Power Electronics - Iii [FIRST ed.] 8184317212, 9788184317213

Citation preview

0 I

Strictly According to the Revised Syllabus of

Rajasthan Technical University (RTU ~ 2006 Course)

Power Electronics • Ill [SEEl] Semester - V (Electrical Engineering)

Dr. J. S. Chitode M. E. (Electronics), Ph.D. Fonnerty Professor & Head, Department of Electronics Engineering Bharati Vidyapeeth University College of Engineering, Pune

Price Rs. 195/-

I

Visit us at : www.vtubooks.com

@)

~

8

Tuchnical Publications Pune (1)

flif1[1\i1i1111111111H YG1U- 8NL-LW8F

t

Power Electronics - Ill ISBN 9788184317213 All rights reserved with Technical Publications. No port of this book should be reproduced in any form, Electronic, Mechanical, Photocopy or any information storage and retrieval system without prior permission In writing, from Technical Publications, Pvne.

Published by : Tuchnic.al Publications Pune• #1 , Amit Residency, • 12, Sn.n1w1, Pfth, Punc . 411 030, lndi..

Printer : AlatOTPrintm Sr.110. , W3,Slnhl,e,d Ro.cl, ~

. -411 0-41

Table of Contents Chapter-~1 ·Power Semlconductor'Devic~~~~~ ·

(1 -1) to (1--70)

1. 1 fntroduction . .... ..... . .. .. . ... .............. .. ... ..... ... . ... ..... ..... .... . ... ... .. . .... .. .... . .... 1 - 1 1.1.1 Applications of Power Electronics • . . . • . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 2

1.1.2 Advantages and Disadvantages of Power Electronic Controllers . . . . . . . . . . . . . . . . . . 1 - 2 1.1.3 Block Diagram of Power Electronic Controller. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 3

1.2 Types of Power E1ectronic Converters................................................ 1 - 4 1.2.1 AC to DC Converters . . . . . . . . • . . . . . . . . . . . . . . . . . • . . . . . . . . . • . . . . . • . . . • . . . . . 1 - 4 1.2.2 DC to AC Converters Inverters . . . . . . . . . • . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . • 1 - 4

1.2.3 DC to DC Converters (Choppers} . ... .•. . . • . .. ...... . ..... . . ... ....... . .. . .. 1- 5 12.4 AC to AC Converters

cloconverters • . . • . . . •• . . • • . . . . . . . . . • . . . . . • . . . • . . . . • 1 - 5

1.2.5 AC Regulators . • . . . . . . . . . . . . . . . . . • . . . . . . • . . . • . . . . . . . . . . • . . . . . • . . . . . . . . . 1 - 5 1.3 Power Semiconductor Devices ............................................................ 1 - 6

1.4 Power Diodes ....................................................................................... 1 - 6

1.4.1 Structure of Power Diode ••• ••••••••••••••••....•. . ..•...... . ............• 1- 7 1.4.21-V Characteristics . . . .. . ....... . . .. ............. . .. . .. . .•.. .. ... . ..... . . 1 - 7 1.4.3 Switching Characteristics of Diodes . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . • . . • . . . 1 - 8

1.4.4 Types of Diodes . . . . . . • . . . . . . . . . . • . . • . . . . . . • . . . . . . . . . . . . • . . . . . . . . . . . . . . . 1 - 9 1.4.4.1 General Purpose Diodes • • . . • • • . . . • . . . . • . . . . . • . . . 1 - 9 1.4.4.2 Fast Recovery Diodes (Switching Diodes) . . . . . . . . . . . . . . . • . • 1 -10 1.4.4.3 Schottky Diodes . . • . . . . . . • . . • . • . . . . . . . . • . . . . 1 • 10

1.4.5 Applk:ations of Power Dkldes. . . . • • . • . . . • . . . • . . • . . . . . . . . • • . • . . . . • • . . • • . . . • 1 - 12 1.4.6 Specifications and Ratings of Power Diodes ...•. . ...... ..... . .....•.... . . .. . 1 -12 1.5 Silicon Controlled Rectifier (SCR) ...................................................... 1 - 12 1.5.1 Construction of SCR • .......... .. ..... ..... .. ... ....•... .•.. .. •. . . . ... . . 1 - 13 1.5.2 Merits, Demerits and Applications of SCR . •...•........•..............••. . . 1 -14 1.6 SCR Characteristics and Modes of Operation .................................. 1 - 14 1.6.1 Reverse Blocking ~e.. . .... . . .. .... . .. . . .. ... . ... . .. . .. .. .... ... •...• 1 -14 1.6.2 Forward Blocking Mode ... . .....• . ..... .. ..•........ • . ... • ... •. ... . . .... 1 - 15 1.6.3 Forward Conduction Mode .. .. . ............ • . . . .. ........•........... . . . 1 -16

1.6.4 Latching and Holding currents . .... . .. . .... . . . . ... . . ................ . . .. . 1 -19

Copynghted ma nal

1.6.4;1 Latching current (IL)

1 -19

1.6.4.2 Holding Current(~).

1 -20

1.6.4.3 Comparison (Difference) between Holding and Latching Currents . .

1 • 20

1.6.5 Two Transistor Model of SCR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 26

1.7 SCR Tum-on and Tum-off ................................................................. 1 - 28 1.7.1 DifferentWaystoTum-ontheSCR ............. . . ......................... 1-28 1.7.2 Tum-on Dynamic Characteristics .. . .......... . ........•...•. ... •...•.... .. 1 - 29 1.7.3 SCR Tum-off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 30 1.7.4 Tum~ff Dynamic Characteristics . ........ ..... .. ... . ... ... . .. .. ... ..... . .. 1 - 30 1.7.5 Inverter Grade and Converter Grade SCRs ..

1- 32 1.8 SCR Gate Characteristics ................................................................. 1 - 32 ft •• ,

•• •

ft . . . ,

•••• ft . . . . . . . II II.

1.8.1 Pulsed Gate Drive . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . • 1 - 33 1.8.2 Requirement of Gate Drive . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 33

1.9 SCR Ratings .................................................................................... 1 - 34 1.9.1 Current Ratings . • . . . • . . . . . • . . . . . . • . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . 1 - 34 1.9.2 Voltage Ratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 35 1.10 Thyristor Types ................................................................................ 1 - 36 1.10.1 Gate Tum-off Thyristor (GTO) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . • . . . • . . . . 1 - 36 1.10.1.1 Structure of GTO . . . . . . . . . . . . . 1 -36 1.10.1.2 Characteristics of GTO . . • . . . . . . . . . . . . . . . . . . . . . . 1 - 38 1.10.1.3 Advantages, Limitations and Applications of GTO . . . . . . . . . . . . . . 1 - 38

1.10.2 light Activated SCR (LASCR) . . . . . . • . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 39 1.10.3 Reverse Conducting Thyristor (RCT) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . 1 - 39

1.10.4 Triac (Bidirectional Triode Thyristors)..................... . ................ 1 -40 1.10.4.1 V-1 Characteristics ofTrfac . . . . . . .

1 - 41

1.10.4.2 Operating Modes of Triac • . . . . • . . . . . . . . . . . • . . . • . . 1 - 41

1.10.4.3 Merits, Limitations and Applications of Triac . . . . . . . . . . . . . • . . 1 -43

1.10.5 MOS Controlled Thyristor (MCD .... . . . ...... ... ...... ........ .......•. .. 1 - 44 1.10.5.1 Internal Structure of MCT .

. . . .

1 - 44

1.10.5.2Tum-on and Tum-off MCT. . . . . . .

1-45

1.10.5.3 Characteristics of MCT . . . . . . . . . .

1- 46 1-46

1.10.5.4 Advantages and Disadvantages of MCT . . .

1.11 Power BJT ....................................................................................... 1 - 45· 1.11 .1 Structureof BJT . . . ... . .. . ... ..... . ... . . .. .. . . ..... . .. ... ... . . .... .... 1 -47 1.11.2 Steady State Characteristics of BJT. .. . . .. . . .. . . .. . . .. . .. . .. . . . .. .. . . . .. . . 1 - 47

1.11 .3 Safe Operating Area (SOA) of BJT . . . . . . . . . . . . . . • . . . . . . . . . . . . • . . . • . . . . . . . 1 - 49 1.11.4 SWitchlng Characteristics of BJT .. . .. .. . .. .. . .. .. .. . . .. . .. . .. . .. . . .. .. .. . 1 - 50

Copynghtcd ma nal

1.11.5 Merits, Demerits and Applications of BJT........ •. . ..... ... . . ..•.. . •. .. . . . . 1 - 51 1.12 Power MOSFET .............................................................................. 1 - 52 1.12.1 Structure of MOSFETs. . . . . . . . . . . . . . . . . .. . . . . . . . . . .. . .. . . . . .. . .. • . . .. .. 1 - 53 1.12.2 Steady State {V-1) Characteristics of MOSFETs • ..• . .. .•.. . .•...•.•.... • .... 1 - 54 1.12.3 Switching Characteristics of MOSFET . . • . . • . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 55 1.12.4 Merits, Demerits and Applications of MOSFETs . . . . . . . . . . . . . . . • . . . . . . . . . . . . . 1 - 57 1.1 3 IGBT................................................................................................ 1 - 58 1.13.1 Structure of IGBT . . . . . . . . . . . . . . . . . . . • . . . . . • . . . . . . . . . . . . . . • • . . . . . • . . . . • 1 - 58 1.13.1.1 Punch through IGBT . . . • . • • . . . . . . . .

1-59

1.13.1.2 Non-punch through IGBT . . . . . . . . . . . • . •

1-59

1.13.1.3 Operation of IGBT . .. . . . • . . . ..

1- 60

1.13.1.4 Latchup in IGBT. . . • . . . . • • . . . . . . . • • . . • .

• . • . 1- 62

1.13.1.5 Body-Source-Short and its Reason • • . • • • . • . . . . • • . . • . . . 1- 62

1.13.2 Safe Operating Area (SOA) of IGBT . . . .. .•..• . ...... .•. . ..•. . •.. . . . ... . • . 1 - 63 1.13.3 Steady State (V-1) Characteristics of IGBT. . . . . . . . . . . . . . . . . . . . . . • . . . . . . • . . . . 1 - 64 1.13.4 Switching Characteristics of IGBT . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . 1 - 64 1.13.5 Merits, Demerits and Applications of IGBT. . .. .. . .................... .. ..... 1 - 66 1.13.6 Protection Circuits for IGBT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 66 1.13.6.1 Gate Overvoltage Protection . . . . . . . . . . . . . . . . . . • . . 1 - 66 1.13.6.2 Overcurrent Protection . . . . . • . . . . . . . . . . . . • . . . . . . 1 - 67 1.13.6.3 Snubber Circuits for IGBT . . . . . . . . . . • . . . . . . . . . . . . .

1 - 67

1.13.7 Comparison of Power Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . 1 • 69

Chapter• 2 Drive and Protection Circuits for Power.Devices

(2 • 1) to (2 ·. 66)

2.1 Firin Circ.uits for the· SCR ................................................................... 2 - 1 2.1.1 Features of Arin Circuits . . . . . . . . . • . . . . . . . . . . . . . . . . . • . . . . . . • . . . . . . . . • . . . • 2 - 2

2.1.2 R-Firing Circuit. . . . ...... . ...... . . .... .. . .. . .. .. ...... ... . .. .. . .. .... . .. 2 - 2 2.1.3 RC Firing Circuit . . .. . . ... . .. .. . .. .. .. . . .. . ... .. ...... ... ......... . .. ... 2 - 3 2.1.4 Full Wave RC Firing Circuit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . 2 - 5 2.1.5 UJT Triggering Circuit . . ................ . ....... . .......... . ....... . ... .. 2 - 7 2.1.6 Pedestal Circuit with Cosine Modified Ramp . . . . . . . . . . . . . . . . . • . . . . . . . . • . . . . . . 2 - 10 2.1.7 SCR I TRIAC Triggering using Diac . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 19 2.1.7.1 V-l Characteristics of Dlac . . . . . . . . . . . . . . . . . • . . . . . • 2 - 20 2.1.7.2 DIAC-TRJAC Phase Control . . . . . . . . . . . . . . . . . . . . . . . 2 - 20

2.2 Drive Circuits for BJT ......................................................................... 2 - 21 2.2.1 Base Drive Control During Tum-on . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . • . . . . 2 - 22 2.2.2 Base Drive Control During Tum-off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 23 2.2.3 Proportional Base Controj . . . • . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . • • . . . . . . . . . . . 2 - 24

Copyr ghted ma nal

2.2.4 Anti-saturation Control . . . . . . . . • . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 24 2.2.5 Typical Driver Circuit for Power BJT. . . . . . .. . . . .. . . . .. . .. .. . .. . . •• . .•....•.. 2 - 26

2.3 Drive Circuit for MOSFET .................................................................. 2 - 27 2.4 Driver Circuit for IGBT and MOSFET................................................. 2 - 27

2.5 Isolation of Gate and Base Drives ..................................... ............... 2 - 28 2.5.1 Necessityof Isolation . • . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . • . . . . . . . . . . . 2 - 28 2.5.2 Isolation using Pulse Transformer ..... . .. .. . . ... . . . . ... . .. . . . . •. . .... . .... 2 - 29 2.5.3 Isolation using Optocoupters . .. . .. .. . .. . . . . . .. . .. . . .. . . .. . . .. • . . .. . .. .. . . 2 - 30

2.6 Protection Circuits ............................................................... ............... 2 - 31 2.7 Protection Against

d¾tand Overvoltages........................ ............... 2 - 32

2.7.1 Snubber Circuits (Tum-?ff Snubber) ..... . . • . .. .. ... .... . .•.. . .•. . . . . . . . . . . 2 - 32

2.7.2 Metal Oxide Varistors {MOVs) . . .. . ......... ... ...... . ........•.. . . ... . . .. 2 • 36 2.7.3 Improving dv/dt Rating with the Help of Cathode Short . ..... . .... . .•. . .... . .. .. 2- 37

2.8

¾t Protection with the Help of Inductor (Tum-on Snubber) ............ 2 - 38

2.9 Overcurrent Protection ...................................................................... 2 - 47 2.9.1 FuS8' ... .. , . .. . ....... . ....... . . .... . ...... .... . ... .. . : ... .. . . . . ....... 2 - 47 2.9.2 Semiconductor Fuses.... . ..... . •. .. .•. . ........ ... .....•.. . •......... . . 2 - 48

2.10 Cooling of Semiconductor Devices .................................................. 2 - 49 2.10.1 Concept of Thermal Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 • 49 2.10.2 Thermal Model of a Power Device • . ..... . ....... . ............•...... . ...• 2 - 50

2.11 Types of Cooling ............................................................... ............... 2 - 55

2.11.1 Natural Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 • 55 2J 1.2 Forced Air Cooling • . . . . . . • . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . • . • . . . . . . . . . 2 - 56 2.1 1.3 Liquid Cooling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . 2 - 56

2.11.4 Vapour Phase Cooling . . . . . . . . . . • . . . . . . . • . . . . . . . . . . . . . . . • . . . . • . . . . • . . . . 2 - 58 2.12 Series and Parallel Operation of Thyristors ..................................... 2 - 59 2.12.1 Necessity of Series and Parallel Connection ..... . . . . . ... . . .. . .. . ...... . .... 2 - 59 2.12.2 Series Connection of Thyristors . . • . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . 2 - 59 2.12.2.1 ProblemsEncountered inSeriesConnection . . . . . .

. . . . . 2 - 59

2.12.2.2 Equalizing Components . . . . . . . . . . . . . . . . . . . . . . . . 2 - 60 2.12.2.3 Dynamic Equalization Circuit. . . . . . . . . . . . . . • . • . . . . . . 2- 61

2.12.3 Parallel Connection of Thyristors ....... .. • . .. • . . ••.. . . .. .. ..• • . . •. .. .• .. . 2 - 65 2.12.3.1 Problems Occurred In Parallel Connection . . . . . . . . . . • . . . . . . 2 - 65

2.12.3.2 Equalizing Arrangements . . . . . . . . . . . . . . . . . . . . . . . . 2 • 65

~tfl pter • 3 Single and Three Phase AC/DC Converters 3.1 Introduction ................ ......................................................................... 3 - 1

Copyr ght

ma nal

3.1.1 Principle of AC/DC Conversion (Control'led Rectifier) . . • . . . . . . . . . • . . . . . . . . . . . . . . 3 - 1 3.1.2 Concept of Commutation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . 3 - 2

3.2 Single Phase Half Wave and Full Converter....................................... 3 - 4 3.2.1 Single Phase Half Wave Controlled Rectifier with Resistive Load . . . . . . . . . . . . . . . . . . 3 • 4 3.2.2 Half Wave Coqtrolled Rectifier with RL Load .• .• . . ...... . . ... •..•..•.•• • , ••• , • 3- 6 3.2.3 Half Wave Controlled Rectifier with Freewheeling Diode .. . ..... : ......•. .. . ..... 3 - 8 3.2.4 Single Phase Full Wave Converter with Resistive Load. . . . . . . . . . . . . . . . . . . . . . . . . 3 - 15

3.2.5 Single Phase Full Wave Converter with Inductive Load Circuit Diagram . . ... . .....• 3 -16 3.3 Single Ph ase Semi converters (Half Bridge Converter) ..................... 3 .. 18 3.3.1 Circuit Diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . 3 - 18 3.3.2 Working with Resistive Load . ....... . ... .. ... . ....... ... . .... .. .. . ... . . .. 3 -19 3.3.3 Working with Inductive (R-l) Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . • . . • 3 - 22 3.3.3.1 Continuous Current Mode . . . . . • . . . . . . . . 3-22 3.3.3.2 Discontinuous Current Mode . . . . . . . . . . . . • • . . . . . . . . . 3 - 24 3.3.3.3 Continuous and Ripplefree Current for Large Inductive Load . . . . . . . . . . . 3 - 25

3.3.4 Asymmetrical Half Bridge Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 35 3.3.4.1 Operation wi1h Resistive load. . . . . . . . . . . . . . . 3- 35 3.3.4.2 Operation of Asymmetrical Half Bridge Converter with Level Load. . . . . . • . . 3 - 37 3.3.4.3 Comparison of Symmetrical and Asymmetrical Configurations . . . . . . . . . . 3- 39

3.4 Single Phase Full Converters ........................................................... 3 - 39 3.4.1 Working with Resistive Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 40

3.4.2 Working with Inductive Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 • 44 3.4.2.1 Continuous Load Current . . . . . . . . . . . . . . 3 • 44 3.4.2.2 Olscontlnuous Load Current . • . . . . . . . . • • • • • • , • , • • • , 3 - 45 3.4.2.3 Continuous and Rlpplefree Current for Large Inductive Load . . . . . . . . . . . 3 - 46

3.4.3 Inversion in 1tt, Full Converter ........ . .............. ..........

•

n

3- 61

.. . .. ... .

3.4.4 Comparison of Half Controlled and Full Controlled Rectifiers • . . . . . . . . • • . . . . . . . . . 3 - 63

3.5 Three Phase Half Wave Converters ................................ :................ 3 - 64 3.5.1 Operation with Resistive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 64 3.5.2.Operation with Inductive Load . ... ................. ..... . ........ . .•. ... .• 3 - 73

3.6 Three Phase Semiconverters ........................................................... 3 - 78 3.6.1 Operation with Resistive Load . . . . . . . . • . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . 3 - 78 3.6.2 Operation with Inductive Load . . • . . . . . . . . . . . . . . . . . . • . . . . . . . . . . .. . . . . . . . . . . . 3 ~ 84

3.7 Three Phase Full Converters ............................................................. 3 - 87 3.7.1 Operation with Resistive Load ............... . .... . .. .. .... . ... . .......... 3 - 87 3.7.2 0

ra•tion with Hi hi Inductive Load . . . . . .. •. .•.... • .. . .. . . . ....... . .•. . .• 3 - 93

3.7.3 Comparison between 3q> and 1ct, Converters ............................ ... 3 • 101

(viii)

'

,

"" ..

......

;.

,. .

~~

..~... 1'

;fi't., •

St

,r:1

Copyr ghted ma nal

3.8 Effect of Source Inductance ..........................: .................................. 3 - 101 3.8.1 Effect of Source Inductance in 1~ Full Converter.. .. . .. ... .. • .. . . ... .... . ... . 3 - 102 3.9 Single Phase Dual Converters ......................................................... 3 - 106 3.9.1 Ckculating Cunent and Non-circulating Current Dual Converters....... . •. ..•... 3 -109 3.9.2 Comparison between Circulating and Non-circulating Current Modes ... . ......... 3- 109 3.1 0 Three Phase Dual Converters ...................................................... 3 - 11 1 3.10.1 CirculatingCurrentMode . ... . . . . . .... ... ... . ...... •... • .. ... ..•. . .. • . . 3-111 3.10.2 Non-circulating Current Mode . . . ...... . ....•.... ... .. .. .. .. .......... .. 3-115 3.11 Power Factor Im rovement.. .......................................................... 3 - 115 3.11.1 Extinction AA le Control . . . . . . . . • . . • • . . . • . . • . . • . . . . • . . • • . . . . • . . . • . . • . . . 3 - 116 3.11.2 Symmetrical Angle Control (SAC) • . . . . . . • . . . • . . . . . . • • . . • . . . . . . . . . • . . . . . . 3 - 118 3.11.3 Putse Width Modulation (PWM) .... . ........ •..•. . .. ... .. ... .• ...•...... 3-120 3.11.4 Sinusoidal Pulse Width Modulation . .. . .....•..•....•........... . .. . : •... 3 - 121

Summa ........ ........................................................................................ 3 - 123

.

. c~,,._

~~

. ).

4.1 Introduction ......................................................................................... 4 - 1 4.2 SteP:down Chopper ............................................................................. 4 - 2 4.2.1 Chopper Control Techniques ... .. . . .. . . . .• .. . ....... .. . .. ... . ... . •. ..• . .•• 4 - 7 4.3 Step-down Chopper with RL Load .................................................... 4 - 11 4.3.1 Continuous Load Current . .. . .. . .......... . .. . .... . .• .. . ... . .. . ......... 4-12 4.3.2 Discontinuous Load Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 • 13 4.4 Step-up Chopper............................................................................... 4 - 27 4.4.1 Use of step-up Operation for Energy Transfer ... . ..• . .. .•. . ......••... . ..... 4 - 34 4.5 Chopper Classification ...................................................................... 4 - 36 4.5.1 Class A Chopper . • . . . . . . . . . . • . . . . . . . . . . • . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . 4 - 36 1

4.5.2 Class B Chopper .•. . .....•..•... .. ... . ................... .. .......... . 4 • 37

4.5.3 Class C Chopper •.. .... . . . ..... . ... .. . ..... ...... . .. .. ... ..... .. ...... 4 - 38 4.5.4 Class D Chopper . . . . . . . . . . . . . . . . . . . • . . . • . . . . . . . . . . . . . . . . . . . . . . . • . . . • . . 4 - 40 4.5.5 Class E Ch r Four Quadrant Cho r . •.... . ......•. . .•. ... .. .. . ..•.. . 4 - 41 4 .6 Application1s of Choppers ................................................................... 4 - 44

4.7 Switching Mode Regulators (SMPS).................................................. 4 - 45 4.7.1 Classification of SMPS .. . . ... ..... . . . , . . .. . .. . .... . .•. . ... ... ••..• . ..• .. 4- 46 4.• Nonisolated Converters ..................................................................... 4 - 47 4.8.1 Buck Regulators . . . .... . . . ........... . ....... . ............ . ... . .. .. •.. 4 - 47 4.8.2 Boost Regulators ........ . .. . .. . ....... . .. . . . . . . .. .. . . .. . . ... .. .. . .. . .. 4 • 53 4.8.3 Buck Boost Regulators . .. ..... . .•. . . . .. . .. .... . ...... ... . ...... . . ....• . 4 - 57

4.8.4 Cuk Regulators .• ... •. ... .... . ......... . ..... . .. ... . . . ...... . .. .. •... . . 4 - 60

Copyrighted mat r al

4.9 Isolated Converters ............................................................................. 4 - 65 4.9.1 Flybadk Converter·.... ... ....... . . .. . .. ....... . . . .. .. ........ .. ...... . . 4 - 65 4.9.2 Forward Converter . .. . .. . .....·..... . .. . .. .. .. . . .. . . . .. . ....... .. . ...... 4 - 70 4.10 Advantages, Disadvantages and Applications ................................. 4 - 73 4.10.1 .Advantages.......•..•..•......•...•... •. .•. . .•.. .. ... .•.•..• ..... .• . 4 - 73

4.102 Disaclvantag.es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 73 4.10.3 Appllcations of SMPS •. .. ... . .. . .. . ... . .. .. ............. . . ....... .. . . .. 4 - 73 4.10.4 Comparison between Linear Switched Mode and Resonant Converters . .. ..... 4 - 74 Summary.................................................................................................. 4 - 74

,-A tmmn

-RelatloDI

Abbreviations and Symbols A AC B C D

Anode AC supply current or voltage Base or B-phase voltage Collector or capacitor Diode DFW Freewheeling diode E Ba.ck emf of the motor, Emitter Eg ,eg Back emf of the motor f Frequency fnpplt Ripple frequency G Gate HF Harmonic factor Supply current Is,~ Peak or maximum current Im ICI i, Capacitor current or charging current Io, io Output current Inductance curr,ent iL Ig I ig Gate current Base current 'B Anode to cathode current /AK Average current rating (SCR) Ir RMS current rating (SCR) ITR 1ripple Ripple current 1FW Freewheeling CWTent Average current l(av) RMS current lrm.s Circulating current ic1, K Cathode L, L, Inductance, or current limiting reactor PF Power factor Pa, P; Output, input power R Resistance or R-phase voltage T Period of the waveform, transistor orSCR On or conduction period of switch Ton

Tum-on time of switch t, tajf Circuit tum-off time Tum-off time of switch tq Off period of switch Toff Rise time t, Spread time tp Delay time td Reverse recovery time ~ Gate recovery time tgr Vs, Vs Supply voltage Peak value of supply phase voltage Vm Forward break-over voltage VBO Reverse break-down voltage VBR Supply or load DC voltage Vdc Vo, Vo Output voltage RMS voltage Vrms Average voltage v(av) Gate voltage Vg Base voltage VB v, Capacitor voltage Vp, VP Peak voltage in UJT or supply V 88 , V Bias voltage or UJT supply voltage Vripple Ripple voltage v R, Vy, v 8 3$ supply voltages Anode to cathode voltage VAi< Intrinsic standoff ratio 11 dv Rate of change of voltage dt tan I

di

dt

Rate of change of current

0

Duty cycle Pi or half cycle period of supply Triggering angle or firing angle Angular frequency Extinction angle

7t

a ro J}

(xii)

Power Semiconductor Devices

Objectives • Principle of operation of power converter . Necessity, Input, output and controls. • Applfcattons of power electontcs. Advantages and disadvantages. • ClassJftcatton of power converters depending upon input and output. • Introduction to power semiconductor devices, their Important rat1ngs. • Charaderistics of power devices.

1.1 Introduction •

Power electronics is one of the important branch of electronics and electrical engineering. It deals with conversion and control of electric energy. We know that AC voltage and current of fixed frequency is available from mains. This supply cannot be used always directly. For example computer needs SMPS (Switched Mode Power Supply) for its working.

•

Fig. 1.1.1 shows the basic functioning of power electronic system. The electric Electric Power energy energy in one form is given at the input Electric energy electronic in the The power electronic system converts the In one sys,tem another electric energy in the other form. For form form Control& example, the input may be AC and the conversion output can be DC. We know that such conversion is performed by rectifier. Thus Fig. 1.1.1 Basic inputs/outputs of rectifier is a power electronic system. power electronic systems

•

The power electronic system thus performs conversion of electric energy. It also controls the amount of el~c energy to be given to the output.

•

The word puwer means high amplitudes of current and voltages. (1 • 1)

Power Electronics -111

1-2

Power Semiconductor Devices

1.1.1 App,lications of Power Electronics We will briefly present the various applications where power electronic systems are used. There are numet;qus applications coming up every day in power electronics. Some of them are mentioned below : 1. Uninterruptible power supplies and stand by power supplies (emergency power supplies) for critical loads such as computers, medical equipments etc. 2. Power control in resistance welding, induction heating, electrolysis, process

industry etc. 3. Power conversion for HVOC and HVAC transmission systems. 4. Speed control of motors which are used in trat tion drives, textile mills, rolling mills, cranes, lifts, compressors, pumps etc. 5. Solid state power compensators, static contactors, transformer tap changers etc.

6. High voltage supplies for electrostatic precipitators, and x-ray generators, etc. 7. Power supplies for communication systems, telephone exchanges, satellite

systems etc. These are some of the important applications of power electronics. 1.1 .2 Advantages and Disadvantages of Power Electronic Controllers

The control and conversion of electric power is performed with the help of power electronic controllers. Thus power electronic systems consists of controllers. The power electronic controllers are also called as power electronic converters. The power electronic controllers have following advantages : 1. Fast dynamic response due to static de~ices. 2. High efficiency of conversion due to

low losses in electronic devices.

3. Compact size and light weight of the V 80 , rather it is turned-on.

1.6.3 Forward Conduction Mode When the SCR is forward biased, then it can go into forward conduction by following techniques : i) When VAK > VBo ii) When gate drive is applied

iii) When dv exceeds permissible value dt

iv) When gate cathode junction is exposed to light Here note that the SCR can go in the forward conduction mode only if it is in the forward blocking mode earlier. (I) When V AK >Vso The SCR is driven into forward conduction mode when anode to cathode voltage (VAK) exceeds the forward break-over voltage (V80 ). The SCR is said to have turned-on when it operates in forward conduction mode. When V A K > V 80 , the SCR is driven in forward conduction even if gate is open. From Fig. 1.6.3, it is clear that junction / 2 is reverse biased during forward blocking mode (VAK < V 80 ). When V AK exceeds v8 0 , the avalanche break-down of junction / 2 takes place even if gate drive is not applied. Hence heavy current starts flowing through the SCR and anode to cathode voltage falls to very small value. This is shown in Fig. 1.6.1. The dotted line

1 -17

Power Electronics - Ill

Load

+

V

Fig. 1.6.4 Use of SCR in forward conduction

Power Semiconductor Devlc.s

(...........) indicates switching of SCR · from forward blocking state (i.e. OFF) to forward conduction state (i.e. ON). The anode to cathode current of the SCR is only limited by the load. Fig. 1.6.4 shows such situation : When the SCR conducts in the forward conduction mode, it is said to have turned 'ON'. 'T he anode to cathode voltage is less than 2 volts. This voltage is normally neglected in calculations. Then the current through the load and SCR will be,

V

1AK = Load

... {1.6.1)

Thus the SCR current is only limited by the load, once the SCR turns 'on'. (II) When gate drive Is applled

A positive gate to cathode signal is applied whenever the SCR is to be driven into forward conduction mode (ON state). This is also called gate triggering of the SQl. Such situation is shown by the typical circuit of Fig. 1.6.5. The SCR is in forward blocking mode when p gate drive is not applied. When the Load positive gate to cathode voltage is applied, current flows from gate to cathode. 1bis current adds to the + forward leakage current. Hence V + avalanche break-down of junction J2 takes place . at lower anode to cathode voltage also. Thus SCR is K driven into forward conduction Fig. 1.6.5 Gata triggering la used to tum.on mode (ON state) even if VAK -V80 . This is shown in Fig. 1.6.1. At these voltages, the leakage current is so high, that internal regenerative starts in the device. dv

dt

SCR can be thought of as a capacitor in the forward biased state. When the anode-cathode voltage changes rapidly, leakage current thought the device increases due to internal capacitor. The leads to tum-on of the SCR.

Power Electronics - Ill

1-29

Power Semiconductor Devices

4. Light

SCR can be turned on by light, when it falls on gate cathode junction of the SCR light induces electronic hole pairs and it helps to increases leakage current.

5. High temperature

SCR tu.ms on due to increased temperature. At higher temperature, there are more electron-hole pairs across junctions. This inverses the leakage current and the SCR turns on.

1.7.2 Tum-on Dynamic Characteristics Fig. 1.7.1 shows the current and voltage of the SCR during tum-on. The gate pulse is applied at t = 0. During the delay time (td ), the anode current rises very slowly and flows only near the narrow region of the gate. Observe that anode to cathode voltage doesnot reduce during td . It remains to the forward blocking value. During the rise time (tr), the anode current increases rapidly and anode to cathode voltage falls rapidly. The high voltage and current are present in the SCR. Hence large dissipation takes place in the SCR.

Fig. 1.7.1 Dynamic characteristics of SCR durtog tum-on

This power dissipation is called switching loss of the SCR. The current starts spreading in the remaining area of the SCR. During the spread time (tp)., the conduction spreads over the complete cross-section of the SCR. The anode current Copy ght

ma r al

Power Electronics - Ill

1 -30

Power Semiconductor Devices

reaches to its maximum value. And the anode to cathode voltage falls to lowest value (i.e. less than 2V). The dissipation in the SCR is also reduced. The turn on time (t011 ) of the SCR is given as total of td ,t, and tp . Thus, t 0 n = td +t, +tp

The turn on time can be defined as,

The turn-on time of the SCR is defined as the time from initiation of gate drive to the time when anode current reaches to its full value. The turn-on ti.me of the SCRs is about 1 to 3 microseconds. The tum-on time can be effectively reduced by applying higher values of gate currents. Because of high gate currents, more electron-holes are injected near junction J2 . Hence avalanche break-down of / 2 takes place fast. Therefore anode current rises fast. Thus effective turn-on time is reduced. To turn-on the SCR, the gate pulse is thus sufficient.

1.7.3 SCR Tum-off We know that SCR can be turned-off, when its forward current falls below holding current. The can be done by two methods :i) Natural commutation and ii) forced commutation.

type of turn-off, the supply voltage becomes zero or negative, Hence SCR is reverse biased. Therefore it is turned-off.

i) Natural Commutation : In this

ii) Forced commutation : When the supply voltage is DC, then external commutation

component are used to tum-off the SCR. The commutation components apply reverse bias across the SCR temporarily or pass impulse of negative current. Therefore SCR turns-off.

1.7.4 Turn-off Dynamic Characteristics Fig. 1.7.2 shows the SCR current and voltage during turn-off. The SCRs are not turned off by gate. They need external circuit for turn-off. These circuits are called commutation circuits. These commutation circuits has to hold negative voltage across the SCR during tum-off period. The SCR is said to be turned-off when it regains forward blocking capability after forward conduction. In the above fiigure observe that anode current falls and then it becomes negative. The negative pulse of current .flows through the SCR for short period. During the conducting state, the SCR is flooded with carriers and it acts as short circuit. The negative anode current flows through the SCR till all these carriers are removed. Then junctions J1 and J3 achleve their forward blocking state. The time required for this is called reverse recovery time (trr ). At the end of trr, reverse voltage appears across the SCR and anode current becomes zero. This is shown in Fig. 1.7.2. But still, the SCR is not turned-on. The commutation circuit has to

Power Electronics -111

1 • 31

Power Semiconductor Devices

Fig. 1.7.2 Dynamic characteristics of SCR during tum-off

hold negative voltage across the SCR for gate recovery time

(tgr )- During this time, the

excess carriers near junction J2 are recombined. If negative voltage is removed by commutation circuit before tg, , then SCR may tum-on again due to these excess carrier near junction J2 • Because they act like gate drive to the SCR. Hence the tum-off is complete at the end of gate recovery time. The SCR regains its forward blocking capability. The negative voltage imposed by commutation circuit can be removed at the end of tgr . The turn-off time (tq) of the SCR is the total time required by re~~ recovery and gate recovery. i.e., tq

=

t" +tgr

The turn-off time can be defined as follows : The turn-off time of the SCR is the time required to achieve forward blocking capability

after commutation is initiated. The tum-off time of the SCR varies from 5 to 200 microseconds. The tum-off time of the commutation circuit is called circuit turn-off time (tc)· And hence circuit turn-off time must be greater than the turn-off time of the SCR (t, > tq).

Power Electronics -111

1 - 32

Power Semiconductor Devices

1.7.5 Inverter Grade and Converter Grade SCRs Inverter grade SCRs

The SCRs which have tum-off time less than 25 µs ar,e called inverter grade SCRs. Such SCRs are used in inverters, choppers etc. Converter grade SCRs

The SCRs having larger turn-off times

(tq> 25 µ s) are called converter grade SCRs.

Such SCRs are used in controlled rectifiers, AC voltage controllers etc.

Review Questions 1. Explain the turn-on and turn-off dynamic characteristics of the SCR. 2. Define the following : (i) turn-mi Hw (ii) turn.off time (iii) convmer grtW SCR (ro) Invmer grtW SCR

1.8 SCR Gate Characteristics In the previous section we studied V-1 characteristics of SCR Now we will have a closer look towards gate characteristics of the SCR. Fig. 1.8.1 shows the gate trigger characteristics. Gate voltage

vo \ \

\ \

\

Reliable tum-on

--

0

Ig(max)

Gate current

lg

Fig. 1.8.1 Gate trigger characteristics

1 -33

Powar Electronics -111

Power Semiconductor DevlcN

The gate voltage is plotted with respect to gate current in the above characteristics. I (max} is the maximum gate current that can flow through the SCR without Jamagmg it. Similarly v g (max) is the maximum gate voltage to be applied. Similarly v g( min) and I g( min) are mmimum gate voltage and current, below which SCR will not be turned-on. Hence to tum-on the SCR successfully the gate current and voltage should be 1g(min) < l g < Ig(max) and

vg (min)
Vc~th), the collector current starts increasing. Tum-on delay, t d(on) is the delay when gate drive is applied and i, starts increasing. When ic increases to its full value, collector emitter voltage starts falling. 'tr;' is the rise time of collector and tfa is the fall time of voltage. Thus, tum-on time of IGBT is, ton

= td( on}

+ tr; + tfa

POW9r a.ctronlcs - Ill

1-65

Power Semiconductor Devices

Fig. 1.13.10 Switching characteristics of IGBT

The tum-off of the IGBT is initiated by reducing the gate voltage. When gate voltage falls to the value equal to vest, vcE starts rising. vc51 is the voltage where IGBT comes out of saturation. Tum-off delay, td(off) is the delay time when gate voltage i~, reduced and Ve£ starts increasing. When vcE reaches to supply voltage, ic starts reducing. ic reduces fast till vcs reach.es to vc~th)· This fast decay in ic is basically due to internal MOSFET. Then Vcs goes to zero and becomes negative. But ic keeps on flowing for some time. This is internal current This current flows due to stored carriers in the drift region. Hence, tum-off time of IGBT is higher than IGBT. The tum-off time of IGBT will be,

err

toff = td(ojf) + trv + tfit

Here,

+ tfi2

trv is voltage rise time tfi1 is MOSFET current~ time. tfi2 is

BJT current fall time.

al

Power Electronics -111

1 -66

Power Semiconductor Devices

1.13.5 Merits, Demerits and Applications of IGBT Merits of IGBT

i. Voltage controlled device. Hence drive circuit is very simple. ii. On-state losses are reduced.

iii Switching frequencies are higher than thyristors. iv. No commutation circuits are required. v. Gate have full control over the operation of IGBT. vi.

IGBTs have approximately flat t.emperature coefficient.

Demerits of IGBT

i. -IGBTs have static charge problems. ii. IGBTs are costlier than BJTs and MOSFETs. Applications of IGBTs

i. AC motor drives, i.e. inverters. ii. OC to DC power supplies, Le choppers.

w. UPS systems. iv. Harmonic compensators.

1.13.6 Protection Circuits for IGBT IGBT can be protected against,

i) Gate overvoltage protection ii) Overcurrent protection iii) Snubber circuits.

1.13.6.1 Gate OVervoltage Protection •

Fig. 1.13.11 shows the circuit diagram of gate overvoltage protection. 'This circuit consists of two zener diode connected in series back to back.

•

Normally the gate overvoltage is ± 20 V. The two zener diodes conduct when overvoltage occurs between gate and source.

•

The breakdown voltage of the zener diodes can be adjusted according to Gate-source breakdown voltage of IGBT.

Drain (Collecto r)

Source (Emitter)

Fig. 1.13.11 Gata overvoltage protection

Power Electronics - Ill

1 -67

Power Semiconductor Devices

1.13.6.2 Overcurrent Protection •

The drain overcurrent of IGBT is continuously monitored. If overcurrent is detected, then drive of IGBT is disabled. This is normally incorporated in drive circuit of IGBT. Fig. 1.13.12 shows drive circuit ofIGBT with overcurrent protection. + ♦

, Over·

:

I I

current protediol I

:

I

I

ca'QMl: I

I I I • • --•••--•••

I

I

OYertwrwtl pl'04edicn cil'NI

I

'II I

''\

~ , ---t \

,

't'

Gate oveMJbge pnMdion

Fig. 1.13.12 Overcurrent protection

Operation

•

In this circuit, the turn on/ off signal is given through optocoupler.

•

The overcurrent protection circuit receivers drive signal from comparator and gives it to gate of IGBT through non-pnp pair of Bffs.

•

The collector voltage of IGBT is sensed through diode D oc , Normally this

T-

diode is forward biased, since collector voltage is very small.. •

If there is overcurrent, then collector voltage increases and diode D oc is

reverse biased. This condition is sensed by overcurrent protection circuit it simply blocks the drive given to gate of IGBT.

affd

1.13.6.3 Snubber Circuits for IGBT Prupose : ....

To 1e nsure that IGBT always operates in its safe operating area at the time of tum-on and tum-off.

If IGBT does not operate in its SOA, then it can be damaged. Thus snubber circuits protect IGBT. There are two types of snubbers : i) Tum-off snubber and ii) Turn-on snubber.

1 -68

Power Electronics - Ill

Power Semiconductor Devices

Tum-off snubber

Fig. 1.13.13 shows the circuit diagram, of tum-off snubber. Operation

•

Tum-off snubber is necessary to limit the voltage across collector-emitter when IGBT turns-off.

•

The load current flows through diode D and capacitor C. This changing of capacitor limits the voltage vcE at the time of tum-off.

•

The resistance 'R' is used to limit the discharge current of capacitor when IGBT turns-on.

•

Fig. 1.13.14 shows the switching trajectory 'Of IGBT for various values of

load

R

G

Fig. 1.13.13 Tum-off snubber

RBSOA

-------------------, C =0 0

C. •

Here Co 11= 0.71

Power Electronlcs - Ill

From equation 2.1.7,

Drive and Protection Circuits for Power Devices

2 -17

T = R,C In ( l

~

11

) J I

Putting values in above equation, 2ox10-3 = 10xl03 xCln (

...

1

_1_ 71)

C = 0.16i f.l F

the I values of charging circuit components if tht line

Example 2.1.7 : Find

synchronised U]T circuit can be operated to get delay angles of 200 to 1600. Assume suitable data if required. :> Solution : The given data is,

20°~a ~60° i.e. 0.349

~(X

~

2.793

Let the operating frequency be f = 50 Hz.

...

c.o

= 2nf = 2n xSO = 100 n

The firing angle is given by equation 2.1.8 as, a = c.oR, C Let T\ = 0.65 and for a 1

In( l ~ ,,)

= 0.349,

0.349 = 100n x R , 1C1ln( l

Rc1C 1

-1_65)

= l.058xl0-3

Similarly with a 2 = 2.793, 2793

...

R, 2C2

= 100n x R,2C2ln( l -1.6S) = 8.468x10-3

Let C = C1 R, 1

=

=C2 = 0.5 µ f .

Hence,

1.ossx10-3 = 2.116 Jdl 0.Sxl0-6

8.468xl0-3 = 16.936 kn R,2 = 0.Sxl0~

,

.

Thus the resistance R, should be varied from 2116 Jdl to 16.936 kO. Hence a variable resistance of 20 kQ can be used.

Pow.r Electronics - Ill

Drive and Protection Circuits for Power Devices

2-18

'l

, . . Example 2.1.8 : An UfT used in a reuwztion oscillator circuit is having = 0.7, Vv = 1 V and the supply voltage to the circuit is 15 V. Design the suitable values of R and C given that the frequency of oscillation is 1 kHz. Peak current is 1 mA and valley current is 8 mA. Solution :

The given data is, VBB

= 15 V

Tl = 0.7

Vv

=

1,

Iv

= 8 mA

f = 1000 Hz

Ip = 1 mA The peak voltage is given as, VP = 'lVBB

+ VD

Let VD = 0.8 and putting other values, VP

=

0.7 x 15 + 0.8

= 11.3 V

The period of oscillation is given by equation 2.1.7 as, T = RcC ln ( l

With T =

l

= ~ 1

~ 'l)

and putting·T1=0.7,

1 1 = R cC ln ( _ _ ) 1000 1 07

...

RcC

Let C

Re

= 8.3058 X 10-4 = 0.3 µ F. Hence, =

8.3058 X 10- 4 = 2.768 lcil 0.3x10- 6

Now let us determine maximum and minimum values of R, . From equation 2.1.7, R c:(m.ax) =

Vaa -VP __ 15 - 11.3 __ "7, 3, ,00 Ip 1 x10-3

n w

POWltr Electronics - Ill

Drive and Protection Circuits tor Power Devices

2 -19

And from equation 2.1.15,

=

15-1 = 1750 .Q 8 x10- 3

Thus the calculated value of R, = 2.768 kn lies in the angle of 1750 Q to 3700 n.

is connected across a· 20 V zener. The valley tmd peak point voltages are 1 V and 15 V respectively The intrinsic stand-off ratio is 0.75. It operates at a frequency of 1200 Hz. Find the charging capacitor if

,_. Example 2.1.9 : An UfT triggering circuit

R = 5.6 k'2. Solution : The given data is,

Tl

= = =

0.75

f

=

1200 Hz

VBB Vv

20 V 1,

R, = 5.6

VP= 15

Hence T

1 = -f1 = 1200

kQ

The period of oscillation of UJT relaxation oscillator is given by equation 2.1.7 as

T=

R, C

In( 1 ~ Tl)

Putting values in above equation, ~ 12

...

= 5.6xl0 3 xCxln (

1

_1_ 75)

C ;:;; 0.107 µF

2.1 .7 SCR / TRIAC Triggering using Diac The diac is the two terminal and four layer device. It is mainly used for triggering triacs. Fig. 2.1.15 {a) shows the symbols of diac. The symbol shown in Fig. 2.1.15 (b) has arrows in both the directions. This means it conducts in either directions. The terminals are not named. It can be used in any direction. It is low power triggering device. There is no control terminal on the diac.

Power Electronics • Ill

Drive and Protection Circuits for Power Devices

2-20

(a)

(b)

Fig. 2.1 .15 Twp different symbols used for dlac

2.1. 7.1 V•I Characteristics of Dlac Fig. 2.1.16 shows the V-1 characteristics of diac. The current through the diac is plotted with respect to voltage across it. The diac remains off till voltage is less than ± Vijo· When voltage exceeds V80 , diac turns 'on' and conducts heavily. Observe that the characteristic is exactly similar for positive as well as negative values of voltage and current. Once the diac turns on, the voltage across it drops to negligible value.

----- ------------

-Veo ....

-~ . ---_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-..,~'--- - - - - . - - - - - - - - V

--------- ----

Vso

----

t

Diac turns on at this voltage

Fig. 2.1.16 V-1 characteristic of diac

2.1.7.2 DIAC-TRIAC Phase Control LOAD - - - - - -- R

C

Fig. 2.1.17 Basic triac triggering using diac

Fig. 2.1.17 shows the basic triac triggering circuit using diac. In this diagram, when capacitor 'C' charges above the breakover voltage (V8 o ) of diac, it starts conducting. Hence capacitor voltage is applied to triac

Power Electronics - Ill

2-21

Drive and Protection Circuits for Power Devices

gate and it turns-on. The charging of the capacitor can be controlled by the variable ~ o r 'R'. Which indirectly controls the firing delay of triac.

Review Questions 1. Explain RC firing circuit for the SCR. Draw the nectsSary waveforms. 2. With the necessary circuit diagram and waveforms explain the operation of

UJT triggering circuit.

3. How diac am be used to trigger triac ?

Unsolved Examples UJT trigger circuit is used to fire a PNPN deuice.

It is supplied from a source across the SCR to be triggerf!d through a 10 V untr. The valley and~ point voltages are found to be 1 V and

1. A

7 V rtspedivtly. Calculate the intrinsic stand-off ratio of UJT and frequency of relaxation oscillator if R = lkO. and C = 1 µ F. {Ans. : Tl = 0.6~ f = 1033 Hz]

2. An SCR is to be triggered using a relaxation oscillator, which has an UIT with '1 = 0.7, IP = 2 µA ,VP = 16.5 volts. Normal leakage current with emitter open is 3 mA, Vv = 1 volts, Iv = 6 mA, R9i 8l = 5.5 kn. Triggering frequency is 100 Hz. With C = 0.1 µ F design the UJT relaxation osa1lator. [Ans. : R, = 83 ~ R 1 = 921.35 0. R 2 = 245.31 Q]

2.2 Drive Circuits for BJT Following points are to be remembered when designing the base drive circuit for transistor. 1. BJT is a current controlled device. 2. Power BJT is used as on/off switch in power converters. 3. Power BJT operates in saturation and cutoff when used as a switch. 4. Sufficient base current is required to drive BJT in saturation. 5. Amount of carrier injected in base region determine storage time of BJT. 6. Storage time determines turn-on and turn-off times of BJT. 7. There should be mechanism to control the amount of saturation so as to control storage tim.e.

There are most important points about BJTs.

Power Electronics - Ill

2-22

Drive and Protection Circuits for Power Devices

2.2.1 Base Drive Control During Tum-on Fig. 2.2.1 shows the base drive circuit for tum-on of a transistor. Due to this circuit the 't0 n ' time is reduced.

Fig. 2.2.1 Drive circuit to make base current high at beginning of

ion

As shown in this circuit, when base drive V8 is applied, the capacitor C1 acts as a short. Hence R2 is virtually by passes. Therefore an initial value of base current is only

limited by R1 and it is given as, I B (nonl-1 ,,.-,

_ Vs - V BE R1

-

This heavy base current drives transistor into saturation for quick tum-on. Once the transistor is turned on, there is no need of such large base CUl'rent. This is taken care-off by R2C1 circuit. The capacitor C1 starts charging and base current starts falling. This is shown in Fig. 2.2.2. Observe that there is peaking of base current at the

beginning of tum-on. Then the current reduces to, IB

_

V B -VBE

-

R1 +R2

Fig. 2.2.2 Base current peaking at the beginning of tum-on •

To turn-off _the transistor, base voltage is made zero. Therefor,e capacitor voltage appears as negative voltage across base-emitter. Hence suddenly base current is reversed as shown in Fig. 2.2.2. This current slowly decays to zero after the stored charge in base region is removed. The capacitor C1 then discharges through R2• This discharge time constant is, 't 2 = R2C1 Copyr ghted ma r al

Drive and Protection Circuits for Power Devices

2-23

Power Electronics - Ill

The charging time constant of the capacitor is, R1R2C1 't1

= Rt+ R2

The Ton period of the transistor must be at least five times of t 1 . Similarly the T0 ff period must be five times of 't 2 . The switching frequency of the BJT will then be,

h s -

i T on +Toff

We know that, > S'tt and Toff(min) > S't2 This gives the maximum switching frequency of, Ton(min)

fs,ma,

=

1

Ton(min) + Toff(min) 1

1

2.2.2 Base Drive Control During Tum-off For quick tum-off, the stored charge in the base region must be removed fast. This can be done by applying negative base drive for tum-off. Due to - Vs, the base-emitter voltag;e becomes negative during tum-off. The voltage on capacitor C1 also adds to this negative voltage. Therefore the negative pea.king of base current (see Fig. 2.2.3) at the beginning of turn-off is increased. This removes the stored charge fast, and the transistor is turned-off quickly. C1

o,

Fig. 2.2.3 Base drive circuit for positive and negative peaking of base current

It is possible to design the different tum-on and tum-off circuits for base current peaking. ·Fig. 2.2.3 shows one of such circuit. In this circuit, R 2C1 is used for positive peaking of base current as discussed earlier. The diode D 2 acts as short when Vs is positive. Capacitor C2 charges to the polarity as shown in above figure when V8 is positive. Normally R3 is much higher than R2, hence base current flows through R2 . The base voltage Vs is made negative to tum-off the transistor. Therefore the voltage

Power Electronics • Ill

2-24

Drive and Protection Circuits for Power Devices

of C2 imposes additional negative input to the base. The negative base current flows through C2, R1 and V 8 . Thus R3C2 becomes effective at the time of turn-off. No current flows through R 2C1 dwing tum-off since diode D 1 is reverse biased. Once the base current is zero, then C2 discharges through R3• Similarly C1 discharges through R2• Thus it is possible to design separate peaking characteristics for tum-on and tum-off.

2.2.3 Proportional Base Control The collector current changes as the load changes. ff the load is reduced, then collector current reduces. Then the base current should also be reduced to avoid excess carries m the base region. Hence proportional base control can be used. Fig. 2.2.4 shows the circuit diagram of proportional base drive.

R

C

~

+

rL Fig. 2.2.4 Proportional base drive control

A short duration pulse is applied to turn-on the transistor. The transistor turns-on and collector current starts flowing. The collector current passes through the coil which is magnetically coupled to base coil. Hence collector current induces the current in the base coil also. This current acts as base drive to the BJT. The RC time constant determines the duration of the pulse to be applied externally to drive BJT into saturation. The turns ratio must be,

N2 _ 1c_f:Ua_J3 N1 -

Ia

---r;;-

The base current then varies according to variations in the collector current. The transistor can be turned off by applying negative pulse through RC circuit. This makes base current negative and BJT tu.ms-off.

2.2.4 Anti-satu1ration Control We know that excess base current increases storage time of the BJT. Hence the turn-off time inaeases. Such excess or heavy base drive is called hard saturation. Therefore transistor must be operated in soft saturation. This means base must be given the carriers which are sufficient to drive the transistor in just saturation (quasi saturation).

Power Electronlcs - Ill

2-25

Drive and Protection Circuits for Power Devices

Fig. 2.2.5 shows, the circuit diagram to achieve quasi saturation. In this circuit the base drive is applied at terminals a-b. We can write following equation for loop consisting of a - D 1 - B - E- b,

+

+ +

Base

Vee

drive

bo----------------Flg. 2.2.5 Base drive with anti-saturation diode Similarly for loop a - Das - C - Ewe can write, Va,-c = VDas + VcE

Hence equating the two equations, VD1

+ VBE = VDas + VcE

Normally v Das= v01. Hence above equation becomes, VcE

=

VBE

1his shows that the collector-emitter voltage will be equal to base-emitter voltage. When BJT turns on, the base-emitter voltage is nearly 0.7 and collector-emitter saturation voltage is 0.3. Because of anti-saturation diode ( v Das ) , the collector-emitter voltage is 'l'aised to v 8 Et i.e. 0.7 V. Hence the BJT is no longer in saturation. It is just above saturation. This effect takes place because of anti-saturation diode v Das . Since BJT is above saturation, there are no excess carriers in base and its storage time is reduced. This reduces 1t 0 ff' and hen.ce switching time. The collector-emitter voltage can be further increased by putting additional diode in series of D 1. Disadvantage of antl-eaturatlon diode

The collector~tter voltage is increased. This increases the on-state losses in BJT.

Power Electronics - Ill

Drive and ProtJction Circuits for Power Devices

2-26

2.2.5 Typical Driver Circuit for Power BJT Fig. 2.2.6 shows the typical driver circuit for power consideration all the earlier requirements.

BJT that

takes into

+

+

Power BJT

Fig. 2.2.6 Typical driver circuit for Power BJT

• •

The drive from control circuit is given to driver circuit through optocoupler. The signal is amplified, buffered by the comparator and then given to ov8 current protection circuit. The collector voltage of main transistor is sensed through diode Doc. If the collector current increases, then collector voltage will rise. 'T itis rise is sensed by over current protection circuit to disable the drive of BJT.

•

The output from over current protection circuit is given to pair of pnp-npn transistors T1 and T2 . These transistors provide the required base curent to power BJT to drive it in saturation.

•

Note that the collector voltage of power Bff is monitored through diode DAS· This provide anti-saturation control. It is called backer's clamp. It ensures that Bff always operates in quasi-saturation.

•

Transistor T 1 provides large positive voltage and current to drive Bff in saturation of quasi-saturation. and transistor T2 provides large negative voltage and current for faster tum-off of BJT.

•

The base-emitter wires of power inductance.

BJT are twisted to

Review Questions 1. Give dijferm,t circuit configurations far base drive of B]T. 2. Draw the drive circuit far B]T. What is tmti-saturation control ?

minimize stray

Drive and Protection Circuits

Power Electronics -111

2-27

for Power Devices

2.3 Drive Circuit for MOSFET The gate drive circuit for MOSFET should satisfy the following requirements : (i) The gate-source input capacitance should be charged quickly. (ii) MOSFET turns on when gate-source input capacitance is charged to sufficient

level. (iii) To turn-off MOSFET quickly, the negative gate current should be sufficiently

high to discharge gate-source input capacitance. Fig. 2.3.1 shows the gate drive as per above requirements. The gate drive is applied across the terminals a-b. Initially the resistance R1 is bypassed by C1 and full drive voltage is applied to the gate. This charges the gate-source capacitance quickly. As the capacitor C1 charges, the gate current reduces. Once the MOSFET is turned on required gate current is very small. When MOSFET is to be turned off, the voltage vH is made zero. This applies capacitor voltage across gate-source in negative direction. Therefore charge on the gate-source capacitance is removed quickly. C1 then discharges through R1. The resistance R2 provides additional discharge path for gate-source capacitance.

c,

b _ _ _ _ ___.__ _____......__ _ ___,

Fig. 2.3.1 Gate drive circuit

Review Questions 1. Explain the typical gate drive circuit for MOSFET. 2. Dnlw the gr,te drive circuit for MOSFF.T.

2.4 Driver Circuit for IGBT and MOSFET Fig. 2.4.1 shows the driver circuit for IGBT which uses IR 2125 IC. •

Here IR 2125 is the high voltage, fast switching MOS gate driver with single floating gate driver channel This IC can be used to drive N-channel power MOSFET or IGBT. .

Drive and Protection Circuits for Power Devices

2 - 28

Power Electronics - Ill

1 Vee

, I

I t

'

\

--

R

2 JN

IN

8

Vs

Out

or -®

7

2

I

I I ;

1 3 Err

-

6

Cs

2

Rs

5

4 Vss

5

Vs

--

Load

Fig. 2.4.1 Driver circuit for IGBT

•

Over current flowing through the IGBT is detected through Rs and Cs terminal of the IC.

•

The error pin of the IC indicates fault conditions.

2.5 Isolation of Gate and Base Drives 2.5.1 Necessity of Isolation We know that driver circuits operate at very low power levels: Normally the signal levels are 3 to 12 volts. Sometimes digital circuits and miaoprocessors are also used in the triggering circuits. The gate and base drives are connected to power devices which operate at high power levels. Fig. 2.5.1 shows this situation. Observe that collector

C t - - t

- - -

- -,

I

I I

,

I

Trigger circuit

t

._

,v8 = 5 v

E

, I

L---- ---"

Fig. 2.5.1 Control / power levels

Power Electronics -111

2-29

Drive and Protection Circuits for Power Devices

of BJT can have voltages of 200 V. But base is connected to trigger circuit that have voltages of 5 V. If BJT is damaged and collector-base gets shorted, then high voltage will get connected to mgger circuit'. This will damage the mgger circuit also. This means mgger circuit is damage due to device damage. Therefore there must be some electric isolation between control and power circuit. There is one more reason for isolation. Consider that the mgger circuit is deriving the two devices as shown in Fig. 2.5.2. Here observe that T 1 is given the drive between a-b. And T 2 is given the drive between c-d. The mgger circuit must isolate the two drives. If there is no electric isolation, the points b ' and 'd' may be shorted due to common ground of the trigger circuit. Isolation can be obtained with the help of pulse transformers and optocouplers.

d o----------

Fig. 2.5.2 Isolation of grounds

2.5.2 Isolation using Pulse Tranafonner Pulse transformer has one primary and one or more secondary windings. It is normally used for pulsed mode of triggering Fig. 2.5.3 shows the isolation using pulse transformer.

I I I

I I

JUUt

I I

Pulsed drive

o----'

I I .___ ___..._ _ _ _ _ _ _____.

Pulse transfonner

Fig. 2..5.3 Electric Isolation using pulse transformer

, ,.

Drive and Protection Circuits for Power Devices

2-30

Power Electronics - Ill

In the above circuit, observe that triggering circuit is electrically isolated from BJT. Hence if there is any electric damage to BJT, there will be no effect on triggering circuit. Advantages

i) Pulse transformer does not need external power for its operation. ii) It is very simple to use.

Disadvantages

i) Pulse transformer saturates at low frequencies hence it can be used only for high frequencies. ii) Due to ma,gnetic coupling, the signal is distorted.

2.5.3 Isolation using Optocouplers Optocoupler consists of a pair of infrared LED and phototransistor Fig. 2.5.4 shows the symbol of optocoupler. When the signal is applied to the infrared LED, it turns on. It's light falls on phototransistor. Therefore phototransistor also starts conducting. There is no electric connection between LED and phototransistor. r- -- -- ----- - ----- --,

I

I

Input side

Output side

L----------------Fig. 2.5.4 Optocoupler

Fig. 2.5.4 shows the triggering circuit that uses optocoupler. In this circuit the mggering pulses are given to the input {LED) of optocoupler. When 'Vg' is positive, LED turns-on. It's light falls on phototransistor. Hence it turns on Therefore base of T 1 is connected to zero volts through phototransistor. Due to this, T 1 turn&-on. Therefore the voltage Vee is applied to gate of the MOSFET. Hence MOSFET turns on. When Vg = 0, the LED ~ f f, therefore phototransistor also tum-off. Therefore base drive of T1 goes to Vee and it tum-off. When T1 turns off, MOSFET gate voltage becomes z.ero. Therefore MOSFET turns-off. Thus gate drive circuit using optocoupler works.

Power Electronics - Ill

2 -31

Drive and Protection Circuits for Power Devices

+ 0-----'11/VV'\r-~

F1lg. 2.5.5 MOSFET triggering circuit using optocoupler Advantages

1) Very good response at low frequencies. 2) Compact and cheaper optocoupler devices are available. Disadvantages

1) Optocoupler need, external biasing voltage for their operation. 2) High frequency response is poor. Applications

Inverters, SMPS, Choppers, AC motor drives use optocouplers.

Review Question 1. WhAt is tht necessity of isouition ? How it is implemented ?

2.6 Protection Circuits Due to unpredicted changes in the load and supply voltages, fault conditions are arised. Under these conditions, the voltage or current ratings of the power devices are exceeded. Hence there is a possibility of damage to the power device. Such damage can be avoided by using protection circuits. Following are the possible faults and required protection circuits :

Power Electronics -111

Drive and Protection Circuits for Power Devices

2-32

i) High rate of change of voltage across the power device is called normally turns-on the thyristor and damages BJTs. The value of

d¾f This

d¾t can be

effectively reduced with the help of snubbers (RC circuit) and Metal oxide varistors (MOV). ii) High rate of change of current in the device is called

d½r The

spread of

current inside the device requires some time. If sufficient time is not given for current spread, then localized hot spot is created in side the device. The device is damaged due to this fault. The t problem normally occurs in thyristors.

d½

This problem can be avoided by putting the current limiting: inductor in series with the device. The inductor limits rapid changes in the current. iii) Overcurrent normally occurs due to variations in the load. Overcurrent causes t

excessive heating of the device and it leads to damage. Fuses are normally used to limit overcurrents.

/4t and Overvoltages

2.7 Protection Against dv

2. 7.1 Snubber Circuits (Tum-off Snubber) The transient overvoltages can switch on the thyristor. In some cases the thyristor

can be damaged due to these transient voltages. These transient voltages are very common when the converter is having inductive loads. The thyristors can be protected against transient voltages by a RC network as shown in Fig. 2.7.1. lbis RC network is connected. in parallel across the thyristor. It is called snubber circuit. The resistance has the value of few hundred ohms. Whenever there is a large spike or voltage transient across the thyristor, it is absorbed by the ( urrPnt 11rn.-,::) RC circuit. The RC circuit :1 RC j1(.;tJI R J 'l'J :•a:15 ":' ; l!(snubber) acts as a lowpass filter 1,~ryr·no :.1y for this voltage transient. The ,C ~.l~.obor resistance has normally low value C ·so that the transient is absorbed by the capacitor quickly. Thus the thyristor is protected against voltage transients. The RC Fig. 2.7.1 A snubber (RC) network Is used for transient voltage protection snubber circuit is very commonly used for protection of thyristors against transient voltages (high frequency voltage spikes).

7

'

Drive and !Protection Circuits for Power .... Devices

2-33

Power Electronics - Ill

dv dt ,D

R

I

f

C

also generates large voltage

transients. These rapid voltage variations can also, be suppressed by snubber circuit. The capacitor acts as a short for these dv variations. The dt

dv Fig. 2.7.2 Snubber is used for dt protection

snubber can be made more effective by connecting a diode across the resistance as shown in Fig. 2.7.2

In case of voltage transient, the current flows through diode and capacitor. The capacitor acts as a short for the voltage transient. Thus it is suppressed. When thyristor turns-on, the capacitor discharges through resistance R. The R, C and diode snubber is more commonly used

because it is very effective for

!:

and other voltage transients.

Design of snubber

The value of capacitor is given as, 2

C = _!_ l( 0.564 Vm lJ 2L

dv

... (2.7.1)

dt Here Vm is the peak value of supply voltage dv . .a.L. • .bl dv is ule pemuss1 e -d . dt t

-

Lis the source inductance.

And resistance is given as,

R=2cri

... (2.7.2)

Here a is the damping factor. It's value is normally taken as 0.65. ,. . Example 2.7.1 : Calculate the required parameters for snubber circuit to provide dv

dt

protection to a SCR used in single phase bridge converter. The SCR has a maximum ddv ~ t capal:rility of 60 V I µ sec. 'fhe. input line to line voltage has a peak value of 425 volts and the source inductance is 0.2 mH.

Drive and Protection Circuits for Power Devices

2-34

Power Electronics - Ill

Solution : Givan :

dv

dt

= 60V

/~ c

L = 0.2 mH V111 = 425 V

From equation 2.7.1 2

_!_(l 0.564 Vm lj =

C =

dv dt

2L

=

1 ( 0.564 x 425 xl0- 6 ) 3 2 x 0.2xl060

2

0.04 µF

In the above equation observe that we have multiplied numerator by 10-6 inside 60 the brackets. It comes from 60 V I µs, i.e. to be substituted in equation for C. l x lO- 6

Let the damping factor be CJ = 0.65. From equation 2.7.2

R = 2 CJ

L

C = 2 x 0.65

02x 10-J 0.04 xl0- 6

R = 92 .Q

Fen: the circuit shown in Fig. 2.7.3, the thyristor is operated at

,_. Example 2.7.2 :

2 kHz. The required

!~

s

is 100 V/]'S. The

discharge current is to be limited to + V 1 = 200 V _

100 A. Detennine

R=SO

Values of R5 and C5 ii) Snubber loss iii) Power rating of R5 Load and stray inductances are negligible. i)

(Nov.-2007, 10 Marks)

L

Fig. 2.7.3 Circuit of example 2.7.2

Solution : Given :

Load resistance, Frequency,

R = 5'2

f \1s

= 2 kHz = 200 V

C

Power Electronics ,. Ill

2 - 35

Drive and Protection Circuits for Power Devices

dv = 100 V/µs dt Iro = 100 A

L = 0 i) To obtain values of R5 and C5

Rs limits the discharge current through T 1. From Fig. 2.7.3, T1 - Rs -Cs forms a loop when T1 turns-on. Prior to tum-on of T1, C5 charges to 200 V. Fig. 2.7.4

Rg

shows this situation. From this figure we can write, ves = Rs Im

Rs

=

ves

I ro

= Vs

Iro

= 200 = 2 Q 100

Fig. 2.7.4 Path of lro

The charging current of Cs can be expressed by I..

Example 2.8.1 : Design the snubber circuit elements Rs and Cs connected across the

45 A I µ s. An inductance dt L = 0.1 H and a resistance R converter is used for HVDC transmission system and ;s operated from 3 4> 25 kV supply. Thyristor each of 1600 V/16 A are available. The forward leakage cu"ent difference of the device is 35 mA. The string efficiency can be assumed to be 85 % and ~ Q max = 25 µC.

,_. Example 2.1 2.2 :

a) Determine the number of devices to be connected in series. b) Equalizing cm_nponents

r

,~ Power Electronics -111

2-64

Drive and Protection Circuits for Power Devices

Solution :- Given data ,f,

V1me = 25 kV

...

V, = PIV = ~ x V line = ~x25 kV= 35.35 kV. VDt = 1600 V

A ~ = 25µC Alo = 35 mA String efficiency, Tl= 0.85 a)To obtain number of devices

Form equation 2.12.8,

11 =

Vs nVDt

...

= n

Vs _ 35. 35xl0 3 flVDt - 0.85x1600

= 26 b) To obtain equallzlng components

Value of R is given by equation 2.12.3 as, R = nV01 -Vs (n-1)61 0

26x1600-35.35x10 3 = (26-1) 35 xio-3 = 7.142 kQ

Value of C1 is given by equation 2.12.5 as, ' (n-l)AQ

Ct = nVDt

-Vs

= 0.1 µF

(26-1)25xl0-6 = 26x1600-35.35x10 3

Power Electronics -111

2-65

Drive and Protection Circuits for Power Devices

2.12.3 Parallel Connection of Thyristors We know that parallel connection of thyristo~ is used to cater higher current demands.

2.12.3.1 Problems Occurred In Parallel Connection The devices used in parallel connection do not have exactly similar characteristics. The thyristor carrying higher current will have more power dissipation. This will increase its temperature and reduce the internal resistance. Therefore the current further increases. This process continues till thyristor damages.

2.12.3.2 Equalizing Arrangements Heat sharing can be done by using common heat sinks for all the devices. This will maintain all the devices at same temperature. Static Cu"nit Sharing Fig. 2.12.5 shows the static current sharing arrangement. Two resistors connected in series with the thyristors.

R1 and R2 try to equalize the currents through T1 and T2 • But power dissipation in R1 and R 2 is very high. Fig. 2.12.5 Static current sharing

Dynamic Cunent Sharing

Fig. 2.12.6 shows an arrangement for dynamic current sharing. The inductors are placed in series with thyristors. These inductors are magnetically coupled. But they are connected in opposite direction. Hence if current in T1 tries to increase, then a voltage of opposite polarity will be induced in L2 . This increases the current in T 2 . Thus current balance is maintained. Fig. 2.12.6 Dynamic current sharing

□□□

(2 - 66)

Single and Three Phase AC/DC Converters

Objectives • Principle of con.trolled redi/ication. • Single phase and 3 phase conuerters. • Half wave and full waue converters.

• Bridge converters

•

--r--+

semiconuerter

L...__. full bridge conuerter

• Resistiue, inductive and motor (RLE) loads on converters. • Continuous and discontinuous output current operation and its effects. • Inverting operation (power flow from load to source) in case of full conuerters.

• Effects of feedback diode and freewheeling operation. • Harmonic analysis of converters.

• Effed of source inductance.

3.1 Introduction 3.1.1 Principle of AC/DC Conversion (Controlled Rectifier) •

Controlled rectifiers are basically AC to OC converters. The power transferred to the load is controlled by controlling triggering angle of the devices. Fig. 3.1.1 shows this operation.

+

AC

Controlled

supply

rectifier

Load

ex

Control circuit

Fig. 3.1.1 Principle of operation of a controlled rectifier (3 • 1)

Single and Three Phase

Power Electronics -111

3-2

AC/DC Converters

•

The triggering angle 'a' of the devices is controlledc,y the control circuit.

•

The input to the controlled rectifier is normally AC mains. The output of the controlled rectifier is adjustable DC voltage. Hence the power transferred across the load is regulated.

Applications :

The controlled rectifiers are used in battery chargers, OC drives, . DC power supplies etc. The controlled rectifiers can be single phase or three phase depending upon the load power requirement.

3.1.2 Concept of Commutation •

Definition : Commutation is the collective operation, which turns of the conducting SCR.

•

Commutation requires external conditions to be imposed in such a way that either current through SCR is reduced below holding current or voltage across it is reversed.

•

There are two types of commutation techniques. Commutation

Natural

Forced

Current commutation

Voltage rommutalion

Fig. 3.1.2

•

Forced commutation : It requires external components to store energy and it is used to apply reverse voltage across the SCR or reduce anode current below holding current of the SCR to turn it off.

•

Current commutation : The SCR is turned off by reducing its anode current below holding current.

•

Voltage commutation : The SCR is turned off by applying large reverse voltage across it.

•

Principle of line commutation

The natural commutation does not need any external components. It uses supply (mains) voltage for turning off the SCR. Hence it is also called as line commutation.

'

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3.3

• Explanation Fig. 3.1.3 shows the circuit using natural commutation. It is basically half wave rectifier. The mains AC supply is applied to the input. The SCR is triggered in the positive half cycle at a . Since the SCR is forward biased, it Mains AC starts conducting and load current i 0 supply rv R starts flowing. The waveforms of currents and voltages are shown in Fig. 3.1.4. Since the load is resistive, Fig. 3.1.3 A hatf wave rectifier uses natural commutation to tum off SCR

.

1

Vo

0

=R-

Hence the shape of the output current is same as output voltage. Observe that the output current is b e~sically SCR current. At '1t' the supply voltage is zero. Hence current through SCR becomes zero. Therefore the SCR turns off. The supply voltage is then negative. This voltage appears across the SCRs and it does not conduct. Thus natural commutation takes place without any external components. Here note that natural commutation takes place only when the supply voltage is AC. Thus the controlled rectifiers use natural commutation.

Fig. 3.1 A Wavefonns of half wave controlled rectifier to Illustrate natural commutation

Copyright

ma r al

Power Electronics -111

Single and Three Phase AC/DC Converters

3-4

Review Questions 1. Explain the principle of n11tural commutation. How it is used in controlled rectifiers ?

2. Explain the basic principle of phase controlled operation.

3.2 Single Phase Half Wave and Full Converter 3.2.1 Single Phase Half Wave Controlled Rectifier with Resistive Load The principle of phase controlled operation can be explained with the help of half wave controlled rectifier shown in Fig. 3.2.1. The secondary of the transformer is connected to resistive load through thyristor or SCR T1. The primary of the transformer is connected to the mains supply. In the positive cycle of the supply, T1 is forward biased. T 1 is niggered at an angle a. This is also called as triggering or firing delay angle. T 1 conducts and secondary (i.e. supply) voltage is applied to the load. + . 'o r, Current iO starts flowing through the load. R The output current and voltage waveforms 11 are shown in Fig. 3.2.2. Since the load current is given as,

resistive, output

is

-

io -

Fig. 3.2.1 Half wave controlled rectifier with R-load.

Vo 1f

Hence the shape of output current waveform is same as output voltage waveform. At 1t supply voltage drops to zero. Hence current i O £lowing through T 1 becomes zero and it turns off. In the negative half cycle of the supply T1 is reverse biased and it does not conduct. There is only one pulse of v O during one cycle of the supply. Hence ripple frequency of the output voltage is, /ripple = 50 Hz i.e. supply frequency Mathematical analysis

The average value of output voltage is given as,

v o(av)

=

1 T

T

J Vo (cot) d cot 0

The period of one pulse of v O (cot) can be considered as T = 2 1t • And v O (cot) = Vm sin oo t from a to 1t. For rest of the period v O (cot) = 0. Hence above equation can be written as,

Vo (av)

=

1 2

1t

7t

J Vm

•

sm wtdwt

a

= Vm [ - cos

2n

cot]1t a

Power Electronics - m

3-5

Single and Three Phase AC/DC Converters

Fig. 3.2.2 Phase control principle as applied to half wave controlled rectifier

...

(1 + cos a)

vo (av) = ~;

... (3.2.1)

The power transferred to the load will be,

_ Po(av) -

vi

o(av) R

Thus the output average voltage and power delivered by the controlled rectifier can be controlled by phase control (i.e. a). The phase control in converters means to control the delay (or triggering) angle a.

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-6

3.2.2 Half Wave Controlled Rectifier with RL Load is +

T1

io R

vs rv

Vo

L

Now let us study the operation of single phase half wave controlled rectifier for inductive {RL) load. Normally motors are inductive load. L is the armature or field coil inductance and R is the resistance of these coils. Fig. 3.2.3 shows the circuit diagram of half wave controlled rectifier with RL load.

The SCR will be forward biased in the positive half cycle of the supply. Hence SCR is applied with the firing pulses in the positive half cycle. The waveforms are shown in Fig. 3.2.4. Fig. 3.2.4(a) shows the supply voltage and Fig. 3.2.4(b) shows the firing pulses to the SCR. Fig. 3.2.3 Half wave controlled rectifier with RL load

Fig. 3.2.4 Waveforms of half wave controlled ractffler for RL load

Copyrighted rn

r al

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-7

When the SCR is triggered, the supply voltage appears across load. We normally neglect small voltage drop in SCR. Hence v O = v5 when SCR is conducting. This is shown in Fig. 3.2.4(c). Observe that output voltage is same as supply voltage after a. Because of the RL load, output current starts increasing slowly from zero. The shape of i 0 depends upon values of Rand L. At 7t, the supply voltage becomes zero and i 0 is maximum. Due to negative supply voltage after n, SCR tries to tum-off. But energy

stored in the load inductance generates the v~ltage L di O • 1his induced voltage

dt

forward biases the SCR and maintains it in conduction. This is shown in Fig. 3.2.S. The basic property of inductance is that it opposes change in current. At n, the current i O is maximum. As SCR tries to tum-off due to negative supply voltage, the output current i O tries to go to zero. Such change in i O is opposed by load inductance. Hence the energy stored in an inductance tries to maintain i0 • To maintain the flow of i 0 , inductance generates the voltage

Ld~;

with

polarity as shown in Fig. 3.2.5. This voltage is higher than negative supply voltage. Hence T 1 is forward biased and it remains in conduction. The output current and supply current flow in the same loop. Hence i O = is all the time. The waveform of i 0 is shown in Fig. 3.2.4(,j) and i5 is shown in Fig. 3.2.4(e). After 1t, i 0 (i.e. i5 ) flows against the supply. Hence energy is

Vs~

+

Fig. 3.2.5 SCR conducts due to Inductance voltage after 1t

consumed in the supply. i 0 flows due to load inductance energy. In other words, the inductance energy is partially fed to the mains and to the load it self. Therefore energy stored in inductance goes on reducing. Hence i O also goes on reducing as ~own in Fig. 3.2.4(d). At~ the energy stored in the inductance is finished. Hence i0 goes to zero. Therefore T 1 turns-off. In Fig. 3.2.4(c) observe that v 0 is negative from 7t to 13. Because T1 conducts from 7t to J1 Hence whenever T1 conducts v 0 = Vs The SCR is triggered again at 21t +a. Hence output voltage remains zero from ti to 2n +a. Output current as well as supply current are also zero from ~ to 21t + a. At 2n +a, T 1 is triggered again and the cycle repeats. Here i 0 goes to zero at p. Hence this 'is called discontinuous conduction.

Derive an expression for average value of output volt.age for 1 war,e controlled rectifier with RL load.

,_., Example 3.2.1 :

t

half

Solution : For discontinuous conduction, the output voltage waveform is shown in Fig. 3.2.4(c). The output voltage waveform repeats at the period of T = 21t . The average value is given as,

Power Electronics - Ill

Single and Three Phase ACIDC Converters

3-8

IT

Jv

= T

... (3.2.2)

0 (cot)drot

0

In Fig. 3.2.4 observe that,

v 0 (rot) =

v 5 = Vm sin mt

from a top from O to a and P to 2 7t

{0

Hence equation 3.2.2 can be written as,

1 p Vo(av) =

...

=

7t

2

JVm sinrot drot a

V 2

; (coscx - cos~)

... (3.2.3)

This is an expression for average value of output voltage.

3.2.3 Half Wave Controlled Rectifier with Freewheeling Diode Now let us consider the half wave controlled rectifier with freewheeling diode across the RL load. This circuit diagram is shown in Fig. 3.2.6.

+ R

The SCR is triggered at firing angle of a in positive half cycle of supply. Hence L v O = v5 • The waveform of v O is shown in Fig. 3.2.7(c). Observe that from a to 7t , v0 Fig. 3.2.6 Freewheeling diode in half is same as supply voltage Vs • The wave controlled rectifier freewheeling diode (D FW) is reverse biased, hence it does not conduct. The output CWTent f 0 increases from zero as shown in Fig. 3.2.7(d). This is shown in equivalent circuit-I in Fig. 3.2.7. See Fig. 3.2.7 on next page.

, I I I I

,,..----.... ... o \ t .

1

I I I

I

I

f iFW \,

i

llo

I

_____

I I

I

:

I

After n, the supply voltage becomes negative. Hence SCR tries to tum-off. Therefore i O mes to go to zero. Observe that i O is maximum at 7t • But the load inductance does not allow i O to go to zero. The energy stored in inductance generates the voltage L di O with polarity as shown in

dt

.,✓

Fig. 3.2.8. Fig. 3.2.8 Freewheeling action In half wave controlled rectifier

Power Electronics - Ill

3-9

--...

'

I

I

I

I I I

'I

Single and Three Phase AC/DC Converters

,-, ;

1.0

\

R

\ I

I

'----------

___ .,II

Equivalent circuit - I

R

I

I

L

\

..

I

I

, ,'

L

EquJvalent circuit • n

Fig. 3.2.7 Wavefonns of half wave converter with freewheeling diode

The induced inductance voltage forward biases freewheeling diode as well as SCR. But freewheeling diode (DFW) is more forward biased. Hence it starts conducting. Therefore T1 turns-off. The output current now flows through the freewheeling diode. In above figure observe that i O = i FW when freewheeling diode conducts. He!e i FW is

Single and Three Phase AC/DC Converters

3-10

Power Electronics -111

freewheeling current. Fig. 3.2.7(d) and (e) shown that i0 = i FW when freewheeling diode conducts. The freewheeling current flows only due to energy stored in the load inductance. The output current flows in the load itself. Thus inductance energy is supplied back to the load itself. This process is called freewheeling. If load energy is fed back to the supply (mains), then it is called feedback. The energy of inductance goes on decreasing after 7t • Hence i 0 also goes on reducing. At~ the inductance energy is finished. Hence i 0 becomes zero at ft Thus freewheeling diode conducts from 7t to t3. The output is shorted due to freewheeling diode. Hence v O = 0 whenever freewheeling diode conducts. This is shown in Fig. 3.2.7(c) also. During freewheeling T1 is off. Hence no supply current flows. Therefore is = 0 during freewheeling period. T1 conducts from ex to 1t . Hence i0 = is from ex to 1t as shown in Fig. 3.2.7.

Derive an expression for average value of output voltage for 1 41 half wave controlled rectifier for RL load and freewheeling diode.

, . Example 3.2.2 :

Solution :

Fig. 3.2.7(c) shows the output voltage waveform. From this we can write

1

Vs

Vo =

{

= Vm sin rot

0

from ex ton from O to ex and 1t to 2 1t

The period of v O is 2 7t . The average value is given as, Vo(av)

=

1 T

T

Jv 0

1 n

0 (rot) drot = 1t 2

f Vm sm. rot drot (X

= Vm [-cos oot] 21t V o(av) =

V.

: (1 + cos ex] 2

... (3.2.4)

Here note that average output voltage is same as that of resistive load given by equation 3.2.1. 'This is because output voltage waveforms are same in both the cases.

> •

A single phase half wave controlled rectifier is used to supply power 230 V, 50 Hz supply at a firing angle of 30". Calculate - i) Average output volta~e ii) Effective output voltage iii) Average load current. Example 3.2.3 : to 10 '2 load from

Solution :

The given data is, R = 10 !l, V5 = 230 V

1t ex= 30°= 6

i) To obtain average output voltage

The load is resistive. For this load

Vo(av)

V o(av)

is given by equation 3.2.1 as

Single and Three Phase

Power Electronics -111

AC/DC Converters

3 -11

=

V: ; {l + cosa) 2

1t)

= 2301t.fi. ( 1 +cos 2 6 = 96.6 V

ii) To obtain effective output voltage Vo(nns)

The rms value is given as,

Vo(rms) =

1

l [! T

T

]2

v;(rot) drot

From the output voltage waveform of Fig. 3.2.2 we can write,

Va(nns) = [ 2~

1

!

2

VJ sin rot

drotr

r 1

cos2rot

drot]

1

= Vm

2

[l- a+ sin2a]2 1t

21t

... (3.2.5)

This is an expression for effective rms value of half wave controlled rectifier. Putting values in above equation,

Vo(rms) =

230-.fi. 2

= 160.27 V

Single and Thrae Phase AC/DC Converters

3-12

Power Electronics - Ill

iii) To obtain average load cunent Io(av) The Io(av) can be calculated as, Io(av) =

Vo(av)

R

= 96.6 =9 .66 A 10 A single phase half wave converter is operated from a 120 V, 50 Hz supply and the load resistance R = 10 n If the average output voltage is 25 % of the maximum possible average output voltage calculate i) Delay angle a ii) The rms and average output currents iii) The nns and average thyristor currents iv) The input power factor.

, . . Example 3.2.4 :

Solution : Given data

Supply voltage Vs

= 120; hence V m = ..fi. x 120 = 169.7 V, Load resistance R = 10 n

Average output voltage

Vo = 25 % of Vo maximum

i) To obtain delay angle ex

The average output voltage of half wave converter is given by equation 3.2.1 as, V o(av)

= ~; (1 + cosa)

Vo will be maximum at ex = 0. Hence above equation will be, V

o('10)m11X

V: V. = _!!!..(1 + cos O' = ---1!!. 27t

=

1697 7t

'

7t

= 54 V

It is given that the average output voltage is 25 % of its maximum v_ alue. i.e.,

V o(av) = 25 % of Vo(ar,)mu = 0.25 x 54 = 13.5 V

Consider again equation 3.2.1, ~.

V: Vo(av) =

; (1 + cosex) 2

Putting values in above equation, 13.5 =

1697 2n (1 + cos a)

Solving above equation for a, a = 2.09 radians= 1200

,

Power Electronlcs - Ill

ii)

Single and Three-.phase AC/DC Converters

3 -13

To obtain rms and average output currents

Average value of output current is given as, Vo(av)

R =

1:05 = 1.35 A

The rms value of output voltage is given by equation 3.2.5 for half wave converter. i.e., 1

= Vm [l - a.+ sin 2a]2 2 1t 21t

= 1697 2

[t -209 + 1t

1

sin(2 X 209)] 21t

2 ,..

= 37.718 V

Hence rms output current will be,

Io[1

f

2

cos nrot drot -

a

cos nwt drot]

n+a

I 7t 2

2

cos

a

2

i

Power Electronics - Ill

HF=

...

Single and Three Phase AC/DC Converters

3-33

n(n-a) _ 2

' 8cos

~

1

... (3.3.21)

This is an expression for harmonic factor of supply current. Iv) Current distortion factor (CDF)

The current distortion factor (CDF) is given as, CDP =

Isl I s(nns)

2..fi. I o(av) ex --n--cos

2

:::

Io(av) ~ 2..fi. cos~

=

... (3.3.22)

.Jn(n -ex)

,,_. Example 3.3.5 : For a 1 cl> half controlled bridge having continuous and ripple free current, obtain, i) Active power and ii) Reactive Power.

Solution : I) Active power

Active power is given as, Pactive = Vs l s1 cos 41>1

= V: •

2

..fi. Io(av) cos a co{- a,.

s

=

✓2½ I«av) 1t

21

2

1t

2cos

2

a

2

Vmlo(av)

.. .(3.3.23)

= ---(l+cosa) 1t

II) Reactive power

Reactive power is given as, Preactive

=

Vs Is1 sin cl>1

::: V: . 2.J2 I o(av) cos 5

1t

since ~1 ::: -~ 2

a a) sin(-

2

2

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-34 .J'i.Vs l o(av) . a a = - - - - - 2sm - cos7t 2 2 V ml o(av) .

= - - - -sma

. .. (3.3.24)

1t

The negative sign indicates that power is reactive. Comments

i) Active power is consumed by the load. ii) Reactive power is not consumed by the load. Hence its sign is negative.

iii) Reactive power fluctuates between load and source. iv) Total powie r includes active as well as reactive power. , . . Example 3.3.6 : Single phase semiconverter is operated frum 120 V, 60 Hz supply. The load cu.rrent with an average value of I a is continuous with negligible ripple

content. Turns ratio of transformer is unity. The delay angle a) Harmonic factor of input current. c) Input power factor Solution :

b) The displacement factor

The given data is, Vs = 120 V

I o(av) = Ia a =

1t

3

a) Harmonic factor is given by equation 3.3.21 as,

HF =

1t(1t

- a) - 1

8cos 2 ~ Putting the values in above equation, 1

HF =

21(21-i) 2

8cos (

- 1 2

"f)

= 0.3108 or 31.08 %

b) The displacement factor is given by equation 3.3.19 as, a

OF = cos-

2

a=;.

Calculate-

Power Electronics -111

Single and Three Phase AC/DC Converters

3-35

=

cos( nf3)

= 0.866

c) The input power factor is given by equation 3.3.20 as, PF =

(l

cos 2 2

= 0.827 lagging

3.3.4 Asymmetrical Half Bridge Converter

3.3.4.1 Operation with Resistive Load Fig. 3.3.12 shows the circuit diagram of asymmetrical half bridge converter. Observe that both the SCRs appear on the same link. +

Load

I

Vo

I Fig. 3.3.12 Single phase controlled ntetlfler

When T1 is triggered, current flows through T1 and D 1 . T2 is triggered in the negative half cycle. Then current flows through T2 and D2, Fig. 3.3.13 (See on next page) shows the waveforms of half bridge converter given in Fig. 3.3.12. These wavefo~ are shown for resistive load and a= ~ - Observe that

the output current waveform is similar to output voltage. Since T1 and 0 1 conduct simultaneously their current waveform is same. Similarly, the current waveform of T2 and 0 2 is same.

Single and Three Phase Power Electronics - Ill

3-36

AC/DC Converters

Fig. 3.3.13 Waveforms of half bridge converter of Fig. 3.3.12

Since the above output voltage is same as that of single phase semiconverter, the rms and average values of output voltage will be, V.

Vo(av) = 2!.(1 + cos a) 1t

Power Electronics :- Ill

Single and Three Phase AC/DC Converters

3-37

3.3.4.2 Operation of Asymmetrical Half Bridge Converter with Level Load With the similar circuit diagram of Fig. 3.3.12 but for highly inductive load, the operation of asymmetrical converter will be as follows. Mode -

I (a s c.o t

s 1t )

SCR T1 is triggered in this mode. Hence load current flows through T1D1. The waveforms are shown in Fig. 3.3.14. (See Fig. 3.3.14-on next page.) Mode - II (n

s c.ot s n +a)

In this mode, the supply voltage becomes zero at 1t. Hence T 1 is no more forward biased. But due to highly inductive load, the constant current is maintained to flow. This load current flows through D 1 -D 2 . The equivalent circuit is shown in Fig. 3.3.14. Thus the freewheeling action takes place through D1 - D 2 and supply current as well as output voltage are zero. Mode - 111.(n +as c.ot S 2n)

SCR T 2 is triggered at 1t + a . Since T2 is more forward biased due to supply voltage, it starts conducting. The load current now flows through T 2 - D 2 • The equivalent circuits - ill in Fig. 3.3.14 shows the current flow. Mode - IV (2n scot S 2n +a)

At 2n, the supply voltage becomes zero. Therefore T2 turns--off. But due to heavy inductive load, the current continuous to flow. This current now flows through D 1 -D2 since they are more forward biased compared to T2 -D2 . At 21t +a, SCR T1 is triggered again and mode-I starts. Thus the cycle repeats. Mathematical analysis

Observe that the waveform of output voltage is same as that of semiconverter. Hence the rms and average values of its output voltage are, Vo(av) =

V: : (1 + cos a) 1

and

V~=l

=-{i[1t-a+½sm1nf

Power Electronics • Ill

r •

I

output 1

h

·• f Cumlflt ,c,(av) l

..

T

.

·l

t.r

Single and Three Phase AC/DC Converters

3-38

I l I q J{r j

±;1

, 1 l

;

:•

11 I

1--rf

•,

l

'

ri. '

. II

I

l

1·

1.

'.

I

• ·

l

'i

t,

' ' '

1

►

! ;

I i

•

_.

,~• ·

r It : •

!1

I ,,

t

I ' :

•

t JI

11t 1

;ii~!

11-1

o, Equivalent Circuit • I

Equivalent Cirwit • II

Equivalent Circuit - II

Equtvalent C.-cuil ·

r-,/

Ag. 3.3.14 Waveforms of asymmetrical half controlled bridge convrert&r for level load

Single and Three Phase AC/DC Converters

3 -39

Power Electronics -111

3.3A.3 Comparison of Symmetrical and Asymmetrical Configurations Table 3.3.1 shows the comparison between symmetrical and asymmetrical configurations of half controlled bridge. Sr. No.

A■ymmetrical

Syrnn:--trtcal configuration

configuration

1.

One SCR is connected on each link.

Both the SCRs are connected on single link.

2.

SCRs can be driven with common cathode.

SCRs must have Isolated cathodes.

3. 4.

5.

Freewheeling ·takes place through on diode

Freewheeling takes place through both the

and on SCR.

SCRs.

Average curli8nts of SCR and diodes are same.

Average currents of diodes are higher than

SCR and diodes conduct for equal durations.

SCRs conduct for shorter duration compared to diodes.

SCR.

Table 3.3.1 Comparison of symmetrical and asymmetrical configuration

Review Questions 1. Explain the working of 1 ♦ semiconvertu with the help of waveforms for resistive load and inductive loads.

2. How freewheeling is present inherently in the semiconverters ?

Unsolved Example 1. Derive ,m expression for rms and average values of output volt.age for 1 ♦ semiamverter.

3.4 Slngle · Phase Full Converters Fig. 3.4.1 shows the block diagram of 1♦ full bridge converter. It contains four SCRs T1, T2, T3 and T4. The conduction of all these SCRs is controlled. Hence this is called full converter. The input to this converter is 1♦ AC supply. The ,o utput is controllable DC. The full bridge converter is mainly used for speed control of DC motors.

+

1. 230 V, 50 Hz AC supply

l l

Vo

io Lo"ad

Fig. 3.4.1 Circuit diagram of 'If full converter

Single and Three Phase AC/DC Converters

3-40

Power Electronics - Ill

3.4.1 Working with Resistive Load

..,-,t

1

Is

Vg

I --~--•----+-..

R

T

Fig. 3.4.2 Conduction of T1 and T2 In positive half cycle of the supply. Dotted line shows path of current flow

1,

Let us consider the working of bridge (Full) converter with resistive load. In the positive half cycle of the supply SCRs T 1 and T2 are triggered at firing angle a. Hence current starts flowing through the load. The equivalent circuit for this operation is shown in Fig. 3.4.2. ·

It is clear from Fig. 3.4.2 that, when T 1 and T2 conducts, VO

and,

= V5

(i.e. supply voltage)

... (3.4.1)

... (3.4.2)

Fig. 3.4.3 shows the waveforms of this circuit. Observe that load voltage is same as supply voltage from a to 1t. Since the load is resistive, waveforms of ~ 0 and i O are same. The supply current is and i 0 are in the same direction hence is =i0 • Ti and T2 tum off when supply voltage becomes zero at n. In the negative half cycle T 3 and T4 are triggered at 1t +a. Fig. 3.4.4 shows the equivalent circuit when T3 and T4 conduct. In the adjacent figure observe that supply current i5 and load current i O flow through the same loop. But directions of i5 and i0 are opposite h ~

The supply current waveform is also shown in Fig. 3.4.3. T 3 and T4 tum off when supply voltage becomes zero at 2n . At 2n +a, T1 and T 2 are triggered again and the cycle repeats. 1

Power Electronics • Ill

3-41

Single and Three Phase AC/DC Converters

Fig. 3.4.3 Waveforms of full bridge converter for resistive load

Power Electronlcs - Ill

Single and Three Phase AC/DC Converters

3-42

....................

.

- .·····--· ...: '

•...

+

•·

i --

~. . . . . . . . .

'

l'

:'

.

. . .J . . . . . . . . . . .

.

~

Vo

j

~

'' '' ... . . .

I +

• • • • • • • • • • • • - 4......... .

~

I R

t

. ''

r,

>•

Fig. 3.4.4 Conduction of T3 In negative half cycle of the supply. Dottad line shows current path

For the 1, fully controlled lmdge converter having load of 'R' determine the following : i) Average output voltage Vo(av) ii) RMS output voltage Vo(rms) If supply voltage is 230 V, 50 Hz and firing angle is 60°, determine average output voltage.

Example 3.4.1 :

Solution : Compare the waveforms of lq, semiconverter given in Fig. 3.3.3 and that

of

1♦

full converter given in Fig. 3.4.3. The waveforms are exactly same. Thus the

operation of semiconverter and full converter is exactly same for resistive load. Hence

their average and RMS output voltages are also same. Hence from equation 33.5, the average output voltage of full converter is, Vo(av)

(1 + cosa)

= ;

... (3.4.3)

The supply voltage is 230 V. Hence, Vs= 230V and

·

Vm=230.J2

ex = 60°

Putting values in equation 3.4.3, Vo(av) =

23o.J2 7t (1 + cos 60°)

= 155.3 V

From equation 3.3.6, RMS output voltage of full converter is, 2

1. 2a]2

V o(rms) = Vm [ 7t -a+ sm 211 2

>•

1

... (3.4.4)

For the circuit shown in Fig. 3.4.5, find the current through 100 .n load, if the SCRs are triggered at 300 delay. Supply voltage is 200 V, 50 Hz. Example 3.4.2 :

Power Electronics - Ill

3-44

Single and Three Phase AC/DC Converters

3.4.2 Working with Inductive Load The inductive load means resistance and inductance in the load . Such loads are DC motors. Because of the inductive (R-L) load, the load current shape is changed. Hence operation of the full bridge converter can be discussed into three modes : i) Continuous load current. ii) Continuous and ripple free current for large inductive load.

iii) Discontinuous load current.

3.4.2.1 Continuous Load Current

(2006)

In the continuous load current, the load or output current iO fl.ows continuously. The waveforms are shown in Fig. 3.4.7.

I

Fig. 3.4.7 Wavefonns of 1' full converter for Inductive load having continuous load current •

Cop ght

ma

al

Power Electronics - Ill

3 -45

Single and Three Phase AC/DC Converters

As shown in the waveforms of Fig. 3.4.7, T1 and T2 conduct from a to 7t . The nature of the load current depends upon values of R and L in the inductive load. Because of the inductance, i 0 keeps on increasing and becomes maximum at 7t . At 7t , the supply voltage reverses but SCRs T1 and T2 does not twn off. This is because, the load inductance does not allow the current i O to go to zero instantly. The load inductance generates a large voltage L dio. dt

r-----, :r, l, if -f •1 1

- ~J Vs " '

+

4

!

l

I

I

◄T2

;

,

R

Vo

1-L

L dio dt

+

I_ -

Fig. 3.4.8 Conduction of T1 and T2 from 7t to 7t + a due ·t o fnductance voltage

This voltage forward biases T 1 and T2 as shown in Fig. 3.4.8. In this figure observe that the load current flows against the supply voltage. The energy stored in the load inductance is supplied partially to the mains supply and to the load itself. Hence this is also called as feedback operation. The output voltage is negative from 7t to 1t + o: since supply voltage is negative. But the load current keeps on reducing.

At 7t + a, SCRs T 3 and T 4 are triggered. The load current starts increasing. The load current remains continuous in the load. The similar operation repeats. The ripple in the load current reduces as the load inductance is increased.

3.4.2.2 Discontinuous Load Current

[2006)

Fig. 3.4.9 sho,w s the circuit diagram of 1~ full converter for discontjnuous load current with RLE load. Two cases are possible as explained next.

Case - I : J3 >n The extinction angle JJ is greater than 1t. Fig. 3.4.10 (b) shows the waveform of output current. Fig. 3..4.10 (c) shows the

+

output voltage wavefo rm. Here observe L

Fig. 3.4.9 1~ full converter with RLE load

that SCRs T1-T2 conduct from a to~Hence output voltage is same as supply voltage from ex to ~- This is shown by equivalent circuit - I in Fig. 3.4.10. The load current becomes zero at t3 and it remains zero till 7t +a:. Thus from f3 to 1t +ex there is no conduction and all the SCRs are off.

Powel' Electronics • Ill

3-46

Single and Three Phase AC/DC Converters

This is shown by equivalent circuit - II in Fig. 3.4.10. Hence battery voltage 'E' appears across the output terminals. Thus, v O = E for ~ to 7t +a. At 7t +a, T 3 -T4 are triggered. Since T 3 -T4 are forward biased due to supply voltage, they start conducting. Again supply voltage appears across the output terminals. This is shown in equivalent circuit - min Fig. 3.4.10. Fig. 3.4.10 (d) shows the supply current. Casa • II :

~


i

L I

-

I I

I

.l

w

v_

1-

J -

-

i

(t)

-· -

I-

-1 = 1-

--

.... -

--

I

t

I

I


..

If a freewheeling diode

is added across the highly inductive load in 14' full converter, derive an expression for average load voltage. Example 3.4.5 :

We know that freewheeling action does not take place in lei> full converter inherently. In the positive half cycle, T1 and T2 conduct from a to 1t as usual. But from 7t to 7t +a freewheeling diode starts conducting. This is shown in Fig. 3.4.15. The freewheeling diode is more forward biased compared to T1 and T2 . Hence freewheeling diode conducts. The freewheeling diode is connected across the output v 0 • Hence v O = 0 during freewheeling. The energy stored in the load inductance is circulated back in the load itself. Fig. 3.4.16 shows the waveforms of this operation. The output voltage becomes zero in the freewheeling periods. Compare the load voltage waveform of Fig. 3.4.15 with that of 141 full _c onverter with resistive load (Fig. 3.4.3). They are same. Hence the average load voltage can be obtained from equation 3.4.3. i.e., Solution :

Vo(av) = ;

(1 + cosa)

... (3.4.10) ,_.,__, I I

l1 !i ___ ' I I

! I

DFW

I

Vo !

I I

I

IFW

R

L dio

I

..

I

I

Fig. 3.4.15 FreewheeUng diode conducts from

io

ff

-,

+

dt

f

to ff + a due to Inductive load

Power Electronics - Ill

Single a'nd Three Phase AC/DC Converters

3-55

The average load current I o(av) or I a is given as,

1o(av) =

Vo(av) R

Putting the values of R =10 .Q and Vo(av) = 146.42 volts, 146.42

Io(av) =

10

= 14.64 A

The supply currie nt waveform will be a square wave as shown in Fig. 3.4.11. The amplitude of the square wave will be Io(av) i.e. 14.64 A. ,. . Example 3.4.7 :

For a 1~ full converter having highly inductive load derive ·the

folluwing: i) Fourier series for supply current ii) Fundamental component of supply current iii) RMS value of supply current Solution : I) To determine Fourier series

The general expression for Fourier series is given as,

i5

( (IJ)

t)

= I5 (av) +

L- cn

sin ( n ro t + 9n)

n=l

where

Cn

= -Ja2n +b2n

and

~n

= tan-1

Here,

an = T

( :: )

2 T

f i (rot) cos nrotdrot 5

0

27t

= /n

f i (rot) cos n cot d ro t 5

0

From the supply current waveform of Fig. 3.4.11 we can write,

2

an = 2~

[ffrr.(av) cosnmt d mt+ 7''( -l•(av)) a

=

n+a

fcos n ro t d mt - 2n+a fcos n

I 0 ( ) [n+a 1tav

a

1t+a

]

ID

t d ro t

COStt(J)

t d (J) 1]

Single and Three Phase AC/DC Converters

3 - 56

Power Electronics - Ill 2

Io(av) nn

= - - - sin n a [ cos n 1t -

-4 Io(av)

=

{

1]

---'--- sin n a

for n = 1,3,5, ...... .

0

for

n 1t

... (3.4.11)

n=0,2,4,., .....

T

Similarly,

Ji

bn = ;

(wt) sin nwt dwt

5

0

=

2n

2

Ji

1t

5

2

(rot) sin

nwt drot

0

From supply current of Fig. 3.4.11,

! ["1'\(ao) sin nw Id wt+ "f''( -/•(av) sin nw dwt)] 2

bn =

=

I

2

a

I 0( ) 1tav

n+a

[c+tt J sin n rot d rot - 2n+tt J sin n rot d wt] a

= =

n+a

21o(av) n 1t cos na (1- cos n 1t]

r

1 o(av) cos n a n1t

0

Hence

Cn

=

=

= And ♦n = tan-1

:n

for

n =1,3, 5, ... ...

for

n =0,2,4,6, ......

... (3.4.12)

.Ja2n +b2n

{(4 In~w) 4

Io(av) n 7t

r

r 1

[ sin 2 n a +cos 2 n a]

for

n =l, 3, 5, ......

... (3.4.13)

= -n a from equation 3.4.11 and equation 3.4.12

n

Thus

... (3.4.14)

Single and Three Phase Power Electronics - Ill

AC/DC Converters

3-57

The average value of supply current is zero. i.e. Js(av) = 0. 1his is clear from

Fig. 3.4.11. Therefore Fourier series is,

i5

(

O>

t)

=

4 1o(av) ~ -~ nn n= 1, 3,5,~.

sin (nrot-n a)

... (3.4.15)

II) Fundamental component of the supply current

The fundamental component of the supply current is given as, C1 1st = ✓ 2 41 o(av) From equation 3.4.13, C 1 - - - with n =1. Hence above equation will be, 1t

4 I o(av)

Isl

2...fi. I o(av)

1

= ------'-' x - = -----11. ...fi. 1t

iii) To obtain nns value of supply current

The

rms vm::(:~v~[;,!I;

(rot) d ro

r

... (3.4.16)

2

From supply current waveform of Fig. 3.4.11, 1

Is(=) =

I Is(nns)

...

>•

Example 3.4.8 :

=

{i.t[T1:c../0>l+:IJ-1.(..>)' drotr Io(~)

For a

I

1♦

... (3.4.17)

full converter having highly inductive load, derive the

folltJWing : i)

Displacement factor (DF)

ii) Supply power factor (PF) iii) Harmonic factor (HF) iv) Current distortion factor (CDF) Solution : i) Displacement factor

The displacement factor (OF) is given as, DF = cos +J. From eqy_ation 3.4.14

♦n

= -n a ;

... (3.4.18) Hence

♦1

= - a.

Single and Three Phase Power Electronics - Ill , . . Example 3.4.9 : current obtain, i)

3-59

AC/DC Converters

For a 1 q, fully controlled bridge having continuous and ripplefree Active power and ii) Reactive power.

Solution : I) Active power

Active power is given as, Pactive = Vs l s1 cos«1>1

2./i Iii.av) = Vs ·----cos(-a), 1t

since , 1 =-a

Ji. Vs 1..tav)

=

2 · - - - " ' - cos(a)

=

- - - - COSCl 1t

7t

2Vm1~nv)

... (3.4.23)

II) Reactive power

Reactive power is given as, P,eactive = Vs l s1 sin 4>1 = V.s ·

2.fi. I

~ 7t

av)

sin(-a)

.fi. Vs l~av) . = -2· - - - - sma 1t

2Vm l ~av) . = - ----sm cx.

. •. (3.4.24)

1t

The negative sign indicates that the power is reactive. Comment

Compare the reactive powers of full converter and half converter. They are as follows : V,n I o(av) •

P,eactive(HCB) =

---smcx

P,eadive (FCB) =

- - - - -smcx.

7t

2 V,n I o(av)

.

1t

From above two equations we have, · P,eactive(FCB) = 2 x P,eactive (HCB)

•

Thus half controlled bridge draws SO % reactive power compared to that of full controlled bridge.

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-62

Fig. 3.4.18 Inversion in 1q, full converters .,

Supply

Fig. 3.4.19 Inverting operation in 1t full converter Triggering •"flle

Moct. of operation

Powar flow

Output voltage

a< 900

Rectification

Source to load

Vo( av} positive

a= 900

None

None

vo(av) ;: 0

a> 900

Inversion

l oad to source

Vo(av) negative

Table 3.4.2 Operating modes of 1' full converter

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-63

A single phase fully controlled bridge rectifier has an AC voltage of 230 nns applied to it. If it is to act as an inverter with a DC source of 150 V, estin:,,ate the trigger angle delay.

, . . Example 3.4.11 :

In Fig. 3.4.18 we have seen that the DC voltage source is connected at the output of converter. The converter then acts as an inverter. Since the conduction is continuous the output average voltage will be, 2Vm V o(av) = - - cosa ... (3.4.25) . Solution :

1t

Here the voltage drop across the load is not given. Hence we can assume zero voltage drop in the load. Hence the DC source voltage becomes as average output voltage. i.e., Vo(av)

= - 150 V

Here 150 V is the DC source voltage and it is negative as shown in Fig. 3.4.18. Supply voltage is, V5 = 230 V Hence Vm = ..fi. x 230 V Putting values in equation 3.4.21, 2 X ..fi. X 2'30 - 150 = - - - -coscx 7t

... ...

coscx = - 0.724383 ex = 136.41°

3.4.4 Comparison of Half Controlled and Full Controlled Rectifiers Now let us oompare the hall controlled and fully controlled bridge rectifiers. Table 3.4.3 shows this comparison. Sr.

Half controlled converter

Fully controlled converter

.No.

1.

This consists of half number of SCRa and half number of diodes.

Thia consists of all the SCRs as controlled devices.

2.

This operates In only one quadrant.

This can operate in two quadrants.

3.

Output voltage is always positive.

Output voltage can be negative in case of Inductive loads.

4.

Inherent freewheeling action is present.

Extemal freewheeling diode is to be connected for freeWheellna.

5.

Power factor is better.

Power factor is poor than half converter.

6.

Inversion is not possible.

Inversion is possible.

7.

Used for battery chargers, lighting and heater control.

Used for DC motor drives.

Table 3.4.3 Comparison of half and fully controlled bridges

Power Electronics - Ill

3-65

Single and Three Phase AC/DC Converters

(i.e. 1200). In the waveforms of Fig. 3.5.3, SCRs are triggered at firing angle of ex = 300.

For each phase a = 0 is actually : i.e. 300 from zero crossing of that phase. As shown in the output voltage waveform, SCR T1 conducts from (:

+a) to

1t .

Fig. 3.5.3 Waveforms of 3 ~ half wave c,onverter for a. = 300 for resistive load

Single and Three Phase

3-66

Power Electronics - Ill

AC/DC Converters

Fig. 3.S.4 shows an equivalent circuit when SCR T1 conducts. T2 and T 3 are 2 reverse biased because of T 1 conduction. The Y phase has zero crossing at ; . For T2 ,

2

a =0 is at ; + : =

5:.

And T2 is triggered at

5 : + a. In the waveforms of

2 Fig. 3.5.3, observe that each SCR conducts for the maximum duration of ; (i.e. 1200). The output current is given as, i o = vRO

for resistive load

N

Fig. 3.5.4 Conduction of T1 In 3 ♦ half wave converter. Dotted line shows current path.

Hence the shape of the output current is same as output voltage. 2 The period of each pulse in the output voltage waveform is ;

. (i.e. 1200). The

supply has the frequency of 50 Hz. There are three pulses of output voltage during one cycle of supply voltage. Hence ripple frequency of output voltage is three times of supply frequency. i.e., /ripple = 3 xso = 150 Hz Sometimes this converter is also called as 3-pulse converter. · , . . Example 3.5.1 :

Derive an equation for average output voltage nf 3 ♦ half

wave

converter. OR

With the help of circuit diagram and relevant wavefonn, describe the operation of 3 ♦ half wave 1converter and hence derive expressions for i) Vde ii) Vdm and iii) V rms (Assume load being resistive.) Solution : I) To obtain average value of output voltage, Vo(av) or Vde

Let the volta:g e of the R-phase be represented as, VR = Vm sin wt

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-67

Here Vm is the peak value of R-phase voltage. In Fig. 3.5.3, observe that T 1 conducts from (: +a) to

(5;: +a} It

is the one

2 period of the load ripple. Thus T = ; . Average valae of load voltage is given as,

1

T

f

V0 (cro) = T

V0

(rot) d rot

... (3.5.1)

0

2 Here T = ; :: +a) to (

and supply voltage VR

=Vm sin co t

appears across output from

5;: +a) . Hence above equation becomes, S1t+a 6

1

-() f 27t ff

3

.

Vm sin rot drot

6 +a

3 V: Sn+a = - ._ m _ [- cos rot] 6

2n

=

~+a

6

3 J3 V:

m cos a

2n

... (3.5.2)

Note that the above equation is valid only for a S 30°. Now let us consider the case when a = 600 (a> 30'>). The last waveform in Fig. 3.5.3 is drawn for a =60°. SCR T 1 conducts from (: + a) to 7t. The output voltage remains zero from

(5;: +a} At (5;: +a) SCR T

2

1t

to

is triggered. At 'n' T 1 turns off since supply voltage VR

becomes negative. Putting the values in equation 3.5.1 we get, 1 Vo(av) = Zn -

1C

J

•

Vm sin rotdrot

1t

3 ~a

... (3.5.3)

Single and Three Phase

3-68

Power Electronics - Ill

AC/DC Converters

Note that this equation is valid only for a > 3Cf, i.e. ·discontinuous operation for resistive load. ii) To obtain peak value of output voltage (Vdm)

Peak value of output will be obtained when a = 0. i.e.,

vdm

= vo(av)l et=o =

3J3vm 0 27t cos

- 3-./3vm -

2n

iii) To obtain rms value of output voltage Vo(nns) RMS value is given as,

.sn

1 6 +a

JV,; sin 2 cot dcot

1t

2

-

1

2

7t

3 6+a 1

=

3VJ 21t

57t+a 6

J 1-cos2cot dcot

2

2

ft

6+a

1llis is the required expression for rms value of output. Above equation is valid for continuous operation, ie., a S 300.

half wave converter is operated from a 3♦ wye connected 220 V, 60 Hz supply and load resistance R = 10 n If the average output voltage is 25 % of the maximum possible average output voltage, calculate a) Delay angle b) RMS and average output currents.

, . . Example 3.5.2 : The 3 q»

Solution : a) To obtain delay angle (a)

The output average voltage for a s 30 is given by equation 3.5.2 as, V o(av) =

3-./3vm i1t

cos a

Hence maximum value of average output voltage will be,

3-69

Power Electronics - Ill Vo(av)max

Single and Three Phase AC/DC Converters

= 'J../3Vm

21t

Here Vm is the peak value of phase voltage. The line voltage is given as 220 V. Hence,

·_ Vune _ 220 V ph - .J3 - ...fj And peak value of phase voltage will be,

Vm = ..fi.Vph

=

..fi. x

2_; = 179.63 V

Hence maximum average output voltage will be, V~av)ma.x

=

'J../3Vm 21t

=

'J../3 X 179·63 = 148.55 V 21t

Average output required is 25 % of the maximum.

...

Vo(av)

= 0.25 x Vo(av)max

=

0.25 x 14855 = 37.138 V

Now let us ealculate the output voltage for a = 30 using equation 3.5.2 i.e., Va(av) =

=

3J3 ;~79.63 cos (30°) 128.65 V

Here observe that 25 % output is less than output for a = 30°. Hence delay angle must be greater than 30°. For resistive load, output current is discontiinuous for a~ 30. Hence w~ have to use equation 3.5.3 for discontinuous operation. i.e., Vc(w)

=

3~ [1+cos(: +a)]

Putting values in above equation,

37.138 = 3x;:·6.,[1+cos(: +a)]

...

a = 94.54° or 1.65 radians.

Thus the d~y angle is more than 30°, is required to get 25 % of maximum output

Single and Three Ptaase Power Electronlcs • Ill

AC/DC Converters

3-72

The rms value of this converter for a > 30° is given by equation 3.5.4 as, 1

V,crn~) = {

3:f [5:-a+; sm(; +2a)]}2 1

83 = {Jx(~:· > [5:-118156+½sm(; +2x118156)]}2 2

= 94.7447 V

RMS output current will be, I

o(rms)

=

Vo(rms) R

= 94 .7447 = 947 A 10

.

iii) Average and nns thyristor currents There are three SCRs. They conduct equally. Hence average SCR current will be, I T(av) =

I o(av)

3

= 7.0225 = 2 34 A 3

.

.

And therms SCR current will be,

1T(rms)

I o(.rms) :;;;

9.47

../3 :;;; ../3 :;;; 5.467 A.

Iv) Rectification efficiency

Rectification efficiency is given as,

11 = =

DC power in load _ V o(av) Io(av) rmspowerinload - Vo(rms)Io(rms)

70.225 X 7.0225 947447 X 9.47

= 0.5496 or 54.96 % v) TUF

The rms input line current is same as the thyristor rms current, i.e. 5.467 A. Therefore VA of one phase will be, VA of one phase :;;; Vph x l ph

= ~ X 5.467 = 656.525

...

VA of 3 cj> = 3 x 656 = 1969.577

Single and Three Phase

Power Electronics - Ill

3-74

AC/DC Converters

Fig. 3.5.6 Waveforms of 3+ half wave converter for highly Inductive load

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3. 79 BY

RY

RB

YB Fig. 3.6.2 Phasor diagram showing the relationship between phase and line voltages of 3' supply Whenat-:)

Here Vm is the peak value of the phase voltage. Putting above expressions in equation 3.6.2, Sn -6 +a

1t

-

Vo(au)

!)

.J3 Vm

=2

sin (

ffit+:) dOl t +

6+a

=

=

!)

[

.J3 v., sin ( Olf- :) dOlt

2

3

1/m a

3

.J;_ftvm {[-cos (Olt+

sin (

r

(J)f + :)dOll+ s;

sin (

Olt-:) dOltl

:)]!+a +[- cos (Olf- :)]ta} 6

2

I

Single and Three Phase

Power Electronics • Ill

=

3../3 V.

7t m

2

(1 + cos

AC/DC Converters

3-82

a)

... (3.6.4)

This is the expression for average output voltage for a S 60. When a = 60

_Fig. 3.6.3 (d) shows the output voltage waveform for a voltage waveform is just continuous. When T1 is triggered at

= 6CJ'. Observe that the a= 1€lr, the line voltage

RB is applied across the load. T 1 and D 2 conducts from (: +a) to (

s:

+a} Similarly

5

next SCR T 3 is triggered at ( : +a) and T 3 D 4 conduct. The current waveform will be similar to voltage waveform for resistive load. The average output voltage is given by equation 3.6.4, since voltage waveform is continuous. When a > 60

Fig. 3.6.3 (e) shows the output voltage waveform for a =90°. SCR T 1 is triggered at

(:+a} T

1

and D 2 conducts and line voltage RB is applied across the load. In the 7

waveform observe that, output voltage becomes zero at : . In Fig. 3.6.3 (b) observe 7 that line voltage RB becom es zero at :. Hence SCR T 1 is turned off. Since T 3 is not

triggered, the output voltage becomes zero. At

(5: +a)

T3 is biggered and T 3 D4

conducts. Line voltage YR is applied across the load. Thus for a> 60, the output voltage is djscontinuous. Since the load is resistive, the current is also discontinuous. The current waveform will be similar to voltage waveform.

>*

Derive an expression for average c,utput voltage c,J ~ semicooverter having resistive load Jc,r a > 60°. Example 3.6.2 :

Fig. 3.6.3 (e) shows the waveform of output voltage for a = 9(11 (i.e. a > 60j. Observe that the period of ripple cycle is Solution :

5 2 T = ( : +a) - (: +a)= ; 7

When T1 -D2 conducts voltage VRB is applied acr~ the load &om (: +a) to :. Hence v O = VRB from

7t

6

+ ex to

71t 6

Power Electronics - Ill

3-85

Single and Three Phase AC/DC Converters

Fig. 3.6.5 Waveforms of 3q semlconverter for highly Inductive load

Single and Three Phase Power Electronics - Ill

AC/DC Converters

3-86

= 0, but the output current i0 [Fig. 3.6.5 (e)] is not 7.ero. The output current waveform is continuous and ripplefree. The load inductance generates a very large

V RB

voltage L dio to maintain i 0 continuous. This situation is shown in Fig. 3.6.6.

dt

,, I

--------------jFW

I

I

R

-

,------, +

I

I I

D

FW

_______.....

I I I I

I I

I

i0

!

R

I

!

1

I

V0

I I I I

1 I

I

I

1

,

f I I

f

-

L dj0

+ dt

, ______ ,,I -

I

Fig. 3.6.6 Freewheeling action In 3+ semlconverter. Dottad fines shows freewheeling cumtnt paths

The freewheeling diode DFW is forward biased by the load inductance voltage

Ld~:. The freewheeling current iFW is basically i

0•

Thus the energy stored in the load

inductance is fed back to the load itself. Fig. 3.6.S (g) shows the waveform of 1 freewheeling current. SCR T 1 turns off at : as soon as &eewheeling diode starts ronducting. This is because &eewheeling diode is more forward biased compared to

T1 and D1 after

1 : . If extra freewheeling diode DFW is not connected, ther,

freewheeling current flows through T1 and D 4 • This is shown by 'thin dotted line' in Fig. 3.6.6. Thus freewheeling action is inherent in 3t semiconverter.

If we neglect the drop of freewheeling diode, then output voltage during freewheeling period is z.ero. As shown in Fig. 3.6.S (f), no supply cunent flows during freewheeling period. Comp~ the output voltage waveform of Fig. 3.6.3 (e) (resistive load) and Fig. 3.6.5 (d) (inductive load). Both the waveforms are same.

,. . Example 3.8..4 : Derive an aprtSSion for average output voltage of 3+ semiconverter having highly inductivt load. Solution : We have seen that the output voltage waveforms of

3+ semiconverter are

same for resistive as well as inductive loads. Hence their average values are also same. Hence from equation 3.6.5 we have,

Power Electronics - Ill

Single and Three Phase AC/DC Converters

3-87

... (3.6.6} This is the expression for average output voltage for resistive as well as inductive loads and O~a ~ff.

Review Question 1. &plain

~

worldng of 3+ semic.onvmer with the help of uxroeforms.

Unsolved Examples 1. A 3♦ semiconr,ierler is operated from 3♦ 230 V, 50 Hz supply. The load is 10 ohms in series with barge mroothing inductor. De.tennine output voltage and current if triggering angle is 60". (Ans. : vo{1111) = 403.S V I lo(av) = 40.3S A)

2. DerhJe an expression for llf1ffllgt wlue of output voltage for 3+ semiconverter.

3.7 Three Phase Full Converters Three phase half converters operate only in first quadrant of v O - i 0 • The output voltage v O is always positive for resistive as well as inductive loads. The output current iO is also always positive. Hence 3 ♦ semiconverter operates in first quadrant only. Three phase full converters can operate in two quadrants. The output voltage of 3 ♦ full converter can be positive as well as negative. It uses six SCRs as shown in Fig. 3.7.1.

+

Ro--.,__-1

Yo---..1 Bo---..1

load

'------t

'--__,I

Fig. 3.7.1

---1

3♦

full converter

3.7.1 Operation with Resistive Load Let us consider the operation of 3 ♦ full converter having resistive load. Fig. 3.7.2 shows the waveforms of 3 ♦ full converter having resistive load. Fig. 3.7.2 (a) shows the supply phase voltages R, Y and B. Fig. 3.7.2 (b) shows the supply line voltages. These supply voltage waveforms are drawn according to the phasor diagram shown in •

Powar Electronics -111

3-89

Single and Three Phase AC/DC Converters

Fig. 3 •.7.2 Wavefonns of~ full converter having resistive load

Single and Three Phase 3 - 91

Power Electronics •- Ill

ACID~ ~converters

For the 3 cf) full converter having resistive load find the following : i) Ripple frequency /ripple . ii) Output average voltage V0 ( av) .

,. . Example 3.7.1 :

Solution :

i) To determine ripple frequency :

In the load voltage waveforms of Fig. 3.7.2 observe-,hat line voltage

applied across the load from (~ +

VRY

is

a) to ( ~ + a} i.e. ~. Similarly other line voltages

are also applied across the loadfor the period of ; . The load voltage has the ripple period of ; . In one cycle of supply, six such ripple cycles are present in output waveform. Hence ripple frequency must be six times of the supply frequency. i.e., /ripple

= 6 xSO = 300 Hz

Thus ripple frequency of 3cf) full converter is higher than 3C? semiconverter. This is true for all the firing angles. Ii) To determine output average voltage

We know that average value is given as,

1 Vo(av) = T

T

f

... (3.7.1)

v 0 (wt) dwt

0

Case -1 cx~W

Consider the case when a S 600 i.e. for continuous output voltage waveform. Fig. 3.7.2 (d) shows the output voltage waveform for a =3Cf. Consider the pericd

(: +a) to (; + a) when voltage

VRY is applied across the load . This period is,

From Fig. 3.7.2 (b) we can write an equation for line voltage VRY as, VRY =

.f3 Vm sin

(wt+:)

Vm is peak value of the phase voltage. Putting the above values in equation 3.7.1,

....

Slnglla and Three Phase AC/DC Converters

3-92

Power Electronics - Ill

1 ft / 3

=

3../3 V. . ft m

[

- cos

(

(I)

1t)]i+a

t + 6 ~+a 6

...

... (3.7.2)

Case II : a > 60°

Now consider the case when a > 6ft'. For this firing angle, the output voltage waveform is discontinuous. Fig. 3.7.2 (f) shows the output voltage waveform for a = 900. Putting the integration limits and VRY in equation 3.7.1 for this case. 5ft

Vo(av) =

11

~

3

T

"3Vmsin(wt+~) dwt

!:+a 6

=

3 ~

Vm

[-cos (wt+:)]!6+a ... (3.7.3)

This is the equation for average output voltage for a > 600.

For a 3 , fully controlled SCR bridge converter operating from 400 V, 3 phase AC supply, calculate the average DC output voltage for a firing angle of 45°.

,_. Example 3.7 .2 :

Solution :

The given data is, Vune Cl

=

400 V (rms)

= 45°

Hence rms value of phase voltage is, 400

Vph

= ../?,

Single and,fhree Phase

Power Electronics -111

3 .93

AC/DC Converters

Hence peak value of phase voltage is, Vm = .J2Vph =

.J2. ~ V

Here firing angle is 45°. Hence the conduction will be continuous for resistive as well as inductive load. Therefore the average DC output is given by equation 3.7.2 i.e.,

'3J3vm

Vo(av) = - - - cos a 7t

Putting values in above equation ,

3/3x./2.- 400

11

3_ cos 45° _ _ _ _./3_ 7t

= 382 volts

3.7.2 Operation with Highly Inductive Load Let us consider the operation of 3 t full converter with highly inductive load. The output current will be continuous and ripplefree. In the waveforms of Fig. 3.7.2, observe that voltage waveform is continuous till a =6CJ'. But with inductive load, voltage waveform is continuous for any value of a. Fig. 3.7.5 shows the waveforms of 34» full converter for highly inductive load. Fig. 3.7.5 (c) shows the output voltage waveform for a = &. Observe that this waveform is same as that of resistive load shown in Fig. 3.7.2 (e). Fig. 3.7.5 (d) shows the continuous and ripplefree output current. Fig. 3.7.5 (e) shows supply phase current waveforms i R, iy and i 8 . Observe that the R-phase current is positive whenever T1 conducts and it is negative whenever T 4 conducts. All the three current waveforms are of the same nature (quasi square

wave) having 1200 phase shift with respect to each other. Fig. 3.7.S (£) shows the output voltage waveform for a = 900. The waveform goes negative for same period, because of inductive load. The load inductance generates a large voltage to maintain the load current in the same direction. Hence SCRs continue to conduct and load voltage becomes negative occasionally. Note that there is no freewheeling in full converter. I

3 . 95

Power Electronics - Ill

Single and Three Phase AC/DC Converters

,_., Example 3.7.3 : Derive an expression for aoerage and r.m.s. output voltage of 3+ full

converter having highly inductive load.

(2004)

In the Fig. 3.7.3 observe that output voltage waveform is continuous for complete range of a . Hence single expression can be drived. · In Fig. 3.7.3 (c), observe Solution :

that one ripple period of output voltage can be, T

=(~+_a)-(:+a)=;

During this period line voltage VRY is applied across the load. From Fig. 3.7.3 (b), VRY

will be, VRY :::;;

✓3 Vm sin ( c.ot+ :)

Here Vm is the peak value of supply phase voltage. I) Average output voltage

The average output voltage is given as, IT Vo(av) = T v o ( co t) do> t

f 0

...

... (3.7.4)

This equation holds for complete range of a . II) RMS output voltage

The rms output voltage is given as,

!

l T

V o(rms) = [ T

1

]2

v;(rot) do>t

Single and Three Phase

Power Electronics - Ill

AC/DC Converters

3-96 1

2

1t

=

1

2+a

7t

1t

I

v~y (cot) dcot

3 6+a

1

=

2

1+a 1-cos(2cot+ ;)

!·

J

3Vn~

dcot

2

ff

6+a I 1t

=

ff

f dcot- 2+a f cos ( 2cot + 31t) dcot

2

2+a 9v.2 m 27t

1t

7(

6 +a

•

6 +a

1 1t

2+a

=

9V. 2

-1!!.

21t

( 2cot + }" sin\ )

~+a

[rot] 2

~+a

2

-

2

6

n

-+a 6 1

= {

:f [~+a- ~ -a-;[sin(n+2a+ ;)-sm(; +2a+ ;)]]}2

9

1

9

= { :: [

;-½(-"3cos2a]]}2

Power Electronics - Ill

3-98

Single and Three Phase AC/DC Converters

Since output voltage is 50 % of its maximum value,

Vo(av) = 0.5 Vo(av)max

= 0.5 x 280.68 = 140.34 V

Consider

the formula for output voltage, V o(av) =

if3v,,, 7t

cosa

Putting values in this equation, 3.J3x1697 140.34 = ----cosa 1t

a

=

60°

II) Average and rms output currents

Average output current is given as, T ~(av)

=

V

'\av) R

=140.34 = 14·034 A 10

To obtain rms current, we have to obtain rms output voltage, for a ~ 60, the output voltage waveform is continuous as shown in Fig. 3.7.2 (d). Consider the period from

(~+a) to (~+a) when voltage

VRY

is applied arrows

the load. 1rus period is,

From Fig. 3.7.2 (b) we can write an equation for line voltage VRY as, V RY = J3Vm

sin(rot+ : )

Here Vm is the peak valve of phase voltage. RMS value is given as, 1

Vo(1- v~~> )r = We kn.o w that Vo(av)

L

= oV

5

or

Vo(av) Vs

1

= 6 and T =f

where f is the frequency of

buck regulator. Hence above equation can be written as, .

1L (,ripple)

=

6V5 (1-6)

Lf

... (4.8.7)

Observe that this equation is same as that of equation 4.8.1. Using Kirchhoff's law at the +ve point of load,

ii = ic +io The ripple in iO is very very small and hence it can be neglected. Hence from above equation, Ripple in iL = Ripple in (ic +i 0 ) 1L(ripple)

=

1C(ripple)

DC/DC eo·nvartars

4-52

Power Electronics - Ill

The average value of capacitor current during period

6T

2

+

(1 - 6) T

2

, i.e.

T . .

2

1S

(~ capacitor current of Fig. 4.8.4),

.

1C(ripple)

l c(av)

= =

Since

ic(ripple)

I c(av)

2x2 ic(ripple)

4

above equation becomes,

= i_L(ripple),

=

1L (ripple)

... (4.8.8)

4

The capacitor voltage is given as vc (t) = vc (t) - vc (t = O)

=

~ Jic dt + vc (t = (JJ

~ f ic

dt

Over the half cycle period, vc (t) - vc (t = 0) will give ripple voltage of capacitor. l T/ 2_ Renee, _vc(ripple) = . C 'c dt

J 0

Equation 4.8.8 gives average value of ic. Hence above equation becomes, Vc(ripple)

_!_ = C

f

T / 2·

l L(ripple ) dt

4

0 ·

= _!_ l L(ripple) C 4

f dt

T/2

0

= ..!_. i L(r ipple) • T C

4

2

Putting the value of i L(ripple) from equation 4.8.7, Vc(ripple)

= _!_ _6V5 (1-6). T C 4 X Lf 2

Since T =

l,

above equation becomes,

_ 6V5 (1-6) vc(ripple) -

BLC /2

.... (4.8.9)

This is the required expression for ripple voltage of capacitor. In Fig. 4.8.3, observe that capacitor voltage is same as output voltage. Hence above equation gives ripple voltage in the output.

DC/DC Converters

4-53

Power Etectronlc:9 - Ill

A buck regulator has a input voltage of 12 V,rei,uired output voltage of 5 V and peak ripple voltage is 20 mV. The switching frequency is 25 kHz. If the peak to peak ripple current of the inductor is limited to 0.8 A, determine : (2003, 8 Marks] i) Duty cycle ii) The filter inductance iii) The filter capacitance

, . . Example 4.8.2 :

Solution :

Given data

Vom>O sin(m1t / n) 7t /

e-x dx = l o

(C-35)

(C-36a)

0

l(a+l) = fa{(i)

where

J

X

2n

I (½)

✓7t

rm

=

I (n)

= (n-1) ! if n is an integer

-a:x2

e

(C-36b)

1;

=

dx _ 1 . 3 . 5 .... (2n-1)

0

- - - - -- 2n+1 a"

(C-36c)

l

•1 e-112:r.2-+-b,x d x = J'ne b2/(4ri2) ,a> 0

-j

a

e -u cos (bx) dx

=

j e-ar sin (bx) dx = 0

f• e-) 0

2 x

(C-37)

(C-38)

a>0

(C-39)

b , a>0 2 2 a +b

(C-40)

2

a

a +b

0

a

(C-36d)

2

,

✓7t e - b2 / V.

cos (bx) dx = - - - -, a> 0 2a

(C-41)

Important Laplace Transform Relations The Laplace transform of a function f(t) is given as, F(s) =

-fJ(t) e- dt st

0-

Sr.No.

Name

1.

Shifting theorem

2.

Differentiation theorem

3.

Convolution therom

5.

Complex translation

6.

Differentiation by s

8.

el [/(t - t0)] = e-sto/(s)

~ f(t)

,.f

dt

=

sF(s)- /(0- ) f (s)

Integration theorem

4.

7.

Property

.t fof(t)dt = ,.f

s

[/1 (t) • !2(t)] = f 1(s) •F2(s) .f [e ,.f

01

f(t) ] = F(s - a)

[t f(t)] =

J(o+) =

Initial value therom

,limf(t) ......

Final value therom

=

d

- - F(s) ds

lim f(t) =lim[sF(s)]

t-+0+

lim[sF(s)]

s-+0

Table 8.1 Properties of Laplace Transfonn

(B • 1)

s ➔-

Power Electronics -111 Sr. No

Name of the function

1.

Exponential function

Appendlx-B

B-2 Laplace transform pair

eat E

1

.1

) --

s-a

2.

Unit step function

Au(t) E .1 >A s

3.

Ramp function

tu(t) E .1

)

1

s2

4.

Impulse function

5.

Sine wave

A sinro0t E .1

)

Aw 0 s2 +wi

6.

Cosine wave

A cosro 0 t ~ ..f

~

A·s 2 s +w~

7.

Damped sine wave

8.

Damped cosine wave

e - ol COS (J)t E

9.

Hyperbolic sine wave

e - at

sin hbt E .f

)

10.

Hyperbolic cosine wave

e-at

coshbt E .1

)

B(t) E .1 )1

e -ot

(J)

sin rot E .1

)

.1

)

2 (s+a) +w 2

s+a (s +a)2 + ro2 b

2 (s+a) -b 2

s+a 2 (s+n) -b 2

Table B.2 Important Laplace Transfonn Pairs

B-3

Power Electronics - Ill Name 0£

Element in time

element

dom~

Appendix - 8

Impedance - voltage

Admittance current

source representation in source representation in

's' domain

's' domain

f

+

v(t)

-

l(s)

+

Vs)

•► 1

:•-

V(s)

•• R

Resistance Y(t): R ~I)

V(s) • R l(s)

1 l(s) • - .V(s) R

- V t ♦ l(s) I 11sl V(s)

Y(t)•~

di

'

V(s)

!-

Inductance

n ~IJ

Uc

Capacitance v(t) .

bl

I

~tJ di+ Y(O-)

~ •o--- - -....

+

V(s) = SL l(s)- U(O-)

l(s).,.

T~1-♦•c,,

l(s)

♦

SCV(s)

!V(s) ,.

- v(O--) ♦

s

1 (s)+ Y(O-) s

~

.!.. V(s) • ~ SL s

Vs)

l

t •CV(s)

==•C

f

C Y(O-)

l(s) ■ sC V(s)-C Y(O--)

Table B.3 Laplace Transformations of Basic Elements

□□□

(B-4)

Power Electronics • III ~ Chapterwlse University Questions with Answen

(P-1)

CD Q.1

Power Semiconductor Devices

Write short notes on following :

[PU : Dec.-2.004, 12 Marks]

i) Repeatitive and non-repeatitive ratings

of SCR.

ii) Fast recovery and schottky diodes. Ans. : i) Refer section 1.9. ii) Refer section 1.4.4. Q.2

Explain the two transistor analogy for an SCR and derive an expression for the anode current in terms of the current gains and leakage currents of the transistors. [PU : May-2000, 2002, 2003, 2006, 2008, 6 Marks ; May-1001, Dec.-2001, 2003, 2006, 8 Marks ; Dec.-2004, 2008, 6 Marks)

Ans. : Refer section 1.6.5. Q.3

Draw the forward and reverse characteristics latching and holding currents.

of

a triac. Clearly indicating the [PU: Dec.-2.003, 4 Marks]

Ans. : Refer section 1.10.4.1. Q.4

A SCR is connected in seria with a RL load and is fed from a 115 V , 60 Hz AC supply. The load resistance is 25 n and load inductance is 0.25 H. If a firing pulse of 60 µs is applied at a firing angle of 45°, what is the maximum permissible latching current of the SCR to ensure turn-on. (PU : May-2.003, 6 Marks]

Ans. : Fig. 1 shows the circuit diagram. The input voltage is,

¼

=

l(t)

vm sinoot

= lls.J2sin45° = 115

R=250

V5 = Vmslnd = 115 "2 sin 45° rv = 115 V ·

L

Thus the input voltage at the time of turning on is 115 V. Fig. 1 RL circuit

Current through RL circuit is given as, V i(t) =

. -tR R(1-e L)

115 25

-t~

= -(1-e 0.25)

= 4.6(1-e-tOOt) (P - 2)

..

= 0.25 H

Power Electronics - Ill

Power Semiconductor Devlen

P-3

Since the firing pulse is of 60 µsec, the current at the end of firing pulse is, i(t)

=

~

4.6(1 - e-lOO x(i()x lO

)

with t

= 60 µsec

= 27.5 mA

Thus the maximum latching current must be 27.5 m.A to ensure tum-on.

Write short notes on : fast recovery diodes.

Q.S

(PU: May-2002, 8 Marks]

Ans. : Refer section 1.4.4. What are different stress demands on power devices ? How we am achieve higher blocking voltage and higher dv/dt capability for SCRs.

Q.6

[PU : May-2002, 5 Marks; May-2008, 6 ~ ]

Ans. : Power devices have following stress demands :

i) They should have higher blocking voltages. ii) They should have very high switching frequencies. ill) They should have very small power dissipation.

iv) The dv / dt and di/dt capabilities should be very high. To achieve higher blocklng voltage

•

Higher blocking voltages can be obtained by lightly doping the n 1 layer and making it wide.

•

Connecting the SCRs in series, the blocking capacity can be further i n ~.

To achieve higher dv/dt capabllfty

•

The displacement current can be reduced with the help of cathode shorts.

•

Due to cathode shorts, the displacement current is diverted (intercepted) in the device. It 'doesnot flow across the gate - cathode junction.

Q.7

Write short notes on : Modes of operation of triac and triac applications along with diJU: as a tri~ dnJice. [PU : Dec.- ~ 8 Marks]

Ans. : Refer section 1.10.4.2 and section 1.14.7..2 gives diac as a trigger device for triac.

!

Q.8

Write short note on : Surgt current and

Ji 2t ratings of SCR. [PU : Dec.-2001, 8 Marks]

Ans. : Refer section 1.9. Q.9

What are the advantages of IGBT over pqwer MOSFET and pqwer B]T? [PU: Dec.-2001, 6 Marks]

Ans. :

Refer section 1.13.5.

Power Semiconductor Devices

Power Electronics - Ill

Q.10

Write short notes on : MCI'

[PU : May-2001, 5 Marks]

Ans. : Refer section 1.10.5. Q.11

Draw the vertical cross-section and 1-V characteristics of an IGBT. [PU : May-2001, 4 Marksl

Ans. : Refer sections 1.13.1 and 1.13.3. Q.12

Write a short note on IGBT as power device.

Ans. :

Refer section 1.13.

Q.13

Describe following ratings as applicable to SCR. i) Surge current rating

2

ii) i t rating

iii)

[PU : Dec.-2000, S Marks]

!!

rating iv) ~; rating.

[PU : Dec.-2000, 10 Marks; Dec.-2006, 2007, 8 Marks]

Ans. : Refer section 1.9. Q.14

Draw forward and reverse characteristics of SCR. Show Ii, I", V80 and V8 R on the characteristic. [PU : May-2000, 4 Marks]

Ans. :

Refer Fig. 1.6.1.

Q.15

Explain the shorted emitter structure to improve the dv/dt rating of SCRs. [PU : May-2004, 2005, 4 Marks]

Ans. :

Refer section 1.19.3.

Q.16

What do you understand by safe operating area (SOA) of a power semiconductor

device ? Draw the forward biased SOA of the IGBT and explain how it is superior to that of power B]T. [PU : Dec.-2004, 8 Marks; May-2006, 6 Marks] Ans. :

Refer section 1.13.2.

Q.17

Draw the vertical cross-section and input and output characteristics of an IGBT. [PU : Dec.-2004 8 Marks]

Ans. :

Refer sections 1.13.1 and 1.13.3.

Q.18

Draw the vertical cross-section of a power MOSFET and explain the following : i)

ii)

Reason for Nbcdy-source-short" in MOSFET structure. Presence of i-"1tegral reverse diode in the structure. [PU: May-2004, 8 Marks; Dec.-2oos,·10 Marks]

Ans. :

Refer section 1.12.1.

Q.19

Draw the forward and reverse characteristics of a mac clearly indicating the latching and holding currents. [PU : Dec.-2003, 4 Marks)

Ans. :

Refer section 1.10.4.1.

Power Electronics - Ill

Power Semiconductor Devices

P-5

Q.20

Draw the typical isolated gate drive circuit for a MOSFET and explain its operation. [PU : Dec.-2002, 8 Marks; May-2003, Dec.-2003, 6 Marks)

Ans. :

Refer section 1.17.

Q.21

What is the principal difference between the forward biased SOAs of i) IGBT and BJT ii) IGBT and power MOSFET. [PU : May-2003, 4 Marks]

Ana. :

Refer sections 1.11.S and 1.13.2.

Q.22

Draw the vertical cross-section and forward biased and reverse biased SOA of a IGBT. [PU : May-2003, 6 Marks)

Ans. :

Refer sections 1.13.1 and 1.13.2.

Q.23

Fast recov'ery and schottky diodes.

Ans. :

Refer sections 1.4.4.2 and 1.4.4.3.

Q.24

Draw the vertical cross-section and VI characteristics of IGBT. Also explain latchup in IGBT and how to avoid it ? [PU : Dec.-2006, 8 Marks]

[PU : May-2001, 5 Marks)

Ans. : Refer section 1.13.1 for vertical cross-section of IGBT. VI characteristics are given is section 1.13.3. Refer section 1.13.1.4 for latchup and how to avoid latchup in

IGBT. Q.25

Draw the vertical cross-section of an IGBT and explain the reason for the body-source-short in the IGBT structure ? [PU : May-2006, 6 Marks]

Ans. : Refer section 1.13.1 for vertical cross-section and section 1.13.1.5 for body-source-short reason. Q.26

Why is a high-frequency pulse train preferred for gating SCRs as compared to DC

triggering.

[PU : May-2006, 4 Marks]

Ans. : Refer section 1.8.1. Q.27

A SCR is connected in series with RL load and is fed from a 120 V, 60 Hz AC supply. The load resistance is 15 n and load is 0.75 ~ - What is the maximum allowable latching current of the SCR if the gate trigger circuit output pulse is of 100 µs duration at a delay angle of 45". [PU : May-2006, 8 Marks)

Ana. : Fig. 2 show the circuit diagram.

,.

i(t)

R = 150

= 120 {2 sin 45°

L = 0.75 H

= 120V

Fig. 2

Power Electronics - Ill

P-6

Power Semiconductor Devices

Current through RL circuit is given as,

R -tRJ L

i(t) = v ( 1-e

... = 16 mA.

Thus at end of 100 µs trigger pulse, SCR current will reach to 16 mA. Hence latching current must be at least 16 mA to keep the device in ON condition. Thus,

Q.28

IL = 16 mA. The Thevenin equivalent of an IGBT gate drive circuit is a DC source of 10 V in series with a resistance R. The IGBT parameters are Cgs = 100 pF, Cgd = 150 PF and V GS = 3 V. Calculate the Niue of R so that the turn-en delay, i.e. time taken for VGS to rise from zero to VGS(TH) is 5 ns. [PU : May-2006, 6 Marks]

Ans. : Fig. 3 shows the Thevenin equivalent circuit.

r:n T R

v, =10V

VGi)TC =c., 11 c,,

Fig. 3

Here 'C' is the parallel combination of Cgs and Cgd hence, c = cgs + cgd = 1000 pF + 150 pF = 1150 pF

The voltage across capacitor will be gate-source voltage. Hence V GS = v g (1 - e -t I RC ) Here we have to determine value of 'R' for t = 5 nsec, VGS = VGS(Th) = 3 V, C = 1150 pF and Vg = 10 V. Hence, 2 · l - e-Sxto-9 /(Rxlt50x!O-l )) 3 =

io(

0.3

...

0.7

R

= l - e-4.3478/ R = e-4.3478/ R = 1219 il='.a.2 .Q

Power Electronics - Ill

Q.29

P-7

Power Semiconductor Devices

The gate-cathode characteristic of a triac is given by Vg = 2+5 I,- A triggering pulse train with an amplitude of 10 V, on period of 10 µs is applied to the gate through a 10 n series resistor. Calculate ( i) Peck gate power (ii) Triggering frequency to obtain an werage gate power of 0.5 W. [PU : Dec-2006, 6 Marks]

Ans. : Refer example 1.10.1. Q.30

Give the constructional details of a SCR. Sketch its schematic diagram and the circuit symbol.

[PU : May-2007, 8 Marks]

Ana. : Re.fer section 1.5.1.

Describe the different modes of operation of a thyristor with the help of its staHc V-1 characteristics. [PU : May-2007, 4 Marks]

Q.31

I

Ans. : Refer section 1.6.

Explain in detail the following current ratings of SCR.

Q.32

i) I 2t rating

ii) di/dt rating.

[PU: May72007, 5 Marks]

Ans. : Refer answer of Q.13 (ii) and (iii). Q.33

Give the comparison between power diodes and thyristors. [PU: May-2007, 5 Marks]

Ans. : Following table lists the comparison between power diodes and thyristors. Sr. No.

Thyristors

Pow.r Diodes

1.

These devices are not controlled.

These devices are controlled by gate.

2.

These devices nonnally have one junction

These devices have more than one junction.

3.

These are used for rectification, freewheeling and feedback.

These are used for controlled rectification AC regulation, inversion and DC-DC conversion.

4.

These are used as protection for thyristors.

Thyrsltors are main power conversion devices.

5.

The diodes of this type are fast recovery diodes, achottky diodes, etc.

Q.34

SCR, TRIAC, GTO are the devic;ea of

thyristor family.

Draw vertical structure of power MOSFET. Explain its operation. Compare MOSFET with other power devices. [PU : Dec.-2007, 8 Marks]

Ans. : Refer section 1.12.1 and Table 1.13.1.

Q.35

Why MOSFET is used at high frequency applications ?

[PU : Dec.-2007, 4 Marks)

Ans. : MOSPETs are preferred at high frequency applications because

Power Electronics -111

P-8

Power Semlcronductor Devices

i) Switching times of MOSFETs are very small. Hence they tum-off and turn-on fast. This makes them suitable to operate at high frequencies. ii) MOSFETs have insulated gate. Hence driver circuits for MOSFETs are very

simple. They can be easily designed for high frequency operation.

Justify-parallel operation of MOSFETS can be done more easily as compare to thyristors. [PU : Dec.-2007, 4 Marks]

Q.36

Ans. : True • MOSFETs have positive temperature coefficient, hence their paralleling is easy. H the current increasing in a particular MOSFET; then due to losses its temperature will rise. This rise in temperature will increase internal resistance of the

MOSFET. Due to increased internal resistance, the current through the MOSFET will reduce. Thus the current is balanced due to positive temperature coefficient. Parallel operation of SCRs is difficult as compared to MOSFETs since, SCRs have negative coefficient of temperature. This doesnot help the internal adjustment of equal current through SCRs. In fact, external equalizing components are required for paralleling of SCR.s.

Draw the vertical cross-section of a power MOSFET and explain the following :

Q.37

Reason for Hbody-source-short" in MOSFET structure ii) Presence of integral reverse diode in the structure [PU : May-2.008, 10 Marks] iii) SOA of power MOSFET. i)

Ans. : For (i) and (ii) refer answer of Q.18.

iii) SOA of power MOSFET : Pig. 8 shows the SOA of power MOSFET.

Three MOSFET,

factors

decide

the

SOA

of

power

i) Maximum drain current (I 0 M) ii) Internal junction temperature (T;max) iii) Breakdown voltage (BV 055 )

BVoss

log Vos

Fig. 4 SOA of power MOSFET

BJT.

•

The MOSFETs doesnot have any second break

= 21tf = 21t x 60 = 377 rad/sec.

a = wRcC /n(1 ~ 11) For a = 20° i.e. 0.349 radians, 0.349 = 377 x Re x 0.1 x 10-6 In( l

-1.t:>3)

• Re = 9312.5 n For a = 1600 i.e. 2.792 radians, 2.792 = 377 x Re x 0.1 x 10-6 In(

1

-1.

63

)

RC = 74.486 lci2

Q.3

Write short note on snubber circuits.

Ans. :

Refer section 2.7.1.

Q.4

Draw the circuit diagram of a line synchronized ramp and pedestal UJT triggering circuit for SCRs and explain its operation with the help of relevant waveforms.

[PU : Dec.-2003, 2004, 6 Marks]

[PU : Dec.-2003, 2006, 8 Marks; Dec.-2008, 10 Marks]

Ans. : Refer section 2.1.5. Q.5

What are the limitations of RC triggering for an SCR ? with the help of circuit diagram and waveforms explain U]T trigger method for controlling firing angle of an SCR. Enumerate its advantages as high frequency carrier gating. [PU : May-2002, 10 Marks]

Ans. : Limitations of RC circuit ., 1. The firing angle depends upon RC time constant. The values of Rand C may change due to temperature.

2. Firing angle can change because of supply voltage fluctuations. 3. The gate firing circuit do not have any isolation from power circuit.

4. The RC firing circuit is not suitable for feedback control applications. Refer section 1.14.5 for UJT triggering circuit. Q.6

A relaxation oscillator using U]T is to be used for triggering on SCR. The UJT has following data : 11 = 0.63, IP = 1 mA, Vv :::: 2 V, Iv :::: 4 mA,. R BB = 8 k{2, supply voltage :::: 30 V, normal leakage current with emitter open. = 2 mA. Assuming VO = 0 V for C = 0.47 µF. Calculate the tJalues af range of R, R 1 and R 2 for changing firing angle from 20° to 160° [PU : May-2002, 8 Marks]

Ans. : This example is similar to that of Q.2. Refer the same for obtaining the answer.

P - 11 Drive and Protection Circuits for Pow.r Devices ·

Power Electronics - Ill

Write short notes on : Cooling methods of power deuices.[PU : Dec.-2001, 8 Marks]

Q.7

Ans. : Refer section 211.

With the help of circuit diagram and relevant wavefonns explain UJT trigger ramp and pedestal contro.l method of SCR. Give its advantages.

Q.8

[PU : D~c.-2001, 10 Marks; May-2001, 8 Marks)

Ans. : Refer section 2.1.5. Calculate ,the values of snubber components R and C in Fig. 1 to protect SCR from reapplied dv/dt, if dv/dt rating of SCR is 100 V,1.lsec. [PU : May-2001, S Marks; Dec.-2008, 6 Marks]

Q.9

250

0.1 mH

500V

+ C

Fig. 1

Ans. : Given : L

vm dv dt C

R

=

0.1 mH

= 500 V

100

= 100 V/µsec = ~ V/ µsec 10

=

_!_[0564 Vm] 2L

dv

2

I dt

r

=

1 [ 0.564 X 50() X lQ-6 2 X 0.1 X 10-3 l()()

= =

0.04 µF

2ai 0.1 x10-3

= 2 x 0.65 - - --

0.04 xl0-6

assuming a = 0.65.

= 65 .Q Out of this, 25, .Q is already present in the circuit. Hence R

= 64 -

25

= ~ n.

With the help of circuit diagram and relevant waveforms explain ramp and pedesteal triggering for an SCR. Give its advantages. [PU : Dec.-2000, 10 Marks] Ans. : Refer section 2.1.5, i.e. UJT triggering. Q.10

P-14

Power Electronics -111

Single and Three Phase AC/DC Converters 1

vo(rms)

= vm [1-a+ sin2a]2 2

27t

7t

1

=

7t/6

311.127

sin(

2x:)

2

1--+---'---~

2

7t

27t

= 153.3 V ro(av)

=

Io(rms)

=

= 9220.4 = 4.62 A

vo(av)

R

Vo(rms) = 153.3 = 7 665 A

R

20

.

Po(av) = Vo(av) XI c(av) = 92.4 X 4.62 = 426.888 W

••

Po(rms) = Vc(rms) xio(rms) = 153.3x 7.665 = 1175.045 W i) Rectification efficiency 1l

=

pc(av) Po(rms)

= 426.888 = 0.363

=

Vo(rms)

= 153.3 = 1_66

or 36.3 %

1175.045

ii) Form factor

FF

Vo(av)

92.4

ili) Transformer utilization factor PG(av}

11JF = Transformer VA Po(av)

Po(cro)

•

= Vs(nns) xls(rms) = Vs xi c(rms) since l s(nns) = lo