Chapter 5. The Discontinuous Conduction Mode 5.1. Origin of the discontinuous conduction mode, and mode boundary 5.2. Analysis of the conversion ratio M(D,K) 5.3. Boost converter example 5.4. Summary of results and key points Fundamentals of Power Electronics 1 Chapter 5: Discontinuous conduction mode Introduction to Discontinuous Conduction Mode (DCM) ● ● ● ● Occurs because switching ripple in inductor current or capacitor voltage causes polarity of applied switch current or voltage to reverse, such that the current- or voltage-unidirectional assumptions made in realizing the switch are violated. Commonly occurs in dc-dc converters and rectifiers, having singlequadrant switches. May also occur in converters having two-quadrant switches. Typical example: dc-dc converter operating at light load (small load current). Sometimes, dc-dc converters and rectifiers are purposely designed to operate in DCM at all loads. Properties of converters change radically when DCM is entered: M becomes load-dependent Output impedance is increased Dynamics are altered Control of output voltage may be lost when load is removed Fundamentals of Power Electronics 2 Chapter 5: Discontinuous conduction mode 5.1. Origin of the discontinuous conduction mode, and mode boundary Buck converter example, with single-quadrant switches L Q1 continuous conduction mode (CCM) iL(t) Vg + – D1 C R iD(t) + iL(t) V I ∆iL – Minimum diode current is (I – ∆iL) Dc component I = V/R Current ripple is (V – V) Vg DD'Ts ∆iL = g DTs = 2L 2L 0 conducting devices: DTs Ts D1 Q1 t Q1 iD(t) I ∆iL Note that I depends on load, but ∆iL does not. 0 Fundamentals of Power Electronics 3 DTs Ts t Chapter 5: Discontinuous conduction mode Reduction of load current Increase R, until I = ∆iL CCM-DCM boundary L Q1 iL(t) + iL(t) Vg + – D1 C R V iD(t) – I ∆iL 0 Minimum diode current is (I – ∆iL) Dc component I = V/R Current ripple is (V – V) Vg DD'Ts ∆iL = g DTs = 2L 2L conducting devices: Ts D1 Q1 t Q1 iD(t) Note that I depends on load, but ∆iL does not. Fundamentals of Power Electronics DTs I ∆iL 0 4 DTs Ts t Chapter 5: Discontinuous conduction mode Further reduce load current Increase R some more, such that I < ∆iL L Q1 iL(t) + iL(t) Vg + – D1 Discontinuous conduction mode C R V iD(t) – I Minimum diode current is (I – ∆iL) Dc component I = V/R Current ripple is (V – V) Vg DD'Ts ∆iL = g DTs = 2L 2L 0 conducting devices: t Ts D2Ts D1 D3Ts Q1 X iD(t) Note that I depends on load, but ∆iL does not. The load current continues to be positive and non-zero. Fundamentals of Power Electronics DTs D1Ts Q1 0 DTs Ts t D2Ts 5 Chapter 5: Discontinuous conduction mode Mode boundary I > ∆iL for CCM I < ∆iL for DCM Insert buck converter expressions for I and ∆iL : DVg DD'TsVg < R 2L Simplify: 2L < D' RTs This expression is of the form where Fundamentals of Power Electronics K < K crit(D) for DCM K = 2L and K crit(D) = D' RTs 6 Chapter 5: Discontinuous conduction mode K and Kcrit vs. D for K < 1: 2 for K > 1: K < Kcrit: DCM Kc ( rit D) = 1 1 –D K > Kcrit: CCM 2 K > Kcrit: CCM K = 2L/RTs Kc ( rit D) = 1–D 1 K = 2L/RTs 0 0 0 Fundamentals of Power Electronics 1 D 7 0 1 D Chapter 5: Discontinuous conduction mode Critical load resistance Rcrit Solve Kcrit equation for load resistance R: where Fundamentals of Power Electronics R < Rcrit(D) for CCM for DCM R > Rcrit(D) Rcrit(D) = 2L D'Ts 8 Chapter 5: Discontinuous conduction mode Summary: mode boundary K > K crit(D) K < K crit(D) or or R < Rcrit(D) R > Rcrit(D) for CCM for DCM Table 5.1. CCM-DCM mode boundaries for the buck, boost, and buck-boost converters Converter K crit(D) max ( K crit ) 0≤D≤1 Buck (1 – D) Boost D (1 – D)2 1 4 27 (1 – D)2 1 Buck-boost Fundamentals of Power Electronics 9 R crit(D) 2L (1 – D)T s 2L D (1 – D) 2 T s 2L (1 – D) 2 T s min ( Rcrit ) 0≤D≤1 2 L Ts 27 L 2 Ts 2 L Ts Chapter 5: Discontinuous conduction mode 5.2. Analysis of the conversion ratio M(D,K) Analysis techniques for the discontinuous conduction mode: Inductor volt-second balance vL = 1 Ts Ts vL(t) dt = 0 0 Capacitor charge balance iC = 1 Ts Ts iC(t) dt = 0 0 Small ripple approximation sometimes applies: v(t) ≈ V because ∆v << V i(t) ≈ I is a poor approximation when ∆i > I Converter steady-state equations obtained via charge balance on each capacitor and volt-second balance on each inductor. Use care in applying small ripple approximation. Fundamentals of Power Electronics 10 Chapter 5: Discontinuous conduction mode Example: Analysis of DCM buck converter M(D,K) L iL(t) + vL(t) – subinterval 1 Vg + – + iC(t) C R v(t) – L Q1 iL(t) + iL(t) Vg + – D1 C iD(t) R subinterval 2 V L + vL(t) – Vg + – + iC(t) C R – v(t) – iL(t) L + vL(t) – subinterval 3 Vg + – C + iC(t) R v(t) – Fundamentals of Power Electronics 11 Chapter 5: Discontinuous conduction mode Subinterval 1 iL(t) vL(t) = Vg – v(t) iC(t) = iL(t) – v(t) / R L + vL(t) – Vg Small ripple approximation for v(t) (but not for i(t)!): + – C + iC(t) R v(t) – vL(t) ≈ Vg – V iC(t) ≈ iL(t) – V / R Fundamentals of Power Electronics 12 Chapter 5: Discontinuous conduction mode Subinterval 2 iL(t) vL(t) = – v(t) iC(t) = iL(t) – v(t) / R L + vL(t) – Vg Small ripple approximation for v(t) but not for i(t): + – C + iC(t) R v(t) – vL(t) ≈ – V iC(t) ≈ iL(t) – V / R Fundamentals of Power Electronics 13 Chapter 5: Discontinuous conduction mode Subinterval 3 iL(t) vL = 0, iL = 0 iC(t) = iL(t) – v(t) / R L + vL(t) – Vg Small ripple approximation: + – C + iC(t) R v(t) – vL(t) = 0 iC(t) = – V / R Fundamentals of Power Electronics 14 Chapter 5: Discontinuous conduction mode Inductor volt-second balance vL(t) Vg – V D1Ts D2Ts D3Ts 0 Ts t –V Volt-second balance: vL(t) = D1(Vg – V) + D2( – V) + D3(0) = 0 Solve for V: V = Vg Fundamentals of Power Electronics D1 D1 + D2 note that D2 is unknown 15 Chapter 5: Discontinuous conduction mode Capacitor charge balance L node equation: iL(t) iL(t) = iC(t) + V / R v(t)/R + iC(t) capacitor charge balance: C R iC = 0 v(t) – hence iL = V / R must compute dc component of inductor current and equate to load current (for this buck converter example) Fundamentals of Power Electronics iL(t) Vg – V L <iL> = I 0 –V L DTs D1Ts 16 ipk Ts D2Ts t D3Ts Chapter 5: Discontinuous conduction mode Inductor current waveform peak current: iL(t) Vg – V iL(D1Ts) = i pk = D 1T s L average current: iL = 1 Ts Vg – V L <iL> = I ipk –V L Ts iL(t) dt 0 0 DTs D1Ts Ts D 2T s t D3 Ts triangle area formula: Ts 0 iL(t) dt = 1 i pk (D1 + D2)Ts 2 iL = (Vg – V) equate dc component to dc load current: V = D1Ts (D + D ) (V – V) 1 2 g R 2L D 1T s (D1 + D2) 2L Fundamentals of Power Electronics 17 Chapter 5: Discontinuous conduction mode Solution for V Two equations and two unknowns (V and D2): V = Vg D1 D1 + D2 V = D1Ts (D + D ) (V – V) 1 2 g R 2L (from inductor volt-second balance) (from capacitor charge balance) Eliminate D2 , solve for V : V = 2 Vg 1 + 1 + 4K / D 21 where K = 2L / RTs valid for K < K crit Fundamentals of Power Electronics 18 Chapter 5: Discontinuous conduction mode Buck converter M(D,K) 1.0 K = 0.01 M(D,K) 0.8 K = 0.1 0.6 D K = 0.5 0.4 M= K≥1 0.2 1+ 2 1 + 4K / D 2 for K > K crit for K < K crit 0.0 0.0 0.2 0.4 0.6 0.8 1.0 D Fundamentals of Power Electronics 19 Chapter 5: Discontinuous conduction mode 5.3. Boost converter example D1 i (t) D L i(t) + vL(t) – Vg + iC(t) + – Q1 C R v(t) – Mode boundary: Previous CCM soln: I > ∆iL for CCM I < ∆iL for DCM Fundamentals of Power Electronics I= 20 Vg D' 2 R ∆iL = Vg DTs 2L Chapter 5: Discontinuous conduction mode Mode boundary Vg DTsVg > 2L D' 2R 2L > DD' 2 RTs where 4 Kcrit ( 13 ) = 27 0.15 for CCM Kcrit(D) for CCM 0.1 K > K crit(D) for CCM for DCM K < K crit(D) K = 2L and K crit(D) = DD' 2 RTs 0.05 0 0 0.2 0.4 0.6 0.8 1 D Fundamentals of Power Electronics 21 Chapter 5: Discontinuous conduction mode Mode boundary CCM 0.15 where K > K crit(D) for CCM for DCM K < K crit(D) K = 2L and K crit(D) = DD' 2 RTs DCM K < Kcrit CCM K > Kcrit 0.1 K (D) K crit 0.05 0 0 0.2 0.4 0.6 0.8 1 D Fundamentals of Power Electronics 22 Chapter 5: Discontinuous conduction mode Conversion ratio: DCM boost L i(t) + vL(t) – subinterval 1 Vg + – + iC(t) C R v(t) – i(t) D1 i (t) D L + vL(t) – Vg + – i(t) + iC(t) Q1 C + vL(t) – subinterval 2 R Vg v(t) L + – + iC(t) C R v(t) – – i(t) L + vL(t) – subinterval 3 Vg + – + iC(t) C R v(t) – Fundamentals of Power Electronics 23 Chapter 5: Discontinuous conduction mode Subinterval 1 i(t) vL(t) = Vg iC(t) = – v(t) / R + vL(t) – + – Vg Small ripple approximation for v(t) (but not for i(t)!): + iC(t) C R v(t) – vL(t) ≈ Vg iC(t) ≈ – V / R Fundamentals of Power Electronics L 0 < t < D1Ts 24 Chapter 5: Discontinuous conduction mode Subinterval 2 i(t) vL(t) = Vg – v(t) iC(t) = i(t) – v(t) / R L + vL(t) – Vg + – C Small ripple approximation for v(t) but not for i(t): R v(t) – vL(t) ≈ Vg – V iC(t) ≈ i(t) – V / R Fundamentals of Power Electronics + iC(t) D1Ts < t < (D1 +D2)Ts 25 Chapter 5: Discontinuous conduction mode Subinterval 3 i(t) vL = 0, i = 0 iC(t) = – v(t) / R L + vL(t) – Vg + – C Small ripple approximation: R v(t) – vL(t) = 0 iC(t) = – V / R Fundamentals of Power Electronics + iC(t) (D1 +D2)Ts < t < Ts 26 Chapter 5: Discontinuous conduction mode Inductor volt-second balance vL(t) Vg D1Ts D2Ts D3Ts 0 Ts t Vg – V Volt-second balance: D1Vg + D2(Vg – V) + D3(0) = 0 Solve for V: D + D2 V= 1 Vg D2 Fundamentals of Power Electronics note that D2 is unknown 27 Chapter 5: Discontinuous conduction mode Capacitor charge balance node equation: D1 i (t) D iD(t) = iC(t) + v(t) / R capacitor charge balance: + iC(t) iC = 0 C hence R v(t) – iD = V / R must compute dc component of diode current and equate to load current (for this boost converter example) Fundamentals of Power Electronics 28 Chapter 5: Discontinuous conduction mode Inductor and diode current waveforms i(t) peak current: Vg i pk = DT L 1 s L Vg – V L average diode current: iD = 1 Ts Ts iD(t) dt 0 0 0 triangle area formula: Ts ipk Vg DTs D 1T s iD(t) Ts D2Ts t D3Ts ipk iD(t) dt = 1 i pk D2Ts 2 Vg – V L <iD> 0 DTs D1Ts Fundamentals of Power Electronics 29 Ts D2Ts t D3Ts Chapter 5: Discontinuous conduction mode Equate diode current to load current average diode current: iD V g D 1 D 2T s 1 1 = i DT = Ts 2 pk 2 s 2L equate to dc load current: V g D 1 D 2T s V = R 2L Fundamentals of Power Electronics 30 Chapter 5: Discontinuous conduction mode Solution for V Two equations and two unknowns (V and D2): D + D2 (from inductor volt-second balance) V= 1 Vg D2 V g D 1 D 2T s V = R 2L (from capacitor charge balance) Eliminate D2 , solve for V. From volt-sec balance eqn: Vg D2 = D1 V – Vg Substitute into charge balance eqn, rearrange terms: V 2gD 21 V – VVg – =0 K 2 Fundamentals of Power Electronics 31 Chapter 5: Discontinuous conduction mode Solution for V V 2gD 21 V – VVg – =0 K 2 Use quadratic formula: V = 1± Vg 1 + 4D 21 / K 2 Note that one root leads to positive V, while other leads to negative V. Select positive root: V = M(D ,K) = 1 + 1 Vg where valid for 1 + 4D 21 / K 2 K = 2L / RTs K < Kcrit(D) Transistor duty cycle D = interval 1 duty cycle D1 Fundamentals of Power Electronics 32 Chapter 5: Discontinuous conduction mode Boost converter characteristics 5 0.0 1 M(D,K) K= 4 3 K = 0 K= 2 K≥ 5 0.0 M = .1 1 1–D 1+ for K > K crit 1 + 4D 2 / K 2 for K < K crit 7 4/2 1 Approximate M in DCM: 0 0 0.25 0.5 0.75 1 D Fundamentals of Power Electronics 33 M≈1+ D 2 K Chapter 5: Discontinuous conduction mode Summary of DCM characteristics Table 5.2. S ummary of CCM-DCM characteristics for the buck, boost, and buck-boost converters K crit (D) Converter Buck (1 – D) Boost D (1 – D)2 (1 – D)2 Buck-boost with DCM M(D,K) 2 1 + 1 + 4K / D 2 1 + 1 + 4D 2 / K 2 – D K K = 2L / RT s. Fundamentals of Power Electronics DCM D2(D,K) K M(D,K) D K M(D,K) D K CCM M(D) D 1 1–D – D 1–D DCM occurs for K < K crit . 34 Chapter 5: Discontinuous conduction mode Summary of DCM characteristics DCM M(D,K) st • DCM buck and boost characteristics are asymptotic to M = 1 and to the DCM buck-boost characteristic 1) o Bo ×– t( os o k-b 1 K c Bu • DCM buck-boost characteristic is linear 1 • CCM and DCM characteristics intersect at mode boundary. Actual M follows characteristic having larger magnitude Buck 0 0 0.2 0.4 0.6 0.8 • DCM boost characteristic is nearly linear D Fundamentals of Power Electronics 1 35 Chapter 5: Discontinuous conduction mode Summary of key points 1. The discontinuous conduction mode occurs in converters containing current- or voltage-unidirectional switches, when the inductor current or capacitor voltage ripple is large enough to cause the switch current or voltage to reverse polarity. 2. Conditions for operation in the discontinuous conduction mode can be found by determining when the inductor current or capacitor voltage ripples and dc components cause the switch on-state current or off-state voltage to reverse polarity. 3. The dc conversion ratio M of converters operating in the discontinuous conduction mode can be found by application of the principles of inductor volt-second and capacitor charge balance. Fundamentals of Power Electronics 36 Chapter 5: Discontinuous conduction mode Summary of key points 4. Extra care is required when applying the small-ripple approximation. Some waveforms, such as the output voltage, should have small ripple which can be neglected. Other waveforms, such as one or more inductor currents, may have large ripple that cannot be ignored. 5. The characteristics of a converter changes significantly when the converter enters DCM. The output voltage becomes loaddependent, resulting in an increase in the converter output impedance. Fundamentals of Power Electronics 37 Chapter 5: Discontinuous conduction mode