Switched Model

Abstract

This chapter focuses on methodologies for obtaining the so-called switched model. This model describes basic low-frequency dynamics, as given by the energy accumulation variations, and it captures the switching dynamics of power electronic converters as well.

Problems

Problem 3.1

In the example below, the Ćuk converter has been considered (see Fig. 3.17) where inductors L ₁ and L ₂ are not coupled. The switching function takes two values {0; 1}. Address the following points.

Fig. 3.17

Electrical circuit of Ćuk DC-DC converter

• (a) Write the dynamical equations of the converter with respect to the switching function u.

• (b) Write the circuit model in the bilinear matrix form.

• (c) Draw the equivalent circuit of the converter and emphasize the coupling terms.

Solution

• (a) Let us take the current sense and voltage polarity as indicated in Fig. 3.17. Transistor H and diode D form the switching network; a single switching function u feeds the transistor gate. When u = 1, switch H is turned on and D is polarized inversely, leading to the configuration depicted in Fig. 3.18a. As u = 0, H turns off, diode D enters into conduction and the second configuration is shown in Fig. 3.18b.

  Fig. 3.18

  

  The two possible configurations of the Cúk converter

  Kirchhoff’s laws for the two topologies give

  $$ \begin{array}{ll}u=1:\left\{\begin{array}{l}\overset{\cdotp }{i_{L1}}=E/{L}_1\\ {}\overset{\cdotp }{v_{C1}}={i}_{L2}/{C}_1\\ {}\overset{\cdotp }{i_{L2}}=-{v}_{C1}/{L}_2-{v}_{C2}/{L}_2\\ {}\overset{\cdotp }{v_{C2}}={i}_{L2}/{C}_2-{v}_{C2}/(RC),\end{array}\right.\hfill & u=0:\left\{\begin{array}{l}\overset{\cdotp }{i_{L1}}=E/{L}_1-{v}_{C1}/{L}_1\\ {}\overset{\cdotp }{v_{C1}}={i}_{L1}/{C}_1\\ {}\overset{\cdotp }{i_{L2}}=-{v}_{C2}/{L}_2\\ {}\overset{\cdotp }{v_{C2}}={i}_{L2}/{C}_2-{v}_{C2}/\left(R{C}_2\right).\end{array}\right.\hfill \end{array} $$

  By multiplying the two sets of equations with u and (1–u), respectively, and summing, one obtains

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

  (3.20)

• (b) By denoting by $\mathbf{x}={\left[\begin{array}{cccc}\hfill {i}_{L1}\hfill & \hfill {v}_{C1}\hfill & \hfill {i}_{L2}\hfill & \hfill {v}_{C2}\hfill \end{array}\right]}^T$ the state vector (each of its components describes an energy accumulation), the set of Eq. (3.20) can be rewritten in matrix form as

  $$ \overset{\cdotp }{\mathbf{x}}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill -1/{L}_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 1/{C}_1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1/{L}_2\hfill \\ {}\hfill 0\hfill & \hfill 1/{C}_2\hfill & \hfill 0\hfill & \hfill -1/\left(R{C}_2\right)\hfill \end{array}\right]\cdot \mathbf{x}+\left[\begin{array}{c}\hfill u\cdot {v}_{C1}/{L}_1\hfill \\ {}\hfill u\cdot {i}_{L1}/{C}_1+u\cdot {i}_{L2}/{C}_1\hfill \\ {}\hfill -u\cdot {v}_{C1}/{L}_2\hfill \\ {}\hfill 0\hfill \end{array}\right]+\left[\begin{array}{c}\hfill E/{L}_1\hfill \\ {}\hfill 0\hfill \\ {}\hfill 0\hfill \\ {}\hfill 0\hfill \end{array}\right]. $$

Note that there is a single switching function, therefore p = 1. By processing the second term of the above matrix relation, one obtains

$$ \overset{\cdotp }{\mathbf{x}}=\mathbf{A}\cdot \mathbf{x}+\left(\mathbf{B}\cdot \mathbf{x}+\mathbf{b}\right)\cdot u+\mathbf{d}, $$

with

$$ \left\{\begin{array}{ll}\begin{array}{l}\mathbf{A}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill -1/{L}_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 1/{C}_1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1/{L}_2\hfill \\ {}\hfill 0\hfill & \hfill 1/{C}_2\hfill & \hfill 0\hfill & \hfill -1/\left(R{C}_2\right)\hfill \end{array}\right],\\ {}\mathbf{b}={\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]}^T,\end{array}\hfill & \begin{array}{l}\mathbf{B}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 1/{L}_1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 1/{C}_1\hfill & \hfill 0\hfill & \hfill 1/{C}_1\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill -1/{L}_2\hfill & \hfill 0\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right],\\ {}\mathbf{d}={\left[\begin{array}{cccc}\hfill E/{L}_1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]}^T.\end{array}\hfill \end{array}\right. $$

Equations (3.20) lead to the equivalent circuit in Fig. 3.19. The first equation of (3.20) gives the first circuit, where the dependent voltage source is a function of the inductor current in the second circuit. The second equation gives the second circuit. Note that the two dependent current sources are coupled with variables from the other two circuits (the inductor current from the first circuit and the capacitor voltage from the third one). These coupled sources can be seen as variable-ratio DC ideal transformers. The third circuit is the image of the last two equations in (3.19).

Fig. 3.19

Equivalent circuit of switched model of Cúk converter, presented in Fig. 3.17

Problem 3.2

Let consider the half-bridge voltage inverter (current rectifier) of Fig. 3.20. Switching function u is defined to take the value 1 if switch F ₁ is turned on and the value −1 if switch F ₂ is turned on. Zero indices concern continuous variables, whereas indices a concern alternative variables.

Fig. 3.20

Half-bridge voltage inverter

• (a) Write the system equations by making appear the switching function u.

• (b) Prove that the switched model of the system may be represented by the equivalent circuit from Fig. 3.21.

  Fig. 3.21

  

  Equivalent topologic diagram of inverter from Fig. 3.20

Solution

The evolution of the filtering capacitor voltages is given by

$$ \left\{\begin{array}{l}{C}_0\cdot \overset{\cdotp }{v_{01}}={i}_0+{\scriptscriptstyle \frac{1+u}{2}}\cdot {i}_a\hfill \\ {}{C}_0\cdot \overset{\cdotp }{v_{02}}=-{i}_0+{\scriptscriptstyle \frac{1-u}{2}}\cdot {i}_a.\hfill \end{array}\right. $$

The expression of the alternative voltage is ${v}_a=\frac{1+u}{2}\cdot {v}_{01}+\frac{1-u}{2}\cdot {v}_{02}$. Taking v _0c  = (v ₀₁ + v ₀₂)/2 and knowing that v ₀ = v ₀₁ − v ₀₂ one obtains

$$ \left\{\begin{array}{l}{v}_a={v}_{0c}+\frac{v_0}{2}\cdot u\hfill \\ {}2{C}_0\cdot \overset{\cdotp }{v_{0c}}={i}_a\hfill \\ {}\frac{C_0}{2}\cdot \overset{\cdotp }{v_0}={i}_0+\frac{i_a}{2}\cdot u.\hfill \end{array}\right. $$

(3.21)

Equation (3.21) allows the equivalent circuit of Fig. 3.21 to be built as representation of the switched model. This proof is useful for solving Problem 3.3.

The following problems are left to the reader to solve.

Problem 3.3

Current-source inverter

Let us consider the half-bridge current source inverter presented in Fig. 3.22. Prove that the equivalent diagram shown in Fig. 3.23 corresponds to the circuit given in Fig. 3.22.

Fig. 3.22

Half-bridge current source inverter

Fig. 3.23

Equivalent topological diagram of current source inverter from Fig. 3.22

Problem 3.4

Series-resonance power supply

Let us consider the converter given in Fig. 3.24. Switching functions u ₁ and u ₂ are such that

Fig. 3.24

Series-resonance voltage power supply based on capacitive half-bridges

$$ {u}_1=\left\{\begin{array}{ll}1\hfill & \mathrm{if}\ {F}_1\ \mathrm{is}\ \mathrm{closed}\hfill \\ {}-1\hfill & \mathrm{if}\ {F}_1\ \mathrm{is}\ \mathrm{open}.\hfill \end{array}\right. $$

• (a) Write the observable state model of the system, i.e., of appropriate order, and prove that for this model the equivalent circuit is the one shown in Fig. 3.25.

  Fig. 3.25

  

  The equivalent topologic diagram of the converter in Fig. 3.24

• (b) Find out the value of capacitor C _eq from the equivalent circuit.

Problem 3.5

Buck-boost power stage

Consider the DC-DC converter given in Fig. 3.26.

Fig. 3.26

Buck-boost DC-DC converter

• (a) Establish the switched model of converter in Fig. 3.26 by considering a switching function that takes its values in the set {0;1}, namely, value 1 corresponds to switch H being turned on and value 0 to the same switch being turned off.

• (b) Identify the variables switched by power switches H and D.

• (c) Draw the equivalent diagram.

Problem 3.6

Zeta DC-DC power stage

• (a) Deduce the switched model of the DC-DC converter from Fig. 3.27 by considering a switching function that takes value 1 when switch H is turned on and value 0 when the same switch is turned off. Consider the voltage polarity and current sense as indicated in Fig. 3.27.

  Fig. 3.27

  

  Zeta DC-DC converter

• (b) Identify the variables switched by power switches H and D.

• (c) Draw the equivalent diagram.