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2.4 Cuk converter example C1 L1 Cuk converter, with ideal switch i1 Vg L2 i2 + v1 – 1 + – 2 C2 + v2 R – Cuk converter: practical realization using MOSFET and diode C1 L1 i1 Vg L2 i2 + v1 – + – Q1 D1 C2 + v2 R – Fundamentals of Power Electronics 28 Chapter 2: Principles of steady-state converter analysis Analysis strategy This converter has two inductor currents and two capacitor voltages, that can be expressed as i 1(t) = I 1 + i 1 -ripple(t) i 2(t) = I 2 + i 2 -ripple(t) v1(t) = V1 + v1 -ripple(t) v2(t) = V2 + v2 -ripple(t) To solve the converter in steady state, we want to ﬁnd the dc components I1, I2, V1, and V2, when the ripples are small. Fundamentals of Power Electronics C1 L1 i1 Vg L2 i2 + v1 – 1 + – 2 C2 + v2 R – Strategy: • Apply volt-second balance to each inductor voltage • Apply charge balance to each capacitor current • Simplify using the small ripple approximation • Solve the resulting four equations for the four unknowns I1, I2, V1, and V2. 29 Chapter 2: Principles of steady-state converter analysis Cuk converter circuit with switch in positions 1 and 2 L1 Switch in position 1: MOSFET conducts Capacitor C1 releases energy to output L2 i1 + vL1 – Vg – + – v1 i2 iC1 + vL2 – C1 iC2 C2 + i1 Switch in position 2: diode conducts Capacitor C1 is charged from input Fundamentals of Power Electronics iC1 + vL1 – Vg + – + C1 v1 – 30 v2 R – L2 L1 + + vL2 – i2 iC2 C2 + v2 R – Chapter 2: Principles of steady-state converter analysis Waveforms during subinterval 1 MOSFET conduction interval Inductor voltages and capacitor currents: vL1 = Vg vL2 = – v1 – v2 i C1 = i 2 v i C2 = i 2 – 2 R L1 i1 + vL1 – Vg + – L2 – v1 + iC1 + vL2 – C1 i2 iC2 C2 + v2 R – Small ripple approximation for subinterval 1: vL1 = Vg vL2 = – V1 – V2 i C1 = I 2 V i C2 = I 2 – 2 R Fundamentals of Power Electronics 31 Chapter 2: Principles of steady-state converter analysis Waveforms during subinterval 2 Diode conduction interval Inductor voltages and capacitor currents: vL1 = Vg – v1 vL2 = – v2 i C1 = i 1 v i C2 = i 2 – 2 R i1 L2 L1 iC1 + vL1 – Vg + – + C1 v1 – + vL2 – i2 iC2 C2 + v2 R – Small ripple approximation for subinterval 2: vL1 = Vg – V1 vL2 = – V2 i C1 = I 1 V i C2 = I 2 – 2 R Fundamentals of Power Electronics 32 Chapter 2: Principles of steady-state converter analysis Equate average values to zero The principles of inductor volt-second and capacitor charge balance state that the average values of the periodic inductor voltage and capacitor current waveforms are zero, when the converter operates in steady state. Hence, to determine the steady-state conditions in the converter, let us sketch the inductor voltage and capacitor current waveforms, and equate their average values to zero. Waveforms: Inductor voltage vL1(t) vL1(t) Volt-second balance on L1: Vg DTs vL1 = DVg + D'(Vg – V1) = 0 D'Ts t Vg – V1 Fundamentals of Power Electronics 33 Chapter 2: Principles of steady-state converter analysis Equate average values to zero Inductor L2 voltage vL2(t) – V2 DTs D'Ts t – V1 – V2 vL2 = D( – V1 – V2) + D'( – V2) = 0 Capacitor C1 current iC1(t) Average the waveforms: i C1 = DI 2 + D'I 1 = 0 I1 DTs D'Ts t I2 Fundamentals of Power Electronics 34 Chapter 2: Principles of steady-state converter analysis Equate average values to zero Capacitor current iC2(t) waveform iC2(t) I2 – V2 / R (= 0) DTs D'Ts t i C2 = I 2 – V2 =0 R Note: during both subintervals, the capacitor current iC2 is equal to the difference between the inductor current i2 and the load current V2/R. When ripple is neglected, iC2 is constant and equal to zero. Fundamentals of Power Electronics 35 Chapter 2: Principles of steady-state converter analysis Solve for steady-state inductor currents and capacitor voltages The four equations obtained from volt-sec and charge balance: Solve for the dc capacitor voltages and inductor currents, and express in terms of the known Vg, D, and R: vL1 = DVg + D' Vg – V1 = 0 vL2 = D – V1 – V2 + D' – V2 = 0 Vg D' V2 = – D Vg D' I1 = – D I2 = D' V I2 = 2 = – D R D' V1 = i C1 = DI 2 + D'I 1 = 0 V i C2 = I 2 – 2 = 0 R Fundamentals of Power Electronics 36 D D' Vg R 2 Vg R Chapter 2: Principles of steady-state converter analysis Cuk converter conversion ratio M = V/Vg D 0 0.2 0.4 0.6 0.8 1 0 M(D) -1 -2 -3 -4 V2 M(D) = =– D Vg 1–D -5 Fundamentals of Power Electronics 37 Chapter 2: Principles of steady-state converter analysis Inductor current waveforms Interval 1 slopes, using small ripple approximation: di 1(t) vL1(t) Vg = = L1 L1 dt di 2(t) vL2(t) – V1 – V2 = = L2 L2 dt i1(t) i1 I1 Vg L1 DTs Fundamentals of Power Electronics Ts DTs Interval 2 slopes: di 1(t) vL1(t) Vg – V1 = = L1 L1 dt di 2(t) vL2(t) – V2 = = L2 L2 dt Vg – V1 L1 I2 – V1 – V2 L2 t t Ts – V2 L2 i2 i2(t) 38 Chapter 2: Principles of steady-state converter analysis Capacitor C1 waveform Subinterval 1: dv1(t) i C1(t) I 2 = = C1 C dt 1 v1(t) v1 V1 I2 C1 Subinterval 2: DTs dv1(t) i C1(t) I 1 = = C1 C1 dt Fundamentals of Power Electronics I1 C1 39 Ts t Chapter 2: Principles of steady-state converter analysis Ripple magnitudes Analysis results Use dc converter solution to simplify: VgDTs i1 = 2L 1 VgDTs i2 = 2L 2 VgD 2Ts v1 = 2D'RC 1 VgDTs i1 = 2L 1 V1 + V2 i2 = DTs 2L 2 – I 2DTs v1 = 2C 1 Q: How large is the output voltage ripple? Fundamentals of Power Electronics 40 Chapter 2: Principles of steady-state converter analysis 2.5 Estimating ripple in converters containing two-pole low-pass ﬁlters Buck converter example: Determine output voltage ripple L 1 iL(t) Vg iC(t) 2 + – + C iR(t) vC(t) R – iL(t) Inductor current waveform. iL(DTs) I iL(0) What is the capacitor current? Vg – V L 0 Fundamentals of Power Electronics –V L DTs 41 iL Ts t Chapter 2: Principles of steady-state converter analysis Capacitor current and voltage, buck example iC(t) Total charge q Must not neglect inductor current ripple! If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current ﬂows through the capacitor. Fundamentals of Power Electronics iL t Ts /2 DTs D'Ts vC(t) v V v t 42 Chapter 2: Principles of steady-state converter analysis Estimating capacitor voltage ripple v iC(t) Current iC(t) is positive for half of the switching period. This positive current causes the capacitor voltage vC(t) to increase between its minimum and maximum extrema. During this time, the total charge q is deposited on the capacitor plates, where Total charge q iL t Ts /2 DTs D'Ts vC(t) V v v t Fundamentals of Power Electronics 43 q = C (2 v) (change in charge) = C (change in voltage) Chapter 2: Principles of steady-state converter analysis Estimating capacitor voltage ripple v iC(t) The total charge q is the area of the triangle, as shown: Total charge q iL t q= Ts /2 DTs D'Ts v= Ts 2 i L Ts 8C v v t Fundamentals of Power Electronics iL Eliminate q and solve for v: vC(t) V 1 2 44 Note: in practice, capacitor equivalent series resistance (esr) further increases v. Chapter 2: Principles of steady-state converter analysis Inductor current ripple in two-pole ﬁlters L1 Example: problem 2.9 iT L2 + i1 Vg Q1 + – vC1 C1 + i2 D1 C2 – vL(t) R v – Total flux linkage v t Ts /2 DTs D'Ts can use similar arguments, with = L (2 i) iL(t) I = inductor ﬂux linkages i = inductor volt-seconds i t Fundamentals of Power Electronics 45 Chapter 2: Principles of steady-state converter analysis 2.6 Summary of Key Points 1. The dc component of a converter waveform is given by its average value, or the integral over one switching period, divided by the switching period. Solution of a dc-dc converter to ﬁnd its dc, or steadystate, voltages and currents therefore involves averaging the waveforms. 2. The linear ripple approximation greatly simpliﬁes the analysis. In a welldesigned converter, the switching ripples in the inductor currents and capacitor voltages are small compared to the respective dc components, and can be neglected. 3. The principle of inductor volt-second balance allows determination of the dc voltage components in any switching converter. In steady-state, the average voltage applied to an inductor must be zero. Fundamentals of Power Electronics 46 Chapter 2: Principles of steady-state converter analysis Summary of Chapter 2 4. The principle of capacitor charge balance allows determination of the dc components of the inductor currents in a switching converter. In steadystate, the average current applied to a capacitor must be zero. 5. By knowledge of the slopes of the inductor current and capacitor voltage waveforms, the ac switching ripple magnitudes may be computed. Inductance and capacitance values can then be chosen to obtain desired ripple magnitudes. 6. In converters containing multiple-pole ﬁlters, continuous (nonpulsating) voltages and currents are applied to one or more of the inductors or capacitors. Computation of the ac switching ripple in these elements can be done using capacitor charge and/or inductor ﬂux-linkage arguments, without use of the small-ripple approximation. 7. Converters capable of increasing (boost), decreasing (buck), and inverting the voltage polarity (buck-boost and Cuk) have been described. Converter circuits are explored more fully in a later chapter. Fundamentals of Power Electronics 47 Chapter 2: Principles of steady-state converter analysis