© 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_7

7. Analysis of Current-Mode Controlled 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
Email: eeyfchen@scut.edu.cn

7.1 Introduction

As a single-loop control system, voltage-mode control usually has only one voltage feedback loop [1]. While the current-mode control [2–6], a dual-loop control system, contains both negative current feedback loop and voltage feedback control. The basic DC/DC converter (Buck/Boost/Buck-boost) is a second-order system with two state variables, i.e., the output filter capacitor voltage and the inductor current. Therefore, from the perspective of control theory, the double closed-loop system is the most optimal control strategy.

Compared with the voltage-mode control technology, the current-mode control technology has the advantages of fast dynamic response, automatic overcurrent protection, being suitable for parallel operation of multiple converters [4–6], etc., and has been widely put into use. It is shown in practice that the output ripple of the current-mode controlled converter is usually about 25 mV, which makes it far superior to similar products adopting the voltage-mode control. However, due to the double closed-loop (voltage loop and current loop) feedback and the impossibility of determining duty cycle explicitly, current-mode control increases the difficulty of modeling and analysis [5, 6]. At present, the analysis of this kind of converter is basically based on the average modeling method, assuming that inductor current changes linearly and ripple is small [5–11]. However, the averaged circuit by applying this kind of method, no longer contains the ripple information of the original circuit and fails to predict the stability of the original system [10]. Recently, symbolic analysis has always been a research topic of power electronics. Compared with numerical simulation [12–15], symbolic analysis helps to investigate circuit behavior and produce an analytical model for circuit optimization and control design. In this chapter, the equivalent-small-parameter method introduced in the previous chapters is extended to the steady-state analysis of the closed-loop current-mode controlled converter system. The analytical solutions of the state variables of the closed-loop converter system, expressed with symbols, are obtained containing DC component and ripple. And this kind of application make it is easy to understand the working mechanism of the circuit, and can provide guidance for engineering design, such as selection of filter capacitor and inductance parameters [16, 17].

The following sections of this chapter are arranged as follows. Section 7.2 introduces the basic principles of current-mode control. And Sect. 7.3 proposes the symbolic analysis method, i.e. equivalent-small-parameter analysis method for the closed-loop current-mode controlled converter. Section 7.4 presents The analysis cases are presented in Sect. 7.4 and finally the summary of this analytical methods in this chapter is given in Sect. 7.5.

7.2 The Basic Principle of Constant Frequency Current-Mode Control

The diagram of constant frequency current-mode control scheme is showed as Fig. 7.1, where the module CSP (current signal processor) is used to detect the value of the inductor current or the current of the switching devices, which may be a transient value or an average value. In Fig. 7.1 the symbol vp denotes a ramp compensation signal; the error amplifier (EA) processes the difference between the output voltage and the reference voltage and obtains a control signal vk, forming a voltage feedback outer loop to realize automatic voltage adjustment; the current feedback control constitutes the inner loop to achieve automatic current regulation. The voltage loop and current loop constitute the double closed-loop feedback controlled system.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig1_HTML.gif
Fig. 7.1

Basic diagram of current-mode controlled system

In traditional duty-ratio programmed (see Fig. 1.​1) DC/DC converter, an analog comparator compares the control signal to a ramp signal, whose period and rising slope is constant. When the value of the ramp signal rises to the control voltage, the comparator flips and the state of the switch changes. The ramp signal is independent of the state variables in power stage. Therefore, the duty cycle is only determined by the control signal. When the control signal is independent of the output variable of the circuit, the open-loop control of the system can be realized. In a switching power converter with constant frequency current-mode control, the conduction of the switch is controlled by a clock signal with a constant frequency, and the turn-off is determined by the moment at which the inductor current signal reaches the control signal. Thus, the duty cycle is not only related to the control signal, but also to the initial value of the inductor current and the rising slope. Therefore, the current-mode control itself introduces a feedback loop [18].

The basic principle of current mode control [5] will be illustrated via an example of constant frequency peak current feedback control. Figure 7.2 shows typical waveforms of the duty cycle and control signal without (Fig. 7.2a) and with (Fig. 7.2b) slope compensation respectively. In the figure, m1, m2, mc represent the rising and falling slope of the inductor current and the slope of the ramp compensation signal respectively. In the peak-current-mode control, the detected current value vl = M * iL, where M is the gain of the CSP.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig2_HTML.gif
Fig. 7.2

Typical waveform of current-mode controlled system

Consider the case without slope compensation shown as Fig. 7.2a. Let T denote the switching period. When t = nT, the clock signal sets the trigger (output high level), turning on the main switch of the circuit in the power stage, and thus the inductor current rises linearly. And at the moment of t = nT + dT, i.e., vl (t) = vk(t), the clock signal resets the trigger (output low level). Consequently, the main switch of circuit in the power stage is turned off and the inductor current drops. At the moment of t = (n + 1)T, the next clock signal arrives and the switch is turned on again. It is pointed out in [5] that the constant frequency control has potential instability (sub-harmonic oscillation), as shown as the dotted line in Fig. 7.3. Assume that, in steady state, the rising and falling slope of the inductor current are $$m_{1}$$ and $$m_{2}$$ respectively, and the converter operates in the continuous inductor current mode. Then if the inductor current has a disturbance $$\Delta I_{0}$$ at the beginning of a switching cycle, there will be a disturbance $$\Delta I_{1}$$ at the end of the cycle. From Fig. 7.3, it can be seen that $$\Delta I_{1} = - \left( {\frac{d}{1 - d}} \right)\Delta I_{0}$$. Thus, after n cycles, the disturbance at the nth cycle would be $$\Delta I_{n} = - \left( {\frac{d}{1 - d}} \right)^{n} \Delta I_{0}$$. Obliviously, the error would gradually increases when the duty cycle $$d > 0.5$$, which means that the system is unstable. When the duty cycle $$d = 0.5$$, a typical sub-harmonic oscillation occurs, as shown in Fig. 7.3.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig3_HTML.gif
Fig. 7.3

Waveform of sub-harmonic oscillation

The solution to eliminate sub-harmonic oscillation is to artificially integrate a ramp function, called ramp compensation, into the detected current ripple of the switching converter. As shown in Fig. 7.2b, let the term $$- m_{c}$$ denotes the falling slope of the function, then the disturbance would be $$\Delta I_{n} = - \left( {\frac{{m_{2} - m_{c} }}{{m_{1} + m_{c} }}} \right)^{n} \Delta I_{0}$$. It is obviously that $$\Delta I_{n}$$ decreases gradually for any duty cycle. Therefore, after several cycles, the disturbance in the inductor current will become negligible and the system remains stable. Particularly, if $$m_{2} = m_{c}$$, the disturbance in the inductor current will disappear after one cycle, which not only ensures the stability of the inner loop but also provides the fastest dynamic response for the current-mode control. Therefore, proper selection of the slope of the ramp compensation function can appropriately improve the dynamic response characteristics of the system [5, 18–20].

7.3 Symbolic Analysis of Closed-Loop Current-Mode Controlled Converter System

Figure 7.1 can be simplified as shown in Fig. 7.4a, and the control law shown in Fig. 7.2b is adopted in Fig. 7.1, which is redrawn as Fig. 7.4b. Combining the power stage and voltage feedback error amplifiers in the system as a whole, then the closed-loop system can be expressed as two parts, with a state variable vector $${\mathbf{x}} = [i,v,v_{a} ]^{Tr}$$, in which the symbols i and v are the state variables of power stage and va denotes voltage across the compensation capacitor in error feedback loop. The meaning of the terms vp, vl, m1, m2 and mc are the same as defined earlier. The symbol vk stands for the control signal. For the case of double closed-loop feedback, the control signal vk is equal to the output voltage of the error amplifier (EA), where the input signal of the EA is the difference between the output voltage and the reference voltage, which is compensated and amplified by the EA by taking some kinds of adjustment method (such as proportional-integral algorithm). In this case, the control signal is related to the output voltage, thus introducing a voltage feedback loop. And when vk is a constant, i.e. vk = Iref, there is only a current feedback loop in the system, which is called single-loop current-mode control, or open-loop current-mode control. In this case, the term Iref represents a reference current signal, and the control signal vk is independent of the output voltage.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig4_HTML.gif
Fig. 7.4

Simplified block diagram of current-controlled system

In the simplified symbolic analysis below, three assumptions, which are also the basic assumptions made by the state-space averaging method in fact, would be proposed as follows.
  1. (i)

    The waveform of inductor current is linear;

     
  2. (ii)

    The control voltage changes slowly during one switching cycle, i.e. $$v_{k} (t) = \bar{v}_{k}$$;

     
  3. (iii)

    The ripple of output voltage is small compared to the steady-state value.

     

7.3.1 Expression of the Duty Cycle for Closed-Loop CMC System

7.3.1.1 Accurate Solution of Duty Cycle

According to the waveform in Fig. 7.4b, the following equation can be deduced as
$$v_{k} (dT) = v_{l} (dT) + m_{c} dT$$
(7.1)
where $$v_{l} (t) = M \cdot i_{L}$$, M is the gain of CSP. According to the discussion in Chap. 5, the control signal vk(t) generally can be expressed as a linear function of the whole system state variables, i.e.
$$v_{k} (t) = k_{0} + {\mathbf{k}}_{1} \cdot {\mathbf{x}}$$
(7.2)
where $$k_{0} ,{\mathbf{k}}_{1}$$ are constants associated with the circuit parameters, here the term $$k_{0}$$ is a scalar, and k1 is a row vector. If x is an column vector with $$n \times 1$$ dimensions, then $${\mathbf{k}}_{1}$$ is row vector with the size of $$1 \times n$$ dimensions. Substituting (7.2) into (7.1) gives the expression for the duty cycle, i.e.,
$$d = \frac{{k_{0} }}{{m_{c} T}} + \frac{{{\mathbf{k}}_{1} }}{{m_{c} T}}\left. {{\mathbf{x}}(t)} \right|_{t = dT} - \frac{M}{{m_{c} T}}\left. {i_{L} (t)} \right|_{t = dT}$$
(7.3)
Obviously, the Eq. (7.3) can be expressed as a linear function of state variables as
$$d = K_{0} + K_{1} \cdot {\mathbf{x}}(dT)$$
(7.4)

The coefficients $$K_{0} ,{\mathbf{K}}_{1}$$ are determined according to the specific circuit. The duty cycle is expressed in the same way as the voltage-mode-controlled PWM converter. Thus, the analysis method for duty cycle, is also the same. The detailed solution can be found in Chap. 5 or the example below.

According to the analysis before, this modeling method doesn’t require any assumption of the original system. Yet, most analysis methods for current-mode controlled converter depend on the geometric relationship of the inductor current waveform to establish the equation with duty cycle. Assumptions of linear waveform and small ripple are needed. Therefore, the proposed symbolic analysis method for closed-loop current-mode controlled converter system is more general.

What’s more, when the feedback doesn’t contain the ramp compensation, the duty cycle cannot be expressed as an explicit function using this method, which is the limitation of this method. Then the following simplified method can be adopted. Since the ramp compensation is usually applied, the method is applicable for the actual circuit.

7.3.1.2 Simplified Solution of Duty Cycle

When the three assumptions (1), (2) and (3) are satisfied, the duty cycle can be obtained by adopting the following simplified method. According to the waveform in Fig. 7.4b, the following relationship can be deduced.
$$\bar{v}_{l} = \bar{v}_{k} - \tfrac{1}{2}\,m_{1} dT - m_{c} dT$$
(7.5)
where $$\bar{v}_{l} = M \cdot \bar{i}_{L}$$ is the average value of inductor current in a switching period. Equation (7.5) is the mathematical model of the duty cycle for most current-mode controlled converters. With the assumptions of linear waveform and small ripple, the term $$m_{1}$$ can generally be expressed as a function of the input voltage and the average value of voltage across filter capacitor in power stage. Assumed that the input voltage Vs remains constant, $$m_{1}$$ can be expressed as
$$m_{1} = g'(\bar{v}_{o} )$$
(7.6)
Substituting (7.2) and (7.6) into (7.5), one can get the expression of the duty cycle as shown in Eq. (7.7), which is a function of mean values of the state variables.
$$d = \frac{{k_{0} + {\mathbf{k}}_{{\mathbf{1}}} {\bar{\mathbf{x}}}(dT) - M \cdot \bar{i}_{L} (dT)}}{{(\tfrac{1}{2}m_{1} + m_{c} ) \cdot T}} = g({\bar{\mathbf{x}}})$$
(7.7)
When the equivalent-small-parameter method is put into use, due to the correction of the DC component by the high-order terms, $${\bar{\mathbf{x}}}$$ may contains both the zero-order component and the DC component of the higher-order correction term. According to the analysis in Chap. 5, suppose that the steady-state periodic solution x is an n-dimensional vector, i.e., x = [x(1), x(2), ... x(n)]Tr, it can be expanded into series as follows.
$${\mathbf{x}} = {\mathbf{x}}_{0} + \varepsilon {\mathbf{x}}_{1} + \varepsilon^{2} {\mathbf{x}}_{2} + \cdots$$
(7.8)
And then the average state variable vector can also expanded as
$${\bar{\mathbf{x}}} = {\bar{\mathbf{x}}}_{0} + \varepsilon {\bar{\mathbf{x}}}_{1} + \varepsilon^{2} {\bar{\mathbf{x}}}_{2} + \cdots$$
(7.9)
where $${\mathbf{x}}_{i}$$ is the ith-order small parameter of x, $$\left| {{\mathbf{x}}_{i} } \right| \ll {\mathbf{x}}_{0}$$. Similarly, the duty cycle can be expanded into
$$d = d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots$$
(7.10)
Then combining with (7.7) leads to
$$d_{0} + \varepsilon d_{1} + \varepsilon^{2} d_{2} + \cdots = g({\bar{\mathbf{x}}}_{0} + \varepsilon {\bar{\mathbf{x}}}_{1} + \varepsilon^{2} {\bar{\mathbf{x}}}_{2} + \cdots )$$
(7.11)
Because $$\left| {\bar{\mathbf{x}}_{i} } \right| \ll {\mathbf{x}}_{0}$$, then according to the linear approximation of the Tailor Series expansion, the increment of d caused by $${\bar{\mathbf{x}}}_{i}$$ can be approximated as
$$\Delta d = d_{i} = \left( {\left. {\frac{\partial g}{{\partial \bar{x_{i}}^{(1)} }}} \right|_{{\bar{\mathbf{x}} = \bar{\mathbf {x}}_{0} + \varepsilon \bar{\mathbf{x}}_{1} + \cdots + \varepsilon^{i - 1} \bar{\mathbf{x}}_{i - 1} }} } \right) \cdot \bar {x}_{i}^{(1)} + \cdots + \left( {\left. {\frac{\partial g}{{\partial \bar{x_{i}}^{(n)} }}} \right|_{{\bar{\mathbf{x}} = \bar{\mathbf{x}}_{0} + \varepsilon \bar{\mathbf{x}}_{1} + \cdots + \varepsilon^{i - 1} \bar{\mathbf{x}}_{i - 1} }} } \right) \cdot \bar {x}_{i}^{(n)}$$
(7.12)

Thus as can be seen from (7.12), $$d_{1}$$ is the increment caused by the DC component in the first-order correction, and $$d_{2}$$ is the increment caused by the DC component in the second-order correction, and so on.

7.3.2 Solution of the Closed-Loop Equation of the CMC Converter

Similar to the voltage-mode controlled converter, the state equation of the current-mode controlled system can still be expressed as
$$G_{0} (p){\mathbf{x}} + G_{1} (p){\mathbf{f}} = {\mathbf{u}}$$
(7.13)
where the nonlinear function
$$f = \delta (t)({\mathbf{x}} + {\mathbf{e}})$$
(7.14)
The meaning of each item in (7.13) and (7.14) is the same as that of in the voltage-mode controlled converter, where x is the state variable of the entire closed-loop system. The switching function $$\delta (t) = 1(0)$$ represent the on/off state of the main switch. If the switch has two states during one cycle, then the switching function can be defined as
$$\delta (t) = \left\{ {\begin{array}{*{20}l} 1 \hfill & {0 < t \le d(t)T} \hfill \\ 0 \hfill & {d(t)T < t \le T} \hfill \\ \end{array} } \right.$$
(7.15)
where T is the switch period, $$d(t)$$ is the duty cycle, decided by (7.3) or (7.7) in the current-mode controlled converters.

Regardless of whether the duty cycle is an exact solution or a simplified solution, the duty cycle can be expressed as the sum of the main term and the small corrections in both cases, so the steady-state solution process of the current-mode controlled converter is exactly the same as that of the voltage-mode controlled converter. The only difference is that the solution of the duty cycle is different.

In order to facilitate the analysis, equivalent-small-parameter method is adopted to obtain the first three components (zero-order, first-order, second-order) of the steady-state periodic solution of the closed-loop system, as shown in (7.16). And for simplicity, in the analysis of the closed-loop current-mode controlled system, the constant vector is set to be zero, i.e., e = 0 in (7.14) is taken into account.
$$\left\{ {\begin{array}{*{20}l} {G_{0} (0){\mathbf{a}}_{00} + G_{1} (0)d_{0} {\mathbf{a}}_{00} = {\mathbf{u}}} \hfill & {({\text{a}})} \hfill \\ {[G{}_{0}(j\omega ) + G_{1} (j\omega ) \cdot d_{0} ] {\mathbf{a}}_{11} = - G_{1} (j\omega ) [(b_{11} + b_{10} ) {\mathbf{a}}_{00} }] \hfill & {({\text{b}})} \hfill \\ {[G_{0} (j2\omega ) + G_{1} (j2\omega )d_{0} ]{\mathbf{a}}_{22} = - G_{1} (j2\omega )[(b_{21} + b_{20} ){\mathbf{a}}_{00} + (b_{11} + b_{10} ){\mathbf{a}}_{11} + b_{30} {\bar{\mathbf{a}}}_{11} ]} \hfill & {({\text{c}})} \hfill \\ {[G_{0} (j3\omega ) + G_{1} (j3\omega )d_{0} ]{\mathbf{a}}_{32} = - G_{1} (j3\omega ) [(b_{31} + b_{30} ){\mathbf{a}}_{00} + b_{10} {\mathbf{a}}_{22} + b_{20} {\mathbf{a}}_{11} ]} \hfill & {({\text{d}})} \hfill \\ {[G_{0} (0) + G_{1} (0)d_{0} ] {\mathbf{a}}_{02} = - G_{1} (0)[(b_{11} + b_{10} ){\bar{\mathbf{a}}}_{11} + (\bar{b}_{11} + \bar{b}_{10} ){\mathbf{a}}_{11} + (d_{1} + d_{2} ){\mathbf{a}}_{00} ]} \hfill & {({{\text{e}}})} \hfill \\ \end{array} } \right.$$
(7.16)
In (7.16) the coefficient $$b{}_{mi}$$ can be determined by the following equations as
$$\begin{aligned} b_{{mi}} & = \tfrac{1}{2}(\alpha _{{mi}} - j\beta _{{mi}} )\quad (m = 1,2, \ldots ,i = 0,1,2, \ldots ) \\ \alpha _{{m0}} & = \frac{{\sin 2m\pi d_{0} }}{{m\pi }},\;\;\beta _{{m0}} = \frac{{1 - \cos 2m\pi d_{0} }}{{m\pi }} \\ \alpha _{{mi}} & \approx 2d_{i} \cos 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon ^{{i - 1}} d_{{i - 1}} )\quad \quad (i = 1,2, \ldots ) \\ \beta _{{mi}} & \approx 2d_{i} \sin 2m\pi (d_{0} + \varepsilon d_{1} + \cdots + \varepsilon ^{{i - 1}} d_{{i - 1}} )\quad \quad (i = 1,2, \ldots ) \\ \end{aligned}$$
(7.17)
where $$d_{0} ,d_{1} ,d_{2} \cdots d_{i}$$ represent the main component, the first- and second-order,…, and the ith-order corrections of the duty cycle respectively. They can be approximately obtained according to the specific duty cycle equation.

In the following content, the application of ESPM in analysis of closed-loop current-mode controlled converter system will be illustrated with examples.

7.4 Examples

7.4.1 Double-Loop Current-Mode Controlled Boost Converter

The typical closed-loop current-mode controlled Boost converter is shown in Fig. 7.5, and the parameters of the PCS (power circuit stage) the FN (feedback network) are set respectively as listed in Table 7.1, the switching cycle T = 1/fs.
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Fig. 7.5

Closed-loop current-mode controlled Boost regulator

Table 7.1

Circuit parameters of the dual-loop CMC-Boost converter

Parameters

Values

Parameters

Values

Input voltage E

28 V

Reference voltage VR

1.8 V

Switching frequency fs

25 kHz

Resistance R1

47.5 kΩ

Inductance L

195 μH

Resistance R2

2.5 kΩ

Capacitance C

2000 μF

Resistance Ra

72.2 kΩ

Load resistance R

11.2 Ω

Feedback capacitance Ca

0.23 μF

Current coefficient M

0.081

Peak ramp voltage Vp

0.25 V

According to the principle of the Boost converter with CCM operation, one can get the state differential equation describing the system as shown in (7.13), where the state variable vector is chosen as $${\mathbf{x}} = [i_{L} ,v_{o} ,v_{a} ]^{Tr}$$, and the square coefficient matrices are determined by
$$G_{1} (p) = \left[ {\begin{array}{*{20}c} p & {\tfrac{1}{L}} & 0 \\ { - \tfrac{1}{C}} & {p + \tfrac{1}{RC}} & 0 \\ 0 & { - \tfrac{1}{{R_{1} C_{a} }}} & p \\ \end{array} } \right]\,\,{\text{and}}\,\,G_{2} (p) = \left[ {\begin{array}{*{20}c} 0 & { - \tfrac{1}{L}} & 0 \\ {\tfrac{1}{C}} & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} } \right]$$
(7.18a)
The nonlinear vector function f and the input vector u are given by
$$f = \delta {\mathbf{x}}\;{\text{and}}\;{\mathbf{u}} = \left[ {\begin{array}{*{20}c} {\tfrac{E}{L}} & 0 & {\tfrac{ - 1}{{C_{a} }}(\tfrac{1}{{R_{1} }} + \tfrac{1}{{R_{2} }}} \\ \end{array} )V_{R} } \right]^{Tr}$$
(7.18b)
The control signal vk(t) can be derived as
$$v_{k} (t) = \left( {1 + \frac{{R_{a} }}{{R_{1} }} + \frac{{R_{a} }}{{R_{2} }}} \right)V_{R} - \frac{{R_{a} }}{{R_{1} }}v_{0} - v_{a}$$
(7.19)

7.4.1.1 Symbolic Analysis with Simplified Solution of Duty Cycle

Due to the smoothing effect of the error amplifier, it can be considered that the control signal changes little during one cycle, that is, the assumption (ii) can be satisfied. In the simplified solution, the duty cycle can be deduced by Eq. (7.5), where the rising slope of the inductor current and the slope of the ramp signal are determined by
$$m_{1} = E/L,\quad m_{c} = v_{p} /T$$
(7.20)
Introducing (7.20) into (7.5) gives the expression of duty cycle as shown in (7.4), in which the coefficients are given by
$$\left\{ {\begin{array}{*{20}l} {K_{0} = \frac{1}{{F_{m} }}(1 + \frac{{R_{a} }}{{R_{1} }} + \frac{{R_{a} }}{{R_{2} }})V_{R} } \hfill \\ {{\mathbf{K}}_{1} = \left[ {\begin{array}{*{20}c} {K_{11} } & {K_{12} } & {K_{13} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\frac{ - M}{{F_{m} }}} & {\frac{{ - R_{a} }}{{R_{1} F_{m} }}} & {\frac{ - 1}{{F_{m} }}} \\ \end{array} } \right]} \hfill \\ \end{array} } \right.$$
(7.21)

Here the symbol $$F_{m} = (\tfrac{1}{2}\,m_{1} + m_{c} )T$$.

According to above discussion, the duty cycle d can be expanded into the series shown in (7.10). Then based on the Eq. (7.12), the following equations for determining d0 and di (i = 1, 2….) can be can be obtained as
$$\begin{array}{*{20}l} {\left\{ {\begin{array}{*{20}l} {d_{0} = K_{0} +{\mathbf {K}}_{1} \bar{\mathbf{x}}_{0} } \hfill \\ {d_{1} = {\mathbf{K}}_{1} \bar{\mathbf{x}}_{1} } \hfill \\ {d_{2} = {\mathbf{K}}_{1} \bar{\mathbf{x}}_{2} } \hfill \\ \end{array} } \right.} \hfill \\ {\quad \cdots } \hfill \\ \end{array}$$
(7.22)
where $${\bar{\mathbf{x}}}_{i}$$ is the average value of the ith correction of x during one switching cycle.
  1. (1)

    Solution of the Main Term

     
Based on the principle of the ESPM, the main term (DC component) can be assumed as
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = [\begin{array}{*{20}c} {I_{00} } & {V_{00} } & {V_{a00} } \\ \end{array} ]^{Tr}$$
(7.23)
According to (7.16a) and (7.22), the following equation for solutions of a00 and d0 can be obtained.
$$G_{0} (0){\mathbf{a}}_{00} + G_{1} (0)d_{0} {\mathbf{a}}_{00} = {\mathbf{u}}$$
(7.24a)
$$d_{0} = K_{0} + {\mathbf{K}}_{1} {\mathbf{a}}_{00}$$
(7.24b)
Then, the steady-state solution can be deduced from (7.24a) and (7.24b), as shown follows.
$$\begin{aligned} V_{{00}} & = \left( {1 + \frac{{R_{1} }}{{R_{2} }}} \right)V_{R} ,\;d_{0} = 1 - \frac{E}{{V_{{00}} }},\;\;I_{{00}} = \frac{{V_{{00}} }}{{R(1 - d_{0} )}} \\ V_{{a00}} & = F_{m} K_{0} - \frac{{R_{a} }}{{R_{1} }}V_{{00}} - M \cdot I_{{00}} \\ \end{aligned}$$
(7.25)
When the circuit parameters are introduced, the values of a00 and d0 are given by
$$a_{00} = [\begin{array}{*{20}c} {4.1327} & {36} & {0.8817]^{Tr} } \\ \end{array} ,d_{0} = 0.2222.$$
(7.26)
  1. (2)

    Solution of First-order Correction

     
The first-order correction term can be assumed as
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} {{e}}^{j\tau } + c.c,\,{\mathbf{a}}_{11} = [\begin{array}{*{20}c} {I_{11} } & {V_{11} } & {V_{a11} } \\ \end{array} ]^{Tr}$$
(7.27)
Since the first-order correction only contains the fundamental, its average value during one switching period is zero, and the duty cycle increment caused by it is also zero, i.e., $${\bar{\mathbf{x}}}_{1} = 0$$ and $$d_{1} = 0$$. Then according to (7.17), the coefficient $$b_{m1} = 0$$ (m = 1, 2, 3…) can be determined, and thus the solution of a11 can be derived by the equation as shown in (7.28).
$$[G{}_{0}(j\omega ) + G_{1} (j\omega ) \cdot d_{0} ] \cdot {\mathbf{a}}_{11} = - G_{1} (j\omega ) \cdot b_{10} \cdot {\mathbf{a}}_{00}$$
(7.28)
By solving the above matrix equation, the symbolic expressions for each element of the column vector a11 can be obtained as
$$\begin{aligned} V_{11} & = \frac{{b_{10} \cdot [j\omega L \cdot I_{00} - (1 - d_{0} ) \cdot V_{00} ]}}{{\omega^{2} LC - (1 - d_{0} )^{2} - j\omega L/R}} \\ I_{11} & = \frac{{b_{10} \cdot V_{00} - (1 - d_{0} ) \cdot V_{11} }}{j\omega L} \\ V_{a11} & = \frac{{V_{11} }}{{j\omega R_{1} C_{a} }} \\ \end{aligned}$$
(7.29)
  1. (3)

    Solution of Second-order Correction

     
Based on the ESPM, the second-order correction term can be assumed as
$${\mathbf{x}}_{2} = {\mathbf{a}}_{02} + {\mathbf{a}}_{22} {{e}}^{j2\tau } + {\mathbf{a}}_{32} {{e}}^{j3\tau } + c.c$$
(7.30)
Since the second-order correction contains the DC component, the second- and the third-order harmonics, its average value during one switching period should be the DC values, i.e., $${\bar{\mathbf{x}}}_{2} = {\mathbf{a}}_{02} ,$$ and then the duty cycle increment caused by it would be $$d_{2} = {\mathbf{K}}_{1} {\mathbf{a}}_{02}$$. Thus according to (7.16) and (7.22), the equations for the solution of the second-order correction can be obtained as
$$[G_{0} (j2\omega ) + G_{1} (j2\omega )d_{0} ]{\mathbf{a}}_{22} = - G_{1} (j2\omega )[b_{20} {\mathbf{a}}_{00} + b_{10} {\mathbf{a}}_{11} + b_{30} {\bar{\mathbf{a}}}_{11} ]$$
(7.31a)
$$[G_{0} (j3\omega ) + G_{1} (j3\omega )d_{0} ]{\mathbf{a}}_{32} = - G_{1} (j3\omega ) [b_{30} {\mathbf{a}}_{00} + b_{10} {\mathbf{a}}_{22} + b_{20} {\mathbf{a}}_{11} ]$$
(7.31b)
$$[G_{0} (0) + G_{1} (0)d_{0} + G_{1} (0)a_{00} \cdot {\mathbf{K}}_{1} ]{\mathbf{a}}_{02} = - G_{1} (0)[b_{10} {\bar{\mathbf{a}}}_{11} + \bar{b}_{10} {\mathbf{a}}_{11} ]$$
(7.31c)
where the coefficient $$b_{m1} = 0$$ (m = 1, 2, 3…) can be determined by (7.17), and thus the solutions of a22, a32 and a02 can be derived by the above equations shown in (7.31).
It is worth noting that all the equations in (7.31) are linear. Then by solving the equations in (7.31) one by one, the symbolic expressions for each element of the column vectors a22, a32 and a02 can be obtained as
$$\begin{array}{*{20}l} {\begin{array}{*{20}l} {V_{22} \approx \frac{{j2\omega L \cdot \Delta_{I22} - (1 - d_{0} ) \cdot \Delta_{V22} }}{{4\omega^{2} LC - (1 - d_{0} )^{2} }}} \hfill \\ {V_{32} \approx \frac{{j3\omega L \cdot \Delta_{I32} - (1 - d_{0} ) \cdot \Delta_{V32} }}{{9\omega^{2} LC - (1 - d_{0} )^{2} }}} \hfill \\ {V_{02} \approx 0} \hfill \\ \end{array} } \hfill & {\begin{array}{*{20}l} {I_{22} = \frac{{\Delta_{V22} - (1 - d_{0} ) \cdot V_{22} }}{j2\omega L}} \hfill \\ {I_{32} = \frac{{\Delta_{V32} - (1 - d_{0} ) \cdot V_{32} }}{j3\omega L}} \hfill \\ {I_{02} \approx \frac{{\Delta_{V02} + \Delta_{I02} }}{{1 - d_{0} }}} \hfill \\ \end{array} } \hfill & {\begin{array}{*{20}l} {V_{a22} = \frac{{V_{22} }}{{j2\omega R_{1} C_{a} }}} \hfill \\ {V_{a32} = \frac{{V_{32} }}{{j3\omega R_{1} C_{a} }}} \hfill \\ {V_{a32} = \frac{{\Delta_{V02} - K_{11} I_{02} }}{{K_{13} }}} \hfill \\ \end{array} } \hfill \\ \end{array}$$
(7.32)
where
$$\begin{array}{*{20}l} {\begin{array}{*{20}l} {\Delta_{I22} = b_{20} I_{00} + b_{10} I_{11} + b_{30} \bar{I}_{11} } \hfill \\ {\Delta {}_{I32} = b_{30} I_{00} + b_{20} I_{11} + b_{10} I_{22} } \hfill \\ {\Delta_{I02} = b_{10} \bar{I}_{11} + \bar{b}_{10} I_{11} \, } \hfill \\ \end{array} } \hfill & {\begin{array}{*{20}l} {\Delta_{V22} = b_{20} V_{00} + b_{10} V_{11} + b_{30} \bar{V}_{11} } \hfill \\ {\Delta_{V32} = b_{30} V_{00} + b_{20} V_{11} + b_{10} V_{22} } \hfill \\ {\Delta_{V02} = b_{10} \bar{V}_{11} + \bar{b}_{10} V_{11} } \hfill \\ \end{array} } \hfill \\ \end{array}$$
When the circuit parameters are introduced into (7.25), (7.29) and (7.32), the steady-state periodic solution (DC component + first three harmonics) of the state variable in the main circuit can be finally obtained as
$$\begin{aligned} i_{L} & = 4.1327 - 0.3093\cos \tau + 0.3684\sin \tau - 0.1814\cos 2\tau + 0.0320\sin 2\tau \\ & \quad - 0.0624\cos 3\tau + 0.0005\sin 3\tau \\ v_{o} & = 36.0000 + 0.0025\cos \tau - 0.0049\sin \tau + 0.0020\cos 2\tau - 0.0005\sin 2\tau \\ & \quad + 0.0007\cos 3\tau + 0.0005\sin 3\tau \\ \end{aligned}$$
(7.33)
where $$\tau = \omega t,\omega = 2\pi f_{s}$$.
Comparison between the DC components from (7.33), which is obtained based on the symbolic ESPM with simplified duty cycle solution, and that of numerical simulation and Pspice simulation are shown in Table 7.2.
Table 7.2

Comparison of DC component of the CCM CMC-Boost resulted from ESPM, numeric and Pspice5 simulation

Variables

Symbolic

Numerical

Pspice

Inductor current

$$I_{dc} = 4.1327$$

$$I_{dc} = 4.1982$$

$$I_{dc} = 4.3052$$

Output voltage

$$V_{dc} = 36.0000$$

$$V_{dc} = 36.2386$$

$$V_{dc} = 36.2237$$

Duty cycle

$$d = 0.222229$$

$$d = 0.222229$$

$$d = 0.2429$$

And the comparison of the state variable ripples resulted from ESPM (dashed line) and Pspice simulation (solid line) would be shown in Fig. 7.6e, f, where the dashed lines from ESPM are the calculation of the sum of the first three harmonic components in the steady-state period solutions shown in (7.32).
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig6_HTML.gif
Fig. 7.6

Ripple comparisons of the CCM CMC-Boost: a and b for the ESPM with accurate duty cycle and the numerical simulation, c and d for the same ESPM and Pspice, e and f for the ESPM with simplified duty cycle and Pspice

7.4.1.2 Symbolic Analysis with Accurate Solution of Duty Cycle

The time-varying differential equations of closed-loop system are still expressed as (7.13) and (7.18). And the duty cycle can be expressed as the linear function of the state vector as shown in (7.4), and rewritten as $$d = K_{0} + {\mathbf{K}}_{1} \cdot {\mathbf{x}}(dT)$$, where the coefficients can be determined by the following equations according to the Eq. (7.3), i.e.,
$$\begin{aligned} K_{0} & = \frac{1}{{V_{p} }}\left( {1 + \frac{{R_{a} }}{{R_{1} }} + \frac{{R_{a} }}{{R_{2} }}} \right)V_{R} \\ {\mathbf{K}}_{1} & = \left[ {\begin{array}{*{20}c} {K_{{11}} } & {K_{{12}} } & {K_{{13}} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\frac{{ - M}}{{V_{p} }}} & {\frac{{ - R_{a} }}{{R_{1} V_{p} }}} & {\frac{{ - 1}}{{V_{p} }}} \\ \end{array} } \right] \\ \end{aligned}$$
(7.34)
where $$V_{p}$$ is the peak-to-peak value of ramp compensation signal.
As can be seen from (7.22), the zero-order component d0 of duty cycle is determined by the average value of inductor current. But actually the duty cycle is determined by the peak value of inductor current according to the peak current-mode control law. It means that when the accurate solution of duty cycle method is adopted, more components are shifted to the high-order equation to be solved. This will slow down the convergence speed of the ESP algorithm, resulting in the need to iterate more times when using ESPM to obtain a steady-state periodic solution with sufficient accuracy. In order to improve this situation, the improved double iteration algorithm mentioned in Chap. 5 is adopted. Then steady-state periodic solution, including DC component and the first three harmonics, in main circuit, can be finally obtained as
$$\begin{aligned} i_{L} & = 4.1326 - 0.3093\cos \tau + 0.3684\sin \tau - 0.1814\cos 2\tau + 0.0320\sin 2\tau \\ & \quad - 0.0623\cos 3\tau + 0.0005\sin 3\tau \\ v_{o} & = 35.9999 + 0.0025\cos \tau - 0.0049\sin \tau + 0.0020\cos 2\tau - 0.0005\sin 2\tau \\ & \quad + 0.0007os3\tau + 0.0005\sin 3\tau \\ \end{aligned}$$
(7.35)

Compare (7.33) and (7.35), it can be found that the AC components are almost the same except a small difference in DC component. The main term of duty cycle obtained in this case is $$d_{0} = 0.2222226$$, which is very close to those of accurate solution and numerical simulation, as shown in Table 7.2.

As for the ripples of the main circuit state variables during a steady-state period, the comparisons among different methods are shown in Fig. 7.6, where the solid line is for the numerical simulation or the Pspice simulation, and the dashed line is for the ESPM, which is the calculation of the sum of the first three harmonic components in the steady-state period solutions shown in (7.33) or (7.35). Figure 7.6a, c and e show the ripple waveform of inductor current, and the Fig. 7.6b, d and f show the ripple waveform of the output voltage. Figure 7.6a, b show the comparison between the ESPM algorithm under accurate solution of duty cycle and the numerical simulation, Fig. 7.6c, d show the comparison between the same ESPM algorithm and the Pspice simulation. And the comparison of the ESPM algorithm under simplified solution of duty cycle and the Pspice simulation is shown in Fig. 7.6e, f. It can be seen from all the figures that the results from the ESPM and those from simulations agree quite well with each other, regardless of whether the duty cycle is an exact solution or a simplified solution.

Therefore, even with a simplified solution for the duty cycle, a steady-state periodic solution with sufficient accuracy can be obtained using the ESPM. Since the assumptions (i)–(iii) mentioned in Sect. 7.3 can be satisfied generally in the actual circuit, the simplified method is often applied.

7.4.2 Single-Loop Current-Mode Controlled Buck Converter

The typical single-loop current-mode controlled Buck converter is shown in Fig. 7.7, and the parameters of the PCS (power circuit stage) the FN (feedback network) are set respectively as follows, i.e., E = 24 V, L = 200 μH, C = 47 μF and R = 2 Ω for the PCS; the reference current Iref = 6 A for the FN, and the switching frequency is chosen as fs = 10 kHz, and the switching cycle T = 1/fs.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig7_HTML.gif
Fig. 7.7

Current-mode controlled Buck regulator

According to the principle of the Buck converter with CCM operation, one can get the square state differential equation describing the system as shown in (7.36), where the state variable vector is chosen as $${\mathbf{x}} = [i_{L} ,v_{o} ]^{Tr}$$, and the square coefficient matrix is determined by (7.37)
$$G_{0} (p){\mathbf{x}} + {\mathbf{f}} = 0$$
(7.36)
$$G_{0} (p) = \left[ {\begin{array}{*{20}c} p & {\tfrac{1}{L}} \\ { - \tfrac{1}{C}} & {p + \tfrac{1}{RC}} \\ \end{array} } \right]$$
(7.37)
The nonlinear vector function f and the input vector u are given by
$${\mathbf{f}} = \delta {\mathbf{u}}\,{\text{and}}\,{\mathbf{u}} = \left[ {\begin{array}{*{20}c} { - \tfrac{E}{L}} & 0 \\ \end{array} } \right]^{Tr}$$
(7.38)
The solution of duty cycle is determined using simplified method mentioned in Sect. 7.4.1.1. According to Fig. 7.7, the control signal is a constant, i.e., vk = Iref. The duty cycle can be obtained by (7.5) as
$$d = \frac{{I_{\text{ref}} - \bar{i}_{l} }}{{m_{1} T/2}}$$
(7.39)
where the rising slope of the inductor current $$m_{1} = \frac{{E - \bar{v}_{o} }}{L},$$ then the duty cycle in (7.39) can be expressed as the function of the mean value of state variable, i.e.
$$d = \frac{{2Lf_{s} (I_{c} - \bar{i}_{L} )}}{{E - \bar{v}_{0} }} = g(\bar{i}_{L} ,\bar{v}_{o} )$$
(7.40)
According to equations from (7.10) to (7.12), the duty cycle in (7.40) can also be expanded into the sum of main item and small corrections. Thus, the state Eq. (7.36) can be solve by the same method with ESPM, the main term equation and each correction are shown as follows.
  1. (1)

    Solution of the Main term

     
Based on the principle of the ESPM, the main term (DC component) can be assumed as
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = \left[ {\begin{array}{*{20}c} {I_{00} } & {V_{00} } \\ \end{array} } \right]^{Tr}$$
(7.41)
where a00 can be determined by the following equations as
$$\left\{ {\begin{array}{*{20}l} {G_{1} (0){\mathbf{a}}_{00} + d_{0} {\mathbf{u}} = 0} \hfill \\ {d_{0} = \frac{{2Lf_{s} (I_{\text{ref}} - I_{00} )}}{{(E - V{}_{00})}}} \hfill \\ \end{array} } \right.$$
(7.42)

It can be seen from (7.42), for current-mode controlled Buck converter, that the solution of the main oscillation component is nonlinear (the main oscillation equation of voltage-mode controlled Buck converter is linear). Equation (7.42) can be solved by the symbolic method or numerical method. It may have multiple solutions, so it is necessary to determine which solution is true according to the situation of the actual converter. In this example, the solution can be obtained by numerical method as

$$d_{0} = 0.3542,I_{00} = 4.251,V_{00} = 8.5020.$$
  1. (2)

    Solution of the First-order Term

     
The first-order correction term can be still assumed as
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} {{e}}^{j\tau } + c.c$$
(7.43)
Since the first-order correction only contains the fundamental, its average value during one switching period is zero, and the duty cycle increment caused by it is also zero, i.e.,$${\bar{\mathbf{x}}}_{1} = 0$$ and $$d_{1} = 0$$. Then according to (7.17), the coefficient $$b_{m1} = 0$$ (m = 1, 2, 3…) can be determined, and thus the solution of a11 can be derived by the equation as shown in (7.44).
$$G{}_{0}(j\omega ){\mathbf{a}}_{11} = - b_{10} \cdot {\mathbf{u}}$$
(7.44)
Obviously, the solution of $${\mathbf{a}}_{11}$$ is linear. By solving the above matrix equation, the symbolic expressions for each element of the column vector a11 can be obtained as
$$\begin{aligned} V_{{11}} & = \frac{{b_{{10}} \cdot E}}{{1 + j\omega L/R - \omega ^{2} LC}} \\ I_{{11}} & = \left( {j\omega C + \frac{1}{R}} \right)V_{{11}} \\ \end{aligned}$$
(7.45)
  1. (3)

    Solution of the Second-order Term

     
Based on the ESPM, the second-order correction term for the Buck converter can be assumed as
$${\mathbf{x}}_{2} = {\mathbf{a}}_{22} {{e}}^{j2\tau } + {\mathbf{a}}_{32} {{e}}^{j3\tau } + c.c$$
(7.46)
It should be noted that, unlike the Boost converter, the second-order correction x2 of the Buck converter does not contain the DC component, thus its average value during one switching period is zero, and the duty cycle increment caused by x2 is also zero, i.e.,$${\bar{\mathbf{x}}}_{2} = 0$$ and $$d_{2} = 0$$. Then the solutions of a22 and a32 can be derived by the equation as shown in (7.47).
$$G_{0} (j2\omega ){\mathbf{a}}_{22} = - b_{20} {\mathbf{u}}$$
(7.47a)
$$G_{0} (j3\omega ){\mathbf{a}}_{32} = - b_{30} {\mathbf{u}}$$
(7.47b)
As can be seen in (7.47) that both of the equations are linear, and the coefficients b10, b20 and b30 can be determined by Eq. (7.17). Then by solving the equations in (7.47) one by one, the symbolic expressions for each element of the column vectors a22 and a32 can be obtained as
$$\left\{ {\begin{array}{*{20}l} {V_{22} = \frac{{b_{20} \cdot E}}{{1 + j2\omega L/R - 4\omega^{2} LC}}} \hfill \\ {I_{22} = (j2\omega C + \frac{1}{R})V_{22} } \hfill \\ \end{array} } \right.\,{\text{and}}\,\left\{ {\begin{array}{*{20}l} {V_{32} = \frac{{b_{30} \cdot E}}{{1 + j3\omega L/R - 9\omega^{2} LC}}} \hfill \\ {I_{32} = (j3\omega C + \frac{1}{R})V_{32} } \hfill \\ \end{array} } \right.$$
(7.48)
Then the steady-state periodic solution of the state variables, including DC component and the first three harmonics, can be finally obtained as
$$\begin{aligned} i_{L} & = 4.2510 - 0.6915\cos \tau - 0.4923\cos 2\tau - 0.2684\cos 3\tau \\ v_{o} & = 8.5020 - 0.4083\cos \tau - 0.0478\cos 2\tau - 0.0013\cos 3\tau \\ \end{aligned}$$
(7.49)
where $$\tau = \omega t,\omega = 2\pi f_{s}$$.
The DC solutions obtained by the ESPM and Pspice simulation are listed in Table 7.3. Obviously, the difference of the DC values from the two methods is quite small.
Table 7.3

Comparison of DC component of the CMC-Buck resulted from ESPM and Pspice5

Variables

Symbolic

Pspice

Inductor current

Idc = 4.1250 A

Idc = 4.1272 A

Output voltage

Vdc = 8.5020 V

Vdc = 8.3388 V

Duty cycle

d0 = 0.3542

d0 = 0.3475

As for the ripples of the main circuit state variables during a steady-state period, the comparisons between the two methods are shown in Fig. 7.8, where the solid line is from the Pspice simulation, and the dashed line is from the ESPM, which is the calculation of the sum of the first three harmonic components in the steady-state period solutions shown in (7.49). It can be seen from Fig. 7.8 that the results from the ESPM show good accordance with those from simulations.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig8_HTML.gif
Fig. 7.8

Comparison of ripple of the CMC-Buck between the ESPM and Pspice simulation, a inductor current, b output voltage

7.5 Steady-State Analysis of CMC-Boost in DCM Operation

Discontinuous current mode is a common operating modes in DC/DC converter. When applying traditional State-space Averaging method to analysis PWM converter in DCM, the reduced order model, which do not contain inductor current, will be obtained. But in some applications, such as PFC (Power Factor Correction) circuit, the inductor current is the ultimate control target, the disappearance of inductor current in the model is not desirable. In the previous section, the ESPM is applied to analyze the current-mode controlled CCM-operated Boost converter. In this section, the analysis method for voltage-mode controlled DCM-operated converters is further extended to the current-mode-controlled converters with DCM operation. Also the obtained steady-state solutions of DC component and ripples of state variables of this type of converter can be expressed in symbolic form.

When using the equivalent-small-parameters method, the analysis for DCM-operated converters of current-mode-control is the same as that of voltage-mode control. The only difference is that the duty ratio is determined differently. Therefore, the related analysis formulas in Chap. 6 will be directly referred to below, and for the convenience of description, some of them are re-listed as follows.

7.5.1 Description of the CMC Converter in DCM

The waveform of inductor current in DCM is shown in Fig. 7.9. According to discussion in Chap. 6, PWM converter operating in DCM can be described with a time-varying differential equation shown in (6.​1), here we rewrite it as.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig9_HTML.gif
Fig. 7.9

The waveform of inductor current in DCM

$${\mathbf{G}}_{0} (p){\mathbf{x}} + {\mathbf{G}}_{1} (p) \cdot {\mathbf{f}}^{(1)} ({\mathbf{x}}) + {\mathbf{G}}_{3} (p) \cdot {\mathbf{f}}^{(2)} ({\mathbf{x}}) = {\mathbf{u}}$$
(7.50)
And the nonlinear vector functions are defined as
$${\mathbf{f}}^{(1)} = \delta_{1} ({\mathbf{x}} + {\mathbf{e}}^{(1)} )$$
(7.51)
$${\mathbf{f}}^{(2)} = \delta_{3} ({\mathbf{x}} + {\mathbf{e}}^{(2)} )$$
(7.52)
where the meaning of the square matrices Gi(p) (i = 0, 1, 3) are similar to those in Chap. 6, and e(1), e(2) are constant vectors. Switching functions δ1, δ3 are defined as:
$$\delta_{1} (t) = \left\{ {\begin{array}{*{20}l} 1 \hfill & {0 < t \le d_{1} (t)T} \hfill \\ 0 \hfill & {d_{1} (t)T < t \le T} \hfill \\ \end{array} } \right.$$
(7.53)
$$\delta_{3} (t) = \left\{ {\begin{array}{*{20}l} 0 \hfill & {0 < t \le [1 - d_{3} (t)]T} \hfill \\ 1 \hfill & {[1 - d_{3} (t)]T < t \le T} \hfill \\ \end{array} } \right.$$
(7.54)
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 Chaps. 5 and 6, when using the ESPM to solve Eq. (7.50), the following iterative equations can be obtained.

$$\begin{array}{*{20}c} {\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} (p){\mathbf{x}}_{0} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{0m}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{f}}_{0m}^{(2)} = {\mathbf{u}}} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{1} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{1m}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{f}}_{1m}^{(2)} = - [{\mathbf{G}}_{1} (p){\mathbf{R}}_{1}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{R}}_{1}^{(2)} ]} \hfill \\ {{\mathbf{G}}_{0} (p){\mathbf{x}}_{2} + {\mathbf{G}}_{1} (p){\mathbf{f}}_{2m}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{f}}_{2m}^{(2)} = - [{\mathbf{G}}_{{\mathbf{1}}} (p){\mathbf{R}}_{2}^{(1)} + {\mathbf{G}}_{3} (p){\mathbf{R}}_{2}^{(2)} ]} \hfill \\ \end{array} } \right.} \\ { \cdots } \\ \end{array}$$
(7.55)
  1. (1)

    Determination of the duty cycle d1 and d3

     
The on/off state of switching transistor is determined by feedback compensation network. According to Chap. 6, the duty cycle d1 can be expressed as a linear function of the system’s state variables, namely:
$$d_{1} (t) = K_{0} + {\mathbf{K}}_{1} {\mathbf{x}}(d_{1} T)$$
(7.56)
Let ts = d1T represents the moment the switch acts, and d1 expand into the following series according to Chaps. 5 and 6, i.e.,
$$d_{1} = d_{10} + \varepsilon d_{11} + \varepsilon^{2} d_{12} + \cdots \varepsilon^{i} d_{1i} + \cdots$$
(7.57)
where each term d1i is determined by
$$\left\{ {\begin{array}{*{20}l} {d{}_{10} = K_{0} + {\mathbf{K}}_{1} {\mathbf{x}}_{0} } \hfill \\ {d_{11} = {\mathbf{K}}_{1} {\mathbf{x}}_{1} (t_{s0} )} \hfill \\ {d_{12} = {\mathbf{K}}_{1} \left. {\frac{{\partial {\mathbf{x}}_{1} }}{\partial t}} \right|_{{t = t_{s0} }} d_{11} T + {\mathbf{K}}_{1} {\mathbf{x}}_{2} (t_{s1} )} \hfill \\ \cdots \hfill \\ \end{array} } \right.$$
(7.58)
And K0, K1 can be determined by the specific circuit, and
$$\begin{aligned} & t_{s0} = d_{10} T \\ & t_{s1} = (d_{10} + \varepsilon d_{11} )T \\ & t_{s2} = (d_{10} + \varepsilon d_{11} + \varepsilon^{2} d_{12} )T \\ & \cdots \\ \end{aligned}$$
(7.59)
where tsi (i = 0,1,2,…) represent the moment at which the switch acts when the steady-state periodic solution only contains the terms of $${\mathbf{x}}_{0} ,\,{\mathbf{x}}_{1} ,\,{\mathbf{x}}_{2} , \ldots {\mathbf{x}}_{i}$$.
Assuming the rising slope and the dropping slop of the inductor current are m1 and m2 respectively, then According to Fig. 7.9 and the discussion in Chap. 6, the duty cycle d3 for DCM operation can be deduced, i.e.,
$$d_{3} = 1 - d_{1} - \frac{{\bar{i}_{L} - (md_{1}^{2} + I_{0} d_{1} )}}{{I_{0} + md_{1} }}$$
(7.60)
where I0 denotes the initial value of the inductor current at the beginning of each switching cycle T, and $$\bar{i}_{L}$$ means the average during one cycle, m = m1T/2.
Generally for most converters, I0 = 0, and when taking into account the circuit’s parasitic parameters, m1 and m2 usually are functions of the average input and the output, then d3 can be expressed as a function of the input, the average state variables $${\bar{\mathbf{x}}}$$ and d1, i.e.,
$$d_{3} = h_{1} (E,d_{1} ,{\bar{\mathbf{x}}})$$
(7.61)
According to the series expansion of $${\bar{\mathbf{x}}}$$ and d1, d3 is be transformed into the series expansion as follows.
$$d_{3} = d_{30} + \varepsilon d_{31} + \varepsilon^{2} d_{32} + \cdots$$
(7.62)
in which each $$d_{3i}$$ (i = 1,2,….) can be approximated by
$$\begin{aligned} d_{{30}} & = h_{1} (E,d_{{10}} ,\bar{{\mathbf{x}}} _{0} ) \\ d_{{31}} & = \left( {\left. {\frac{{\partial h_{1} }}{{\partial d_{1} }}} \right|_{\begin{subarray}{l} d_{1} = d_{{10}} \\ \bar{{\mathbf{x}}} = \bar{{\mathbf{x}}} _{0} \end{subarray} } } \right) \cdot d_{{11}} + \left( {\left. {\frac{{\partial h_{1} }}{{\partial \bar{\mathbf{x}}}}} \right|_{\begin{subarray}{l} d_{1} = d_{{10}} \\ \bar{{\mathbf{x}}} = \bar{{\mathbf{x}}} _{{0}} \end{subarray} } } \right) \cdot \bar{{\mathbf{x}}} _{1} = H_{{11}}^{1} \cdot d_{{11}} + H_{{12}}^{1} \cdot \bar{{\mathbf{x}}} _{{{1}}} \\ d_{{32}} & = \left( {\left. {\frac{{\partial h_{1} }}{{\partial d_{1} }}} \right|_{\begin{subarray}{l} d_{1} = d_{{10}} + d_{{11}} \\ \bar{{\mathbf{x}}} = \bar{{\mathbf{x}}} _{0} + \bar{{\mathbf{x}}} _{{{1}}} \end{subarray} } } \right) \cdot d_{{12}} + \left( {\left. {\frac{{\partial h_{1} }}{{\partial \bar{\mathbf{x}}}}} \right|_{\begin{subarray}{l} d_{1} = d_{{10}} + d_{{11}} \\ \bar{{\mathbf{x}}} = \bar{{\mathbf{x}}} _{{{0}}} + \bar{{\mathbf{x}}} _{{{1}}} \end{subarray} } } \right) \cdot \bar{{\mathbf{x}}} _{2} = H_{{21}}^{1} \cdot d_{{12}} + H_{{22}}^{1} \cdot \bar{{\mathbf{x}}} _{2} \\ \ldots \\ \end{aligned}$$
(7.63)
  1. (2)

    Determination of Switching functions δ1 and δ3

     
According to the analysis above, d3 and d1 have the same form of series expansion. Thus, δ1 and δ3 can be transformed into series expansion in the same way.
$$\delta^{(k)} = \delta_{0}^{(k)} + \varepsilon \delta_{1}^{(k)} + \varepsilon^{2} \delta_{2}^{(k)} + \cdots$$
(7.64)
Here, k = 1,3. According the Fourier series expansion as
$$\delta^{(k)} = b_{0}^{(k)} + \sum\limits_{m = 1}^{\infty } {[b_{m}^{(k)} \cdot e^{jm\tau } + \bar{b}_{m}^{(k)} } \cdot e^{ - jm\tau } ]$$
(7.65)
where $$\tau = 2\pi f_{s} t$$. Then each term $$\delta_{i}^{(k)}$$ can be chosen as
$$\begin{aligned} \delta_{0}^{(k)} & = d_{k0} + b_{10}^{(k)} {{e}}^{j\tau } + c.c\quad , \\ \delta_{1}^{(k)} & = d_{k1} + b_{11}^{(k)} {{e}}^{j\tau } + b_{20}^{(k)} {{e}}^{j2\tau } + b_{30}^{(k)} {{e}}^{j3\tau } + c.c\quad , \\ \delta_{2}^{(k)} & = d_{k2} + b_{12}^{(k)} {{e}}^{j\tau } + b_{21}^{(k)} {{e}}^{j2\tau } + b_{31}^{(k)} {{e}}^{j3\tau } + b_{40}^{(k)} {{e}}^{j4\tau } \, + b_{50}^{(k)} {{e}}^{j5\tau } + c.c\quad , \\ & \quad \cdots \\ \end{aligned}$$
And the coefficient $$b_{mi}^{(k)}$$ is determined by
$$\left\{ {\begin{array}{*{20}l} {b_{mi}^{(k)} = \tfrac{1}{2}(\alpha_{mi}^{(k)} - j\beta_{mi}^{(k)} )\quad \quad (m = 1,2, \cdots ,i = 0,1,2, \cdots )} \hfill \\ {\alpha_{m0}^{(k)} = \frac{{\sin 2m\pi d_{k0} }}{m\pi } ,\,\, \beta_{m0}^{(k)} = \pm \frac{{1 - \cos 2m\pi d_{k0} }}{m\pi }} \hfill \\ {\alpha_{mi}^{(k)} \approx 2d_{ki} \cos 2m\pi (d_{k0} + \varepsilon d_{k1} + \cdots + \varepsilon^{i - 1} d_{k(i - 1)} )} \hfill \\ {\beta_{mi}^{(k)} \approx \pm 2d_{i} \sin 2m\pi (d_{k0} + \varepsilon d_{k1} + \cdots + \varepsilon^{i - 1} d_{k(i - 1)} )} \hfill \\ {(i = 1,2, \cdots )} \hfill \\ \end{array} } \right.$$
(7.66)

7.5.2 Steady-State Solution of the CMC Converter with DCM Operation

For the current-mode control converter shown in Fig. 7.5, when the load resistor is chosen greater than 72.53Ω, i.e. $$R > 72.53\Omega$$, the circuit will operate in DCM. Choosing the load resistance $$R = 112\Omega$$, and all the other circuit parameters are the same as those in Table 7.1.

The differential equation to describe the system is shown in Eq. (7.50), in which the constant vectors are $${\mathbf{e}}^{(1)} = 0$$ and $${\mathbf{e}}^{(3)} = [\begin{array}{*{20}c} 0 & { - E} & {0]^{{T_{r} }} } \\ \end{array}$$, the square matrices $$G_{0} (p)$$ and $$G_{1} (p)$$ are the same as those of CCM operation, $$G_{3} (p) = G_{1} (p)$$, they are rewritten as follows.
$${\mathbf{G}}_{1} (p) = {\mathbf{G}}_{3} (p) = \left[ {\begin{array}{*{20}c} p & {\tfrac{1}{L}} & 0 \\ { - \tfrac{1}{C}} & {p + \tfrac{1}{RC}} & 0 \\ 0 & { - \tfrac{1}{{R_{1} C_{a} }}} & p \\ \end{array} } \right]\,{\text{and}}\,G_{2} (p) = \left[ {\begin{array}{*{20}c} 0 & { - \tfrac{1}{L}} & 0 \\ {\tfrac{1}{C}} & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} } \right]$$
(7.67)
The nonlinear vector functions are given by (7.51) and (7.52), and the input vector u are given by
$${\mathbf{u}} = \left[ {\begin{array}{*{20}c} {\tfrac{E}{L}} & 0 & {\tfrac{ - 1}{{C_{a} }}(\tfrac{1}{{R_{1} }} + \tfrac{1}{{R_{2} }}} \\ \end{array} )V_{R} } \right]^{Tr}$$
(7.68)
The duty cycle $$d_{1}$$ is the linear function of state vector, can be expressed as Eq. (7.56), and here it is solved by the simplified method introduced in Sect. 7.4.1, i.e.,
$$d_{1} (t) = K_{0} + {\mathbf{K}}_{1} {\bar{\mathbf{x}}}(d_{1} T)$$
(7.69)
In which the terms K0 and K1 can be determined by the Eq. (7.21), and then d1 is expanded in series as shown in (7.57), where the term d1i is determined by
$$\begin{array}{*{20}l} {\left\{ {\begin{array}{*{20}l} {d_{10} = K_{0} + {\mathbf{K}}_{1} {\bar{\mathbf{x}}}_{0} } \hfill \\ {d_{11} = {\mathbf{K}}_{1} {\bar{\mathbf{x}}}_{1} } \hfill \\ {d_{12} = {\mathbf{K}}_{1} {\bar{\mathbf{x}}}_{2} } \hfill \\ \end{array} } \right.} \hfill \\ {\quad \cdots } \hfill \\ \end{array}$$
(7.70)
Then according to Eq. (7.60), d3 is determined by:
$$d_{3} = 1 - {\mathbf{K}}_{11} {\bar{\mathbf{x}}}/(md_{1} )$$
(7.71)
where $${\mathbf{K}}_{11} = [\begin{array}{*{20}c} 1 & 0 & {0]} \\ \end{array} ,m = (E \cdot T)/(2L)$$.
The duty cycle d3 can be expanded as the series shown in (7.62), in which each term is defined by according to (7.63), i.e.,
$$\begin{aligned} d_{30} & = 1 - {\mathbf{K}}_{11} {\bar{\mathbf{x}}}_{0} /(md_{10} ) \\ d_{31} & = \frac{{{\mathbf{K}}_{11} {\bar{\mathbf{x}}}_{0} }}{{(md_{10} )^{2} }} \cdot d_{11} - \frac{{{\mathbf{K}}_{11} }}{{(md_{10} )}} \cdot {\bar{\mathbf{x}}}_{1} = H_{11}^{1} \cdot d_{11} + H_{12}^{1} \cdot {\bar{\mathbf{x}}}_{{\mathbf{1}}} \\ d_{32} & = \frac{{{\mathbf{K}}_{11} ({\bar{\mathbf{x}}}_{0} + {\bar{\mathbf{x}}}_{0} )}}{{m^{2} (d_{10} + d_{11} )^{2} }} \cdot d_{12} - \frac{{{\mathbf{K}}_{11} }}{{m(d_{10} + d_{11} )}} \cdot {\bar{\mathbf{x}}}_{2} = H_{21}^{1} \cdot d_{12} + H_{22}^{1} \cdot {\bar{\mathbf{x}}}_{2} \\ & \cdots \cdots \\ \end{aligned}$$
(7.72)
  1. (1)

    Solution of the Main Component

     
First, the main wave x0 to the steady-state value of state variable x is chosen as
$${\mathbf{x}}_{0} = {\mathbf{a}}_{00} = [\begin{array}{*{20}c} {I_{00} } & {V_{00} } & {V_{f00} } \\ \end{array} ]^{Tr}$$
(7.73)
It can be obtained by combining the following equations.
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{{\mathbf{0}}} (0){\mathbf{a}}_{00} + {\mathbf{G}}_{1} (0)d_{10} {\mathbf{a}}_{00} + {\mathbf{G}}_{3} (0)d_{30} ({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} ) = {\mathbf{u}}} \hfill \\ {d_{10} = K_{0} + {\mathbf{K}}_{1} {\mathbf{a}}_{00} } \hfill \\ {d_{30} = 1 - {\mathbf{K}}_{11} {\mathbf{a}}_{00} /(md_{10} )} \hfill \\ \end{array} } \right.$$
(7.74)
Obviously, the Eq. (7.73) is nonlinear and need to be solved by symbolic method or numerical method.
  1. (2)

    Solution of the First-order Correction

     
The first-order correction can be assumed as
$${\mathbf{x}}_{1} = {\mathbf{a}}_{11} {{e}}^{j\tau } + {\bar{\mathbf{a}}}_{11} {{e}}^{ - j\tau }$$
(7.75)
where the amplitude coefficient a11 can be found by solving the following equation by applying the ESPM.
$$\begin{aligned} & [{\mathbf{G}}{}_{0}(j\omega ) + {\mathbf{G}}_{1} (j\omega )d_{10} + {\mathbf{G}}_{3} (j\omega )d_{30} ]{\mathbf{a}}_{11} \\ & = - {\mathbf{G}}_{1} (j\omega )(b_{11}^{(1)} + b_{10}^{(1)} ){\mathbf{a}}_{00} - {\mathbf{G}}_{3} (j\omega )[(b_{11}^{(3)} + b_{10}^{(3)} )({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} )] \\ \end{aligned}$$
(7.76)
The Eq. (7.76) is linear. Let $${\mathbf{G}}_{1} = {\mathbf{G}}{}_{0}(j\omega ) + {\mathbf{G}}_{1} (j\omega )d_{10} + {\mathbf{G}}_{3} (j\omega )d_{30}$$, the solution for (7.76) can be obtained by
$${\mathbf{a}}_{11} = - {\mathbf{G}}^{ - 1} \{ {\mathbf{G}}_{1} (j\omega )(b_{11}^{(1)} + b_{10}^{(1)} ){\mathbf{a}}_{00} - {\mathbf{G}}_{3} (j\omega )[(b_{11}^{(3)} + b_{10}^{(3)} )({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} )]\}$$
(7.77)

Here, the superscript “−1” indicates the inversion matrix.

Furthermore, we can define the following symbols as
$$\begin{aligned} & \tau_{s0} = 2\pi d_{10} ,\sigma_{s0} = 2\pi d_{30} ,{\mathbf{A}}_{12} = - {\mathbf{G}}^{ - 1} {\mathbf{G}}_{1} (j\omega ),{\mathbf{A}}_{13} = - {\mathbf{G}}^{ - 1} {\mathbf{G}}_{3} (j\omega ) \\ & {\mathbf{B}}_{10} = {\mathbf{A}}_{12} {\mathbf{b}}_{10}^{(1)} \cdot {{e}}^{{j\tau_{s0} }} ,{\mathbf{Q}}_{10} = {\mathbf{A}}_{13} b_{10}^{(3)} \cdot {{e}}^{{j\sigma_{s0} }} ,{\mathbf{F}} = {\mathbf{H}}_{11}^{1} \cdot {{e}}^{{j(\tau_{s0} + \sigma_{s0} )}} \\ & {\mathbf{A}}_{1} = 1 - {\mathbf{K}}_{1} [({\mathbf{A}}_{12} + {\bar{\mathbf{A}}}_{12} ){\mathbf{a}}_{00} + ({\mathbf{A}}_{13} {\mathbf{F}} + {\bar{\mathbf{A}}}_{13} {\bar{\mathbf{F}}})({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} )] \\ & {\mathbf{B}}_{1} = {\mathbf{K}}_{1} [({\mathbf{B}}_{10} + {\bar{\mathbf{B}}}_{10} ){\mathbf{a}}_{00} + ({\mathbf{Q}}_{10} + {\bar{\mathbf{Q}}}_{10} )({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} )], \\ \end{aligned}$$
Then the first-order correction d11 of the duty cycle d1 can be obtained as
$$d_{11} = {\mathbf{B}}_{{\mathbf{1}}} /{\mathbf{A}}_{1}$$
(7.78)
  1. (3)

    Solution of the Second-order Correction

     
Next, according to process of the ESPM, the second-order correction can be assumed as
$${\mathbf{x}}_{2} = {\mathbf{a}}_{02} + ({\mathbf{a}}_{22} {{e}}^{j2\tau } + {\mathbf{a}}_{32} {{e}}^{j3\tau } + c.c)$$
(7.79)
And then based on the third equation in (7.55), the equations with DC component, the second-order harmonic and the third-order harmonic in the second-order correction can be obtained:
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{a}}_{22} = - {\mathbf{G}}_{2}^{ - 1} \cdot ({\mathbf{B}}_{22} + {\mathbf{Q}}_{22} )} \hfill \\ {{\mathbf{a}}_{32} = - {\mathbf{G}}_{3}^{ - 1} \cdot ({\mathbf{B}}_{32} + {\mathbf{Q}}_{32} )} \hfill \\ {{\mathbf{G}}_{0} a_{02} + {\mathbf{G}}_{1} (0){\mathbf{a}}_{00} d_{12} + {\mathbf{G}}_{3} (0)({\mathbf{a}}_{00} + e^{(3)} )d_{32} = - ({\mathbf{B}}_{02} + {\mathbf{Q}}_{02} )} \hfill \\ \end{array} } \right.$$
(7.80)
where the symbolic expressions are defined as follows.
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{2} = {\mathbf{G}}_{0} (j2\omega ) + {\mathbf{G}}_{1} (j2\omega ) \cdot d_{10} + {\mathbf{G}}_{3} (j2\omega ) \cdot d_{30} } \hfill \\ {{\mathbf{B}}_{22} = {\mathbf{G}}_{1} (j2\omega )[(b_{10}^{(1)} + b_{11}^{(1)} ){\mathbf{a}}_{11} + (b_{20}^{(1)} + b_{21}^{(1)} ){\mathbf{a}}_{00} + b_{30}^{(1)} {\bar{\mathbf{a}}}_{11} ]} \hfill \\ {{\mathbf{Q}}_{22} = {\mathbf{G}}_{3} (j2\omega )[(b_{10}^{(3)} + b_{11}^{(3)} ){\mathbf{a}}_{11} + (b_{20}^{(3)} + b_{21}^{(3)} )({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} ) \, + b_{30}^{(3)} {\bar{\mathbf{a}}}_{11} ]} \hfill \\ \end{array} } \right.$$
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{3} = {\mathbf{G}}_{0} (j3\omega ) + {\mathbf{G}}_{1} (j3\omega ) \cdot d_{10} + {\mathbf{G}}_{3} (j3\omega ) \cdot d_{30} } \hfill \\ {{\mathbf{B}}_{32} = {\mathbf{G}}_{1} (j3\omega )[b_{10}^{(1)} a_{22} + b_{20}^{(1)} {\mathbf{a}}_{11} + (b_{30}^{(1)} + b_{31}^{(1)} ){\mathbf{a}}_{00} ]} \hfill \\ {{\mathbf{Q}}_{32} = {\mathbf{G}}_{3} (j3\omega )[b_{10}^{(3)} a_{22} + b_{20}^{(3)} {\mathbf{a}}_{11} + (b_{30}^{(3)} + b_{31}^{(3)} )(a_{00} + {\mathbf{e}}^{(3)} )]} \hfill \\ \end{array} } \right.$$
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{G}}_{0} = G_{0} (0) + {\mathbf{G}}_{1} (0)d_{10} + {\mathbf{G}}_{3} (0)d_{30} } \hfill \\ {{\mathbf{B}}_{02} = {\mathbf{G}}_{1} (0)[(b_{10}^{(1)} + b_{11}^{(1)} ){\bar{\mathbf{a}}}_{11} + (\bar{b}_{10}^{(1)} + \bar{b}_{11}^{(1)} ){\mathbf{a}}_{11} + d_{11} {\mathbf{a}}_{00} ]} \hfill \\ {{\mathbf{Q}}_{02} = {\mathbf{G}}_{3} (0)[(b_{10}^{(3)} + b_{11}^{(3)} ){\bar{\mathbf{a}}}_{11} + (\bar{b}_{10}^{(3)} + \bar{b}_{11}^{(3)} ){\mathbf{a}}_{11} + d_{31} ({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} )]} \hfill \\ \end{array} } \right.$$
It can be seen from Eq. (7.80) that the solutions of a22 and a32 are linear, yet the solution of a02 is nonlinear. Assume that
$$\left\{ {\begin{array}{*{20}l} {{\mathbf{P}}_{11} = {\mathbf{a}}_{11} e^{{j\tau_{s0} }} ,{\mathbf{P}}_{22} = {\mathbf{a}}_{22} e^{{j\tau_{s1} }} ,{\mathbf{P}}_{32} = {\mathbf{a}}_{32} e^{{j\tau_{s1} }} ,} \hfill \\ {(\tau_{s0} = 2\pi d_{10} ,\tau_{s1} = 2\pi (d_{10} + d_{11} ))} \hfill \\ {{\mathbf{M}} = - {\mathbf{K}}_{1} \cdot 4\pi d_{11} \text{Im} ({\mathbf{P}}_{11} ) + 2{\mathbf{K}}_{1} [\text{Re} ({\mathbf{P}}_{22} ) + \text{Re} ({\mathbf{P}}_{32} )]} \hfill \\ \end{array} } \right.$$
(7.81)
in which $$\text{Im} ( \cdot ),\text{Re} ( \cdot )$$ represent the imaginary part and the real part of conjugate complex respectively.

According to Eq. (7.70) and (7.72), $$d_{12} = K_{1} {\mathbf{a}}_{02} ,\,d_{32} = H_{21}^{1} \cdot d_{12} + H_{22}^{1} \cdot {\mathbf{a}}_{02}$$. Then introducing d12 and d32 into the equation for solving a02 in (7.80) gives $${\mathbf{a}}_{02} = {\mathbf{B}}_{2} /{\mathbf{A}}_{2}$$,

in which
$$\begin{aligned} {\mathbf{A}}_{2} & = {\mathbf{G}}_{0} + {\mathbf{G}}_{10} (0){\mathbf{a}}_{00} {\mathbf{K}}_{1} - {\mathbf{G}}_{3} (0)({\mathbf{a}}_{00} + {\mathbf{e}}^{(3)} )(H_{21}^{1} {\mathbf{K}}_{1} + H_{22}^{1} ) \\ {\mathbf{B}}_{2} & = {\mathbf{B}}_{02} + {\mathbf{Q}}_{02} - {\mathbf{G}}_{1} (0){\mathbf{a}}_{00} M - {\mathbf{G}}_{3} (0)({\mathbf{a}}_{00} + e^{(3)} )H_{2} M \\ \end{aligned}$$
Finally, with all the circuit parameters being introduced in, the steady-state periodic solution of state variable in main circuit, including DC component and the first three harmonics, can be obtained as
$$\begin{aligned} i_{L} & = 0.4249 - 0.1939\cos \tau + 0.4153\sin \tau - 0.1308\cos 2\tau \\ & \quad + 0.0852\sin 2\tau \, - 0.0616\cos 3\tau - 0.0195\sin 3\tau \\ v_{o} & = 36.0132 - 0.0008\cos \tau - 0.0013\sin \tau + 0.0002\cos 2\tau \\ & \quad { - }0.0002\sin 2\tau \, + 0.00008\cos 3\tau - 0.00008\sin 3\tau \\ \end{aligned}$$
(7.82)
The DC components and duty cycles d1 and d3 from the symbolic ESPM with simplified duty cycle solution, and that of Pspice simulation are shown in Table 7.4, and the comparison of the ripples of the state variables in the main circuit between the ESPM and the Pspice simulation is shown in Fig. 7.10.
../images/419194_1_En_7_Chapter/419194_1_En_7_Fig10_HTML.gif
Fig. 7.10

The comparison of ripples of the DCM CMC-Boost from ESPM (dashed line) and Pspice (solid line)

It can be seen from Fig. 7.10 and Table 7.4 that both the DC components and the AC components from the ESPM are consistent with those from simulation.
Table 7.4

Comparison of DC component of the DCM CMC-Boost resulted from ESPM and Pspice

Variables

Symbolic

Pspice

Inductor current

Idc = 0.4249 A

Idc = 0.4203 A

Output voltage

Vdc = 36.0132 V

Vdc = 36.0031 V

Duty cycle

d1 = 0.1876

d3 = 0.1570

d1 = 0.19

d3 = 0.21

7.6 Summary

Current-mode control scheme has been widely used in practice because it has more advantages than duty cycle control scheme. However, due to the double closed loops (current loop and voltage loop) introduced in current-mode control, the duty cycle cannot be expressed explicitly, and thus it is difficult to model and analyze current-mode controlled system. At present, the analysis of current-mode control converter basically adopts the average modelling method based on inductor current waveform analysis. Its premise is that the waveform of inductor current is linear and the assumption of small ripples can be satisfied, and there is no correlation to the analysis of the ripple of the state variable.

For the analysis of closed-loop converter system, the determination of duty cycle is critical. When using the equivalent-small-parameter method (ESPM) to analysis the current-mode controlled converters, two methods for determining the duty cycle are presented in this paper, one is the accurate solution of duty cycle, and the other is the simplified solution of duty cycle. The former would be found wider applicability, as it does not need to satisfy the assumptions of small ripples and linear waveform with inductor current. The results are all analytical expressions, which can help to understand the working mechanism of the circuit. The examples and simulation results confirm that the proposed two methods are of high accuracy.

Two methods have the same manner as the ESPM of the duty-cycle controlled PWM converter mentioned in Chaps. 5 and 6. Thus, current-mode controlled, as well as the duty-cycle controlled converter system, can be analyzed uniformly with the same mathematical model using ESPM.