1st course
- 1. Power Electronics 2 (H5CPE2) Dr Christian Klumpner Power Electronics, Machines and Control Group School of Electrical and Electronic Engineering, UoN Tower Building, 508 email: christian.klumpner@nottingham.ac.uk Module webpage: www.eee.nott.ac.uk/teaching/h5cpe2
- 2. Introduction Line-side Converter Motor-side Converter AC DC DC AC DC-link Typical AC/DC/AC power conversion (Adjustable Speed Drive) • AC/DC converter (Rectifier): fixed voltage&frequency to DC voltage • DC/AC converter (Inverter): DC voltage to variable voltage/variable frequency Why AC/DC/AC? Electrolytic capacitors (compact and cheap), only unipolar devices available (transistors) Operation of rectifier, stress on devices ($), design of filter ($), operation of inverter
- 3. Introduction Pre requisites Circuit theory and electronics at first year undergraduate level, knowledge of switching regulators and single phase rectification (controlled and uncontrolled) such as that provided by module H5BPE1. Aims and objectives of the module The aim of this module is to provide an in depth knowledge of power electronics at a level suitable for final year undergraduate students. Since power electronics is a rapidly growing subject the course tries to reflect this by covering the well established and widely used technologies (such as three phase rectification) as well as more recent developments such as resonant converters. The increasing importance of power quality is also addressed and various high power factor utility interface circuits are discussed. Inverter circuits employing pulse width modulation (PWM) are studied due to their very widespread use in variable speed drives and power supply systems. High power (multi-level) converter structures are then discussed. Throughout the course, emphasis is placed on circuits and their applications rather than on the technology of power switching devices.
- 4. Lecture course syllabus Lecture TOPIC 1 Introduction to the course, review of 3-phase supplies and the associated waveforms. 2-3 3-phase uncontrolled (diode) rectifiers. Basic mode of operation and waveforms. Concept and importance of power factor, displacement factor and distortion factor applied to power electronic equipment. 4-5 Overlap in diode rectifiers, waveforms and calculations. Introduction to thyristor characteristics. 6-7 3-Phase controlled rectification, waveforms and calculations, effect of overlap. Power factor calculations. Inversion. 8-9 Smoothing circuits. Capacitive smoothing, waveforms and analysis. Inductive smoothing, waveforms and analysis, discontinuous current. Multiple converter circuits and HVDC. 10-12 Resonant converters, review of hard switching, introduction to soft switching and different types of resonant switches and converters. Forward converter employing zero voltage switching, analysis and waveforms. 13-15 Single phase inverters, the H-bridge circuit and its operation, applications, quasi-square wave and PWM techniques for voltage and frequency control, typical frequency spectra, relationship between AC and DC side harmonics. 16-17 3-phase PWM inverters, High power (multi-level) converter structures. 18-20 High power factor utility interface circuits, single switch boost converter with input current wave shaping. PWM rectifiers (pulse converters), control strategies.
- 5. Recommendations Booklist There are no essential books for this course. However, the following book is excellent and covers most of the material in this course and the second year power electronics course. POWER ELECTRONICS: Converters, Applications and Design (2-ed) by Mohan, Undeland and Robbins, Wiley publishing Another book worth looking at for power electronics in general, rather than specifically this course is: ELEMENTS OF POWER ELECTRONICS, by Philip T Krein, Oxford University Press - familiarize yourself with emergency exits (fire alarm) in the building - don’t get late (not more than 5 minutes) into the classroom - switch off mobile phones - attend to the course equipped with a ruler, 4 or more colored pens/markers - if you have a computer at home, install a simulation pack (PSPICE, Simcad)
- 6. Review of 3-phase supplies (1) Why sinusoidal voltage? Behavior of passive components Resistor Inductor Capacitor v 1 dv i= i = ∫ v ×dt i=C R L dt Proportional Integrative Derivative Rectangular Voltage: Rectangular Current Triangular Current Pulse Current Production, transport & distribution system = Resistors + Inductors + Capacitors We need to preserve the voltage waveform
- 7. Review of 3-phase supplies (2) We need a supply voltage waveform which preserves its shape when is derivated or integrated ⇒ sinusoidal Resistor Inductor Capacitor Behavior of passive components v 1 dv i= R i= L ∫ v ×dt i=C dt Proportional Integrative Derivative Sinusoidal Voltage E i=− cos ( ω t ) = i = C ×E ×ω ×cos(ω t ) E × ( ωt ) sin ωL v = E ×sin ( ω t ) i= π R E π = C ×E ×ω ×sin(ω t − ) = sin ω t + ÷ 2 ωL 2 Sinusoidal Current Sinusoidal Current Sinusoidal Current
- 8. Review of 3-phase supplies (3) Assume a “STAR” connected supply In practice, the 3 voltage sources represent the voltages generated by 3 coils (physically displaced by 120O from each other) in an AC rotating machine (Alternator) Line A A Phasor diagram Phase Line to VCA Neutral voltage line VAN voltage VAB VCN N VBN VBC C B “Line to line” voltage often called “line voltage”
- 9. Review of 3-phase supplies (4) Assuming the peak phase voltage is E (a convention used throughout the course) then: VAN = E sin(ωt ) B lags A by 120O, C lags B VBN = E sin(ωt − 2π / 3) by 120O etc VCN = E sin(ωt − 4π / 3) = E sin(ωt + 2π / 3) This is for “phase sequence” A-B-C, A-C-B is also possible – we will always assume A-B-C Drawing a phasor diagram and converting back to time functions, it is easy to show that the line voltages are given by: VAB = 3E sin(ωt + π / 6) VBC = 3E sin(ωt − π / 2) VCA = 3E sin(ωt + 5π / 6) 3-phase supplies are specified using the RMS line voltage. Hence “a 415V, 50Hz, 3-phase system” means: 3E = 415 V, ω = 100π 2
- 10. Review of 3-phase supplies (5) Why three-phase voltage systems (120O displaced)? E ×I p = v ×i = E ×sin ( ω t ) ×I sin ( ω t + ϕ ) = sin ( 2ω t + ϕ ) + cos(ϕ ) 2 Displacement angle = 0O Displacement angle = 90O Necessity to deliver - smooth power (require less filtering) - smooth torque in a motor (less mechanical stress, noise)