Classical Averaged Model

Abstract

This chapter deals with methodologies of obtaining the so-called averaged model, which focuses on capturing the low-frequency behavior of power electronic converters while neglecting high-frequency variations due to circuit switching. This appears to be a natural action, as every converter employs filters in order to limit the ripple of various variables. The result is a continuous-time model, one which is easier to handle by classical analysis and control formalisms.

Problems

Problems 4.1, 4.2, and 4.3 are given with solutions. Problems 4.4, 4.5, and 4.6 are left as exercises.

Problem 4.1

Flyback converter

Let us consider the flyback converter in Fig. 4.24a, allowing isolated non-inverting boost topology via a transformer with ratio n.

Fig. 4.24

Flyback converter: (a) electrical circuit; (b) electrical circuit including model of the transformer

It is supposed that the transformer has negligible primary and secondary winding resistances and leakage inductances, whereas it has nonzero core reluctance. These assumptions lead to transformer modeling as a “two-winding inductor”, i.e., represented by a magnetizing inductance referred to the primary winding coupled with an ideal transformer, as in Fig. 4.24b (Erikson and Maksimović 2001). It is required to address the following points.

• (a) Obtain the averaged model and the corresponding equivalent diagram by using Algorithm 4.1.

• (b) Compute the steady-state model.

• (c) Considering that input voltage E varies (representing a disturbance input), deduce the small-signal state-space model and draw the associated equivalent diagram.

• (d) Using the previously obtained small-signal model, get expressions for the transfer functions representing the influence from the duty ratio (control input) and the input voltage E (disturbance input) to the state variables.

Solution

(a) The averaged model is deduced from the switched model. This latter results from an analysis of the circuit operation; it has two configurations: switch H turned on and diode D blocked (switching function u takes value 1), and switch H turned off and diode D conducting (switching function u takes value 0) (see Fig. 4.25).

Fig. 4.25

Two configurations of flyback converter, corresponding respectively to the two states of the switch network composed of H and D: (a) H turned on and D blocked (u = 1); (b) H turned off and D in conduction (u = 0)

When the diode is blocked (u = 1), the transformer’s primary and secondary currents are zero. When switch H is turned off (u = 0) the magnetizing inductor is decoupled from the input source, the primary voltage is v _C /n and the secondary current is i _L /n. Hence, the system has two state variables, current i _L and voltage v _C , whose dynamics are described by the following relations:

$$ \begin{array}{ll}u=1:\left\{\begin{array}{l}L\frac{d{i}_L}{ dt}=E\\ {}C\frac{d{v}_C}{ dt}=-\frac{v_C}{R}\end{array}\right.\hfill & \kern2em u=0:\left\{\begin{array}{l}L\frac{d{i}_L}{ dt}=-\frac{v_C}{n}\\ {}C\frac{d{v}_C}{ dt}=\frac{i_L}{n}-\frac{v_C}{R}.\end{array}\right.\hfill \end{array} $$

(4.41)

Equations (4.41) can be expressed by a single set, which represents the switched model of the flyback converter:

$$ \left\{\begin{array}{l}L\cdot \overset{\cdotp }{i_L}=-\left(1-u\right)\cdot \frac{v_C}{n}+u\cdot E\hfill \\ {}C\cdot \overset{\cdotp }{v_C}=\left(1-u\right)\cdot \frac{i_L}{n}-\frac{v_C}{R}.\hfill \end{array}\right. $$

(4.42)

The averaged model results directly from model (4.42) by replacing switching function u by its average, i.e., the duty ratio denoted by α. The equivalent diagram of the averaged model can be drawn as shown in Fig. 4.26. The three subcircuits are linked by two couplings, each of which acts as an AC + DC transformer, with ratio α (input voltage side) and (1 − α)/n (load side).

Fig. 4.26

Equivalent diagram of flyback converter averaged model

(b) The steady-state model results by zeroing the derivatives in Eq. (4.42). Equilibrium values of the state variables are denoted by subscript e:

$$ \begin{array}{cc}\hfill {v}_{Ce}=n\cdot \frac{E_e\cdot {\upalpha}_e}{1-{\upalpha}_e},\hfill & \hfill {i}_{Le}={n}^2\cdot \frac{\upalpha_e}{{\left(1-{\upalpha}_e\right)}^2}\cdot \frac{E_e}{R}\hfill \end{array}. $$

(4.43)

The first equation of (4.43) shows that the output voltage has expression similar to that of the buck-boost case, except it is positive and contains a supplementary multiplying factor n.

(c) The small-signal model may result from perturbation and linearization. Let us replace the perturbed input and state variables into Eq. (4.42): $\upalpha ={\upalpha}_e+\tilde{\upalpha}$, $E={E}_e+\tilde{E}$, ${i}_L={i}_{Le}+\tilde{i_L}$ and ${v}_C={v}_{Ce}+\tilde{v_C}$. By using (4.43) and neglecting products of small variations, one obtains, after some simple algebra:

$$ \left\{\begin{array}{ll}L\cdot \overset{\cdotp }{\tilde{i_L}}=\hfill & \kern-5.9em -\frac{1-{\upalpha}_e}{n}\cdot \tilde{v_C}+\left(\frac{v_{Ce}}{n}+E\right)\cdot \tilde{\upalpha}+{\upalpha}_e\cdot \tilde{E}\hfill \\ {}C\cdot \overset{\cdotp }{\tilde{v_C}}=\frac{1-{\upalpha}_e}{n}\cdot \tilde{i_L}\hfill & \kern0.5em -\frac{1}{ RC}\cdot \tilde{v_C}\kern1em -\frac{i_{Le}}{n}\cdot \tilde{\upalpha}.\hfill \end{array}\right. $$

(4.44)

These lead to the small-signal AC equivalent diagram in Fig. 4.27.

Fig. 4.27

Small-signal AC equivalent diagram of flyback converter

Equations (4.44) allow state-space matrix representation as in Eq. (4.45), emphasizing the small-signal state vector $\mathbf{x}={\left[\begin{array}{cc}\hfill \tilde{i_L}\hfill & \hfill \tilde{v_C}\hfill \end{array}\right]}^T$, the small-signal input vector $\mathbf{u}={\left[\begin{array}{cc}\hfill \tilde{\upalpha}\hfill & \hfill \tilde{E}\hfill \end{array}\right]}^T$, the state matrix A and the input matrix B:

$$ \underset{\overset{\cdotp }{\mathbf{x}}}{\underbrace{\left[\begin{array}{c}\hfill \overset{\cdotp }{\tilde{i_L}}\hfill \\ {}\hfill \overset{\cdotp }{\tilde{v_C}}\hfill \end{array}\right]}}=\underset{\mathbf{A}}{\underbrace{\left[\begin{array}{cc}\hfill 0\hfill & \hfill -\frac{1-{\upalpha}_e}{ nL}\hfill \\ {}\hfill \frac{1-{\upalpha}_e}{ nC}\hfill & \hfill -\frac{1}{ RC}\hfill \end{array}\right]}}\cdot \underset{\mathbf{x}}{\underbrace{\left[\begin{array}{c}\hfill \tilde{i_L}\hfill \\ {}\hfill \tilde{v_C}\hfill \end{array}\right]}}+\underset{\mathbf{B}}{\underbrace{\left[\begin{array}{cc}\hfill \frac{1}{L}\cdot \left(\frac{v_{Ce}}{n}+{E}_e\right)\hfill & \hfill \frac{\upalpha_e}{L}\hfill \\ {}\hfill -\frac{i_{Le}}{ nC}\hfill & \hfill 0\hfill \end{array}\right]}}\cdot \underset{\mathbf{u}}{\underbrace{\left[\begin{array}{c}\hfill \tilde{\upalpha}\hfill \\ {}\hfill \tilde{E}\hfill \end{array}\right]}}. $$

(4.45)

(d) Setting the output vector to be identical to the state vector, i.e., y ≡ x, results in output matrix C being the 2 × 2 identity matrix.

Based upon the matrix state representation in (4.45) and on the definition of the output matrix C, one can compute the transfer matrix H(s), containing the transfer functions of the four input-to-output channels. To this end, the following well-known formula can be used:

$$ \mathbf{H}(s)=\mathbf{C}\cdot {\left(s\mathbf{I}-\mathbf{A}\right)}^{-1}\cdot \mathbf{B}. $$

After performing the computation and knowing that element H _ij (s) of matrix H(s) represents the Laplace image of the transfer from input j to output i, the expression of the transfer matrix can be written as

$$ \mathbf{H}(s)=\left[\begin{array}{ll}{H}_{\upalpha \to {i}_L}(s)\hfill & {H}_{E\to {i}_L}(s)\hfill \\ {}{H}_{\upalpha \to {v}_C}(s)\hfill & {H}_{E\to {v}_C}(s)\hfill \end{array}\right], $$

(4.46)

where the transfer functions are

$$ \left\{\begin{array}{ll}{H}_{\upalpha \to {i}_L}(s)={k}_{\upalpha {i}_L}\cdot \frac{T_{\upalpha {i}_L}s+1}{s\left({T}_0s+1\right)}\hfill & {H}_{E\to {i}_L}(s)=\frac{k_{E{i}_L}}{s}\hfill \\ {}{H}_{\upalpha \to {v}_C}(s)={k}_{\upalpha {v}_C}\cdot \frac{1-{T}_{\upalpha {v}_C}s}{s\left({T}_0s+1\right)}\hfill & {H}_{E{v}_C}(s)=\frac{k_{E{v}_C}}{s\left({T}_0s+1\right)},\hfill \end{array}\right. $$

(4.47)

with the different gain and time constant notations standing for

$$ \left\{\begin{array}{ll}{T}_0= RC\hfill & \hfill \\ {}{k}_{\upalpha {i}_L}=\frac{v_{Ce}+n{E}_e+\left(1-{\upalpha}_e\right)\cdot {i}_{Le}/n}{ nL}\hfill & {T}_{\alpha {i}_L}=\frac{ RC\left({v}_{Ce}+n{E}_e\right)}{v_{Ce}+n{E}_e+\left(1-{\upalpha}_e\right)\cdot {i}_{Le}/n}\hfill \\ {}{k}_{\upalpha {v}_C}=\frac{\left(1-{\upalpha}_e\right)\cdot \left({v}_{Ce}/n+{E}_e\right)}{n\cdot \left(L/R\right)}\hfill & {T}_{\upalpha {v}_C}=\frac{i_{Le}L}{\left(1-{\upalpha}_e\right)\cdot \left({v}_{Ce}/n+{E}_e\right)}\hfill \\ {}{k}_{E{i}_L}=\frac{\upalpha_e}{L}\hfill & {k}_{E{v}_C}=\frac{\upalpha_e\left(1-{\upalpha}_e\right)}{n\cdot \left(L/R\right)}.\hfill \end{array}\right. $$

(4.48)

Equations (4.47) and (4.48) indicate that the small-signal model is a linear-parameter-varying one because its parameters depend on the steady-state operating point, as expected. Note also that all the four transfer functions have a pole at the origin. As in the case of the buck-boost converter, the capacitor voltage exhibits nonminimum-phase behavior in response to duty ratio variation, as shown by the associated transfer function ${H}_{\upalpha \to {v}_C}(s)$ having right-half-plane zero (see Eq. (4.47)). Based upon expressions (4.47) of the transfer functions, one can draw the corresponding frequency responses in the form of Bode diagrams.

Problem 4.2

Single-ended primary-inductor converter (SEPIC)

Let us consider the single-ended primary-inductor converter (SEPIC) in Fig. 4.28, allowing noninverting up/down voltage conversion using two uncoupled inductors. The circuit is driven by a single binary switching function and hence has two configurations (Sira-Ramírez and Silva-Ortigoza 2006).

Fig. 4.28

Electrical circuit of SEPIC

The following points must be addressed.

• (a) Obtain the averaged model and the corresponding equivalent diagram.

• (b) Compute the steady-state model.

• (c) Assuming that input voltage E varies (being a disturbance input), deduce the small-signal state-space model and draw the associated equivalent diagram.

• (d) Using the previously obtained small-signal model, get the expression of the transfer function representing the influence from the duty ratio (control input) to the voltage v _C2 as output variable.

Solution

(a) The averaged model is deduced from the switched model, which results from analyzing the two configurations of the circuit: for switch H turned on and diode D blocked (switching function u takes value 1 – Fig. 4.29a), and for switch H turned off and diode D conducting (switching function u takes value 0 – Fig. 4.29b).

Fig. 4.29

The two configurations of the SEPIC power stage, corresponding respectively to the two states of the switch H: (a) H turned on (u = 1); (b) H turned off (u = 0)

The circuit is described by four state variables, the two inductor currents i _L1 and i _L2, and the two capacitor voltages v _C1 and v _C2. Operation of the two configurations can be merged to yield the switched model given in Eq. (4.49).

$$ \left\{\begin{array}{l}{L}_1\overset{\cdotp }{i_{L1}}=-\left(1-u\right)\cdot \left({v}_{C1}+{v}_{C2}\right)+E\hfill \\ {}{C}_1\overset{\cdotp }{v_{C1}}=\left(1-u\right)\cdot {i}_{L1}-u\cdot {i}_{L2}\hfill \\ {}{L}_2\overset{\cdotp }{i_{L2}}=u\cdot {v}_{C1}-\left(1-u\right)\cdot {v}_{C2}\hfill \\ {}{C}_2\overset{\cdotp }{v_{C2}}=\left(1-u\right)\cdot \left({i}_{L1}+{i}_{L2}\right)-\frac{v_{C2}}{R}\hfill \end{array}\right. $$

(4.49)

The averaged model results directly from model (4.49), in which one replaces the switching function u by its average, i.e., the duty ratio denoted by α. The equivalent diagram of the averaged model is shown in Fig. 4.30.

Fig. 4.30

Equivalent diagram of SEPIC averaged model

(b) By zeroing the derivatives in Eq. (4.49) one obtains the algebraic relations allowing computation of the steady-state values of state variables (marked with subscript e) as follows:

$$ \begin{array}{cccc}\hfill {i}_{L1e}={\left(\frac{\upalpha_e}{1-{\upalpha}_e}\right)}^2\cdot \frac{E_e}{R},\hfill & \hfill {v}_{C1e}={E}_e,\hfill & \hfill {i}_{L2e}=\frac{\upalpha_e}{1-{\upalpha}_e}\cdot \frac{E_e}{R},\hfill & \hfill {v}_{C2e}=\frac{\upalpha_e}{1-{\upalpha}_e}\cdot {E}_e\hfill \end{array}. $$

(4.50)

Equations (4.50) show that the steady-state value of v _C1 equals the input voltage E, with the steady-state output voltage v _C2e corresponding to the noninverting buck-boost topology.

(c) The perturb-and-linearize method may be used to get the small-signal model. Thus, the perturbed input and state variables $\upalpha ={\upalpha}_e+\tilde{\upalpha}$, $E={E}_e+\tilde{E}$, ${i}_{L1}={i}_{L1e}+\tilde{i_{L1}}$, ${v}_{C1}={v}_{C1e}+\tilde{v_{C1}}$, ${i}_{L2}={i}_{L2e}+\tilde{i_{L2}}$ and ${v}_{C2}={v}_{C2e}+\tilde{v_{C2}}$ are replaced in Eq. (4.49). After computations, by using Eq. (4.50) and neglecting products of small variations, one obtains Eq. (4.51), whose associated equivalent diagram is given in Fig. 4.31.

Fig. 4.31

Small-signal equivalent diagram of SEPIC

$$ \left\{\begin{array}{c}{L}_1\overset{\cdotp }{\tilde{i_{L1}}}=\begin{array}{llllll}\hfill & \kern1.30em -\left(1-{\upalpha}_e\right)\tilde{v_{C1}}\hfill & \hfill & \kern1.30em -\left(1-{\upalpha}_e\right)\tilde{v_{C2}}\hfill & +\left({v}_{C1e}+{v}_{C2e}\right)\tilde{\upalpha}\hfill & +\tilde{E}\hfill \end{array}\\ {}{C}_1\overset{\cdotp }{\tilde{v_{C1}}}=\begin{array}{llllll}\left(1-{\upalpha}_e\right)\tilde{i_{L1}}\hfill & \hfill & \kern2.5em -{\upalpha}_e\tilde{i_{L2}}\hfill & \hfill & \kern5.75em +\left({i}_{L1e}+{i}_{L2e}\right)\tilde{\upalpha}\hfill & \hfill \end{array}\\ {}{L}_2\overset{\cdotp }{\tilde{i_{L2}}}=\begin{array}{llllll}\hfill & \kern3.75em {\upalpha}_e\tilde{v_{C1}}\hfill & \hfill & \kern4.25em -\left(1-{\upalpha}_e\right)\tilde{v_{C2}}\hfill & +\left({v}_{C1e}+{v}_{C2e}\right)\tilde{\upalpha}\hfill & \hfill \end{array}\\ {}{C}_2\overset{\cdotp }{\tilde{v_{C2}}}=\begin{array}{llllll}\left(1-{\upalpha}_e\right)\tilde{i_{L1}}\hfill & \hfill & \kern1.75em +\left(1-{\upalpha}_e\right)\tilde{i_{L2}}\hfill & -\left(1/R\right)\tilde{v_{C2}}\hfill & +\left({i}_{L1e}+{i}_{L2e}\right)\tilde{\upalpha}\hfill & \hfill \end{array}\end{array}\right. $$

(4.51)

Equations (4.51) can further be put into the matrix form:

$$ \overset{\cdotp }{\mathbf{x}}=\mathbf{A}\cdot \mathbf{x}+\mathbf{B}\cdot \mathbf{u}, $$

(4.52)

where $\mathbf{x}={\left[\begin{array}{cccc}\hfill \tilde{i_{L1}}\hfill & \hfill \tilde{i_{L2}}\hfill & \hfill \tilde{v_{C1}}\hfill & \hfill \tilde{v_{C2}}\hfill \end{array}\right]}^T$ is the small-signal state vector, $\mathbf{u}={\left[\begin{array}{cc}\hfill \tilde{\upalpha}\hfill & \hfill \tilde{E}\hfill \end{array}\right]}^T$ is the small-signal input vector and state matrix A and input matrix B are

$$ \left\{\begin{array}{l}\mathbf{A}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill -\left(1-{\upalpha}_e\right)/{L}_1\hfill & \hfill 0\hfill & \hfill -\left(1-{\upalpha}_e\right)/{L}_1\hfill \\ {}\hfill \left(1-{\upalpha}_e\right)/{C}_1\hfill & \hfill 0\hfill & \hfill -{\upalpha}_e/{C}_1\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill {\upalpha}_e/{L}_2\hfill & \hfill 0\hfill & \hfill -\left(1-{\upalpha}_e\right)/{L}_2\hfill \\ {}\hfill \left(1-{\upalpha}_e\right)/{C}_2\hfill & \hfill 0\hfill & \hfill \left(1-{\upalpha}_e\right)/{C}_2\hfill & \hfill -1/\left(R{C}_2\right)\hfill \end{array}\right]\\ {}\mathbf{B}=\left[\begin{array}{cc}\hfill \left({v}_{C1e}+{v}_{C2e}\right)/{L}_1\hfill & \hfill 1/{L}_1\hfill \\ {}\hfill -\left({i}_{L1e}+{i}_{L2e}\right)/{C}_1\hfill & \hfill 0\hfill \\ {}\hfill \left({v}_{C1e}+{v}_{C2e}\right)/{L}_2\hfill & \hfill 0\hfill \\ {}\hfill -\left({i}_{L1e}+{i}_{L2e}\right)/{C}_2\hfill & \hfill 0\hfill \end{array}\right].\end{array}\right. $$

(4.53)

Relations (4.52) and (4.53) define the SEPIC small-signal state-space model.

(d) Using the matrix state representation in (4.52), the matrix definitions in (4.53) and the definition of the output matrix C as the four-by-four identity matrix, one can compute the transfer matrix H(s) as

$$ \mathbf{H}(s)=\mathbf{C}\cdot {\left(s\mathbf{I}-\mathbf{A}\right)}^{-1}\cdot \mathbf{B}, $$

(4.54)

which contains the transfer functions of the eight input-to-output channels. Applying relation (4.54) requires in this case the computation of the inverse of a four-by-four matrix in analytical form, which is quite difficult. In order to get the expression of the transfer function of the channel from the duty ratio $\tilde{\upalpha}$ (first input) to the output voltage $\tilde{v_{C2}}$ (fourth state variable), the output vector must be set as y ≡ v _C2, hence the output matrix must be set as $\mathbf{C}={\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]}^T$. Therefore, computing only the fourth row of matrix H(s) is sufficient to get the transfer function sought for. A quite laborious but simple computation finally produces the duty-ratio-to-output-voltage transfer function,

$$ {H}_{\upalpha \to {v}_{C2}}(s)=\frac{b_3{s}^3+{b}_2{s}^2+{b}_1s+{b}_0}{s^4+{a}_3{s}^3+{a}_2{s}^2+{a}_1s+{a}_0}, $$

(4.55)

where the different parameters of numerator and denominator are, respectively

$$ \left\{\begin{array}{ll}{b}_3=-\frac{i_{L1e}+{i}_{L2e}}{C_2}\hfill & {b}_2=\left(1-{\upalpha}_e\right)\frac{v_{C1e}+{v}_{C2e}}{L_1{L}_2{C}_1{C}_2}\left({L}_1+{L}_2\right){C}_1\hfill \\ {}{b}_1=-{\upalpha}_e{L}_1\frac{i_{L1e}+{i}_{L2e}}{L_1{L}_2{C}_1{C}_2}\hfill & {b}_0=\left(1-{\upalpha}_e\right)\frac{v_{C1e}+{v}_{C2e}}{L_1{L}_2{C}_1{C}_2},\hfill \end{array}\right. $$

$$ \left\{\begin{array}{ll}{a}_3=\frac{1}{R{C}_2}\hfill & {a}_2=\frac{{\left(1-{\upalpha}_e\right)}^2\left({L}_1{C}_1+{L}_2{C}_2+{L}_2{C}_1\right)+{\upalpha}_e^2{L}_1{C}_2}{L_1{L}_2{C}_1{C}_2}\hfill \\ {}{a}_1=\frac{1}{R}\cdot \frac{\upalpha_e^2{L}_1+{\left(1-{\upalpha}_e\right)}^2{L}_2}{L_1{L}_2{C}_1{C}_2}\hfill & {a}_0=\frac{{\left(1-{\upalpha}_e\right)}^2}{L_1{L}_2{C}_1{C}_2}.\hfill \end{array}\right. $$

Analyzing the numerator of expression (4.55), one can identify the presence of unstable zeros, characterizing the nonminimum-phase behavior of the output voltage in response to duty ratio variations, similar to the buck-boost and flyback converter cases.

Problem 4.3

Nonideal Boost Converter

Let us consider the boost converter in Fig. 4.32, where the inductor is modeled as a pure inductance L = 2 mH and a series resistance R _L  = 0.5 Ω due to copper losses. The output capacitor has C = 100 μF and an equivalent series resistance, R _C  = 0.05 Ω. Both input voltage and load resistor are variant around their respective rated values E = 5 V and R = 10 Ω. It is required to solve the following points.

• (a) Deduce the small-signal state-space model and draw the associated equivalent diagram taking into account that both input voltage E and load resistance R vary (they will be represented as disturbance inputs).

• (b) Get the expressions of the three following transfer functions: from duty ratio to output voltage, from input voltage to output voltage and from load resistance to output voltage, respectively, by using the previously obtained diagram.

• (c) Compute the steady-state model and draw the steady-state characteristics and input–output efficiency curve with respect to the duty ratio.

• (d) Draw the zero-pole diagrams for each of the three above mentioned influence channels using the MATLAB^®-Simulink^® software.

• (e) Simulate the averaged nonlinear (large signal) model and assess the results for small variations around the steady-state operating point corresponding to the full load; make comparisons with the zero-pole diagram.

Fig. 4.32

Nonideal boost power stage

Solution

(a) For the sake of simplicity, in this example the brackets 〈 · 〉₀ will be dropped; therefore, any variable encountered will denote in fact the corresponding average. Following the developments in Sect. 4.5.3 of Chap. 4, the large-signal averaged state-space model may be written as

$$ \left\{\begin{array}{l}L\overset{\cdotp }{i_L}=E-{v}_0\left(1-\upalpha \right)-{R}_L{i}_L\hfill \\ {}C\overset{\cdotp }{v_C}={i}_L\left(1-\upalpha \right)-\frac{v_0}{R}\hfill \\ {}{v}_0=C{R}_C\overset{\cdotp }{v_C}+{v}_C.\hfill \end{array}\right. $$

(4.56)

Next, variables describing the steady-state (equilibrium) operating point and rated values bear the subscript e. With notation α ^′ _e  = 1 − α _e , in the equilibrium point it holds that

$$ \left\{\begin{array}{l}{E}_e-{v}_{0e}\cdot {\upalpha}_e^{\prime }-{R}_L{i}_{Le}=0\hfill \\ {}{i}_{Le}\cdot {\upalpha}_e^{\prime }-\frac{v_{0e}}{R_e}=0\hfill \\ {}{v}_{0e}={v}_{Ce}.\hfill \end{array}\right. $$

(4.57)

In order to obtain the small-signal model around the considered operating point, one must differentiate the model (4.56):

$$ \left\{\begin{array}{l}L\overset{\cdotp }{\tilde{i_L}}={E}_e+\tilde{E}-\left({v}_{0e}+\tilde{v_0}\right)\cdot \left(1-{\upalpha}_e-\tilde{\upalpha}\right)-{R}_L\cdot \left({i}_{Le}+\tilde{i_L}\right)\hfill \\ {}C\overset{\cdotp }{\tilde{v_C}}=\left({i}_{Le}+\tilde{i_L}\right)\cdot \left(1-{\upalpha}_e-\tilde{\upalpha}\right)-\frac{v_{0e}}{R_e}-\frac{\tilde{v_0}}{R_e}+\frac{v_{0e}}{R_e^2}\cdot \tilde{R}\hfill \\ {}{v}_{0e}+\tilde{v_0}=C{R}_C\overset{\cdotp }{\tilde{v_C}}+{v}_{Ce}+\tilde{v_C}.\hfill \end{array}\right. $$

(4.58)

Using relations (4.57) in the system of (4.58) and neglecting small variations, one obtains the small-signal model of the considered boost power stage:

$$ \left\{\begin{array}{l}L\overset{\cdotp }{\tilde{i_L}}=\tilde{E}+{v}_{0e}\cdot \tilde{\upalpha}-{\upalpha}_e^{\prime}\cdot \tilde{v_0}-{R}_L\cdot \tilde{i_L}\hfill \\ {}C\overset{\cdotp }{\tilde{v_C}}=-{i}_{Le}\cdot \tilde{\upalpha}+{\upalpha}_e^{\prime}\cdot \tilde{i_L}-\frac{\tilde{v_0}}{R_e}-\tilde{i_S}\hfill \\ {}\tilde{v_0}=C{R}_C\overset{\cdotp }{\tilde{v_C}}+\tilde{v_C},\hfill \end{array}\right. $$

(4.59)

where the variation of the load current due to the load variation has been denoted by $\tilde{i_S}=-{v}_{0e}/{R}_e^2\cdot \tilde{R}$. The associated equivalent diagram is given in Fig. 4.33.

Fig. 4.33

Small-signal model of the nonideal boost circuit: equivalent diagram

This diagram shows the influence in variations of all the exogenous variables over system output $\tilde{v_0}$: output results from superposition of all input variables. In order to extract a certain transfer function corresponding to one of these influence channels, one must nullify all other input variations.

(b) The duty-ratio-to-output-voltage transfer function is obtained by putting $\tilde{E}=0$ and $\tilde{i_S}=0$ in Eq. (4.59) or in Fig. 4.33. The result is presented in Fig. 4.34a. Further, the inductor and the voltage source in Fig. 4.34a may be pushed through the transformer; the circuit in Fig. 4.34b results, where the new inductor current has been denoted by $\tilde{i}$. Using Kirchhoff’s laws, one solves this circuit by expressing output voltage variations $\tilde{v_0}$ as a function of the duty ratio variations $\tilde{\upalpha}$:

Fig. 4.34

(a) Small-signal model of nonideal boost circuit: duty-ratio-to-output-voltage influence; (b) push of inductor and voltage source through transformer

$$ \left\{\begin{array}{l}\frac{1}{\upalpha_e^{\prime 2}}\left( sL+{R}_L\right)\cdot \tilde{i_L}=\frac{v_{0e}}{\upalpha_e^{\prime }}\cdot \tilde{\upalpha}-\tilde{v_0}\hfill \\ {}\tilde{i_L}={i}_{Le}\cdot \tilde{\upalpha}+\frac{\tilde{v_0}}{R_e\left|\right|\left({R}_C+\frac{1}{ sC}\right)}.\hfill \end{array}\right. $$

(4.60)

Combining Eq. (4.60) one finds that

$$ \tilde{v_0}\cdot \left[\frac{1}{\upalpha_e^{\prime 2}}\cdot \frac{ sL+{R}_L}{R_e\left|\right|\left({R}_C+\frac{1}{ sC}\right)}+1\right]=\tilde{\upalpha}\cdot \left(\frac{v_{0e}}{\upalpha_e^{\prime }}-\frac{ sL+{R}_L}{\upalpha_e^{\prime }}\cdot {i}_{Le}\right). $$

(4.61)

Simple algebra gives the required transfer function, ${H}_{v_0\upalpha}(s)=\frac{\tilde{V_0}(s)}{\tilde{\upalpha}(s)}$:

$$ {H}_{v_0\upalpha}(s)=\frac{R_e\left(C{R}_Cs+1\right)\left({v}_{0e}{\upalpha}_e^{\prime }-{i}_{Le}{R}_L-{i}_{Le} Ls\right)}{ CL\left({R}_e+{R}_C\right){s}^2+\left[{\upalpha}_e^{\prime 2}{R}_e{R}_CC+C{R}_L\left({R}_e+{R}_C\right)+L\right]s+{\upalpha}_e^{\prime 2}{R}_e+{R}_L}. $$

(4.62)

Similarly, the output-current-to-output-voltage transfer function may be obtained by putting $\tilde{E}=0$ and $\tilde{\upalpha}=0$ in Eq. (4.59) or in Fig. 4.33. The result is shown in Fig. 4.35a. Further, the inductor in Fig. 4.35a may be pushed through the transformer, resulting in the circuit in Fig. 4.35b. One aims at expressing the output voltage variations $\tilde{v_0}$ as a function of output current variations $\tilde{i_S}$ and output impedance:

Fig. 4.35

(a) Small-signal model of the nonideal boost circuit: load-current-to-output-voltage influence; (b) push of inductor through transformer

$$ \tilde{v_0}=\tilde{i_S}\cdot \left[{R}_e\left|\right|\left({R}_C+\frac{1}{ sC}\right)\left|\right|\left(s\frac{L}{\upalpha_e^{\prime 2}}+\frac{R_L}{\upalpha_e^{\prime 2}}\right)\right]. $$

(4.63)

Simple algebra gives the required transfer function, ${H}_{v_0{i}_S}(s)=\frac{\tilde{V_0}(s)}{\tilde{I_S}(s)}$:

$$ {H}_{v_0{i}_S}(s)=\frac{R_e\left(C{R}_Cs+1\right)\left({R}_L+ Ls\right)}{ CL\left({R}_e+{R}_C\right){s}^2+\left[{\upalpha}_e^{\prime 2}{R}_e{R}_CC+C{R}_L\left({R}_e+{R}_C\right)+L\right]s+{\upalpha}_e^{\prime 2}{R}_e+{R}_L}. $$

(4.64)

Computation of the input-voltage-to-output-voltage transfer function results similarly and it is left to the reader. Solutions to the questions proposed in (c), (d) and (e) are also left to the reader, as are the solutions to the following problems.

Problem 4.4

Noninverting Buck-boost Converter

The circuit in Fig. 4.36 has a switching network composed of four switches (two transistors and two diodes). As the transistors are operated synchronously with the same binary switching function, u ∈ {0,1}, this switching network leads to two circuit configurations.

Fig. 4.36

Noninverting buck-boost power stage

By taking i _L and v _C (see Fig. 4.30) as state variables

• (a) obtain the switched model (bilinear form);

• (b) obtain the averaged model and the corresponding equivalent diagram;

• (c) compute the steady-state model; draw the static input/output characteristic with respect to the duty ratio;

• (d) using the perturb-and-linearize method, deduce the small-signal state-space model and draw the associated equivalent diagram (E is considered constant);

• (e) using the previously obtained small-signal model, get the expression of the transfer function representing the influence from the duty ratio (control input) and from the input voltage E (disturbance input) to the output voltage;

• (f) analyze the system at point (d) for nonminimum-phase behavior and how its poles and zeroes migrate as the load resistor varies.

Problem 4.5

Watkins–Johnson Converter

Given the circuit in Fig. 4.37, answer the same requirements as in Problem 4.4.

Fig. 4.37

Watkins–Johnson power stage

In addition, it is required to simulate numerically (for example, using MATLAB^®) the switched and the averaged models in the following case: input voltage E = 5 V, inductance L = 5 mH, output capacitor C = 100 μF and output resistor R = 10 Ω. Compare the system behavior for duty ratios larger than 0.5 with the behavior for duty ratios smaller than 0.5. Draw the Bode diagram of the transfer from the duty ratio to the output voltage for α _e  = 0.6 and α _e  = 0.4.

Problem 4.6

Quadratic Buck Converter

Given the circuit in Fig. 4.38, answer the same requirements as in Problem 4.4.

Fig. 4.38

Quadratic buck power stage