Chapter 10 Ac and Dc Equivalent Circuit Modeling of the Discontinuous Conduction Mode Introduction 10.1. DCM Averaged Switch Model 10.2. Small-Signal AC Modeling of the DCM Switch Network 10.3. Generalized Averaged Switch Modeling 10.4. Summary of Key Points Chapter 10: Ac and dc equivalent circuit modeling 1 of the discontinuous conduction mode Fundamentals of Power Electronics
We are missing ac and dc equivalent circuit models for the discontinuous conduction mode DC AC e ( s ) d ( s ) L e 1 : M(D) 1 : M(D) + – + + + + j ( s ) d ( s ) V g R V v ( s ) R v g ( s ) C CCM – – – – + + DCM ? + + ? V g R V v ( s ) R v g ( s ) – – – – Chapter 10: Ac and dc equivalent circuit modeling 2 of the discontinuous conduction mode Fundamentals of Power Electronics
Change in characteristics at the CCM/DCM boundary Steady-state output voltage becomes strongly load-dependent ● Simpler dynamics: one pole and the RHP zero are moved to very high ● frequency, and can normally be ignored Traditionally, boost and buck-boost converters are designed to operate ● in DCM at full load All converters may operate in DCM at light load ● So we need equivalent circuits that model the steady-state and small- signal ac models of converters operating in DCM The averaged switch approach will be employed Chapter 10: Ac and dc equivalent circuit modeling 3 of the discontinuous conduction mode Fundamentals of Power Electronics
10.1 Derivation of DCM averaged switch model: buck-boost example Switch network • Define switch terminal i 1 i 2 quantities v 1 , i 1 , v 2 , i 2 , as + – shown + v 1 v 2 • Let us find the averaged + v g – + C R v quantities 〈 v 1 〉 , 〈 i 1 〉 , 〈 v 2 〉 , – + 〈 i 2 〉 , for operation in DCM, L v L – and determine the – i L relations between them Chapter 10: Ac and dc equivalent circuit modeling 4 of the discontinuous conduction mode Fundamentals of Power Electronics
DCM waveforms i L (t) i 1 (t) i pk i pk Area q 1 v v g L L i 1 ( t ) T s 0 t v L (t) v g – v v g v 1 (t) v 1 ( t ) T s v g 0 0 v i 2 (t) i pk Area q 2 Switch network i 1 i 2 i 2 ( t ) T s + – + v 1 v 2 v g – v v 2 (t) + v g – + C R v v 2 ( t ) T s – – v + L v L 0 – – i L t d 1 T s d 2 T s d 3 T s T s Chapter 10: Ac and dc equivalent circuit modeling 5 of the discontinuous conduction mode Fundamentals of Power Electronics
Basic DCM equations i 1 (t) Peak inductor current: i pk Area q 1 i pk = v g L d 1 T s i 1 ( t ) T s Average inductor voltage: v g – v v 1 (t) T s + d 3 ⋅ 0 v L ( t ) T s = d 1 v g ( t ) T s + d 2 v ( t ) v 1 ( t ) T s v g 0 In DCM, the diode switches off when the i 2 (t) inductor current reaches zero. Hence, i (0) i pk Area q 2 = i ( T s ) = 0 , and the average inductor voltage is zero. This is true even during i 2 ( t ) T s transients. v L ( t ) T s = d 1 ( t ) v g ( t ) T s + d 2 ( t ) v ( t ) T s = 0 v g – v v 2 (t) v 2 ( t ) T s – v Solve for d 2 : 0 v g ( t ) t T s d 2 ( t ) = – d 1 ( t ) d 1 T s d 2 T s d 3 T s T s v ( t ) Chapter 10: Ac and dc equivalent circuit modeling T s 6 of the discontinuous conduction mode Fundamentals of Power Electronics
Average switch network terminal voltages i 1 (t) Average the v 1 ( t ) waveform: i pk Area q 1 T s = d 1 ( t ) ⋅ 0 + d 2 ( t ) v 1 ( t ) v g ( t ) T s – v ( t ) T s + d 3 ( t ) v g ( t ) T s i 1 ( t ) T s Eliminate d 2 and d 3 : v g – v v 1 (t) v 1 ( t ) T s = v g ( t ) T s v 1 ( t ) T s v g Similar analysis for v 2 ( t ) waveform leads to 0 i 2 (t) i pk T s + d 2 ( t ) ⋅ 0 + d 3 ( t ) – v ( t ) Area q 2 v 2 ( t ) T s = d 1 ( t ) v g ( t ) T s – v ( t ) T s i 2 ( t ) T s = – v ( t ) T s v g – v v 2 (t) v 2 ( t ) T s – v 0 t d 1 T s d 2 T s d 3 T s T s Chapter 10: Ac and dc equivalent circuit modeling 7 of the discontinuous conduction mode Fundamentals of Power Electronics
Average switch network terminal currents Average the i 1 ( t ) waveform: i 1 (t) i pk Area q 1 t + Ts = q 1 T s = 1 i 1 ( t ) i 1 ( t ) dt T s T s i 1 ( t ) T s t The integral q 1 is the area under the i 1 ( t ) waveform during first subinterval. Use triangle v g – v v 1 (t) area formula: v 1 ( t ) T s v g t + Ts = 1 0 q 1 = i 1 ( t ) dt 2 d 1 T s i pk i 2 (t) t i pk Area q 2 Eliminate i pk : 2 ( t ) T s T s = d 1 i 1 ( t ) v 1 ( t ) i 2 ( t ) T s 2 L T s Note 〈 i 1 ( t ) 〉 Ts is not equal to d 〈 i L ( t ) 〉 Ts ! v g – v v 2 (t) v 2 ( t ) T s – v Similar analysis for i 2 ( t ) waveform leads to 2 0 v 1 ( t ) 2 ( t ) T s T s = d 1 T s t i 2 ( t ) d 1 T s d 2 T s d 3 T s 2 L v 2 ( t ) T s T s Chapter 10: Ac and dc equivalent circuit modeling 8 of the discontinuous conduction mode Fundamentals of Power Electronics
Input port: Averaged equivalent circuit i 1 ( t ) T s 2 ( t ) T s T s = d 1 i 1 ( t ) v 1 ( t ) + 2 L T s v 1 ( t ) R e ( d 1 ) v 1 ( t ) T s T s i 1 ( t ) T s = R e ( d 1 ) – R e ( d 1 ) = 2 L 2 T s d 1 Chapter 10: Ac and dc equivalent circuit modeling 9 of the discontinuous conduction mode Fundamentals of Power Electronics
Output port: Averaged equivalent circuit i(t) 2 + v 1 ( t ) 2 ( t ) T s T s = d 1 T s i 2 ( t ) 2 L v 2 ( t ) T s p(t) v(t) – 2 v 1 ( t ) T s i 2 ( t ) T s v 2 ( t ) T s = R e ( d 1 ) = p ( t ) T s Chapter 10: Ac and dc equivalent circuit modeling 10 of the discontinuous conduction mode Fundamentals of Power Electronics
The dependent power source i(t) i(t) + v(t)i(t) = p(t) p(t) v(t) – v(t) • Must avoid open- and short-circuit connections of power sources • Power sink: negative p(t) Chapter 10: Ac and dc equivalent circuit modeling 11 of the discontinuous conduction mode Fundamentals of Power Electronics
How the power source arises in lossless two-port networks In a lossless two-port network without internal energy storage: instantaneous input power is equal to instantaneous output power In all but a small number of special cases, the instantaneous power throughput is dependent on the applied external source and load If the instantaneous power depends only on the external elements connected to one port, then the power is not dependent on the characteristics of the elements connected to the other port. The other port becomes a source of power, equal to the power flowing through the first port A power source (or power sink) element is obtained Chapter 10: Ac and dc equivalent circuit modeling 12 of the discontinuous conduction mode Fundamentals of Power Electronics
Properties of power sources Series and parallel connection of power P 1 sources P 1 + P 2 + P 3 P 2 P 3 n 1 : n 2 Reflection of power source through a transformer P 1 P 1 Chapter 10: Ac and dc equivalent circuit modeling 13 of the discontinuous conduction mode Fundamentals of Power Electronics
The loss-free resistor (LFR) i 1 ( t ) T s i 2 ( t ) T s p ( t ) T s + + v 2 ( t ) T s v 1 ( t ) T s R e ( d 1 ) – – A two-port lossless network Input port obeys Ohm’s Law Power entering input port is transferred to output port Chapter 10: Ac and dc equivalent circuit modeling 14 of the discontinuous conduction mode Fundamentals of Power Electronics
Averaged modeling of CCM and DCM switch networks Switch network Averaged switch model i 1 ( t ) T s i 2 ( t ) T s i 1 (t) i 2 (t) 1 : d(t) + + + + CCM v 1 ( t ) T s v 2 ( t ) T s v 1 (t) v 2 (t) – – – – i 1 ( t ) T s i 2 ( t ) T s i 1 (t) i 2 (t) p ( t ) T s + + + + v 2 ( t ) T s v 1 (t) v 2 (t) v 1 ( t ) T s DCM R e ( d 1 ) – – – – Chapter 10: Ac and dc equivalent circuit modeling 15 of the discontinuous conduction mode Fundamentals of Power Electronics
Averaged switch model: buck-boost example Switch network i 1 i 2 + – Original circuit + v 1 v 2 + v g – + C R v – + L v L – – i L i 1 ( t ) T s i 2 ( t ) T s Averaged model p ( t ) T s + – + R e (d) v 2 ( t ) T s v 1 ( t ) T s + v g ( t ) T s – + C R v ( t ) Ts – L – Chapter 10: Ac and dc equivalent circuit modeling 16 of the discontinuous conduction mode Fundamentals of Power Electronics
Solution of averaged model: steady state I 1 Let + L → short circuit P + C → open circuit V g R e (D) R V – – Converter input power: 2 Equate and solve: P = V g 2 P = V g R e = V 2 R e R Converter output power: V P = V 2 R V g = ± R e R Chapter 10: Ac and dc equivalent circuit modeling 17 of the discontinuous conduction mode Fundamentals of Power Electronics
Steady-state LFR solution V R V g = ± is a general result, for any system that can R e be modeled as an LFR. For the buck-boost converter, we have R e ( D ) = 2 L D 2 T s Eliminate R e : D 2 T s R V = – D V g = – 2 L K which agrees with the previous steady-state solution of Chapter 5. Chapter 10: Ac and dc equivalent circuit modeling 18 of the discontinuous conduction mode Fundamentals of Power Electronics
Steady-state LFR solution with ac terminal waveforms i 1 (t) i 2 (t) + p(t) + v g (t) C R v(t) R e – – Converter average input Note that no average power power: flows into capacitor 2 P av = V g , rms R e Equate and solve: Converter average output V rms R V g , rms = power: 2 P av = V rms R e R Chapter 10: Ac and dc equivalent circuit modeling 19 of the discontinuous conduction mode Fundamentals of Power Electronics
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