_{22} \\ \Delta_{V32} &amp; = (b_{31} + b_{30} )V_{00} + b_{20} V_{11} + b_{10} V_{22} \\ \Delta_{I02} &amp; = (b_{11} + b_{10} )\bar{I}_{11} + (\bar{b}_{11} + \bar{b}_{10} )I_{11} + d_{1} I_{00} ) \\ \Delta_{V02} &amp; = (b_{11} + b_{10} )\bar{V}_{11} + (\bar{b}_{11} + \bar{b}_{10} )V_{11} + d_{1} V_{00} ) \\ d^{\prime}_{0} &amp; = 1 - d_{0} ,\;M_{02} = g_{1} V_{00} (1 - g_{2} RI_{00} )/R \\ \end{aligned}$$

The final expression of the steady-state periodic solution can be obtained from Eqs. (5.51) to (5.57), as shown in Eq. (5.48a), in which the steady-state DC solution is $${\mathbf{x}}_{dc} = {\mathbf{a}}_{00} + {\mathbf{a}}_{02}$$, i.e., $$I_{0} = I_{00} + I_{02} \quad ,V_{0} = V_{00} + V_{02}$$.

The results of the steady-state DC solution obtained from the ESPM are compared with those from numerical simulations, as shown in Table 5.2.
Table 5.2

Steady-state solutions of state variables in example 1

Frequency

ESPM

Numerical simulation

fs = 50 kHz

I0 = 0.3325A V0 = 6.9796 V

I0 = 0.3750A V0 = 7.0472 V

fs = 100 kHz

I0 = 0.4253A V0 = 7.7490 V

I0 = 0.4430A V0 = 7.6504 V

fs = 1 MHz

I0 = 0.5032A V0 = 8.3934 V

I0 = 0.5267A V0 = 8.3355 V

For the AC component, the sum of the first three harmonic components of the state variable during one switching cycle can be obtained by using the ESPM, and the results from ESPM (dashed line) are compared with those from numerical simulations (solid line), as shown in Fig. 5.3. The ripples of the inductor current and the capacitor voltage with the converter system operating at the frequency fs = 50 kHz are shown in Fig. 5.3a, b, and those for switching frequency fs = 100 kHz are shown in the Fig. 5.3c, d.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig3_HTML.gif
Fig. 5.3

Steady-state ripples of state variables for Boost with proportional control

It can be seen from the data in Table 5.2 and the waveforms in Fig. 5.3 that the results obtained by the two methods are close, especially at higher frequencies. The difference is smaller, means that the ESPM introduced in this paper is effective. At the same time, it can be seen from Table 5.2 that when the switching frequency is low, the DC component has a certain change, which is called DC offset.

5.4.2 Boost Regulator with Proportional-Integral Control

In this example, the main circuit parameters are the same as those in the Sect. 5.4.1, and the proportional-integral control law is adopted with the feedback circuit being shown in Fig. 5.1b. The circuit parameters of the feedback loop are chosen as follows, i.e., R1 = 0.5 kΩ, R2 = 500 kΩ, Cf = 5 μF, and the reference voltage is set as Vr = 0.4091 V, the parameters of sawtooth voltage are the same as those in Fig. 5.1a, i.e., Vl = 0 and Vu = 1 V. As shown in Table 5.3, the steady-state DC solution obtained from the ESPM is compared with the results from simulations.
Table 5.3

Steady-state solutions of state variables

Frequency

ESPM

Simulation

fs = 50 kHz

I0 = 0.5117A V0 = 8.4618 V

I0 = 0.5011A V0 = 8.2435 V

fs = 100 kHz

I0 = 0.5115A V0 = 8.4617 V

I0 = 0.5166A V0 = 8.3137 V

The fundamental components of the state variables obtained from the ESPM at both switching frequencies, together with the first-order correction d1 of the duty cycle, are shown in Table 5.4.
Table 5.4

Fundamental components of the state variables and the 1-order correction of duty cycle

Frequency

$$I_{11}$$

$$V_{11}$$

$$d_{1}$$

fs = 50 kHz

−0.1617 − 0.0431j

0.0873 + 0.1032j

−0.00002

fs = 100 kHz

−0.0796 − 0.0223j

0.0490 + 0.0336j

−0.000005

For the AC component, the sum of the first three harmonic components of the state variable during one switching cycle can be obtained by using the ESPM, and the results from ESPM (dashed line) are compared with those from numerical simulations (solid line), as shown in Fig. 5.4. The ripples of the inductor current and the capacitor voltage with the converter system operating at the frequency fs = 50 kHz are shown in Fig. 5.4a, b, and those for switching frequency fs = 100 kHz are shown in the Fig. 5.4c, d.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig4_HTML.gif
Fig. 5.4

Steady-state ripples of state variables for Boost with PI control

It can be seen from the data in Tables 5.3 and 5.4, and the ripple waveforms in Figs. 5.3 and 5.4, that when there is a compensation capacitor in the feedback control circuit, the results from the ESPM agree quite well with those from numerical simulation, which means that the ESPM has a higher accuracy. The compensation capacitor makes the ripple have a very small effect on the duty cycle because of its filtering characteristics (see Table 5.4). Thus, the errors introduced by some of the assumptions made during the algorithm derivation are quite small.

5.5 Improvement of the Algorithm

5.5.1 Improved Algorithm for Duty Cycle Correction

According to the basic principle of the equivalent small parameter method, when the term di is solved, the items that are explicitly related to di are moved to the solution equation of higher order correction, so the right side of Eqs. (5.18a)~(5.18d) no longer contains the explicit relation with di, it is the linear equation of di.

If all the di-related terms on the right side of (5.13) are taken into account when solving di, a more accurate expression for di can be obtained, as shown in Eqs. (5.58a)~(5.58d).
$$d{}_{0} = K_{0} + K_{1} \cdot \mathbf{x}_{0}$$
(5.58a)
$$d_{1} = K_{1} \cdot \mathbf{x}_{1} (t_{s0} ) + K_{1} \cdot \left. {\frac{{\partial \mathbf{x}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} d_{1} T$$
(5.58b)
$$d_{2} = K_{1} \cdot \mathbf{x}_{2} (t_{s1} ) + K_{1} \left( {\left. {\frac{{\partial \mathbf{x}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} + \left. {\frac{{\partial \mathbf{x}_{2} }}{\partial t}} \right|_{{t = t_{s1} }} } \right) \cdot d_{2} T$$
(5.58c)
$$\begin{array}{*{20}c} {d_{3} = K_{1} \left( {\left. {\frac{{\partial \mathbf{x}_{1} }}{{\partial t}}} \right|_{{t = t_{{s0}} }} + \left. {\frac{{\partial \mathbf{x}_{2} }}{{\partial t}}} \right|_{{t = t_{{s1}} }} + \left. {\frac{{\partial \mathbf{x}_{3} }}{{\partial t}}} \right|_{{t = t_{{s2}} }} } \right) \cdot d_{3} T + K_{1} \cdot \mathbf{x}_{3} (t_{{s2}} )} \\ \ldots \\ \end{array}$$
(5.58d)
Taking the above correction algorithm, the solution of d1 would be non-linear in (5.42), that is:
$$Ad_{1}^{2} + Bd_{1} + D = 0$$
(5.59)
where the coefficients are determined by the following equations as
$$\begin{aligned} A &amp; = K_{1} \cdot j2\pi (A_{1} - \bar{A}_{1} )a_{00} \\ B &amp; = 1 + K_{1} \cdot j2\pi (A_{1} b_{10} \cdot e^{{j\tau_{s0} }} - \overline{{A_{1} b_{10} }} \cdot e^{{ - j\tau_{s0} }} )a_{00} + K_{1} \cdot (A_{1} + \bar{A}_{1} )a_{00} \\ D &amp; = K_{1} \cdot (A_{1} b_{10} \cdot e^{{j\tau_{s0} }} + \overline{{A_{1} b_{10} }} \cdot e^{{ - j\tau_{s0} }} )a_{00} \\ A_{1} &amp; = [G_{0} (j\omega ) + d_{0} G_{1} (j\omega )]^{ - 1} \cdot G_{1} (j\omega ) \\ \end{aligned}$$

Obviously, according to (5.59), there would be more than one solution for d1, thus the correct value of d1 needs to be selected according to the actual situation of the system.

Similarly, the solutions d2 and a02 in (5.45a) can be modified to be
$$\begin{aligned} a_{02} &amp; = - [G_{0} (0) + d_{0} G_{1} (0) + G_{1} (0) \cdot K_{1} \cdot a_{00} /C_{1} ]^{ - 1} \cdot G_{1} (0)(B_{1} + a_{00} \cdot B_{2} /C_{1} ) \\ d_{2} &amp; = (K_{1} a_{02} + B_{2} )/C_{1} \\ \end{aligned}$$
(5.60)
where the coefficients are determined by
$$\begin{aligned} B_{1} &amp; = (b_{11} + b_{10} )\bar{a}_{11} + (\bar{b}_{11} + \bar{b}_{10} )a_{11} + d_{1} a_{00} \\ B_{2} &amp; = K_{1} \cdot [2R(a_{22} e^{{j\tau_{s1} }} ) + 2R(a_{32} e^{{j\tau_{s1} }} )] \\ C_{1} &amp; = 1 + K_{1} \cdot [4\pi \cdot I(a_{11} e^{{j\tau_{s1} }} ) + 8\pi \cdot I(a_{22} e^{{j\tau_{s1} }} ) + 12\pi \cdot I(a_{32} e^{{j\tau_{s1} }} )] \\ \end{aligned}$$

5.5.2 Correction Algorithm for Series Expansion of the Switching Function δ(t)

In the previous analysis of the state equation of the closed-loop system using the ESPM, it can be found that solving the first-order correction a11 needs to know the coefficients b11 and b10 in the expansion series of the switching function, and the first-order correction d1 of the duty cycle. And in the solution of the second-order correction term, it is necessary to know the values of b20, b30, b21, b31, and d1. As d1 has been obtained during the solution of the first-order correction, thus it can be utilized in the solutions of b20 and b30. Similarly, when solving bk0 and b(k+1)0 (k = 2,4,6…), we can use d0, d1…, d(k−1), which have been obtained in the iterative process. Therefore, in Eq. (5.32b), the expansions of α1and β1 are unchanged, while the others are modified to be
$$\begin{aligned} \alpha _{m} &amp; = \frac{1}{\pi }\int\limits_{0}^{{2\pi (d_{0} + d_{1} + \cdots d_{{m - 1}} )}} {\cos m\tau d\tau + \frac{1}{\pi }} \int\limits_{{2\pi (d_{0} + d_{1} + \cdots d_{{m - 1}} )}}^{{2\pi (d_{0} + d_{1} + \cdots d_{{m - 1}} + \varepsilon ^{m} d_{m} )}} {\cos m\tau d\tau } \\ &amp; \quad + \frac{1}{\pi }\int\limits_{{2\pi (d_{0} + d_{1} + \cdots d_{{m - 1}} + \varepsilon ^{m} d_{m} )}}^{{2\pi (d_{0} + d_{1} + \cdots d_{{m - 1}} + \varepsilon ^{m} d_{m} + \varepsilon ^{{m + 1}} d_{{m + 1}} )}} {\cos m\tau d\tau } + \cdots \\ &amp; = \alpha _{{m0}} + \varepsilon \alpha _{{m1}} + \varepsilon ^{2} \alpha _{{m2}} + \cdots \varepsilon ^{m} \alpha _{{mm}} + \cdots \\ \end{aligned}$$
(5.61)
In which
$$\begin{aligned} \alpha_{m0} &amp; = \frac{{\sin 2m\pi (d_{0} + d_{1} + d_{2} + \cdots + d_{m - 1} )}}{m\pi } \\ \alpha_{m1} &amp; = \alpha_{m2} = \cdots \alpha_{m(m - 1)} = 0 \\ \alpha_{mi} &amp; \approx 2d_{i} \cos 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon^{i - 1} d_{i - 1} ) \\ \end{aligned}$$
Here m =  k or k + 1, and k = 2,4,6…, i =  k, k + 1, k + 2, ….The same method can be done with the expansion of βm, i.e.,
$$\beta_{m} = \beta_{m0} + \varepsilon \beta_{m1} + \varepsilon^{2} \beta_{m2} + \cdots$$
(5.62)
where
$$\begin{aligned} \beta_{m0} &amp; = \frac{{1 - \cos 2m\pi (d_{0} + d_{1} + d_{2} + \cdots + d_{m - 1} )}}{m\pi } \\ \beta_{m1} &amp; = \beta_{m2} = \cdots \beta_{m(m - 1)} = 0 \\ \beta_{mi} &amp; \approx 2d_{i} \sin 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon^{i - 1} d_{i - 1} ) \\ \end{aligned}$$

In this way, the series expansion form of the switching function in the closed-loop converter system is unchanged, and the process of solving is exactly the same as the original algorithm, except that the coefficients of some items would become zero, for example b21 = 0, b31 = 0.

5.5.3 Double Iterative Symbol Algorithm

In order to distinguish the algorithm to be proposed below, we refer to the algorithm of the closed-loop system proposed in Sect. 5.4 as the single-iteration symbol algorithm (or basic algorithm).

When the ESPM is used to solve the state equation of the closed-loop system, the main oscillation equation should be:
$${\mathbf{G}}_{0} (0){\mathbf{a}}_{00} + {\mathbf{G}}_{1} (0)(d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots ){\mathbf{a}}_{00} = {\mathbf{u}}$$
(5.63)

However, it can be found from the expansion of δ in (5.32a) that in the closed-loop solution, a portion of the DC component in (5.63) is shifted to higher order equations for solution because of the duty cycle and the steady-state periodic solution are both expanded into the sum of the main term and the small correction terms. Similarly this is also the case in the solution of other components. Thus, it is possible to achieve a higher accuracy by using single-iteration symbol algorithm to iterate several times.

When the equivalent small-parameter method is adopted, since the influence of the higher harmonics on the duty ratio is small, it can be considered that when solving the steady-state periodic solution of the closed-loop system, the duty ratio has been determined after three iterations, that is, d =  d0 +  d1 +  d2, is considered to be constant. In order to improve the convergence speed, the system can be solved once again by the equivalent small parameter solution method of the open-loop system. This method of solving the closed-loop system of the switching power converter first, after determining the duty ratio, considering the system as an open-loop system, and then solving it again with the ESPM is called double iterative symbol method.

In fact, we find that if d0 is replaced by the term ( d0 +  d1 +  d2) in the expansion of the switching function δ, the first three iterative equations of the closed-loop system are exactly the same as those of the open-loop system. Therefore, it can be set that:
$$D = d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} ,\quad \tau_{s2} = 2\pi (d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} )$$
(5.64)
Here the meaning of $$\varepsilon$$ is the same as before. According to equations from (5.23) to (5.29) on the method of dividing the integral interval, there are:
$$\alpha_{1} = \frac{{\sin 2\pi (D + \varepsilon^{3} d_{3} + \cdots )}}{\pi } \approx \frac{\sin 2\pi D}{\pi } + \varepsilon^{3} \cdot 2d_{3} \cos \tau_{s2} + \cdots$$
(5.65)
$$\beta_{1} = \frac{{1 - \cos 2\pi (D + \varepsilon^{3} d_{3} + \cdots )}}{\pi } \approx \frac{1 - \cos 2\pi D}{\pi } + \varepsilon^{3} \cdot 2d_{3} \sin \tau_{s2} + \cdots$$
(5.66)
Thus the coefficient b1 can be calculated by the following equations as
$$b_{1} \approx \tfrac{1}{2}(\alpha_{1} - j\beta_{1} ) = b_{10} + \varepsilon^{3} d_{3}e^{{-j\tau_{s2}}} + \cdots$$
(5.67)
Similarly, the series expansion of b2 and b3 are given as follows.
$$b_{2} = b_{20} + \varepsilon^{3} d_{3}e^{{-j\tau_{s2}}} + \cdots ,\quad b_{3} = b_{30} + \varepsilon^{3} d_{3}e^{{-j\tau_{s2}}} + \cdots$$
(5.68)
Thus, in the series expansion of switching function $$\delta$$ in (5.32a), the value of the following coefficients is zero, i.e., b11 = b12 = b21 = b22 = b31 = b32 = 0, and the coefficient bm0 is determined by
$$b_{m0} = \tfrac{1}{2}(\alpha_{m0} - j\beta_{m0} )$$
(5.69)
where
$$\alpha_{m0} = \frac{\sin 2m\pi D}{m\pi },\quad \beta_{m0} = \frac{1 - \cos 2m\pi D}{m\pi },\;(m = 1,2,3)$$

It can be seen that the first three terms of the series expansion of $$\delta$$ are exactly the same as those of the open loop, which shows that the joint of the closed-loop solution method and the open-loop solution method are reasonable to determine the steady-state periodic solution of the closed-loop system.

See Chap. 4 for the detailed process of the steady-state periodic solution of the open-loop switching power converter system of the switching power converter by the equivalent small-parameter method. Here we only give the equations for the main component and the first-order and second-order corrections, as shown in Eq. (5.70), which is also available by setting d0 =  D, d1 =  d2 = 0, b11 =  b12 =  b21 =  b31 = 0 in Eqs. (5.37), (5.41a), and (5.45a)~(5.45c).
$$\begin{array}{*{20}l} {G_{0} (0)a_{00} + G_{1} (0)Da_{00} = u} \hfill \\ {[G{}_{0}(j\omega ) + G_{1} (j\omega ) \cdot D] \cdot a_{11} = - G_{1} (j\omega ) \cdot b_{10} a_{00} } \hfill \\ {[G_{0} (j2\omega ) + G_{1} (j2\omega )D]a_{22} = - G_{1} (j2\omega ) \cdot [b_{20} a_{00} + b_{10} a_{11} + b_{30} \bar{a}_{11} ]} \hfill \\ {[G_{0} (j3\omega ) + G_{1} (j3\omega )D]a_{32} = - G_{1} (j3\omega ) \cdot [b_{30} a_{00} + b_{10} a_{22} + b_{20} a_{11} ]} \hfill \\ {[G_{0} (0) + G_{1} (0)D]a_{02} = - G_{1} (0)(b_{10} \bar{a}_{11} + b_{10} a_{11} )} \hfill \\ \end{array}$$
(5.70)

Note that when using the double iterative algorithm, the original closed-loop system needs to be decoupled, that is, the power stage main circuit is separated from the feedback compensation network, and only the main circuit state variables are solved. Since the duty cycle is considered constant at this time, the feedback network has no effect on the steady-state periodic solution of the power stage.

5.5.4 Analysis Example

We analyze the Boost converter in Sect. 5.4.1 by the improved symbolic algorithm and then compare it with other methods. When the converter operates with different frequencies, the results from the four methods, i.e., the basic equivalent small parameter method (B-ESPM), the improved equivalent small parameter method (I-ESPM), the numerical simulation, and the Pspice simulation method, are listed in Table 5.5.
Table 5.5

Simulated results of state variables in different methods

fs

Methods

a0

a1

b1

a2

b2

a3

b3

50 kHz

d0 = 0.4091

d1 = −0.066

d2 = −0.047

B-ESPM

0.3325

−0.2744

0.1387

−0.0587

−0.0379

−0.0013

0.0039

6.9796

0.1179

−0.2186

0.0310

0.0407

−0.0151

−0.0052

I-ESPM

0.3566

−0.1938

0.13640

−0.0649

−0.0196

−0.0040

−0.0095

7.0651

0.0332

−0.1791

0.0372

0.0124

−0.0084

0.0094

Numerical

0.3766

−0.2124

0.1349

−0.0631

−0.0289

0.0018

−0.0071

7.2230

0.0412

−0.1994

0.0351

0.0237

−0.0179

0.0070

Pspice

0.3750

−0.2295

0.1280

−0.0643

−0.0299

−0.0073

−0.0029

7.0472

0.0571

−0.1966

0.0294

0.0269

−0.0145

0.0011

100 kHz

d0 = 0.4091

d1 = −0.032

d2 = −0.022

B-ESPM

0.4253

−0.1483

0.0584

−0.0211

−0.0210

−0.0032

0.0069

7.7490

0.0875

−0.0741

0.0098

0.0191

−0.0014

−0.0064

I-ESPM

0.4287

−0.1286

0.0606

−0.0244

−0.0188

−0.0010

0.0036

7.7465

0.0651

−0.0698

0.0127

0.0153

−0.0029

−0.0030

Numerical

0.4298

−0.1295

0.0613

−0.0247

−0.0207

−0.0009

0.0040

7.7414

0.0641

−0.0732

0.0127

0.0166

−0.0049

−0.0037

Pspice5

0.4430

−0.1442

0.0481

−0.0220

−0.0222

−0.0071

0.0053

7.6504

0.0791

−0.0669

0.0083

0.0169

0.0005

−0.0066

1000 kHz

d0 = 0.4091

d1 = −0.003

d2 = −0.002

B-ESPM

0.5032

−0.0157

0.0048

−0.0013

−0.0020

−0.0008

0.0009

8.3984

0.0107

−0.0036

0.0008

0.0014

0.0005

−0.0007

I-ESPM

0.5023

−0.0155

0.0048

−0.0014

−0.0020

−0.0007

0.0009

8.3856

0.0104

−0.0036

0.0009

0.0014

0.0005

−0.0006

Numerical

0.5049

−0.0156

0.0049

−0.0014

−0.0021

−0.0008

0.0010

8.3750

0.0106

−0.0039

0.0009

0.0014

0.0005

−0.0008

Pspice5

0.5256

−0.0171

0.0031

−0.0012

−0.0019

−0.0008

0.0008

8.3348

0.0117

−0.0035

0.0009

0.0011

0.0008

−0.0008

Note The upper row of each set of data in Table 5.5 represents the inductor current (A), and the lower row represents the output capacitor voltage (V)

The closed-loop Boost converter system in Sect. 5.4.1 is used as an analysis example. The basic algorithm of ESPM, the improved algorithm of ESPM, the numerical and the Pspice simulation methods are adopted respectively to analyze the Boost converter operated with different switching frequencies. The resulted DC component and the first three harmonic components of the state variables are listed in Table 5.5, in which a0 represents the DC component, ai and bi represent amplitudes of the cosine and sine components of the ith (i = 1,2,3) harmonic, respectively (see Eq. (5.48b)).

When the converter operates at the switching frequency of fs = 50 and 100 kHz respectively, the ripples of the state variables during one cycle obtained from the ESP analysis method (dashed line) are compared with those from numerical simulation method (solid line) in Fig. 5.5, in which the results of the ESPM are obtained by calculation of the sum of the first three harmonic components of the state variables during one cycle. Among Fig. 5.5, (a–d) are the results of comparing the improved algorithm with numerical simulation; and (e–h) are the results of the comparison between the original single-iteration algorithm (basic algorithm) and the numerical simulation.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig5_HTML.gif
Fig. 5.5

Comparison of simulated state ripple waveforms between numerical simulations and the ESPM (a–d: with the improved algorithm; e–h: with the original algorithm)

It can be seen from Table 5.5 and the waveforms in Fig. 5.5, that using the improved symbol algorithm to analyze the closed-loop switching power system can get more accurate results.

When the switching frequency is low and the ripple is large, the values of d1 and d2 in the direct ripple feedback control mode are larger. Therefore, when the single-iteration algorithm is used, the obtained DC component would have a larger error. However, this situation will be improved, and more accurate results can be obtained by using the improved symbol algorithm.

It can be seen from the above examples and Table 5.5, that the basic symbol algorithm can get more accurate results when the switching frequency is higher, or when the average feedback control law is used (i.e., the appropriate error compensation capacitor is added in the control loop to smooth the effects of ripples on the duty cycle). Because under these situations, the values of the first-order correction d1 and the second-order d2 are quite small, they have less influence on the duty cycle d.

5.6 Experiments and Verification

5.6.1 Diagram of the Experimental Circuit

A Boost regulator and its feedback-controlled circuit is shown in Fig. 5.6, where $${r_{1}}$$ and $$R_{s}$$ are the parasitic resistance and the current sensing resistor of the inductor respectively. In order to verify that the symbolic Equivalent-Small-Parameters (ESP) analysis method is still applicable in the case of high output ripple, we choose a lower switching frequency and a smaller output capacitance.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig6_HTML.gif
Fig. 5.6

Experimental circuit of Boost regulator

The TL494 PWM control chip is considered for experiments, its detailed internal block diagram, pin configuration and functions, and the operation principle can be found in related power integrated control circuit datasheet. The voltage of Pin-4 is used to limit the maximum width of the pulse, which determines the dead time control of the main circuit. The internal voltage reference $$V_{R} = + 5V$$ is provided by Pin-14. The resistors R3 and R4 set the reference of about 2.5 V on the negative input port (Pin-2) of the error amplifier (EA), and the output voltage of the main circuit is divided by resistors R1 and R2, setting the sampling voltage on the positive input port of the error amplifier (Pin-1). The error amplifier compares and amplifies the difference between the sampled and the reference voltage, and the resulted output error voltage is then compared with the sawtooth wave voltage signal of a constant frequency to generate a pulse with a certain width (the width of the pulse is directly determined by the error voltage). The pulse is shaped and amplified by TL-494 and then output via Pin-8, Pin-9 or Pin 10 and Pin-11 with a certain drive capability to turn on the power switch. Pins 5 and 6 are connected to the capacitor C6 and the resistor R9 respectively, which determine the oscillating frequency of sawtooth wave, i.e., $$f_{s} = 1.2/R_{9} C_{6}\,({\text{kHz}})$$. The oscillation amplitude of the sawtooth wave is given by the relevant datasheet or measured by Pin-5. The maximum and minimum values measured in the experiment are $$v_{u} = 3.12V$$ and $$v_{l} = 0$$ (see Fig. 5.7b below). Pin-3 is the output of the TL-494 internal error amplifier. It is connected to the non-inverting input (Pin-2) of the amplifier with a RC circuit for error compensation and the self-oscillation suppression. To increase the stability of the error-amplifier circuit, the output of the error amplifier is fed back to the inverting input port through a resistor R5.

According to the PWM principle of the TL-494 chip, when the circuit works normally, i.e., the circuit is controlled effectively, that is, the duty ratio is in the unsaturated state. The schematic of the control circuit in Fig. 5.6 can be shown in Fig. 5.7a, where EA indicates the operational amplifier, COMP indicates the comparator, $$R_{2}$$ is the sum of the resistance of resistor $$R_{20}$$ and potentiometer P1 in Fig. 5.6, and C1 is the sum of C11 and C12 in Fig. 5.6.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig7_HTML.gif
Fig. 5.7

a Diagram of feedback controlled circuit, b measured sawtooth ramp and switching signal waveforms

5.6.2 Comparison of Experiment, ESPM and Simulation for Open-Loop System

The input filter capacitor $$C_{0}$$ in Fig. 5.6 has such a large value that it can be replaced with an open circuit when the circuit operates in steady state, which means it does not affect the dynamic characteristics of the circuit, so it was ignored in symbolic analysis. Thus the main circuit is considered to be a second-order circuit. Disconnect the point A in Fig. 5.6, and connect a constant voltage $$V_{A}$$ independent of the output voltage of the main circuit at point A, then the circuit is in open-loop operation. The experimental results are $$V_{A} = 2.46(V)$$, $$V_{R} = 4.97(V)$$; the current sensing resistor of the inductor $$R_{s} = 0.595\,\Omega$$. According to Fig. 5.7, the output voltage $$V_{k}$$ of the error amplifier and the duty cycle $$D_{0}$$ can be calculated as:
$$\begin{aligned} V_{k} &amp; = V_{A} (2R_{5} /R_{3} + 1) - V_{R} \cdot R_{5} /R{}_{3} = 21V_{A} - 10V_{R} = 1.96(V) \\ D_{0} &amp; = 1 - V_{k} /(v_{u} - v_{l} ) = 0.3718 \\ \end{aligned}$$
(5.71)
And the operating frequency of the switch calculated from the circuit parameters is:
$$f_{s0} = 1.0695\,({\text{kHz}})$$
(5.72)
Figure 5.7b shows the measured sawtooth ramp and switching signal waveforms when the circuit is in open-loop operation. The figure shows that the switching period $$T = 961.6\,{{\upmu }\text{s}}$$, $$t_{on} = 368.2\,{{\upmu }\text{s}}$$, thus, the actual switching frequency and duty cycle would be
$$f_{s} = 1.039933\,({\text{kHz}}),\;D = 0.3829$$
(5.73)

From the comparison of Eqs. (5.71)–(5.73), it can be seen that the measured values are close to the theoretically calculated values.

According to the analysis of the Chap. 4, the Boost converter in CCM can be described by the following general vector state variable differential equation:
$${\mathbf{G}}_{0} (p){\mathbf{x}} + {\mathbf{G}}_{1} ({\mathbf{p}}){\mathbf{f}} = {\mathbf{u}}$$
(5.74)
where
$$\begin{aligned} {\mathbf{G}}_{0} (p) &amp; = \left[ {\begin{array}{*{20}c} {p + \tfrac{{r1 + R_{s} }}{L}} &amp; {\tfrac{1}{L}} \\ {\tfrac{ - 1}{C}} &amp; {p + \tfrac{1}{RC}} \\ \end{array} } \right],\quad {\mathbf{G}}_{1} (p) = \left[ {\begin{array}{*{20}c} 0 &amp; {\tfrac{ - 1}{L}} \\ {\tfrac{1}{C}} &amp; 0 \\ \end{array} } \right],\;{\mathbf{x}} = \left[ {\begin{array}{*{20}c} {i_{L} } \\ {v_{o} } \\ \end{array} } \right], \\ {\mathbf{u}} &amp; = \left[ {\begin{array}{*{20}c} {E/L} \\ 0 \\ \end{array} } \right],\quad {\mathbf{f}} = \delta {\mathbf{x}},\;\delta = \left\{ {\begin{array}{*{20}l} 1 \hfill &amp; {0 \le t \le DT} \hfill \\ 0 \hfill &amp; {DT \le t \le T} \hfill \\ \end{array} } \right. \\ \end{aligned}$$
The switching frequency and duty cycle are determined by Eq. (5.73). Use the analysis method in Chap. 4, the approximated steady-state periodic analytic solution of state variables are acquired:
$$\begin{aligned} i_{L} &amp; = 0.8907 - 0.4157\cos \tau + 0.1448\sin \tau - 0.0527\cos 2\tau - 0.0490\sin 2\tau \\ &amp; \quad - 0.0130\cos 3\tau + 0.0200\sin 3\tau \\ v_{o} &amp; = 27.5296 + 1.2535\cos \tau - 1.5518\sin \tau + 0.0862\cos 2\tau + 0.3408\sin 2\tau \\ &amp; \quad - 0.0398\cos 3\tau - 0.1254\sin 3\tau \\ \end{aligned}$$
(5.75)
where $$\tau = \omega t$$, $$\omega = 2\pi f_{s}$$.
Figure 5.8 shows waveforms of capacitor output voltage and switching signal and sensed inductor current signal, Fig. 5.9 shows waveforms of measured state ripples and comparison of state ripples obtained by symbolic analysis (solid line) and Pspice simulation (dotted line), detailed data comparison is shown in Table 5.6.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig8_HTML.gif
Fig. 5.8

Waveforms of capacitor output voltage and switching signal (left) and sensed inductor current signal (right) for open-loop Boost converter

../images/419194_1_En_5_Chapter/419194_1_En_5_Fig9_HTML.gif
Fig. 5.9

Comparison of state ripples for open-loop Boost converter: a and b measured results; c and d ESPM analysis (solid line) and Pspice simulation (dotted line)

Table 5.6

Comparison of symbolic (ESPM) analysis, Pspice simulation and experiment for open-loop Boost converter

E = 18.30 V

DC values

Ripples

$$I_{L} \;(A)$$

$$V_{o} \;(V)$$

$$I_{L} R_{s} \;(V)$$

$$V_{o} \;(V)$$

Max

Min

Amp.

Max

Min

Amp.

ESPM

0.8907

27.5296

0.2390

−0.2493

0.4883

1.7654

−2.4454

4.2108

Pspice5

0.8792

27.2668

0.2616

−0.2783

0.5399

1.6904

−2.8187

4.5090

Exp.

0.9529

26.0

0.228

−0.278

0.506

1.5

−2.64

4.14

5.6.3 Comparison of Experiment, ESPM and Simulation for Closed-Loop System

According to Fig. 5.7 and the analysis method in Chap. 5, the time-variant differential equation of the closed-loop system can still be expressed as (5.74), where
$${\mathbf{G}}_{0} (p) = \left[ {\begin{array}{*{20}c} {p + \tfrac{{r1 + R_{s} }}{L}} &amp; {\tfrac{1}{L}} &amp; 0 &amp; 0 \\ { - \tfrac{1}{C}} &amp; {p + \tfrac{1}{RC}} &amp; 0 &amp; 0 \\ 0 &amp; {\tfrac{{m_{1} m_{2} R_{5} }}{{C_{1} }}} &amp; {p + \tfrac{{m_{1} }}{{C_{1} }}} &amp; { - \tfrac{{m_{1} m_{2} R_{5} }}{{C_{1} }}} \\ 0 &amp; { - \tfrac{1}{{R_{2} C_{2} }}} &amp; 0 &amp; {p + \tfrac{{m_{3} }}{{C_{2} }}} \\ \end{array} } \right],\quad {\mathbf{G}}_{1} (p) = \left[ {\begin{array}{*{20}c} 0 &amp; { - \tfrac{1}{L}} &amp; 0 &amp; 0 \\ {\tfrac{1}{C}} &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ \end{array} } \right],$$
$${\mathbf{x}} = [\begin{array}{*{20}c} {i_{L} } &amp; {v_{o} } &amp; {v_{1} } &amp; {v_{2} } \\ \end{array} ]^{{\prime }} ,\quad {\mathbf{u}} = [\begin{array}{*{20}c} {E/L} &amp; 0 &amp; {V_{R} R_{5} m_{1} /(R_{3} C_{1} )} &amp; {0]^{{\prime }} } \\ \end{array} ,\quad {\mathbf{f}} = \delta {\mathbf{x}},$$
where
$$m_{1} = 1/(R_{5} + R_{6} ),\quad m_{2} = 1/R_{3} + 1/R_{4} ,\quad m_{3} = 1/R_{1} + 1/R_{2}$$
and the switching function is:
$$\delta = \left\{ {\begin{array}{*{20}l} 1 \hfill &amp; {0 \le t \le d(t)T} \hfill \\ 0 \hfill &amp; {d(t)T \le t \le T} \hfill \\ \end{array} } \right.$$
where $$d(t)$$ is the duty cycle, which can be expressed as a linear function of state variable:
$$d = K_{0} + {\mathbf{K}}_{1} {\mathbf{x}}(t_{s} )\;(t_{s} = dT)$$
(5.76)
where
$$K_{0} = 1 + {{R_{5} R_{6} m_{1} V_{R} } \mathord{\left/ {\vphantom {{R_{5} R_{6} m_{1} V_{R} } {(R_{3} v_{u} )}}} \right. \kern-0pt} {(R_{3} v_{u} )}}$$
$${\mathbf{K}}_{1} = [\begin{array}{*{20}c} 0 &amp; {1 + m_{1} m_{2} R_{5} R_{6} )} &amp; { - R_{5} m_{1} } &amp; { - (1 + m_{1} m_{2} R_{5} R_{6} )]/v_{u} } \\ \end{array}$$
The switching frequency of the circuit is the same as that of the open loop. Using the double iterative symbol algorithm in Sect. 5.5.3, the steady-state periodic solution of the inductor current and the output voltage containing the DC component and the first three harmonics can be calculated as Eq. (5.77), where $$\tau = \omega t$$, $$\omega = 2\pi f_{s}$$.
$$\begin{aligned} i_{L} &amp; = 1.5146 - 0.5747\cos \tau - 0.0962\sin \tau - 0.0279\cos 2\tau + 0.0372\sin 2\tau \\ &amp; \quad - 0.0533\cos 3\tau - 0.0212\sin 3\tau \\ v_{o} &amp; = 34.7894 + 3.2657\cos \tau - 0.6497\sin \tau - 0.3595\cos 2\tau - 0.3204\sin 2\tau \\ &amp; \quad + 0.2960\cos 3\tau + 0.1295\sin 3\tau \\ \end{aligned}$$
(5.77)
Figure 5.10 shows waveforms of capacitor output voltage and switching signal and sensed inductor current signal when the circuit is in closed-loop operation, waveforms of measured state ripples and comparison of state ripples obtained by symbolic analysis (solid line) and Pspice simulation (dotted line) are shown in Fig. 5.11, detailed data comparison is shown in Table 5.7. As shown in Fig. 5.10a, the experimentally measured duty cycle is $$d = 0.5467$$, the duty cycle calculated by symbolic analysis method is $$d = 0.5420$$, one can see that the measured result and the analytic calculation are in a good agreement.
../images/419194_1_En_5_Chapter/419194_1_En_5_Fig10_HTML.gif
Fig. 5.10

Waveforms of output voltage and switching signal (left) and sensed inductor current signal (right) for closed-loop Boost converter

../images/419194_1_En_5_Chapter/419194_1_En_5_Fig11_HTML.gif
Fig. 5.11

Comparison of state ripples for closed-loop Boost converter: a and b measured results; c and d symbolic analysis

Table 5.7

Comparison of symbolic (ESPM) analysis and experiment for closed-loop Boost converter

E 18.00 V

DC values

Ripples

$$I_{L} \;(A)$$

$$V_{o} \;(V)$$

$$I_{L} R_{s} \;(V)$$

$$V_{o} \;(V)$$

Max

Min

Amp.

Max

Min

Amp.

ESPM

1.5146

34.7894

0.3733

−0.3921

0.7654

3.2941

−3.9366

7.2247

Exp.

1.7142

33.7

0.318

−0.344

0.662

3.28

−3.96

7.24

5.7 Summary

In this chapter, the equivalent small parameter method (ESP) is applied to the steady-state analysis of the closed-loop converter system with CCM (continuous-conduction-mode) operation. A single-iterative and a double iterative symbol algorithm for analyzing the steady-state solution of the closed-loop system are proposed. It is shown that the equivalent small parameter method can be extended to the steady-state analysis of the closed-loop PWM switching converter system and still possesses the advantages of simple algorithm and high accuracy. The results are all analytical expressions, and from which the working mechanism of the circuit can be easily mastered. And moreover, the analytical expressions of the ripples would have obvious applications in the engineering design of circuit and computer symbolic simulation analysis.

In addition, the principle of the method and the analysis results in this chapter show that: (1) The effect of ripple on the duty cycle of the PWM closed-loop switching power converter system is quite small, especially at a higher switching frequency; (2) The DC offset presented in the system can be suppressed by adding the appropriate integral compensation in the feedback circuit; (3) The data comparison in Table 5.5 shows that using a single-iterative algorithm to analyze the closed-loop system with PI feedback control law has a higher accuracy (the actual application circuit is generally the case).

The comparison between experiments and the results of symbolic analysis further verified that the ESP symbol analysis method has high accuracy for the steady state analysis of open-loop and closed-loop systems of power switching converters even if the output contains large ripples.