Chapter 3. Steady-State Equivalent Circuit Modeling, Losses, and - PowerPoint PPT Presentation

Chapter 3. Steady-State Equivalent Circuit Modeling, Losses, and Efficiency 3.1. The dc transformer model 3.2. Inclusion of inductor copper loss 3.3. Construction of equivalent circuit model 3.4. How to obtain the input port of the model 3.5. Example: inclusion of semiconductor conduction losses in the boost converter model 3.6. Summary of key points 1 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

3.1. The dc transformer model I g I Basic equations of an ideal dc-dc converter: Switching + + Power Power dc-dc V g V P in = P out output input ( η = 100%) – – converter V g I g = V I D V = M ( D ) V g (ideal conversion ratio) I g = M ( D ) I control input These equations are valid in steady-state. During transients, energy storage within filter elements may cause P in ≠ P out 2 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Equivalent circuits corresponding to ideal dc-dc converter equations P in = P out V g I g = V I V = M ( D ) V g I g = M ( D ) I dependent sources Dc transformer I g I I g I 1 : M(D) + + + + Power Power Power Power + V g V M(D) I M(D)V g V g V – output input output input – – – – D control input 3 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

The dc transformer model I g I 1 : M(D) Models basic properties of + + Power Power ideal dc-dc converter: V g V output input – – • conversion of dc voltages and currents, ideally with 100% efficiency D • conversion ratio M controllable via duty cycle control input • Solid line denotes ideal transformer model, capable of passing dc voltages and currents • Time-invariant model (no switching) which can be solved to find dc components of converter waveforms 4 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Example: use of the dc transformer model 1. Original system 3. Push source through transformer M 2 (D)R 1 R 1 Switching + + + + + dc-dc V 1 V g V R M(D)V 1 V R – – – converter – – D 2. Insert dc transformer model 4. Solve circuit R 1 R 1 : M(D) V = M ( D ) V 1 R + M 2 ( D ) R 1 + + + V g V V 1 R – – – 5 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

3.2. Inclusion of inductor copper loss Dc transformer model can be extended, to include converter nonidealities. Example: inductor copper loss (resistance of winding): L R L Insert this inductor model into boost converter circuit: L R L 2 + i 1 + V g C R v – – 6 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Analysis of nonideal boost converter L R L 2 + i 1 + V g C R v – – switch in position 1 switch in position 2 L R L L R L i i + + + v L – + v L – i C i C + + V g C R v C R v V g – – – – 7 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Circuit equations, switch in position 1 L R L Inductor current and i capacitor voltage: + + v L – i C v L ( t ) = V g – i ( t ) R L + V g C R v – i C ( t ) = – v ( t ) / R – Small ripple approximation: v L ( t ) = V g – I R L i C ( t ) = – V / R 8 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Circuit equations, switch in position 2 L R L i + + v L – i C + C R v V g – – v L ( t ) = V g – i ( t ) R L – v ( t ) ≈ V g – I R L – V i C ( t ) = i ( t ) – v ( t ) / R ≈ I – V / R 9 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Inductor voltage and capacitor current waveforms v L (t) Average inductor voltage: V g – IR L Ts v L ( t ) = 1 D' T s DT s v L ( t ) dt T s 0 t = D ( V g – I R L ) + D '( V g – I R L – V ) V g – IR L – V i C (t) Inductor volt-second balance: I – V/R 0 = V g – I R L – D ' V -V/R Average capacitor current: i C ( t ) = D ( – V / R ) + D ' ( I – V / R ) Capacitor charge balance: 0 = D ' I – V / R 10 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for output voltage We now have two 5 R L / R = 0 equations and two 4.5 unknowns: R L / R = 0.01 4 0 = V g – I R L – D ' V 3.5 0 = D ' I – V / R 3 R L / R = 0.02 V / V g Eliminate I and 2.5 solve for V : 2 R L / R = 0.05 V V g = 1 1 1.5 D ' 2 R ) (1 + R L / D ' R L / R = 0.1 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D 11 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

3.3. Construction of equivalent circuit model Results of previous section (derived via inductor volt-sec balance and capacitor charge balance): v L = 0 = V g – I R L – D ' V i C = 0 = D ' I – V / R View these as loop and node equations of the equivalent circuit. Reconstruct an equivalent circuit satisfying these equations 12 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Inductor voltage equation v L = 0 = V g – I R L – D ' V L R L • Derived via Kirchoff’s voltage law, to find the inductor voltage + <v L > – + IR L – during each subinterval = 0 + + V g D' V • Average inductor voltage then – – I set to zero • This is a loop equation: the dc components of voltage around • IR L term: voltage across resistor a loop containing the inductor of value R L having current I sum to zero • D’V term: for now, leave as dependent source 13 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Capacitor current equation i C = 0 = D ' I – V / R node • Derived via Kirchoff’s current V/R law, to find the capacitor + <i C > current during each subinterval = 0 D' I C V R • Average capacitor current then set to zero – • This is a node equation: the dc components of current flowing • V/R term: current through load into a node connected to the resistor of value R having voltage V capacitor sum to zero • D’I term: for now, leave as dependent source 14 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Complete equivalent circuit Dependent sources and transformers The two circuits, drawn together: I 1 R L + + + nV 2 nI 1 V 2 – + + V g D' V D' I V R I – – – – n : 1 The dependent sources are equivalent + I 1 to a D’ : 1 transformer: V 2 R L D' : 1 + I – + • sources have same coefficient V g V R – • reciprocal voltage/current – dependence 15 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Solution of equivalent circuit Converter equivalent circuit R L D' : 1 + I + V g V R – – Refer all elements to transformer Solution for output voltage secondary: using voltage divider formula: R L / D' 2 + D' I V = V g = V g 1 R D ' R + R L D ' R L + V g / D' V R 1 + 2 R – 2 D ' D ' – 16 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for input (inductor) current R L D' : 1 + I + V g V R – – V g = V g 1 I = 2 R + R L 2 R L D ' D ' 1 + 2 R D ' 17 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Solution for converter efficiency R L D' : 1 P in = ( V g ) ( I ) + I + V g V R – P out = ( V ) ( D ' I ) – η = P out P in = ( V ) ( D ' I ) ( V g ) ( I ) = V V g D ' 1 η = R L 1 + 2 R D ' 18 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Efficiency, for various values of R L 100% 0.002 1 η = 90% R L 0.01 1 + 2 R 80% D ' 0.02 70% 0.05 60% η 50% R L /R = 0.1 40% 30% 20% 10% 0% 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 D 19 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

3.4. How to obtain the input port of the model Buck converter example —use procedure of previous section to derive equivalent circuit i g L R L 1 i L + + v L – 2 + V g C v C R – – Average inductor voltage and capacitor current: v L = 0 = DV g – I L R L – V C i C = 0 = I L – V C / R 20 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Construct equivalent circuit as usual v L = 0 = DV g – I L R L – V C i C = 0 = I L – V C / R R L + + <v L > – V C /R <i C > = 0 = 0 + V C R DV g – I L – What happened to the transformer? • Need another equation 21 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Modeling the converter input port Input current waveform i g (t) : i g (t) i L (t) ≈ I L area = DT s I L t 0 0 DT s T s Dc component (average value) of i g (t) is Ts I g = 1 i g ( t ) dt = DI L T s 0 22 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...

Input port equivalent circuit Ts I g = 1 i g ( t ) dt = DI L T s 0 + I g D I L V g – 23 Fundamentals of Power Electronics Chapter 3: Steady-state equivalent circuit modeling, ...