2011_notes

Department of Electrical and Computer Engineering

ELE754 Course Notes

Power Electronics

2011 Edition

Bin Wu and David Xu

ELE 754 Power Electronics 1

Contents 1. Course Outline………………………………………………………………………………. 2

2. Design Projects………………………………………………………………….…………… 5 2.1 Project 1- Microprocessor Controlled dc-dc Converter and dc Motor Drive……………….. 5 Part A Preliminary Study……………………………………………………….. 5 Part B One-quadrant Copper (Buck Converter)………………………………... 7 Part C dc Motor Speed Control………………………………………………….11 Appendix I Connector Pin Assignment……………………………………………… 16 Appendix II Report Writing Guidelines……………………………………………… 17 Appendix III ASCII Code Chart………………………………………………………. 18 Appendix IV Sample Program………………………………………………………… 19 2.2 Project 2 – Microprocessor Controlled SCR Rectifier……………………………………… 24 Part A Zero-crossing Detector and 68HC11 Programming…………………….. 24 Part B Programming……………………………………………………………. 25 Part C Experiments……………………………………………………………... 26 Appendix I Sample Program………………………………………………………… 31 2.3 Project 3 – IGBT Inverter Operation and Induction Motor Speed Control (to be provided through blackboard) 3. Solutions to Selected Problems………………………………………………………….. 33 3.1 dc-dc Switch Mode Converters……………………………………………………………... 33 3.2 Single and Three Phase Diode Rectifiers…………………………………………………… 37 3.3 Single and Three Phase Thyristor Rectifiers………………………………………………... 42 3.4 Switch Mode Inverters……………………………………………………………………… 46 4. Examination Samples……………………………………………………………………… 50


1. Course Outline Course Description A course on microprocessor-controlled solid state converters. Major topics includes: solid state switching devices, dc-dc switch mode converters, diode & thyristor rectifiers, current & voltage source inverters, industry applications and microprocessor programming techniques. Typical control schemes for these converters will also be discussed. Important concepts are illustrated with laboratory design projects. An MC68HC11 microprocessor based MPP board will be used in the projects. Prerequisite All required third year courses. Course Organization This course consists of three hours of lecture and two hours of laboratory per week. Course Material Text "Power Electronics -- Converters, Applications and Design" by N.Mohan, T.Undeland and

W.Robbins, 3rd Edition, published by John Wiley & Sons, Inc., Reference

“Fundamental of Power Electronics, Second Edition” by R.W. Erickson and D. Maksimovic, published by Springer Science+Business Media Inc.

Course Notes

"ELE754 Power Electronics - Course Notes" by B. Wu & D. Xu, 2011 Edition, which can be downloaded from /home/courses/ele754/2011_notes.pdf

Course Evaluation Theoretical component 70%

Mid-term Examination 20% Final Examination 40% Quizzes 10%

Laboratory component 30%

P controlled dc-dc converter and dc motor drive 10% P based thyristor rectifiers 10% P controlled IGBT inverter and induction motor drive 10%


Lecture Topics 1 dc-dc Switch Mode Converters (pp 161-199) 7 hrs 1.1 Introduction 1.2 Buck converters 1.3 One-quadrant chopper 1.4 Two-quadrant chopper 1.5 Review of 68HC11 based MPP board 1.6 Microprocessor control of dc-dc converters 2 Microprocessor Controlled dc Motor Drives (pp 377-398) 5 hrs 2.1 Introduction 2.2 Equivalent circuit of dc motors 2.3 dc motor speed control 2.4 Converters used in the dc motor drives 2.5 Microprocessor control of dc motor drives 3 Diode and Thyristor Rectifiers (pp 79-160) 8 hrs 3.1 Introduction 3.2 Single and three phase diode rectifiers 3.3 Total harmonic distortions and power factor 3.4 Single and three phase thyristor (SCR) rectifiers 3.5 Microprocessor control of thyristor rectifiers 4 Inverters (dc -ac converters) (pp 200-248) 8 hrs 4.1 Introduction 4.2 Single-phase Inverters 4.3 Three-phase IGBT Inverters 4.4 PWM techniques 4.5 Current source Inverters 4.6 Induction Motor Speed Control (pp 399-434) 5 Applications (pp 354-364, 460-504) 6 hrs 5.1 Introduction 5.2 Uninterruptible power supplies (UPS) 5.3 Power supplies 5.4 Motor drives 5.5 Active power filters 5.6 Static var compensators 5.7 Electronic ballasts 6 Design Considerations (pp 667-730) 3 hrs 6.1 Introduction 6.2 Snubber circuit design 6.2 Gate drive circuits 6.3 Heatsink design


Laboratory Schedule

Project

Topic

Week #

Project 1 Microprocessor controlled dc-dc converter and dc motor drive

2-5

Project 2 Microprocessor control of thyristor rectifiers

6-8

Project 3 Microprocessor controlled IGBT inverter and induction motor drive

9-12

Course grade A final letter grade shall be assigned based on the following conversion table. A+ 90% - 100% C+ 63% - 65% A 85% - 89% C 60% - 62% A- 80% - 84% C- 57% - 59% B+ 75% - 79% D+ 54% - 56% B 70% - 74% D 52% - 53% B- 66% - 69% D- 50% - 51% F 00% - 49% In order to achieve a passing grade, the student must achieve an average of at least 50% in both theoretical and laboratory components. Instructor David Xu, PhD Room ENG333, 245 Church Street, Toronto Department of Electrical and Computer Engineering Ryerson University (416) 979-5000 ext: 6075 eMail: [email protected] Counseling Hours Every Tuesday 2-4PM at ENG333. Faculty Course Survey The faculty course survey will be held in November, 2011.


2. Design Projects

2.1 Project 1 Microprocessor Controlled dc-dc Converter and dc Motor Drive Objectives To review the MPP board and 68HC11 assembly language; To become familiar with Power Converter Module (PCM); To study the performance of one-quadrant chopper with an RL load; To investigate the performance of a dc motor drive. Equipment Power Convert Module 68HC11 based MPP board (included in the PCM) Personal computer 120V, 3A dc motor Part A Preliminary study A.1 Start Linux and logged on, to download, compile and run PROJ1.ASM (sample program), follow the following steps. Assume that the working directory is your home directory.

- Start terminal window - /home/student1/xxx>cp /home/couses/ele754/proj1.asm ~/ Copy the sample program PROJ1.ASM

in your home directory. - /home/student1/xxx>as11 proj1.asm –l > proj1.lst Compile the program (which generates a file

“proj1.s19” - Reset the MPP board by pressing the reset button on the PCM front panel - /home/student1/xxx>vt6811 proj1.s19 vt6811 downloads “proj1.s19” into the MPP

board. - > G 6500 Run the program with starting address $6500,

which is specified in the source program. To exit vt6811, press CTRL and C simultaneously.

Note: 1. For your convenience, a hardcopy of the sample program is given in Appendix IV. 2. To create or edit source asm file, use jedit, emacs, pico, joe, xemacs in Linux.


A.2 Observe the PWM waveform at Pin 6 of Port A, and complete the following table using digital oscilloscope.

Duty Cycle Ton / (Ton + Toff)

Keyboard Input

10 %

50 %

90 %

Ton

Toff

Ton

Toff

Ton

Toff

Measured Value

Duty Cycle

A.3 Modified the program such that the frequency of the PWM gating pulse is 1000Hz. And then complete

the following table.

Duty Cycle Ton / (Ton + Toff)

Keyboard Input

10 %

90 %

Ton Toff Ton Toff Measured Value

Duty Cycle

A.4 Display a message ‘Welcome to ELE754 Lab’ on the computer monitor every time when a new duty

cycle value is entered. A.5 Modified the program to limit the duty cycle value entered from keyboard. The minimum value is 10%

and the maximum value 90%.


Part B One-quadrant Chopper (Buck Converter) B.1 Power Converter Module (PCM)

The circuit diagram of Power Converter Module is shown in Fig. 1 for your reference.

Fig. 1 Circuit diagram of Power Converter Module

Fig. 2 shows the circuit diagram of a one-quadrant chopper which uses IGBT as a switching device. A continuous gate signal (PA7 = logic “1") is applied to the SCR gates such that the SCR rectifier performs the same functions as a diode bridge rectifier. As a result, the SCR rectifier provides a near constant dc voltage for the chopper. An RL load is connected to the output terminals (A3 and COM) of the chopper.

Fig. 2 Circuit diagram of the IGBT based chopper


B.2 Power Circuit Connection 1) Make sure the panel switch (SW1) is in OFF position. 2) Connect the output of the SCR rectifier (A1) to the input of the chopper (A2). 3) RL load connection:

- Connect three units of 300mH inductors in parallel (to increase the current rating) - Connect a 100 resistor in series with the inductors.

Now you are ready for programming.

B.3 Programming requirements

1) Use Port A (PA6) to generate a gate signal for the IGBT device Tc. 2) Minimum duty cycle: 10%; Maximum duty cycle: 90%. 3) The value of duty cycle is entered from keyboard. 4) Display the value of duty cycle on LCD. 5) Switching frequency: 500Hz.

B.4 Measurements

- Set duty cycle D to 10%. Make sure that the gate signal at PA6 is correct (i.e., D = 10%, f = 500Hz, and the amplitude of the gate signal is 5V).

- Ask your lab instructor to check your connection before energizing the system. - Close Switch SW1 (ON position).

Warning: The load resistor R will be very hot. Do not touch it. B.4a) RL load (Switching frequency: 500Hz, R = 100 Ohms and L = 100mH)

Use an analog scope and a multimeter for the following measurements.

Table A Duty Cycle

D (%)

10

30

50

70

90

VA2 (V) Chopper Input Voltage

(Voltage at Point A2 with respect to COM)

(Use multimeter, DC)

VA3 (V)

Chopper Output Voltage (Voltage at Point A3 w.r.t.

common COM) (Use multimeter, DC)

VR (V, p-p)

Ripple Voltage (Use oscilloscope)

ME

AS

UR

EM

EN

T

IO (A, p-p)

Ripple Current (Io = VR / R)

(Calculate from above)


VA3 (V)

Chopper Output Voltage Calculation:

VA3 = D VA2

CA

LC

UL

AT

ION

IO (A, p-p) Calculation:

Io = Imax - Imin

(Use formulas in class)

B.4b) RL load (Switching frequency: 1000Hz, R = 100 Ohms and L = 100mH)

Modifier the program such that the switching frequency is 1000 Hz. Complete the following table.

Table B Duty Cycle

D (%)

10

30

50

70

90 VR (V, p-p)


ME

AS

UR

EM

EN

T

IO (A, p-p) Ripple Current (Io = VR / R)

(Calculate from above)

CA

LC

UL

AT

ION

IO (A, p-p) Calculation:

Io = Imax - Imin

(Use formulas in class)

B.4c) RLC load (Switching frequency: 500Hz, R = 100 , L = 100mH and C = 40F. C is in parallel with

R) Modifier the program such that the switching frequency is 500 Hz. Complete the following table.

Table C Duty Cycle

D (%)

10

30

50

70

90


(Voltage at Point A2 w.r.t. common COM)

(Use multimeter)

ME

AS

UR

EM

EN

T

VC (V)

Converter Output Voltage (Voltage across the filter cap C)

(Use multimeter)


VC (V, p-p)


ME

AS

UR

EM

EN

T

IL (A, p-p)

Ripple Current (IL = VR3 / R3)

(Use oscilloscope)

CA

LC

UL

AT

ION

VC (V, p-p)

Ripple Voltage (Use formulas in class)

2.5 Report - Complete the calculations specified in the above tables. - Produce graphs that illustrate the following relationships:

1) Output voltage of the chopper vs. duty cycle D (RL and RLC loads); 2) Io vs. D for the RL load (experimental and calculation results); 3) IL vs. D and VC vs. D for the RLC load.

- Summarize your observations.


Part C dc Motor Speed Control C.1 Programming requirements Make sure that the panel switch (SW1) is in OFF position before you start programming.

1) To avoid excessively high current during motor starting, a ramp function for the duty cycle should be implemented. An example is shown in Fig. 3. At t = 1.0 second, a reference duty cycle (Dref) of 90% is entered from the keyboard. The actual duty cycle of the gate signal (Dg) increases at a certain rate 1%/50ms. At t = 5.0 second, the duty cycle Dg reaches the reference value. As a result, the motor accelerates smoothly and starting current is limited. For the motor speed deceleration, the ramp function has a deceleration rate of 1% per 50 ms.

2) Duty cycle ramp function: 1% / 50ms 3) Minimum duty cycle: 10%. 4) Maximum duty cycle: 90%. 5) Switching frequency: 1000Hz. 6) Use Port A (PA6) to generate a gate signal for the IGBT device TC. 7) Display the value of duty cycle on LCD.

Note: A 50ms delay subroutine is included in the sample program.

2 141210864

100

80

60

40

20

0

Dg (%)

Dref (%)

100

80

60

40

20

0

t (sec)

Fig. 3 A duty cycle ramp function for the dc motor drive

C.2 Power Circuit Connection

1) Make sure the panel switch (SW1) is in OFF position. 2) Connect the output of the SCR rectifier (A1) to the input of the chopper (A2). 3) Motor Connection:

- Connect the shunt field winding of the dc motor to the SCR rectifier output (A1 and COM) which provides a constant dc voltage.

- Connect the armature winding to the chopper output (A3 and COM) through a 150mH inductor (two 300mH inductors in parallel).


C.3 Measurements - Set duty cycle D to 10%. Make sure that the gate signal at PA6 is correct (i.e., D = 10%, f =

1000Hz, and the amplitude of the gate signal is 5V). - Ask your lab instructor to check your connection before energizing the system. - Close Switch SW1 (ON position).

C.3a) dc motor drive with an external inductor (150mH)

Switching frequency: 1000Hz Use an digital scope and a multimeter for the following measurements.

Table D Duty Cycle

D (%)

10

30

50

70

90


(Voltage at A2 w.r.t. COM) (Use multimeter)

VA3 (V)

Chopper Output Voltage (Voltage at A3 w.r.t. COM)

(Use multimeter)

VA3 (V)

Trace the waveform*

Yes

No

No

Yes

No

IO (A, p-p)

Ripple Current (Io = VR3 / R3)

Connect scope probe to TP5 w.r.t to A3, R3 is 1Ohm.**

IO (A, p-p)

Trace the waveform*

Yes

No

No

Yes

No

Operating mode

(Continuous or discontinuous)

Speed (rpm)

(Use tachometer)

* Please follow the following procedures to capture the waveforms. 1) Open web browser (Netscape, Mozilla, Firefox, etc). 2) Type: http://scope/ in the address area. 3) When the scope displayed, right click on the scope image and save the picture as file. 4) Choose proper file name in the dialog. ** In order to trace both the waveforms of VA3 and IO, connect the probes ground to A3. Connect one probe to TP5,

which measures the voltage drop across sample resistor R3 (1Ohm). Connect the other probe to COM, which measures the voltage of COM w.r.t A3. Invert the waveform of this channel.


Vertical: V/div Vertical: V/div

Horizontal: ms/div Horizontal: ms/div

Duty Cycle: % Duty Cycle: %





C.3b) dc motor drive without any external inductor Switching frequency: 1000Hz Use an digital scope and a multimeter for the following measurements.

Table E

Duty Cycle D (%)

10

30

50

70

90



VA3 (V) Chopper Output Voltage


VA3 (V) Trace the waveform*

Yes

No

No

Yes

No

IO (A, p-p) Ripple Current

(Io = VR3 / R3)

IO (A, p-p) Trace the waveform*

Yes

No

No

Yes

No

Operating mode (Continuous or discontinuous)

Speed (rpm)








C.4 Report - Complete the calculations specified in the above tables. - Produce graphs that illustrate the following relationships:

1) Terminal voltage of the armature vs. duty cycle D; 2) IL vs. D; 3) Motor speed vs. D.

- Summarize your observations and draw conclusions.


Appendix I Connector Pin Assignment

Panel Connector 1 (D type)

13654321 3 4 PD5

14

1PD2GNDPA7PD0

25

Panel Connector 2 (D type)

13

14

1Vsyn+

25

Vsyn- -12V+12VGND+5V

Test Board

Purple

PA7

Blue

PA6 PA5 PA4 PA3 PA2 PA1 PA0

Green Yellow Orange Red Brown Black

GND

White


Appendix II Report Writing Guidelines Title page This page includes: the title of the report, author name, for whom and when the report was prepared, course name and laboratory group of the author. Abstract An abstract is a short paragraph summarizing the report. One or two sentences for each of the following items would be appropriate: Purposes (objectives) Methods Observations (figures-of-merit) Conclusions & recommendations An abstract can be executive summary. The reader wants to read a short paragraph and to be able to put the report into context with other relating materials without spending much time reading the whole report. Thus, an abstract requires a careful preparation and is the LAST item to be written in a report. It should be independent and the rest of the report should be written as if an abstract doesn’t exist. Introduction This paragraph explains the initiation of the study, the problem to be investigated, the approach or the method to be employed for the study.

Theory Develop all theoretical formulations or explanation for the performance of the systems under investigation.

Experiments Record all data in several neat tables. Oscilloscope tracings and waveform drawings should be reported in this section. Calculations should also be given to verify the experimental results. Conclusions & Recommendations It is important to make a precise conclusion of the project. List the major results together with short explanations and comments. Recommendations are also necessary in a report. Appendix, Flowchart and Program Listing Include Flowchart(s) for the program you developed. Insert comments/explanations into the program listing where appropriate.


Appendix III ASCII Code Chart hexadecimal

code

Character hexadecimal

code


code


code

Character00 NUL 20 SP 40 @ 60 ` 01 SOH 21 ! 41 A 61 a 02 STX 22 “ 42 B 62 b 03 ETX 23 # 43 C 63 c 04 EOT 24 $ 44 D 64 d 05 ENQ 25 % 45 E 65 e 06 ACK 26 & 46 F 66 f 07 BEL 27 ‘ 47 G 67 g 08 BS 28 ( 48 H 68 h 09 HT 29 ) 49 I 69 I 0A LF 2A * 4A J 6A j 0B VT 2B + 4B K 6B k 0C FF 2C , 4C L 6C l 0D CR 2D - 4D M 6D m 0E SO 2E . 4E N 6E n 0F SI 2F / 4F O 6F o 10 DLE 30 0 50 P 70 p 11 DC1 31 1 51 Q 71 q 12 DC2 32 2 52 R 72 r 13 DC3 33 3 53 S 73 s 14 DC4 34 4 54 T 74 t 15 NAK 35 5 55 U 75 u 16 SYN 36 6 56 V 76 v 17 ETB 37 7 57 W 77 w 18 CAN 38 8 58 X 78 x 19 EM 39 9 59 Y 79 y 1A SUB 3A : 5A Z 7A z 1B ESC 3B ; 5B [ 7B { 1C FS 3C < 5C \ 7C | 1D GS 3D = 5D ] 7D } 1E RS 3E > 5E ^ 7E ~ 1F US 3F ? 5F _ 7F DEL


Appendix IV Sample Program *----------------------------------------------------------------------------* * ELE754 POWER ELECTRONICS * * * * SAMPLE PROGRAM * * FOR * * PROJECT 1 dc-dc CONVERTER AND dc MOTOR DRIVE * * * * File Name: Proj1.asm * * Version: 3 * * * * Please read the following program and try to understand it. Compile * * and run this program. Use an oscilloscope to observe waveforms at * * PA7 (Always high for the SCR rectifier) and PA6 (pulses with a variable * * duty cycle for the chopper) * * * * This sample program will help you to learn * * (1) how to get a number from keyboard; * * (2) how to display information on the computer monitor; * * (3) how to use LCD; * * (4) how to write a subroutine; * * (5) how to write an interrupt service routine; * * (6) how to use Output Compare Register; and * * (7) how to generate a PWM (pulse width modulated) signal. * * * *----------------------------------------------------------------------------* * Define subroutine addresses located in EPROM INCHAR EQU $E55B * Subroutine to get strings from keyboard OUTSTR EQU $E52F * subroutine to display strings on monitor WCTRL EQU $E096 * subroutine to control LCD configuration WDAT EQU $E0A3 * subroutine to display strings on LCD * Define address of HC11 registers OC1M EQU $100C * OC1 Action Mask Register OC1D EQU $100D * OC1 Action Data Register TCNT EQU $100E * Time Counter Register TOC1 EQU $1016 * Output Compare Register 1 TCTL1 EQU $1020 * Timer Control Register 1 TMSK1 EQU $1022 * Timer Interrupt Mask Register 1 TFLG1 EQU $1023 * Timer Interrupt Flag Register 1 PACTL EQU $1026 * Pulse Accumulator Control Resister * Messages to be displayed on PC monitor (CRT) or LCD ORG $6000 CRTMSG1 FCC 'ENTER DUTY CYCLE (%):' FCB $0A * Line Feed FCB $04 * String terminator CRTMSG2 FCC 'DUTY CYCLE RANGE: 5% TO 95%' FCB $0A FCB $04 LCDMSG1 FCC 'DUTY CYCLE: ' * LCD message ENDMSG FCB


LOGIC FCB 0 * Logic level of PWM signal. Logic low = 0 * Reserve RAM locations KB_DUTY RMB 3 * Storage memory for duty cycle from keyboard NUM1 RMB 1 * Number of digits for duty cycle from keyboard DUTY RMB 1 * Decimal value of duty cycle TIME_HI RMB 2 * Number of E-clock cycles for logic high period TIME_LO RMB 2 * Number of E-clock cycles for logic low period BUFFER RMB 2 *--------------------------- MAIN PROGRAM ----------------------------------- * Program initialization ORG $00DF * Interrupt vector for OC1 service routine JMP INT_SVC ORG $6500 LDS #$7FFF * Set stack pointer LDAA #$80 STAA PACTL * Set Bit 7 of Port A as output pin LDAA #$80 * Clear the flag for Output Compare 1 STAA TFLG1 LDAA #$C0 * Bits 6&7 of Port A are affected by OC1 compare STAA OC1M CLR TCTL1 * Clear Timer Control Register 1 LDAA #$80 * Unmask Output Compare Register 1 interrupt STAA TMSK1 MAIN LDX #CRTMSG1 * Display message on PC monitor (CRT) LDAA 0,X JSR OUTSTR JSR GETCHAR * Get strings from keyboard JSR LCD_DSP * Display duty cycle on LCD JSR CONV * Data conversion CLI * Clear interrupt mask JMP MAIN *------------------------- SUBROUTINE GETCHAR ------------------------------- * Get characters from keyboard GETCHAR LDAB #0 STAB NUM1 LDY #KB_DUTY START JSR INCHAR * See the flowchart distributed in class LDAB NUM1 CMPB #0 BEQ NEXT CMPA #$0D * $0D - ASCII code, representing 'Carriage Return' key BEQ CHECK1 NEXT CMPA #$39 * $39 - ASCII code, representing keyboard character '9' BHI ERROR CMPA #$30 * $30 - ASCII code, representing keyboard character '0'


BLO ERROR STAA 0,Y LDAB NUM1 INCB CMPB #3 * Error occurs when the max number of digits is BHS ERROR * larger than two. STAB NUM1 INY BRA START CHECK1 LDAA NUM1 * Check the number of digits CMPA #1 BEQ CHECK2 LDAB KB_DUTY CMPB #$39 BEQ CHECK3 BRA END CHECK2 LDAA KB_DUTY * For a single digit number, CMPA #$35 * the minimum value is 5. BLO ERROR JMP END CHECK3 LDAA KB_DUTY+1 CMPA #$35 * For a two-digit number, the max value is 95. BHI ERROR JMP END ERROR LDX #CRTMSG2 * Display error message on CRT LDAA 0,X JSR OUTSTR BRA GETCHAR END RTS *--------------------------- SUBROUTINE LCD_DSP ----------------------------- * LCD display LCD_DSP LDAA #$01 * LCD initialization JSR WCTRL LDAA #$02 JSR WCTRL LDAA #$38 JSR WCTRL LDAA #$0C JSR WCTRL LDY #LCDMSG1 LOOP CPY #ENDMSG * Display a message on LCD one character by BEQ NEXT2 * another until the end of the message. LDAA 0,Y JSR WDAT INY LDAA #$06 JSR WCTRL BRA LOOP NEXT2 CLR BUFFER * Display duty cycle on LCD LDY #KB_DUTY LOOP1 LDAA 0,Y


JSR WDAT INY INC BUFFER LDAA BUFFER CMPA NUM1 * Check if all the digits are displayed BEQ NEXT3 LDAA #$06 JSR WCTRL BRA LOOP1 NEXT3 LDAA #'%' * Display % JSR WDAT RTS *-------------------------- SUBROUTINE CONV -------------------------------- * Convert duty cycle in hexadecimal to decimal and then number of * E-clock cycles CONV LDAA NUM1 * Check # of digits (duty cycle) CMPA #1 BEQ SINGLE LDAA KB_DUTY * If it has 2 digits, convert the 1st to decimal SUBA #$30 LDAB #10 MUL STAB BUFFER LDAA KB_DUTY+1 * Convert the 2nd digit to decimal SUBA #$30 ADDA BUFFER STAA DUTY BRA CALULAT SINGLE LDAA KB_DUTY * Convert the single digit number to decimal SUBA #$30 STAA DUTY CALULAT LDAA #20 * Convert the duty cycle (decimal) to number of LDAB DUTY * E-clock cycles. MUL STD TIME_HI LDAA #100 SUBA DUTY LDAB #20 MUL STD TIME_LO RTS *---------------------- Subroutine 50ms Delay ------------------------------ DELAY LDY #7142 * (7142 * 7) + 4 = 49998us LOOP2 DEY *takes 4 clock cycles BNE LOOP2 *takes 3 clock cycles RTS *---------------------- INTERRUPT SERVICE ROUTINE (OC1) -------------------- * PWM waveform generation INT_SVC LDD OC1D


ANDA #%01000000 * Test Bit 6 BEQ NEXT5 LDAA #%10000000 * Set PA7 high (for the SCR rectifier) and * * PA6 low (for the IGBT Chopper) STAA OC1D LDD TOC1 * Set the time for the next interrupt ADDD TIME_HI STD TOC1 BRA NEXT6 NEXT5 LDAA #$C0 * Set PA7 high (for the SCR rectifier) and * * PA6 high (for the IGBT Chopper) STAA OC1D LDD TOC1 * Set the time for the next interrupt ADDD TIME_LO STD TOC1 NEXT6 LDAA #$80 * Clear OC1 interrupt flag STAA TFLG1 RTI


2.2 Project 2 Microprocessor Controlled SCR Rectifier Objectives To generate SCR gate signals using 68HC11 based MPP board; and To study the performance of a single phase SCR rectifier. Equipment Power Convert Module 68HC11 based MPP board (included in the PCM) Personal computer 300mH dc reactors 100 resistors Part A Zero-crossing Detector and 68HC11 Programing A.1 Zero-crossing detector

The gate signal for the SCRs used in the single phase half bridge rectifier should be synchronized with the input ac line voltage. Therefore, zero-crossings of the line voltage should be detected. A typical zero-crossing detector is shown in Fig. 2.1. It is composed of a low pass filter, phase shifter and comparator. Adjust Potentiometer P1 such that the phase shift between Vsyn and Vo2 is zero. Record the waveforms of Vsyn & Vo2 and Vsyn & Vo3.

+

-

+

-

+

-Vsyn P1

R3

-12V

+12V

R8

R7

R6

R5

500K

5.1K

2.7K

5.1KPA1

Vo3

2 1

3IC1

IC2

IC3

C12

C11

10K

10K

100K

27K

R1

27K

R2

27K

Low Pass Filter

Phase Shifter

Comparator

0.1F

0.22F

Vo1

Vo2

+

-

Fig. 2.1 Zero-crossing detector (a low pass filter is used to avoid multiple zero-crossings due to noise).


Vsyn and Vo2 Vsyn and Vo3



Part B Programming B.1 Preliminary study

- Study the sample program “Proj2.asm” and make sure you fully understand it. This program can be used to generate SCR gate signals according to the firing angle specified by ALPHA_D in the program. The firing angle is the angle between the zero crossing of Vsyn and the rising angle of the gating signal.

- Compile and run the program. - Measure the phase angle between Vsyn and the gate signal. Confirm that the measured angles are

in a good agreement with ALPHA_D. Note: The minimum delay angle alpha in this program is one degree (1). The program will not work

when the delay angle is zero. B.2 Programming a) Programming requirements

- Minimum delay (firing) angle: 1 - Maximum delay (firing) angle: 165 - The delay angle should be obtained from keyboard and be displayed on LCD; - Two gating pulses should be generated per cycle of the input source voltage; - Use TIC2 (PA1) for input capture; and - Use PA7 to generate the SCR gate signal.

b) Procedures

1) Combine the program you developed in Project 1 with the sample program “Proj2.asm”. 2) Run the program and complete the following table.


Keyboard Input

30 90 150

Measured

3) Modify the program such that only two interrupt service routines are used instead of three in the

sample program. Part C Experiment C.1 Resistive load Power Circuit Connection 1) Make sure the panel switch (SW1) is in OFF position. 2) Load connection:

- Connect two pieces of 100 Ohm resistors in parallel; and - Connect the load to Connectors A1 and COM.

Measurements - Set delay angle to 165. Make sure that the gate signal at PA7 is correct (i.e., = 165, and the

amplitude of the gate signal is 5V). - Ask your lab instructor to check your connection before energizing the system. - Close Switch SW1 (ON position). - Use an digital scope and a multimeter for the following measurements.

Warning: The load resistors will be very hot during experiment. Do not touch them.

TABLE A Delay angle 1 40 80 120 160

VS

(Voltage between TP1 and TP2) (Multimeter: V~, RMS: ac)

VA1, avg (dc) (V)

(Voltage at A1 with respect to COM) Average Value

(Multimeter: V= )

VA1, rms (ac) (V)

RMS Value of ac component only (Multimeter: V~, RMS: ac)

VA1, rms (ac + dc) (V)

RMS Value (Multimeter: V~, RMS: ac + dc)


Verification

V V Vrms dc ac 2 2 +

VA1, avg (dc) (V)

Calculation

Use the equation derived in the lecture class


Calculation

Use the equation derived in the lecture class

Record the rectifier output voltage waveforms using the digital oscilloscope. Delay angles: 40 and 120



Delay angle 40° Delay angle 120°


C.2 RL load Power Circuit Connection 1) Make sure the panel switch (SW1) is in OFF position. 2) RL load connection:

- Connect four units of 300mH inductors in series; - Connect two pieces of 100 Ohm resistors in parallel; and - Connect the inductor (1200mH) in series with the resistor (50 ). - Connect the RL load to Connectors A1 and COM.

Measurements - Ask your lab instructor to check your connection before energizing the system. - Close Switch SW1 (ON position).

Warning: The load resistors will be very hot during experiment. Do not touch them.

TABLE B Delay angle 40

VS

(Voltage between TP1 and TP2) (Multimeter: V~, RMS: ac)

VA1, avg (dc) (V)

(Voltage at A1 with respect to COM) Average Value

(Multimeter: V= )

VA1, rms (ac) (V)

RMS Value of ac component only (Multimeter: V~, RMS: ac)


RMS Value (Multimeter: V~, RMS: ac + dc)

Verification

V V Vrms dc ac 2 2 +

Compare the above measured results with those ( = 40) given in Table A. Make your conclusions. Include your conclusions in the report.


Record the rectifier output voltage waveforms using the digital oscilloscope. Delay angles: 1 and 40




Record the rectifier output current waveforms (Measure the waveform across the load resistor Delay angles: 1 and 40





Record the rectifier input current waveforms (Measure the waveform across sample resistor R1). Delay angles: 1 and 40




Find the thyristor conduction angles with = 1 and 40. C.3 Report

- Complete the calculations specified in the above tables. - Produce graphs that illustrate the following relationships:

1) Output dc (average) voltage of the rectifier (Table A) versus delay angle . Show the curves obtained from the measured and calculated results on the same graph.

2) Output dc (rms) voltage of the rectifier (Table A) versus delay angle . Show the curves obtained from the measured and calculated results on the same graph.

- Summarize your observations. - Attach a copy of your program listing to the report.


Appendix I Sample Program *----------------------------------------------------------------------------* * ELE754 POWER ELECTRONICS * * * * SAMPLE PROGRAM * * FOR * * PROJECT 2 Microprocessor Controlled Thyristor Rectifier * * * * File Name: Proj2.asm * * Version: 3 * * * * Please read the following program and try to understand it. Compile * * and run this program. Use an oscilloscope to observe the waveform at * * PA7 (Gate signal for the SCRs used in the rectifier). * * * * Zero Crossing Detector: No phase shift between its input and output * * Input Capture: PORT A BIT 1 (PA1) * * SCR Gate Signal: PORT A BIT 7 (PA7) * * ALPHA Range: 1 to 165 degrees * *----------------------------------------------------------------------------* * Define addresses of HC11 registers used in this program PORTA EQU $1000 * Port A PACTL EQU $1026 * Pulse Accumulator Control Resistor TIC2 EQU $1012 * Input Capture Register 2 TOC1 EQU $1016 * Output Compare Register 1 TOC2 EQU $1018 * Output Compare Register 2 TCNT EQU $100E * Timer Counter Register (Free running) TCTL2 EQU $1021 * Timer Control Register 1 TMSK1 EQU $1022 * Timer Interrupt Mask Register 1 TFLG1 EQU $1023 * Timer Interrupt Flag Register 1 * Define configurations of input capture and output compare LOW EQU $00 * Logic '0' of the gate signal HIGH EQU $80 * Logic '1' of the gate signal PERIOD EQU 8333 * 8.333ms (half cycle of 60Hz) GAP EQU 400 * 0.4ms ALPHA_D FCB 90 * Delay angle ALPHA in degrees ALPHA FCB * Delay angle ALPHA in # of counts *--------------------------- Main Program --------------------------------* ORG $00E5 * IC2 Interrupt Vector JMP INTER1 ORG $00DF * OC1 Interrupt Vector JMP INTER2 ORG $00DC * OC2 Interrupt Vector JMP INTER3 ORG $6500 LDS #$7FFF


LDAA #$80 * Set Port A bit 7 as an output STAA PACTL LDAA #$0C * Input capture on both rising and falling edges STAA TCTL2 LDAA #$C2 * Clear OC1, OC2 and IC2 interrupt flags STAA TFLG1 LDAA #$C2 * Allow OC1, OC2 and IC2 interrupt requests STAA TMSK1 LDAB ALPHA_D LDAA #45 * Convert delay angle to # of counts MUL STD ALPHA CLI * Interrupt enable LOOP NOP * Wait for interrupt requests JMP LOOP *---------------------- IC2 Interrupt Service Routine ---------------------* INTER1 LDD TIC2 ADDD ALPHA STD TOC1 QUIT LDAA #$02 * Clear input capture STAA TFLG1 RTI *---------------------- OC1 Interrupt Service Routine ---------------------* INTER2 LDAA #HIGH STAA PORTA LDD TCNT ADDD #PERIOD SUBD ALPHA SUBD #GAP STD TOC2 LDAA #$80 * Clear output compare flag STAA TFLG1 RTI *---------------------- OC2 Interrupt Service Routine ---------------------* INTER3 LDAA #LOW STAA PORTA LDAA #$40 * Clear output compare flag STAA TFLG1 RTI


3. Solutions to Selected Problems

3.1 dc-dc Switch-mode Converters Step-down Converters (Chapter 7, Moham)

7-1 VO=5V, Vd=10V to 40V, PO5W, fs=50kHz. Find the minimum inductance to keep the converter in the continuous conduction mode under all conditions. Solution:

For a given load and output voltage, the likelihood that the inductor current will fall to zero is increased by lowering the duty ratio and thus increasing the OFF time. The duty ratio is lowest when Vd=40V. PO/VO=5W/5V=IO=1A

H43.75L

]540[1500002

125.0][

2f

DL ;125.0

40

5

][2

I

5,-7 Eq. from conduction continuousFor

S

O

OdO

OdS

VVI

D

VVLf

D

7-2 Vo=5V, fs=20kHz, L=0.001H, C=470F, Vd=12.6V, Io=0.2A. Find Vo. Solution: Is the circuit operating in the continuous mode?

2.01mV

62o

2

o

OB

10470001.08000,20

56.1251V

8

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397.012.6

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2i 5,-Eq.7 From

LC

VDTIt

VVLf

D

os

ods

7-3 Find the RMS ripple current through L. (optional)

Solution:

-0.0755A

t

TS

DTS (1-D)TS

iL,ripple

0.0755A


sAsAs

DsAdt

di

sAdt

di

dt

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,

0

2,

s

s,

RMS

iti

stti

rippleL

T

rippleL

rippleL

s

7-5 Calculate the ripple voltage in problem 7-2 if the load current is reduced to IOB/2. (optional)

Solution:

1.66mV

O

2

111

o

11d

o

V

4,-problem7in derivedequation the Using

0.281.D ;52.11197.0

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;001.0000,202

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52.1 ;6.12

5

V

V ;0377.0

D

D

D

L

DTV

Us

DD

DAI

sd

o

One-quadrant chopper

+

Vd

_

G iD

D

T iT iO L

R

Ea

+

VO

_

Fig 7-a Circuit diagram of one-quadrant chopper

7-a In the one-quadrant chopper circuit of Fig. 7-a, Vd=600V, Ea=200V, L=4mH, R=1.5,

Ts=4000s, ton=2500s. Show that the output current iO is continuous. Solution:


.continuous is i 0, I Since

3.1333.1785.1

200

1

1

1.5

600I

5.126674000

937.026672500

26675.14

Omin

5.1

937.0

min

45Ae

e

T

t

smH

s

on

7-b For the chopper circuit of problem 7-a, determine: (a) The average output voltage and current, VO and IO. (b) The output current at the instant of IGBT turn-off.

Solution:

1.5

200-

e-1

e-1

1.5

600I

1.5104

5.1104T

9375 0.104

5.1102.5t )(

1.5

200-375

R

V

6000.625V T

tV

continous.iscurrent a,-7Prob. From )(

1.5-

0.9375-

max

3-

3-s

3-

3-on

O

onO

179.9A

116.7A

375V

b

EI

a

aO

7-c In the chopper circuit of Fig.7-a, Vd=600V, Ea=200V, L=1mH, R=1.5, Ts=4000s, and ton=2500s. Show that the output current iO is discontinuous. Solution:

ous.discontinu is i 0, I Since

133415.1

200

1

1

5.1

600I

0.66674000

748.36672500

6675.11

Omin

0.6

748.3

min

-92Ae

e

T

t

smH

s

on

7-d For the chopper circuit of problem 7-c, determine:

(a) The average output voltage and current, VO and IO. (b) The output current at the instant of commutation.

Solution:

s

e

a

3-

75.33.753-

x

on

-3

103.222

} ] )1(200

2006001[e {ln

5.1

10 t

3.751.52.5t

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200-600I (b)

5.1

2009.413I

2004

3.222)-(4600

4

5.2V

3.75-max

O

O

260.4A

142.6A

413.9V

R

EV aO

7-e In the chopper circuit of Fig. 7-a, Vd=600v, Ea=350V, R=0.1, Ts=1800s, and L is of so large a value that the output current may be assumed constant or ripple-free. If the output current is to be IO=100A: (a) Calculate the required value of ton. (b) Sketch to scale the time variations of vG, vO, iO, iD and iT. Solution:

st

V

VT

tA

R

EVIa

on

donaO

O

3-101.08

600

108.1360

3603501001.0V

V V )(

3

O

O

(b)

t

t

t

t

t

v G

v O

i O

i T

i D

100

100

100

600 360

t o n T s


3.2 Single and Three phase Diode Rectifier 5-1 (Chapter 5, Moham)

5-3

2

32

122

3

22

33311

23

21

2d

23

2

12

d

33311d

33311

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cos45cos2

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cos45cos2VP

cos2cos2

cos2sin2cos2

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tItIIi

tVtVtVVv

dd

dd

d

d

d

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sin2cos2

:

111

11

111

1111

tV

tV

ttV

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Note

sinˆˆ

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33.6A 641.82.058397.9-2.058-450.1cosI

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1.4260.884cos

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coscos2L

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b

VV

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1I

ICF

0.9DPFI

IPF

1.0DPF

4.489.0

9.01100100

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9.024

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:analysis F

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s1

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ourierFrom

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32

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I

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34b

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1-3 Table from Therefore,

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s1

sn

s1

1

s1

ds

nd

nd

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dd

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nI

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3.3 Single and Three Phase Thyristor Rectifiers 6-3 (Chapter 6, Moham)

t

t

Vm

vs

vd

Vm

2

2

A1 Vm

+

T1 and T2 on T3 and T4 on

VAt

VAt

VVAV mm

mm

100 v,135

100 v,45

21V 2v

d

d

1d

6-5

707.02cosDPF

90 22

0.9cos

9.02

1cos1

2

2

1

2cos

DPF V :power equatingby calculated becan 180

-180

2cos

1cos 2

tdt sinV 2V

9.0

1

1s1

s

Lor

VV

VVb

V

IVI

IVII

II

DPF

V

VVa

ss

dod

s

dds

ddss

ds

sd

sdo


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shows (c) and (b) ofn Comparisio

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17-6 . 45.0cos9.0

16-6 . 5.0cos

60 ,9.0V

6-6 . cos

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90180

d

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o

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6-6

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d

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5.103%10115V and 0 ,100V

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9.0V 26,-6 Eq.in

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s

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uIV

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uVV

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Lu

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VV

dLLdLL

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ddodd

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63

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44-6 . I6

I :ionApproximat

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62,-6 Eq.by 55-6 Eq.in L R

36-6 . 23

62-6 . 2

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55-6 Eq. 3

cos

1

1

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s

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d

s

6-12

330.77-33.76u

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6

6

dds

Therefore

or

IV

Lu

IL

VV

AV

PI

kWVHVV

d

LL

s

dLLd

d

dd

LL


6-a Consider the 3 full bridge thyristor rectifier shown in Fig.6-19 (page 138). Draw waveforms of VPn , VNn and Vd at =30, 90 and 150.


3.4 Switch Mode Inverters 8-a (Chapter 8, Moham)

The single-phase half-bridge inverter in Fig.8-4 has a resistive load of R=2.4 and the dc input voltage is Vd=48V. Consider square wave operation with device conduction angle of 180 per cycle. Determine (a) the rms output voltage Vo1 at the fundamental frequency; (b) the average output power Po; (c) the peak and average currents of each transistor; and (d) the total voltage harmonic distortion. Solution:

48.06.21

6.2124THD )(

5105.0I

50% cycleDuty

102I )(

240P

241

V )(

6.214845.0

)sin(2

seriesFourier )(

22

1

21

2

avg

Peak

2

o

0

2o

1

5,3,1

o

oo

d

o

T

o

o

n

do

V

VVd

A

ARVc

WRV

VdtvT

b

VV

tnn

Vv

a

8-b Repeat Problem 8-a for a single phase bridge inverter in Fig.8-11.

Solution:

48.0V

THD )(

10205.0I

20I )(

960P

481

V )(

2.43489.0V

)sin(4

v

seriesFourier )(

1

21

2o

avg

Peak

2

o

0

2o

o1

5,3,1o

o

o

d

o

T

o

n

d

V

Vd

A

ARVc

WRV

VdtvT

b

V

tnn

V

a


8-c The single phase full bridge inverter in Fig.8-11 has an RLC load with R=10, L=31.5mH, and C=112F. The inverter frequency fo = 60Hz and dc input voltage Vd = 220V. Consider the square wave operation with device conduction angle of 180 per cycle. (a) Express the instantaneous load current in Fourier series; (b) Calculate the rms load current I1 at the fundamental frequency; (c) Calculate the THD of the load current; (d) Calculate the power absorbed by the load Po and the fundamental power Po1; and (e) Calculate the average current of dc supply Id. Solution:

AVP

WRI

WRI

I

IITHD

III

c

Ab

t

ttt

t

tttt

a

nn

nn

and

nn

n

j

n

j

njnj

do

o

o

o

oo

ooo

o

oo

7.7I (e)

4.1638P

1695P (d)

%18_

02.13I

harmonic.ninth the toupConsider )(

8.1221.18I )(

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17.703773sin17.372.49377sin1.18i

ascurrent load ousinstantane obtain thecan weangles,factor power the todue

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368.2187.1tan

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68.2387.11tan

is voltageharmonicnth for the anglefactor power the

68.2387.1110Z

is voltageharmonicnth for the impedance The

68.23

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is voltageharmonicnth for the reactance capacitive The

87.11105.31602X

is voltageharmonicnth for the reactance inductive The

d

21o1

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29

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o

o

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o

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8-d Repeat Problem 8-c assuming that C = 0. The solution to this problem is not provided. 8-e In the single phase full bridge PWM inverter circuit of Fig.8-11, Vd = 300V, ma =0.8, mf = 39 and

the frequency of the sine modulating wave is 60Hz. The load resistance is 10 and load inductance is 1mH. Bipolar modulation technique is used. Determine (a) switching frequency of each transistor; (b) the rms values of voltage and current at the fundamental frequency; (c) the rms values of 37th and 39th harmonic voltages and currents; and (d) the ratio of 39th harmonic current to fundamental current.

Answers:

575.0II )(

759.9001.06039210

52.173

Z

VI 52.173

2

V818.0V

719.2001.06037210

67.46

Z

VI 67.46

2

V22.0V )(

96.16001.060210

7.169

Z

VI 7.169

2

V8.0V

is lfundamenta of voltagerms the1,-8 Table From )(

23403960 )(

o1o39

L39

o39o39

do39

L37

o37o37

do37

L1

o1o1

do1

d

Aj

V

Aj

Vc

Aj

V

b

HzHzfa sw

8-f Consider a 3 inverter in Fig.8-21. Assume that IGBTs are used as switching devices. The

conduction angle of each IGBT is 180o per cycle. (a) Draw waveforms of IGBT gate signals, line-to-line and line-to-neutral voltages of the

inverter. (b) Derive an expression for the calculation of rms values of line-to-line and line-to-neutral

voltages. (c) Derive an expression for the calculation of rms values of fundamental components of line-

to-line and line-to-neutral voltages.

8-g The three-phase inverter in Fig.8-21 has a wye-connected load of R = 5 and L = 23mH. The inverter frequency is fo= 60Hz and the dc input voltage is Vd = 220V. The conduction angle of each switching device is 1800 per cycle. (a) Express the instantaneous line-to-line voltage vab(t) and line current ia(t) in a Fourier series. (b) Determine the rms line voltage VL ; (c) Calculate the rms phase voltage VP; (d) Calculate the rms line voltage VL1 at the fundamental frequency; (e) Calculate the rms phase voltage at the fundamental frequency, VP1; (f) Determine the total harmonic distortion THD of line current ia; (g) Determine the load power Po; and

(h) Determine the average dc current Id. Solution (continued on next page):


A. 6.72201473220PIcurrent dc average The )(

. 1473591.933Ppower load The

91.92

06.010.013.033.064.014I

iscurrent line rms The current. phase theas same theiscurrent line theloads, connected-For wye )(

053.014

06.010.013.033.064.0THD

harmonic.17th toupConsider )(

. 03.993

V )(

. 53.1712207797.0V )(

. 7.1032204714.0V )(

. 63.1792208165.0V )(

)8837717sin(06.0)5.8737713sin(10.0

)8737711sin(13.0 )3.853777sin(33.0

)4.833775sin(64.0)60377sin(14i

567.8tan67.85tanZ

)30(377t17 sin27.14)30377(13 sin66.18

)30(337t1122.05sin )30(377t7 sin66.34

303775 sin52.4830337 sin58.242v

as written becan t v voltageline -to-line ousinstantane The notes. classyour Refer to )(

od

22o

21

222222

L

21

22222

1p1

L1

p

L

a

122122L

o

oo

ab

ab

h

WRI

A

g

f

VVe

Vd

Vc

Vb

tt

tt

ttt

nnRLnLnR

t

ttt

a

L

L

oo

oo

oo

o

8-h The three phase inverter in Fig.8-21 operates in a PWM mode. A three phase wye-connected load

of R = 10 and L = 2mH is connected to the inverter. The input dc voltage is 240V. The gating pattern is sinusoidally modulated with ma = 1.0 and mf = 21. The fundamental frequency is 100Hz. Determine: (a) the rms values of fundamental voltage and current; (b) the rms values of 5th and 7th harmonic voltages and currents; and (c) the rms values of 19th and 23rd harmonic voltages and currents.

The solution to this problem is not provided.


4. Examination Samples

MIDTERM EXAMINATION

ELE 754 Power Electronics

Ryerson University Department of Electrical and Computer Engineering

Duration: 2 hours

Instructor: Dr. B.Wu Instructions: 1. Answer all four questions.

2. Show details of your derivations/calculations. Marks: Q1. 30% ______________ Q2. 20% ______________ Q3. 30% ______________ Q4. 20% ______________ Student Name: _________________________________ Student Number: _________________________________ Question 1 (30%)


+

R

Ea

Vo

+

-

D

iD

iT iO

-

Vd

G

T

Fig. 1 One quadrant chopper with a dc motor load An IGBT-based one quadrant chopper is used to drive a dc motor, whose armature circuit parameters are represented by the winding resistance R (0.2 Ohms), winding inductance L (0.25mH) and back emf as shown in Fig.1. The input dc voltage of the chopper is 115V and its switching frequency is 500Hz. When the duty cycle of the chopper is 0.4, the motor runs at 1200rpm with its emf of 55V. Answer Questions 1.1 to 1.3. 1.1) The peak value of armature current.

3% 33

1025.12.0

1025.0

R

L , sec108.0500

14.0 3ont

)1( /,

ontadpeako e

R

EVI

64.0/ ont

Ae 8.141)1(2.0

55115 64.0

6.125.1

2

ms

msTs

1.2) The average (dc) armature voltage and armature (dc) current.

3% Ae

eIo 6.144

2.0

55

1

1

2.0

1156.1

64.0

min,

discontinuous mode

3%

64.064.03 1

55

551151ln1025.1 eetx

ms32.1

)874.2ln(1025.1 3

VV dco 7.642

32.1255

2

8.0115,

AI dco 5.48

2.0

557.64,

1.3) Sketch to scale the waveforms of vo and io, indicating the peak values.


3% 115V

Vo

iO141.8V

tx

Question 1 (continued) An external dc choke of 6mH is connected in series with the armature winding of the above mentioned dc motor. The duty cycle is adjusted to 0.5 and the motor runs at 1120rpm. The field winding current remains unchanged. Find: 1.4) The dc component (average value), fundamental component (peak value) and third harmonic (pick

value) in. 3% Vo,dc = ________ 3% Vo1, peak = ________ 3% Vo3, peak = ________ Check: continuous or discontinuous mode?

33

1025.312.0

1025.6

R

L , msmston 0.125.0

032.025.31

0.1/ ont 064.0

25.31

2

ms

msTs

Ae

eIo 25.26

2.0

33.51

1

1

20

115064.0

032.0

min,

continuous mode 33.5111201200

55aE

Vd

Vo

Use Fourier Series to find Vo,dc, Vo1, peak and V03, peak


1.5) The dc component (average value), fundamental component (peak value) and third harmonic (peak value) in io.

3% Io,dc = ________ 3% Io1, peak = ________ 3% Io3, peak = ________ Vo

115V

t

10

...5,3,1

)sin(5.574

2

115

no tn

nV

...5sin64.143sin4.24sin21.735.57 ttt dc 1st 3rd

AI

ALR

VI

AI

peako

peako

dco

414.0)1025.650032(2.0

4.24

73.3)1025.65002(2.0

2.7321.73

82.302.0

33.515.57

232,3

23222,1

,

Question 2 (20%) The input terminals of a single phase full bridge diode rectifier are connected to a single phase ac voltage source rated at 208V and 400Hz with a source inductance of 4%. The rectifier is rated at 20kVA. The load of this rectifier is an ideal dc voltage source of 240V. Answer the following questions: 2.1) Find the value of source inductance in mH. 4% 2.2) Derive an expression for the calculation of output dc current. 4%


2.3) Sketch to scale the waveforms of input voltage, dc voltage, dc current, and the voltage on source 6% inductance. 2.4) Calculate the diode conduction angle per half cycle of the input line frequency. 6% Solutions: 2.1)

HL

mHL

Z

AI

rated

rated

rated

6109.45148.104.0

148.14002

884.2

884.212.72

208

12.72208

15000

vs

id

vL

b p

f

294V

383.2A

use equ (1)

294V - 240V = 54V

2.2) rad

rad

bp

b

187.2

945.067.542082

240sin 1

)1(......................34592080cos2549

)96.2282401.170cos1.294(668.8

)954.0(240954.0coscos2082109.454002

16

equAi

i

i

d

d

d

2.4) f = ? 357.1816.0cos

034592080cos2549

ff

ff

Try f = 2.6 cos + 0.816 = 1.265

Try f = 2.8 cos + 0.816 = 1.342

Try f = 2.82 cos + 0.816 = 1.352 conduction angle: f - = 2.82 – 0.954 = 1.866 rad = 106.9


Question 3 (30%) A three-phase full bridge SCR rectifier without a freewheeling diode is connected to a three phase balanced voltage source of 60Hz, 208V (line to line). Assume the load of the rectifier is a constant current source of 100A and its firing (daley) angle is 60 degrees. Find: 3.1) The average dc output voltage of the rectifier.

6% VVd 4.140cos20835.1 at which is 60

3.2 The total harmonic distortion of the input current and overall input power factor.

12% AItdiI daa 6.81816.02

1 2

0

2

A

II

tItIi

da

da

78.772

1.1

...)(5sin22.0)sin(1.1

1

%312

1

21

2

a

aa

I

IITHD

995.01

12

THDDF , 5.0cos DPF

478.0 DPFDFPF

3.3 Use Fig.2 given below, draw the dc output waveform assuming that the delay angle remains at 60 6% degrees.


3.4 If a freewheeling diode is connected in parallel with the dc current source, draw the dc output waveform assuming that the delay angle remains at 90 degrees

6%

Question 4 (20%) 4.1) List four switching devices that are commonly used in power electronic converters. IGBT, GTO, MOSFET, SCR 4.2) Indicate which of the switching devices you have listed are voltage controlled devices and which

are current controlled devices. IGBT, MOSFET Voltage Controlled Devices

SCR, GTO Current Controlled Devices 4.3) Draw a typical circuit diagram of an IGBT-based two quadrant chopper.

LOAD

Vd

+

- 4.4) Draw a curve which illustrates the relationship between the boundary current and the duty cycle D

of a buck converter. IBD

D 4.5) Write an equation which describes a dynamic relationship between motor speed, electromagnetic

torque and load torque. mLem BTT

dt

dJ

In the following multiple-choice questions, circuit one best answer. 4.6) In a single phase, single diode rectifier with an RL load, the diode conduction angle is

a) grater than 180. b) less than 180. c) equal to 180. d) grater than 120. e) less than 120. f) equal to 120. g) grater than 60. h) less than 60. i) equal to 60. j) none of the above.


4.7) In a three-phase, full SCR bridge rectifier with a three RL load where L is assumed very large, the delay angel is 60 degrees. The SCR conduction angle is a) grater than 180. b) less than 180. c) equal to 180. d) grater than 120. e) less than 120. f) equal to 120. g) grater than 60. h) less than 60. i) equal to 60. j) none of the above.

4.8) In a three-phase, full SCR bridge rectifier with a three R load, the delay angel is 60 degrees. The

SCR conduction angle is a) grater than 180. b) less than 180. c) equal to 180. d) grater than 120. e) less than 120. f) equal to 120. g) grater than 60. h) less than 60. i) equal to 60. j) none of the above.

4.9) The maximum voltage and current ratings of a single IGBT device currently available in the market

are a) 600A, 1000V b) 600A, 2000V c) 600A, 4500V d) 600A, 6500V

e) 1200A, 1000V f) 1200A, 2000V g) 1200A, 4500V h) 1200A, 6500V 4.10) The source inductance between an ideal voltage source and three phase SCR rectifier

a) may be used to assist SCR commutation b) may increase input current THD c) may reduce the dc output voltage d) may increase the dc output voltage d) none of the above.

2011_notes

Documents

interrupt

ton toff

dc motor drive

power convert

igbt device

ac dc

power converter

versus delay