Chapter 18 Low Harmonic Rectifier Modeling and Control 18.1 Modeling losses and efficiency in CCM high-quality rectifiers Expression for controller duty cycle d ( t ) Expression for the dc load current Solution for converter efficiency η Design example 18.2 Controller schemes Average current control Feedforward Current programmed control Hysteretic control Nonlinear carrier control 18.3 Control system modeling Modeling the outer low-bandwidth control system Modeling the inner wide-bandwidth average current controller ECEN5807 Power Electronics 2 1 Chapter 18: Low harmonic rectifier modeling and control
18.1 Modeling losses and efficiency in CCM high-quality rectifiers Objective: extend procedure of Chapter 3, to predict the output voltage, duty cycle variations, and efficiency, of PWM CCM low harmonic rectifiers. Approach: Use the models developed in Chapter 3. Integrate over one ac line cycle to determine steady-state waveforms and average power. Boost example R L DR on V F R L L D 1 i g (t) i g (t) i(t) D' : 1 i(t) + – + + + + v g (t) R v(t) v g (t) Q 1 C R v(t) – – – – Dc-dc boost converter circuit Averaged dc model ECEN5807 Power Electronics 2 2 Chapter 18: Low harmonic rectifier modeling and control
Modeling the ac-dc boost rectifier i g (t) i d (t) i(t) Boost + + i ac (t) R L L D 1 rectifier v ac (t) v g (t) Q 1 C v(t) R circuit – – controller R L d(t) R on V F i g (t) i d (t) i(t) = I d'(t) : 1 Averaged + – + model + C v g (t) R v(t) = V – (large) – ECEN5807 Power Electronics 2 3 Chapter 18: Low harmonic rectifier modeling and control
Boost rectifier waveforms v g (t) i g (t) Typical waveforms 300 10 v g (t) 8 (low frequency components) 200 i g (t) 6 i g ( t ) = v g ( t ) 4 100 R e 2 0 0 0 ° 30 ° 60 ° 90 ° 120 ° 150 ° 180 ° 1 6 d(t) i d (t) 5 0.8 4 0.6 3 i(t) = I 0.4 2 0.2 1 0 0 0 ° 30 ° 60 ° 90 ° 120 ° 150 ° 180 ° 0 ° 30 ° 60 ° 90 ° 120 ° 150 ° 180 ° ω t ECEN5807 Power Electronics 2 4 Chapter 18: Low harmonic rectifier modeling and control
Example: boost rectifier with MOSFET on-resistance i g (t) i(t) = I d'(t) : 1 + i d (t) d(t) R on + C v g (t) R v(t) = V – (large) – Averaged model Inductor dynamics are neglected, a good approximation when the ac line variations are slow compared to the converter natural frequencies ECEN5807 Power Electronics 2 5 Chapter 18: Low harmonic rectifier modeling and control
18.1.1 Expression for controller duty cycle d(t) i g (t) Solve input side of i(t) = I d'(t) : 1 model: + i d (t) d(t) R on i g ( t ) d ( t ) R on = v g ( t ) – d '( t ) v + C v g (t) R v(t) = V – i g ( t ) = v g ( t ) (large) with R e – v g ( t ) = V M sin ω t eliminate i g ( t ) : solve for d ( t ) : v – v g ( t ) v g ( t ) d ( t ) = R e d ( t ) R on = v g ( t ) – d '( t ) v v – v g ( t ) R on R e Again, these expressions neglect converter dynamics, and assume that the converter always operates in CCM. ECEN5807 Power Electronics 2 6 Chapter 18: Low harmonic rectifier modeling and control
18.1.2 Expression for the dc load current i g (t) Solve output side of i(t) = I d'(t) : 1 model, using charge + i d (t) d(t) R on balance on capacitor C : + C I = i d v g (t) R v(t) = V – T ac (large) i d ( t ) = d '( t ) i g ( t ) = d '( t ) v g ( t ) – R e But d’ ( t ) is: hence i d ( t ) can be expressed as v g ( t ) 1 – R on R e 1 – R on d '( t ) = 2 ( t ) R e i d ( t ) = v g v – v g ( t ) R on R e R e v – v g ( t ) R on R e Next, average i d ( t ) over an ac line period, to find the dc load current I . ECEN5807 Power Electronics 2 7 Chapter 18: Low harmonic rectifier modeling and control
Dc load current I Now substitute v g ( t ) = V M sin ω t , and integrate to find 〈 i d ( t ) 〉 Tac : T ac /2 sin 2 ω t 1 – R on R e 2 V M T ac = 2 I = i d dt T ac R e v – V M R on sin ω t R e 0 This can be written in the normalized form sin 2 ω t T ac /2 2 VR e 1 – R on V M I = 2 1 – a sin ω t dt T ac R e 0 a = V M R on with V R e ECEN5807 Power Electronics 2 8 Chapter 18: Low harmonic rectifier modeling and control
Integration By waveform symmetry, we need only integrate from 0 to T ac /4 . Also, make the substitution θ = ω t : sin 2 θ π /2 2 I = V M VR e 1 – R on 2 1 – a sin θ d θ π R e 0 This integral is obtained not only in the boost rectifier, but also in the buck-boost and other rectifier topologies. The solution is sin 2 θ π /2 – 2 a – π + 4 sin – 1 a + 2 cos – 1 a 4 = F ( a ) = 2 1 – a sin θ d θ π a 2 π 1 – a 2 0 • a is typically much smaller than • Result is in closed form unity • a is a measure of the loss resistance relative to R e ECEN5807 Power Electronics 2 9 Chapter 18: Low harmonic rectifier modeling and control
The integral F ( a ) sin 2 θ π /2 – 2 a – π + 4 sin – 1 a + 2 cos – 1 a 4 = F ( a ) = 2 1 – a sin θ d θ π a 2 π 1 – a 2 0 Approximation via 1.15 polynomial: 1.1 F ( a ) ≈ 1 + 0.862 a + 0.78 a 2 1.05 For | a | ≤ 0.15 , this approximate expression is F(a) 1 within 0.1% of the exact 0.95 value. If the a 2 term is omitted, then the accuracy 0.9 drops to ±2% for | a | ≤ 0.15. 0.85 The accuracy of F(a) –0.15 –0.10 –0.05 0.00 0.05 0.10 0.15 coincides with the accuracy of the rectifier efficiency η . a ECEN5807 Power Electronics 2 10 Chapter 18: Low harmonic rectifier modeling and control
18.1.4 Solution for converter efficiency η Converter average input power is 2 T ac = V M P in = p in ( t ) 2 R e Average load power is 2 a = V M R on F ( a ) VR e 1 – R on V M P out = VI = V with 2 R e V R e So the efficiency is η = P out P in = 1 – R on F ( a ) R e Polynomial approximation: 2 η ≈ 1 – R on 1 + 0.862 V M R on R e + 0.78 V M R on R e V V R e ECEN5807 Power Electronics 2 11 Chapter 18: Low harmonic rectifier modeling and control
Boost rectifier efficiency 1 η = P out P in = 1 – R on η F ( a ) R e = 0.05 0.95 / R R e on R on / R e = 0.1 • To obtain high 0.9 efficiency, choose V R on / R e = 0.15 slightly larger than V M 0.85 2 . 0 • Efficiencies in the range = R / e R 90% to 95% can then be on 0.8 obtained, even with R on as high as 0.2 R e 0.75 0.0 0.2 0.4 0.6 0.8 1.0 • Losses other than V M / V MOSFET on-resistance are not included here ECEN5807 Power Electronics 2 12 Chapter 18: Low harmonic rectifier modeling and control
18.1.5 Design example Let us design for a given efficiency. Consider the following specifications: Output voltage 390 V 500 W Output power rms input voltage 120 V Efficiency 95% Assume that losses other than the MOSFET conduction loss are negligible. Average input power is P in = P out η = 500 W 0.95 = 526 W Then the emulated resistance is 2 = (120 V) 2 R e = V g , rms 526 W = 27.4 Ω P in ECEN5807 Power Electronics 2 13 Chapter 18: Low harmonic rectifier modeling and control
Design example V M V = 120 2 V Also, = 0.435 390 V 95% efficiency with V M / V = 0.435 occurs 1 with R on / R e ≈ 0.075. η = 0.05 0.95 / R So we require a R e on MOSFET with on R on / R e = 0.1 resistance of 0.9 R on / R e = 0.15 R on ≤ (0.075) R e 0.85 = (0.075) (27.4 Ω ) = 2 Ω 2 . 0 = R / e R on 0.8 0.75 0.0 0.2 0.4 0.6 0.8 1.0 V M / V ECEN5807 Power Electronics 2 14 Chapter 18: Low harmonic rectifier modeling and control
18.2 Controller schemes Average current control Feedforward Current programmed control Hysteretic control Nonlinear carrier control ECEN5807 Power Electronics 2 15 Chapter 18: Low harmonic rectifier modeling and control
18.2.1 Average current control i g ( t ) + L Boost example + v g ( t ) v ( t ) – Low frequency (average) component – of input current is controlled to follow Gate input voltage driver + – Pulse width modulator v a ( t ) Compensator G c ( s ) ≈ R s 〈 i g ( t ) 〉 T s – + Current reference v r ( t ) ECEN5807 Power Electronics 2 16 Chapter 18: Low harmonic rectifier modeling and control
Use of multiplier to control average power i g (t) As discussed in Chapter + 17, an output voltage feedback loop adjusts the + v g (t) v(t) C – emulated resistance R e such that the rectifier – power equals the dc load power: Gate driver + – 2 P av = V g , rms = P load Pulse width v g (t) R e modulator v a (t) Compensator G c (s) An analog multiplier Multiplier introduces the v ref 1 (t) v err (t) x – k x xy + y dependence of R e v(t) on v ( t ) . v control (t) – G cv (s) + v ref 2 (t) Voltage reference ECEN5807 Power Electronics 2 17 Chapter 18: Low harmonic rectifier modeling and control
18.2.2 Feedforward Feedforward is sometimes i g (t) used to cancel out + disturbances in the input + voltage v g ( t ) . v g (t) v(t) – To maintain a given power – throughput P av , the reference voltage v ref 1 ( t ) should be Gate driver + – v ref 1 ( t ) = P av v g ( t ) R s Pulse width v g (t) 2 modulator V g , rms v a (t) Compensator G c (s) multiplier x v ref 1 (t) Peak xy – z k v + detector V M z 2 y v control (t) – G cv (s) + v ref 2 (t) Voltage reference ECEN5807 Power Electronics 2 18 Chapter 18: Low harmonic rectifier modeling and control
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