Ch18 Low Harmonic Modeling and Control

# Ch18 Low Harmonic Modeling and Control - Chapter 18 Low...

• Notes
• 39

This preview shows pages 1–8. Sign up to view the full content.

ECEN5807 Power Electronics 2 1 Chapter 18: Low harmonic rectifier modeling and control 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

This preview has intentionally blurred sections. Sign up to view the full version.

ECEN5807 Power Electronics 2 2 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 + Q 1 L C R + v(t) D 1 v g (t) i g (t) R L i(t) + R + v(t) v g (t) i g (t) R L i(t) DR on + D' : 1 V F Dc-dc boost converter circuit Averaged dc model
ECEN5807 Power Electronics 2 3 Chapter 18: Low harmonic rectifier modeling and control Modeling the ac-dc boost rectifier R v ac (t) i ac (t) + v g (t) i g (t) + v(t) i d (t) Q 1 L C D 1 controller i(t) R L + R + v(t) = V v g (t) i g (t) R L i(t) = I d(t) R on + d'(t) : 1 V F i d (t) C (large) Boost rectifier circuit Averaged model

This preview has intentionally blurred sections. Sign up to view the full version.

ECEN5807 Power Electronics 2 4 Chapter 18: Low harmonic rectifier modeling and control Boost rectifier waveforms 0 2 4 6 8 10 0 100 200 300 v g (t) v g (t) i g (t) i g (t) 0 ° 30 ° 60 ° 90 ° 120 ° 150 ° 180 ° d(t) 0 0.2 0.4 0.6 0.8 1 0 ° 30 ° 60 ° 90 ° 120 ° 150 ° 180 ° 0 1 2 3 4 5 6 i d (t) i(t) = I ϖ t 0 ° 30 ° 60 ° 90 ° 120 ° 150 ° 180 ° Typical waveforms (low frequency components) i g ( t ) = v g ( t ) R e
ECEN5807 Power Electronics 2 5 Chapter 18: Low harmonic rectifier modeling and control Example: boost rectifier with MOSFET on-resistance + R + v(t) = V v g (t) i g (t) i(t) = I d(t) R on d'(t) : 1 i d (t) C (large) Averaged model Inductor dynamics are neglected, a good approximation when the ac line variations are slow compared to the converter natural frequencies

This preview has intentionally blurred sections. Sign up to view the full version.

ECEN5807 Power Electronics 2 6 Chapter 18: Low harmonic rectifier modeling and control 18.1.1 Expression for controller duty cycle d(t) + R + v(t) = V v g (t) i g (t) i(t) = I d(t) R on d'(t) : 1 i d (t) C (large) Solve input side of model: i g ( t ) d ( t ) R on = v g ( t ) – d '( t ) v with i g ( t ) = v g ( t ) R e eliminate i g ( t ) : v g ( t ) R e d ( t ) R on = v g ( t ) – d '( t ) v v g ( t ) = V M sin ϖ t solve for d ( t ) : d ( t ) = v v g ( t ) 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 7 Chapter 18: Low harmonic rectifier modeling and control 18.1.2 Expression for the dc load current + R + v(t) = V v g (t) i g (t) i(t) = I d(t) R on d'(t) : 1 i d (t) C (large) Solve output side of model, using charge balance on capacitor C : I = i d T ac i d ( t ) = d '( t ) i g ( t ) = d '( t ) v g ( t ) R e But d’ ( t ) is: d '( t ) = v g ( t ) 1 – R on R e v

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '11
• ECEN5807 Power Electronics

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern