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Chapter 02

T
Tha Mike

This chapter discusses principles of steady-state analysis of DC-DC power converters. It introduces inductor volt-second balance and capacitor charge balance, which relate the average inductor voltage and capacitor current to be zero during steady-state. A small ripple approximation is used to simplify analysis by ignoring output voltage ripple. Examples of steady-state analysis of the buck and boost converters are presented using these principles to determine output voltage, inductor current, and capacitor sizing for given ripple levels.

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Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis1 Chapter 2 Principles of Steady-State Converter Analysis 2.1. Introduction 2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation 2.3. Boost converter example 2.4. Cuk converter example 2.5. Estimating the ripple in converters containing two- pole low-pass filters 2.6. Summary of key points
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis2 2.1 Introduction Buck converter SPDT switch changes dc component Switch output voltage waveform complement D′: D′ = 1 - D Duty cycle D: 0 ≤ D ≤ 1 + – R + v(t) – 1 2 + vs(t) – Vg vs(t) Vg DTs D'Ts 0 t0 DTs Ts Switch position: 1 2 1
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis3 Dc component of switch output voltage vs = 1 Ts vs(t) dt 0 Ts vs = 1 Ts (DTsVg) = DVg Fourier analysis: Dc component = average value vs(t) Vg 0 t0 DTs Ts 〈vs〉 = DVgarea = DTsVg
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis4 Insertion of low-pass filter to remove switching harmonics and pass only dc component v ≈ vs = DVg + – L C R + v(t) – 1 2 + vs(t) – Vg Vg 0 0 D V 1
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis5 Three basic dc-dc converters Buck Boost Buck-boost M(D) D 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 M(D) D 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 M(D) D –5 –4 –3 –2 –1 0 0 0.2 0.4 0.6 0.8 1 (a) (b) (c) + – L C R + v – 1 2 + – L C R + v – 1 2 + – L C R + v – 1 2 M(D) = D M(D) = 1 1 – D M(D) = – D 1 – D iL (t) Vg iL (t) Vg iL (t) Vg
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis6 Objectives of this chapter G Develop techniques for easily determining output voltage of an arbitrary converter circuit G Derive the principles of inductor volt-second balance and capacitor charge (amp-second) balance G Introduce the key small ripple approximation G Develop simple methods for selecting filter element values G Illustrate via examples
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis7 2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation Buck converter containing practical low-pass filter Actual output voltage waveform v(t) = V + vripple(t) Actual output voltage waveform, buck converter + – L C R + v(t) – 1 2 iL(t) + vL(t) – iC(t) Vg v(t) t 0 V Actual waveform v(t) = V + vripple(t) dc component V
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis8 The small ripple approximation In a well-designed converter, the output voltage ripple is small. Hence, the waveforms can be easily determined by ignoring the ripple: v(t) ≈ V v(t) = V + vripple(t) v(t) t 0 V Actual waveform v(t) = V + vripple(t) dc component V vripple < V
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis9 Buck converter analysis: inductor current waveform original converter switch in position 2switch in position 1 + – L C R + v(t) – 1 2 iL(t) + vL(t) – iC(t) Vg L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis10 Inductor voltage and current Subinterval 1: switch in position 1 vL = Vg – v(t) Inductor voltage Small ripple approximation: vL ≈ Vg – V Knowing the inductor voltage, we can now find the inductor current via vL(t) = L diL(t) dt Solve for the slope: diL(t) dt = vL(t) L ≈ Vg – V L ⇒ The inductor current changes with an essentially constant slope L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis11 Inductor voltage and current Subinterval 2: switch in position 2 Inductor voltage Small ripple approximation: Knowing the inductor voltage, we can again find the inductor current via vL(t) = L diL(t) dt Solve for the slope: ⇒ The inductor current changes with an essentially constant slope vL(t) = – v(t) vL(t) ≈ – V diL(t) dt ≈ – V L L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis12 Inductor voltage and current waveforms vL(t) = L diL(t) dt vL(t) Vg – V t – V D'TsDTs Switch position: 1 2 1 – V L Vg – V L iL(t) t0 DTs Ts I iL(0) iL(DTs) ∆iL
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis13 Determination of inductor current ripple magnitude (change in iL) = (slope)(length of subinterval) 2∆iL = Vg – V L DTs ∆iL = Vg – V 2L DTs L = Vg – V 2∆iL DTs⇒ – V L Vg – V L iL(t) t0 DTs Ts I iL(0) iL(DTs) ∆iL
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis14 Inductor current waveform during turn-on transient When the converter operates in equilibrium: iL((n + 1)Ts) = iL(nTs) iL(t) t0 DTs Ts iL(0) = 0 iL(nTs) iL (Ts ) 2Ts nTs (n + 1)Ts iL((n + 1)Ts) Vg – v(t) L – v(t) L
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis15 The principle of inductor volt-second balance: Derivation Inductor defining relation: Integrate over one complete switching period: In periodic steady state, the net change in inductor current is zero: Hence, the total area (or volt-seconds) under the inductor voltage waveform is zero whenever the converter operates in steady state. An equivalent form: The average inductor voltage is zero in steady state. vL(t) = L diL(t) dt iL(Ts) – iL(0) = 1 L vL(t) dt 0 Ts 0 = vL(t) dt 0 Ts 0 = 1 Ts vL(t) dt 0 Ts = vL
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis16 Inductor volt-second balance: Buck converter example Inductor voltage waveform, previously derived: Integral of voltage waveform is area of rectangles: λ = vL(t) dt 0 Ts = (Vg – V)(DTs) + ( – V)(D'Ts) Average voltage is vL = λ Ts = D(Vg – V) + D'( – V) Equate to zero and solve for V: 0 = DVg – (D + D')V = DVg – V ⇒ V = DVg vL(t) Vg – V t – V DTs Total area λ
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis17 The principle of capacitor charge balance: Derivation Capacitor defining relation: Integrate over one complete switching period: In periodic steady state, the net change in capacitor voltage is zero: Hence, the total area (or charge) under the capacitor current waveform is zero whenever the converter operates in steady state. The average capacitor current is then zero. iC(t) = C dvC(t) dt vC(Ts) – vC(0) = 1 C iC(t) dt 0 Ts 0 = 1 Ts iC(t) dt 0 Ts = iC
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis18 2.3 Boost converter example Boost converter with ideal switch Realization using power MOSFET and diode + – L C R + v – 1 2 iL(t) Vg iC(t) + vL(t) – + – L C R + v – iL(t) Vg iC(t) + vL(t) – D1 Q1 DTs Ts + –
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis19 Boost converter analysis original converter switch in position 2switch in position 1 + – L C R + v – 1 2 iL(t) Vg iC(t) + vL(t) – C R + v – iC(t) + – L iL(t) Vg + vL(t) – C R + v – iC(t) + – L iL(t) Vg + vL(t) –
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis20 Subinterval 1: switch in position 1 Inductor voltage and capacitor current Small ripple approximation: vL = Vg iC = – v / R vL = Vg iC = – V / R C R + v – iC(t) + – L iL(t) Vg + vL(t) –
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis21 Subinterval 2: switch in position 2 Inductor voltage and capacitor current Small ripple approximation: vL = Vg – v iC = iL – v / R vL = Vg – V iC = I – V / R C R + v – iC(t) + – L iL(t) Vg + vL(t) –
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis22 Inductor voltage and capacitor current waveforms vL(t) Vg – V t DTs Vg D'Ts iC(t) – V/R t DTs I – V/R D'Ts
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis23 Inductor volt-second balance Net volt-seconds applied to inductor over one switching period: vL(t) dt 0 Ts = (Vg) DTs + (Vg – V) D'Ts Equate to zero and collect terms: Vg (D + D') – V D' = 0 Solve for V: V = Vg D' The voltage conversion ratio is therefore M(D) = V Vg = 1 D' = 1 1 – D vL(t) Vg – V t DTs Vg D'Ts
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis24 Conversion ratio M(D) of the boost converterM(D) D 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 M(D) = 1 D' = 1 1 – D
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis25 Determination of inductor current dc component Capacitor charge balance: iC(t) dt 0 Ts = ( – V R ) DTs + (I – V R ) D'Ts Collect terms and equate to zero: – V R (D + D') + I D' = 0 Solve for I: I = V D' R I = Vg D'2 R Eliminate V to express in terms of Vg: iC(t) – V/R t DTs I – V/R D'Ts D 0 2 4 6 8 0 0.2 0.4 0.6 0.8 1 I Vg/R
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis26 Determination of inductor current ripple Inductor current slope during subinterval 1: diL(t) dt = vL(t) L = Vg – V L Inductor current slope during subinterval 2: 2∆iL = Vg L DTs diL(t) dt = vL(t) L = Vg L Change in inductor current during subinterval 1 is (slope) (length of subinterval): Solve for peak ripple: ∆iL = Vg 2L DTs • Choose L such that desired ripple magnitude is obtained Vg – V L Vg L iL(t) t0 DTs Ts I ∆iL
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis27 Determination of capacitor voltage ripple Capacitor voltage slope during subinterval 1: Capacitor voltage slope during subinterval 2: Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval): Solve for peak ripple: • Choose C such that desired voltage ripple magnitude is obtained • In practice, capacitor equivalent series resistance (esr) leads to increased voltage ripple dvC(t) dt = iC(t) C = – V RC dvC(t) dt = iC(t) C = I C – V RC – 2∆v = – V RC DTs ∆v = V 2RC DTs v(t) t0 DTs Ts V ∆v I C – V RC – V RC
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis28 2.4 Cuk converter example + – L1 C2 R + v2 – C1 L2 1 2 + v1 –i1 i2 Vg + – L1 C2 R + v2 – C1 L2 + v1 –i1 i2 D1Q1Vg Cuk converter, with ideal switch Cuk converter: practical realization using MOSFET and diode
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis29 Cuk converter circuit with switch in positions 1 and 2 + – L1 C2 R + v2 – C1 L2 i1 i2 – v1 + iC1 iC2 + vL2 –+ vL1 – Vg + – L1 C2 R + v2 – C1 L2i1 i2 + v1 – iC1 iC2 + vL2 –+ vL1 – Vg Switch in position 1: MOSFET conducts Capacitor C1 releases energy to output Switch in position 2: diode conducts Capacitor C1 is charged from input
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis30 Waveforms during subinterval 1 MOSFET conduction interval + – L1 C2 R + v2 – C1 L2 i1 i2 – v1 + iC1 iC2 + vL2 –+ vL1 – Vg vL1 = Vg vL2 = – v1 – v2 iC1 = i2 iC2 = i2 – v2 R Inductor voltages and capacitor currents: Small ripple approximation for subinterval 1: vL1 = Vg vL2 = – V1 – V2 iC1 = I2 iC2 = I2 – V2 R
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis31 Waveforms during subinterval 2 Diode conduction interval Inductor voltages and capacitor currents: Small ripple approximation for subinterval 2: + – L1 C2 R + v2 – C1 L2i1 i2 + v1 – iC1 iC2 + vL2 –+ vL1 – Vg vL1 = Vg – v1 vL2 = – v2 iC1 = i1 iC2 = i2 – v2 R vL1 = Vg – V1 vL2 = – V2 iC1 = I1 iC2 = I2 – V2 R
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis32 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: vL1(t) Vg – V1 t DTs Vg D'Ts Inductor voltage vL1(t) vL1 = DVg + D'(Vg – V1) = 0 Volt-second balance on L1:
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis33 Equate average values to zero vL2(t) – V1 – V2 t DTs – V2 D'Ts iC1(t) I2 t DTs I1 D'Ts Inductor L2 voltage Capacitor C1 current vL2 = D( – V1 – V2) + D'( – V2) = 0 iC1 = DI2 + D'I1 = 0 Average the waveforms:
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis34 Equate average values to zero iC2(t) I2 – V2 / R (= 0) tDTs D'Ts Capacitor current iC2(t) waveform 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. iC2 = I2 – V2 R = 0
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis35 Cuk converter conversion ratio M = V/VgM(D) D -5 -4 -3 -2 -1 0 0 0.2 0.4 0.6 0.8 1 M(D) = V2 Vg = – D 1 – D
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis36 Inductor current waveforms di1(t) dt = vL1(t) L1 = Vg L1 di2(t) dt = vL2(t) L2 = – V1 – V2 L2 Interval 1 slopes, using small ripple approximation: Interval 2 slopes: di1(t) dt = vL1(t) L1 = Vg – V1 L1 di2(t) dt = vL2(t) L2 = – V2 L2 i1(t) tDTs Ts I1 ∆i1 Vg – V1 L1 Vg L1 – V2 L2 – V1 – V2 L2 i2(t) tDTs Ts I2 ∆i2
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis37 Capacitor C1 waveform dv1(t) dt = iC1(t) C1 = I2 C1 Subinterval 1: Subinterval 2: dv1(t) dt = iC1(t) C1 = I1 C1 I1 C1 I2 C1 v1(t) tDTs Ts V1 ∆v1
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis38 Ripple magnitudes ∆i1 = VgDTs 2L1 ∆i2 = V1 + V2 2L2 DTs ∆v1 = – I2DTs 2C1 Use dc converter solution to simplify: ∆i1 = VgDTs 2L1 ∆i2 = VgDTs 2L2 ∆v1 = VgD2 Ts 2D'RC1 Analysis results Q: How large is the output voltage ripple?
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis39 2.5 Estimating ripple in converters containing two-pole low-pass filters Buck converter example: Determine output voltage ripple Inductor current waveform. What is the capacitor current? + – L C R + vC(t) – 1 2 iC(t) iR(t) iL(t) Vg – V L Vg – V L iL(t) t0 DTs Ts I iL(0) iL(DTs) ∆iL
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis40 Capacitor current and voltage, buck example 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. iC(t) vC(t) t t Total charge q DTs D'Ts Ts /2 V ∆iL ∆v ∆v
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis41 Estimating capacitor voltage ripple ∆v q = C (2∆v) 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 (change in charge) = C (change in voltage) iC(t) vC(t) t t Total charge q DTs D'Ts Ts /2 V ∆iL ∆v ∆v
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis42 Estimating capacitor voltage ripple ∆v The total charge q is the area of the triangle, as shown: q = 1 2 ∆iL Ts 2 Eliminate q and solve for ∆v: ∆v = ∆iL Ts 8 C Note: in practice, capacitor equivalent series resistance (esr) further increases ∆v. iC(t) vC(t) t t Total charge q DTs D'Ts Ts /2 V ∆iL ∆v ∆v
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis43 Inductor current ripple in two-pole filters Example: problem 2.9 can use similar arguments, with λ = L (2∆i) λ = inductor flux linkages = inductor volt-seconds R + v – + – C2 L2L1 C1 + vC1 – i1 iT i2 D1 Q1 Vg vL(t) iL(t) t t Total flux linkage λ DTs D'Ts Ts /2 I ∆v ∆i ∆i
Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis44 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 steady- state, voltages and currents therefore involves averaging the waveforms. 2. The linear ripple approximation greatly simplifies the analysis. In a well- designed 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 Chapter 2: Principles of steady-state converter analysis45 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 steady- state, 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.

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Chapter 02

  • 1. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis1 Chapter 2 Principles of Steady-State Converter Analysis 2.1. Introduction 2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation 2.3. Boost converter example 2.4. Cuk converter example 2.5. Estimating the ripple in converters containing two- pole low-pass filters 2.6. Summary of key points
  • 2. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis2 2.1 Introduction Buck converter SPDT switch changes dc component Switch output voltage waveform complement D′: D′ = 1 - D Duty cycle D: 0 ≤ D ≤ 1 + – R + v(t) – 1 2 + vs(t) – Vg vs(t) Vg DTs D'Ts 0 t0 DTs Ts Switch position: 1 2 1
  • 3. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis3 Dc component of switch output voltage vs = 1 Ts vs(t) dt 0 Ts vs = 1 Ts (DTsVg) = DVg Fourier analysis: Dc component = average value vs(t) Vg 0 t0 DTs Ts 〈vs〉 = DVgarea = DTsVg
  • 4. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis4 Insertion of low-pass filter to remove switching harmonics and pass only dc component v ≈ vs = DVg + – L C R + v(t) – 1 2 + vs(t) – Vg Vg 0 0 D V 1
  • 5. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis5 Three basic dc-dc converters Buck Boost Buck-boost M(D) D 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 M(D) D 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 M(D) D –5 –4 –3 –2 –1 0 0 0.2 0.4 0.6 0.8 1 (a) (b) (c) + – L C R + v – 1 2 + – L C R + v – 1 2 + – L C R + v – 1 2 M(D) = D M(D) = 1 1 – D M(D) = – D 1 – D iL (t) Vg iL (t) Vg iL (t) Vg
  • 6. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis6 Objectives of this chapter G Develop techniques for easily determining output voltage of an arbitrary converter circuit G Derive the principles of inductor volt-second balance and capacitor charge (amp-second) balance G Introduce the key small ripple approximation G Develop simple methods for selecting filter element values G Illustrate via examples
  • 7. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis7 2.2. Inductor volt-second balance, capacitor charge balance, and the small ripple approximation Buck converter containing practical low-pass filter Actual output voltage waveform v(t) = V + vripple(t) Actual output voltage waveform, buck converter + – L C R + v(t) – 1 2 iL(t) + vL(t) – iC(t) Vg v(t) t 0 V Actual waveform v(t) = V + vripple(t) dc component V
  • 8. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis8 The small ripple approximation In a well-designed converter, the output voltage ripple is small. Hence, the waveforms can be easily determined by ignoring the ripple: v(t) ≈ V v(t) = V + vripple(t) v(t) t 0 V Actual waveform v(t) = V + vripple(t) dc component V vripple < V
  • 9. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis9 Buck converter analysis: inductor current waveform original converter switch in position 2switch in position 1 + – L C R + v(t) – 1 2 iL(t) + vL(t) – iC(t) Vg L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg
  • 10. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis10 Inductor voltage and current Subinterval 1: switch in position 1 vL = Vg – v(t) Inductor voltage Small ripple approximation: vL ≈ Vg – V Knowing the inductor voltage, we can now find the inductor current via vL(t) = L diL(t) dt Solve for the slope: diL(t) dt = vL(t) L ≈ Vg – V L ⇒ The inductor current changes with an essentially constant slope L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg
  • 11. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis11 Inductor voltage and current Subinterval 2: switch in position 2 Inductor voltage Small ripple approximation: Knowing the inductor voltage, we can again find the inductor current via vL(t) = L diL(t) dt Solve for the slope: ⇒ The inductor current changes with an essentially constant slope vL(t) = – v(t) vL(t) ≈ – V diL(t) dt ≈ – V L L C R + v(t) – iL(t) + vL(t) – iC(t) + – Vg
  • 12. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis12 Inductor voltage and current waveforms vL(t) = L diL(t) dt vL(t) Vg – V t – V D'TsDTs Switch position: 1 2 1 – V L Vg – V L iL(t) t0 DTs Ts I iL(0) iL(DTs) ∆iL
  • 13. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis13 Determination of inductor current ripple magnitude (change in iL) = (slope)(length of subinterval) 2∆iL = Vg – V L DTs ∆iL = Vg – V 2L DTs L = Vg – V 2∆iL DTs⇒ – V L Vg – V L iL(t) t0 DTs Ts I iL(0) iL(DTs) ∆iL
  • 14. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis14 Inductor current waveform during turn-on transient When the converter operates in equilibrium: iL((n + 1)Ts) = iL(nTs) iL(t) t0 DTs Ts iL(0) = 0 iL(nTs) iL (Ts ) 2Ts nTs (n + 1)Ts iL((n + 1)Ts) Vg – v(t) L – v(t) L
  • 15. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis15 The principle of inductor volt-second balance: Derivation Inductor defining relation: Integrate over one complete switching period: In periodic steady state, the net change in inductor current is zero: Hence, the total area (or volt-seconds) under the inductor voltage waveform is zero whenever the converter operates in steady state. An equivalent form: The average inductor voltage is zero in steady state. vL(t) = L diL(t) dt iL(Ts) – iL(0) = 1 L vL(t) dt 0 Ts 0 = vL(t) dt 0 Ts 0 = 1 Ts vL(t) dt 0 Ts = vL
  • 16. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis16 Inductor volt-second balance: Buck converter example Inductor voltage waveform, previously derived: Integral of voltage waveform is area of rectangles: λ = vL(t) dt 0 Ts = (Vg – V)(DTs) + ( – V)(D'Ts) Average voltage is vL = λ Ts = D(Vg – V) + D'( – V) Equate to zero and solve for V: 0 = DVg – (D + D')V = DVg – V ⇒ V = DVg vL(t) Vg – V t – V DTs Total area λ
  • 17. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis17 The principle of capacitor charge balance: Derivation Capacitor defining relation: Integrate over one complete switching period: In periodic steady state, the net change in capacitor voltage is zero: Hence, the total area (or charge) under the capacitor current waveform is zero whenever the converter operates in steady state. The average capacitor current is then zero. iC(t) = C dvC(t) dt vC(Ts) – vC(0) = 1 C iC(t) dt 0 Ts 0 = 1 Ts iC(t) dt 0 Ts = iC
  • 18. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis18 2.3 Boost converter example Boost converter with ideal switch Realization using power MOSFET and diode + – L C R + v – 1 2 iL(t) Vg iC(t) + vL(t) – + – L C R + v – iL(t) Vg iC(t) + vL(t) – D1 Q1 DTs Ts + –
  • 19. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis19 Boost converter analysis original converter switch in position 2switch in position 1 + – L C R + v – 1 2 iL(t) Vg iC(t) + vL(t) – C R + v – iC(t) + – L iL(t) Vg + vL(t) – C R + v – iC(t) + – L iL(t) Vg + vL(t) –
  • 20. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis20 Subinterval 1: switch in position 1 Inductor voltage and capacitor current Small ripple approximation: vL = Vg iC = – v / R vL = Vg iC = – V / R C R + v – iC(t) + – L iL(t) Vg + vL(t) –
  • 21. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis21 Subinterval 2: switch in position 2 Inductor voltage and capacitor current Small ripple approximation: vL = Vg – v iC = iL – v / R vL = Vg – V iC = I – V / R C R + v – iC(t) + – L iL(t) Vg + vL(t) –
  • 22. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis22 Inductor voltage and capacitor current waveforms vL(t) Vg – V t DTs Vg D'Ts iC(t) – V/R t DTs I – V/R D'Ts
  • 23. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis23 Inductor volt-second balance Net volt-seconds applied to inductor over one switching period: vL(t) dt 0 Ts = (Vg) DTs + (Vg – V) D'Ts Equate to zero and collect terms: Vg (D + D') – V D' = 0 Solve for V: V = Vg D' The voltage conversion ratio is therefore M(D) = V Vg = 1 D' = 1 1 – D vL(t) Vg – V t DTs Vg D'Ts
  • 24. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis24 Conversion ratio M(D) of the boost converterM(D) D 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 M(D) = 1 D' = 1 1 – D
  • 25. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis25 Determination of inductor current dc component Capacitor charge balance: iC(t) dt 0 Ts = ( – V R ) DTs + (I – V R ) D'Ts Collect terms and equate to zero: – V R (D + D') + I D' = 0 Solve for I: I = V D' R I = Vg D'2 R Eliminate V to express in terms of Vg: iC(t) – V/R t DTs I – V/R D'Ts D 0 2 4 6 8 0 0.2 0.4 0.6 0.8 1 I Vg/R
  • 26. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis26 Determination of inductor current ripple Inductor current slope during subinterval 1: diL(t) dt = vL(t) L = Vg – V L Inductor current slope during subinterval 2: 2∆iL = Vg L DTs diL(t) dt = vL(t) L = Vg L Change in inductor current during subinterval 1 is (slope) (length of subinterval): Solve for peak ripple: ∆iL = Vg 2L DTs • Choose L such that desired ripple magnitude is obtained Vg – V L Vg L iL(t) t0 DTs Ts I ∆iL
  • 27. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis27 Determination of capacitor voltage ripple Capacitor voltage slope during subinterval 1: Capacitor voltage slope during subinterval 2: Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval): Solve for peak ripple: • Choose C such that desired voltage ripple magnitude is obtained • In practice, capacitor equivalent series resistance (esr) leads to increased voltage ripple dvC(t) dt = iC(t) C = – V RC dvC(t) dt = iC(t) C = I C – V RC – 2∆v = – V RC DTs ∆v = V 2RC DTs v(t) t0 DTs Ts V ∆v I C – V RC – V RC
  • 28. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis28 2.4 Cuk converter example + – L1 C2 R + v2 – C1 L2 1 2 + v1 –i1 i2 Vg + – L1 C2 R + v2 – C1 L2 + v1 –i1 i2 D1Q1Vg Cuk converter, with ideal switch Cuk converter: practical realization using MOSFET and diode
  • 29. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis29 Cuk converter circuit with switch in positions 1 and 2 + – L1 C2 R + v2 – C1 L2 i1 i2 – v1 + iC1 iC2 + vL2 –+ vL1 – Vg + – L1 C2 R + v2 – C1 L2i1 i2 + v1 – iC1 iC2 + vL2 –+ vL1 – Vg Switch in position 1: MOSFET conducts Capacitor C1 releases energy to output Switch in position 2: diode conducts Capacitor C1 is charged from input
  • 30. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis30 Waveforms during subinterval 1 MOSFET conduction interval + – L1 C2 R + v2 – C1 L2 i1 i2 – v1 + iC1 iC2 + vL2 –+ vL1 – Vg vL1 = Vg vL2 = – v1 – v2 iC1 = i2 iC2 = i2 – v2 R Inductor voltages and capacitor currents: Small ripple approximation for subinterval 1: vL1 = Vg vL2 = – V1 – V2 iC1 = I2 iC2 = I2 – V2 R
  • 31. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis31 Waveforms during subinterval 2 Diode conduction interval Inductor voltages and capacitor currents: Small ripple approximation for subinterval 2: + – L1 C2 R + v2 – C1 L2i1 i2 + v1 – iC1 iC2 + vL2 –+ vL1 – Vg vL1 = Vg – v1 vL2 = – v2 iC1 = i1 iC2 = i2 – v2 R vL1 = Vg – V1 vL2 = – V2 iC1 = I1 iC2 = I2 – V2 R
  • 32. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis32 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: vL1(t) Vg – V1 t DTs Vg D'Ts Inductor voltage vL1(t) vL1 = DVg + D'(Vg – V1) = 0 Volt-second balance on L1:
  • 33. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis33 Equate average values to zero vL2(t) – V1 – V2 t DTs – V2 D'Ts iC1(t) I2 t DTs I1 D'Ts Inductor L2 voltage Capacitor C1 current vL2 = D( – V1 – V2) + D'( – V2) = 0 iC1 = DI2 + D'I1 = 0 Average the waveforms:
  • 34. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis34 Equate average values to zero iC2(t) I2 – V2 / R (= 0) tDTs D'Ts Capacitor current iC2(t) waveform 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. iC2 = I2 – V2 R = 0
  • 35. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis35 Cuk converter conversion ratio M = V/VgM(D) D -5 -4 -3 -2 -1 0 0 0.2 0.4 0.6 0.8 1 M(D) = V2 Vg = – D 1 – D
  • 36. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis36 Inductor current waveforms di1(t) dt = vL1(t) L1 = Vg L1 di2(t) dt = vL2(t) L2 = – V1 – V2 L2 Interval 1 slopes, using small ripple approximation: Interval 2 slopes: di1(t) dt = vL1(t) L1 = Vg – V1 L1 di2(t) dt = vL2(t) L2 = – V2 L2 i1(t) tDTs Ts I1 ∆i1 Vg – V1 L1 Vg L1 – V2 L2 – V1 – V2 L2 i2(t) tDTs Ts I2 ∆i2
  • 37. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis37 Capacitor C1 waveform dv1(t) dt = iC1(t) C1 = I2 C1 Subinterval 1: Subinterval 2: dv1(t) dt = iC1(t) C1 = I1 C1 I1 C1 I2 C1 v1(t) tDTs Ts V1 ∆v1
  • 38. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis38 Ripple magnitudes ∆i1 = VgDTs 2L1 ∆i2 = V1 + V2 2L2 DTs ∆v1 = – I2DTs 2C1 Use dc converter solution to simplify: ∆i1 = VgDTs 2L1 ∆i2 = VgDTs 2L2 ∆v1 = VgD2 Ts 2D'RC1 Analysis results Q: How large is the output voltage ripple?
  • 39. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis39 2.5 Estimating ripple in converters containing two-pole low-pass filters Buck converter example: Determine output voltage ripple Inductor current waveform. What is the capacitor current? + – L C R + vC(t) – 1 2 iC(t) iR(t) iL(t) Vg – V L Vg – V L iL(t) t0 DTs Ts I iL(0) iL(DTs) ∆iL
  • 40. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis40 Capacitor current and voltage, buck example 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. iC(t) vC(t) t t Total charge q DTs D'Ts Ts /2 V ∆iL ∆v ∆v
  • 41. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis41 Estimating capacitor voltage ripple ∆v q = C (2∆v) 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 (change in charge) = C (change in voltage) iC(t) vC(t) t t Total charge q DTs D'Ts Ts /2 V ∆iL ∆v ∆v
  • 42. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis42 Estimating capacitor voltage ripple ∆v The total charge q is the area of the triangle, as shown: q = 1 2 ∆iL Ts 2 Eliminate q and solve for ∆v: ∆v = ∆iL Ts 8 C Note: in practice, capacitor equivalent series resistance (esr) further increases ∆v. iC(t) vC(t) t t Total charge q DTs D'Ts Ts /2 V ∆iL ∆v ∆v
  • 43. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis43 Inductor current ripple in two-pole filters Example: problem 2.9 can use similar arguments, with λ = L (2∆i) λ = inductor flux linkages = inductor volt-seconds R + v – + – C2 L2L1 C1 + vC1 – i1 iT i2 D1 Q1 Vg vL(t) iL(t) t t Total flux linkage λ DTs D'Ts Ts /2 I ∆v ∆i ∆i
  • 44. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis44 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 steady- state, voltages and currents therefore involves averaging the waveforms. 2. The linear ripple approximation greatly simplifies the analysis. In a well- designed 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.
  • 45. Fundamentals of Power Electronics Chapter 2: Principles of steady-state converter analysis45 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 steady- state, 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.