UNIT I Power Semiconductor Devices -...

UNIT I

Power Semiconductor

Devices

www.Vidyarthiplus.com


Fatima Michael College of Engineering & Technology



Introduction

• What are Power Semiconductor Devices (PSD)?

They are devices used as switches or rectifiers in

power electronic circuits

• What is the difference of PSD and low-power

semiconductor device?

Large voltage in the off state

High current capability in the on state






Classification

Fig. 1. The power semiconductor devices family






Important Parameters

• Breakdown voltage.

• On-resistance.

Trade-off between breakdown voltage and

on-resistance.

• Rise and fall times for switching between on

and off states.

• Safe-operating area.






Power MOSFET: Structure

Power MOSFET has much higher current handling capability in

ampere range and drain to source blocking voltage(50-100V)

than other MOSFETs.

Fig.2.Repetitive pattern of the cells

structure in power MOSFET






Power MOSFET: R-V Characteristics

An important parameter of a power MOSFET is on resistance:

, where on S CH D

R R R R= + +( )

CH

n ox GS T

LR

W C V Vµ=

−

Fig. 3. Typical RDS versus ID characteristics of a MOSFET.






Thyristor: Structure • Thyristor is a general class of a four-layer pnpn

semiconducting device.

Fig.4 (a) The basic four-layer pnpn structure.

(b) Two two-transistor equivalent circuit.






Three States:

Reverse Blocking

Forward Blocking

Forward Conducting

Thyristor: I-V Characteristics

Fig.5 The current-voltage

characteristics of the pnpn device.






Applications

Power semiconductor devices have widespread applications:

Automotive

Alternator, Regulator, Ignition, stereo tape

Entertainment

Power supplies, stereo, radio and television

Appliance

Drill motors, Blenders, Mixers, Air conditioners and Heaters






Thyristors

• Most important type of power semiconductor device.

• Have the highest power handling capability.they have a rating of 1200V / 1500A with switching frequencies ranging from 1KHz to 20KHz.






• Is inherently a slow switching device compared to BJT or MOSFET.

• Used as a latching switch that can be turned on by the control terminal but cannot be turned off by the gate.






Different types of Thyristors

• Silicon Controlled Rectifier (SCR).

• TRIAC.

• DIAC.

• Gate Turn-Off Thyristor (GTO).






SCR

Symbol of

Silicon Controlled Rectifier






Structure

µ

µ

µ

µ






Device Operation

Simplified model of a

thyristor






V-I

Characteristics






Effects of gate current






Two Transistor Model of SCR

⇒






( ) ( )

1 1

1 1

1 1

1

1

Considering PNP transistor

of the equivalent circuit,

, , ,

,

1 1

E A C C

CBO CBO B B

B A CBO

I I I I

I I I I

I I I

α α

α

= = =

= =

∴ = − − − − −






( ) ( )

2 2 2

2 2

2 2

2

2

Considering NPN transistor

of the equivalent circuit,

, ,

2

C C B B E K A G

C k CBO

C A G CBO

I I I I I I I I

I I I

I I I I

α

α

= = = = +

= +

= + + − − −






( )

2 1

2 1 2

1 2

From the equivalent circuit,

we see that

1

C B

g CBO CBO

A

I I

I I II

α

α α

∴ =

+ +⇒ =

− +






( )1 2

1 2

Case 1: When 0

1

g

CBO CBO

A

I

I II

α α

=

+=

− +

( )2 1 2

1 2

Case 2: When 0

1

G

g CBO CBO

A

I

I I II

α

α α

≠

+ +=

− +






Turn-on

Characteristics

o n d rt t t= +






Turn-off

Characteristi

c

!!

"

#$% #

"

& '

(

")$ **

) **






Methods of Thyristor Turn-on

• Thermal Turn-on.

• Light.

• High Voltage.

• Gate Current.

• dv/dt.






Thyristor Types

• Phase-control Thyristors (SCR’s).

• Fast-switching Thyristors (SCR’s).

• Gate-turn-off Thyristors (GTOs).

• Bidirectional triode Thyristors (TRIACs).

• Reverse-conducting Thyristors (RCTs).






• Static induction Thyristors (SITHs).

• Light-activated silicon-controlled rectifiers (LASCRs).

• FET controlled Thyristors (FET-CTHs).

• MOS controlled Thyristors (MCTs).






Phase Control Thyristor • These are converter thyristors.

• The turn-off time tq is in the order of 50 to

100µsec.

• Used for low switching frequency.

• Commutation is natural commutation

• On state voltage drop is 1.15V for a 600V device.






• They use amplifying gate thyristor.






Fast Switching

Thyristors • Also called inverter thyristors.

• Used for high speed switching applications.

• Turn-off time tq in the range of 5 to 50µsec.

• On-state voltage drop of typically 1.7V for

2200A, 1800V thyristor.

• High dv/dt and high di/dt rating.






Bidirectional Triode

Thyristors (TRIAC)






Mode-I

Operation

MT2 Positive,

Gate Positive

P1

N1

N2

P2Ig

Ig

MT2 (+)

MT1 ( )−G

V(+)






Mode-II

Operation

MT2 Positive,

Gate Negative

P1

N1

N2N3

P2

Ig

MT2 (+)

MT1 ( )−G

V

Finalconduction

Initialconduction






Mode-III Operation

MT2 Negative,

Gate Positive

P1

N1

N4

N2

P2

Ig

MT2 ( )−

MT1 (+)G(+)






Mode-IV Operation

MT2 Negative,

Gate Negative

P1

N1

N4

P2

Ig

MT2 ( )−

MT1 (+)

N3

G(-)






Triac Characteristics






BJT structure

note: this is a current of electrons (npn case) and so the conventional current flows from collector to emitter.

heavily doped ~ 10^15 provides the carriers

lightly doped ~ 10^8 lightly doped ~ 10^6






BJT characteristics






BJT modes of operation

Mode

EBJ

CBJ

Cutoff Reverse Reverse

Forward

active

Forward Reverse

Reverse

active

Reverse Forward

Saturation Forward Forward






Cutoff: In cutoff, both junctions reverse biased. There is very little current flow, which corresponds to a logical "off", or an open switch.

Forward-active (or simply, active): The emitter-base junction is forward biased and the base-collector junction is reverse biased. Most bipolar transistors are designed to afford the greatest common-emitter current gain, βf in forward-active mode. If this is the case, the collector-emitter current is approximately proportional to the base current, but many times larger, for small base current variations. Reverse-active (or inverse-active or inverted): By reversing the biasing conditions of the forward-active region, a bipolar transistor goes into reverse-active mode. In this mode, the emitter and collector regions switch roles. Since most BJTs are designed to maximise current gain in forward-active mode, the βf in inverted mode is several times smaller. This transistor mode is seldom used. The reverse bias breakdown voltage to the base may be an order of magnitude lower in this region. Saturation: With both junctions forward-biased, a BJT is in saturation mode and facilitates current conduction from the emitter to the collector. This mode corresponds to a logical "on", or a closed switch.

BJT modes of operation






BJT structure (active)

current of electrons for npn transistor – conventional current flows from collector to emitter.

B

C E

IE IC

IB

-

+

VBE VCB

-

+

+ - VCE






• A GATE electrode is placed above (electrically insulated

from) the silicon surface, and is used to control the

resistance between the SOURCE and DRAIN regions

• NMOS: N-channel Metal Oxide Semiconductor

L

• L = channel length

“Metal” (heavily

doped poly-Si)

W • W = channel width

MOSFET

SOURCE

DRAIN

GATE






• Without a gate-to-source voltage applied, no current can

flow between the source and drain regions.

• Above a certain gate-to-source voltage (threshold

voltage VT), a conducting layer of mobile electrons is

formed at the Si surface beneath the oxide. These

electrons can carry current between the source and drain.

N-channel MOSFET

n

p

oxide insulator gate

n

Drain Source

Gate

ID

IG

IS






N-channel vs. P-channel

MOSFETs

• For current to flow, VGS > VT

• Enhancement mode: VT > 0

• Depletion mode: VT < 0

– Transistor is ON when VG=0V

p-type Si

n+ poly-Si

n-type Si

p+ poly-Si

NMOS PMOS

n+ n+ p+ p+

• For current to flow, VGS < VT

• Enhancement mode: VT < 0

• Depletion mode: VT > 0

– Transistor is ON when VG=0V

(“n+” denotes very heavily doped n-type material; “p+” denotes very heavily doped p-type material)






MOSFET Circuit Symbols

p-type Si

n+ poly-Si

NMOS

n+ n+

n-type Si

p+ poly-Si

PMOS

p+ p+

G G

G G

S

S S

S

Body

Body






• The voltage applied to the GATE terminal determines whether current can flow between the SOURCE & DRAIN terminals.

– For an n-channel MOSFET, the SOURCE is biased at a lower potential (often 0 V) than the DRAIN

(Electrons flow from SOURCE to DRAIN when VG > VT)

– For a p-channel MOSFET, the SOURCE is biased at a higher potential (often the supply voltage VDD) than the DRAIN

(Holes flow from SOURCE to DRAIN when VG < VT )

• The BODY terminal is usually connected to a fixed potential.

– For an n-channel MOSFET, the BODY is connected to 0 V

– For a p-channel MOSFET, the BODY is connected to VDD

MOSFET Terminals






VGS

S

semiconductor oxide

G

VDS

+ − + −

D

always zero!

IG

VGS

The gate is insulated from the semiconductor, so there is no significant steady gate current.

IG

NMOSFET IG vs. VGS Characteristic

Consider the current IG (flowing into G) versus VGS :






VGS

S

semiconductor oxide

G

VDS

ID

+ − + −

D

ID

zero if VGS < VT

VDS

Next consider ID (flowing into D) versus VDS, as VGS is varied:

Below “threshold” (VGS < VT): no charge no conduction

Above threshold (VGS > VT): “inversion layer” of electrons appears, so conduction between S and D is possible

VGS > VT

NMOSFET ID vs. VDS Characteristics






The MOSFET as a Controlled Resistor

• The MOSFET behaves as a resistor when VDS is low:

– Drain current ID increases linearly with VDS

– Resistance RDS between SOURCE & DRAIN depends on VGS

• RDS is lowered as VGS increases above VT

NMOSFET Example:

ID

IDS = 0 if VGS < VT

VDS

VGS = 1 V > VT

VGS = 2 V

Inversion charge density Qi(x) = -Cox[VGS-VT-V(x)] where Cox ≡≡≡≡ εεεεox / tox

oxide thickness ≡ tox






ID vs. VDS Characteristics

The MOSFET ID-VDS curve consists of two regions:

1) Resistive or “Triode” Region: 0 < VDS < VGS −−−− VT

2) Saturation Region:

VDS > VGS −−−− VT

( )

oxnn

TGSn

DSAT

Ck

VVL

WkI

µ=′

−′

=

where

2

2

oxnn

DSDS

TGSnD

Ck

VV

VVL

WkI

µ=′

−−′=

where

2

process transconductance parameter

“CUTOFF” region: VG < VT





Part I: Bipolar Power Transistors

The Evolution Of IGBT

• Bipolar Power Transistor Uses Vertical Structure For Maximizing Cross Sectional Area Rather Than Using Planar Structure





Part II: Power MOSFET


• Power MOSFET Uses Vertical Channel Structure Versus The Lateral Channel Devices Used In IC Technology






Lateral MOSFET structure






• Discrete BJT + Discrete Power MOSFET In Darlington Configuration

Part III: BJT(discrete) + Power MOSFET(discrete)





Part IV: BJT(physics) + Power MOSFET(physics) = IGBT


• More Powerful And Innovative Approach Is To Combine Physics Of BJT With The Physics Of MOSFET Within Same Semiconductor Region

• This Approach Is Also Termed Functional Integration Of MOS And Bipolar Physics

• Using This Concept, The Insulated Gate Bipolar Transistor (IGBT) Emerged

• Superior On-State Characteristics, Reasonable Switching Speed And Excellent Safe Operating Area






• IGBT Fabricated Using Vertical Channels (Similar To Both The Power BJT And MOSFET)

Part IV: BJT(physics) + Power MOSFET(physics) = IGBT





Device Operation • Operation Of IGBT Can Be Considered Like A PNP Transistor With Base Drive Current Supplied By The MOSFET






DRIVER CIRCUIT (BASE / GATE)

• Interface between control (low power electronics) and (high power) switch.

• Functions:

– amplifies control signal to a level required to drive power switch

– provides electrical isolation between power switch and logic level

• Complexity of driver varies markedly among switches. MOSFET/IGBT drivers are simple but GTO drivers are very complicated and expensive.






ELECTRICAL ISOLATION FOR DRIVERS

• Isolation is required to prevent damages on the high power switch to propagate back to low power electronics.

• Normally opto-coupler (shown below) or high frequency magnetic materials (as shown in the thyristor case) are used.






ELECTRICAL ISOLATION FOR DRIVERS

• Power semiconductor devices can be categorized into 3

types based on their control input requirements:

a) Current-driven devices – BJTs, MDs, GTOs

b) Voltage-driven devices – MOSFETs, IGBTs, MCTs

c) Pulse-driven devices – SCRs, TRIACs






CURRENT DRIVEN DEVICES (BJT)

• Power BJT devices have low current gain due to constructional consideration, leading current than would normally be expected for a given load or collector current.

• The main problem with this circuit is the slow turn-off time. Many standard driver chips have built-in isolation. For example TLP 250 from Toshiba, HP 3150 from Hewlett-Packard uses opto-coupling isolation.






ELECTRICALLY ISOLATED DRIVE CIRCUITS






EXAMPLE: SIMPLE MOSFET GATE DRIVER

• Note: MOSFET requires VGS =+15V for turn on and 0V to turn off. LM311 is a simple amp with open collector output Q1.

• When B1 is high, Q1 conducts. VGS is pulled to ground. MOSFET is off.

• When B1 is low, Q1 will be off. VGS is pulled to VGG. If VGG is set to +15V, the MOSFET turns on.





UNIT II

PHASE CONTROLLED

CONVERTERS




Phase-Control Converters

Single-Phase

Semiconverter

Three-Phase

Full converter

Dual converter

Semiconverter

Full converter

Dual converter




Semiconverter

..is a one-quadrant converter and it has one polarity

Full converter

..is a two-quadrant converter and the polarity of its

output can be either positive or negative.

However

the output current of full converter has

one polarity only

Dual converter ..can operate in four quadrants ; both the output

voltage and current can be either positive or negative




cos12

sin2

1 mmdc

VttdVV

Average Output Voltage

m

dm

VV Maximum

Output Voltage

cos15.0 dm

dcn

V

VVNormalizing

Output Voltage

2

2sin1

2sin

2

1 22

mmrms

VttdVV

RMS Output Voltage




If the converter has a purely resistive load of R and

the delay angle is , determine (a) the rectification efficiency

(b) the form factor FF (c) the ripple factor RF

and (d) the peak inverse voltage PIV of thyristor T1

2/




%27.203536.0

1592.0

3536.02

22

sin2

1

2

1592.0

2cos1

2sin

2

1

2

2

2

2

2

m

m

rms

dc

mm

rms

mdc

mmdc

V

V

V

V

VV

V

VV

VttdVV






(b) the form factor FF

(c) the ripple factor RF and (d) the peak inverse voltage PIV of thyristor T1

2/

221.21592.0

3536.0

m

m

dc

rms

V

V

V

VFF






(b) the form factor FF

(c) the ripple factor RF

and (d) the peak inverse voltage PIV of thyristor T1

2/

983.11221.21 22 FFRF

mVPIV



www.Vidyarthiplus.com Semiconverter




Single-Phase Semiconverter

ttdVV

VttdVV

mrms

mmdc

22 sin2

2

cos1sin2

2




Single-Phase Semiconverter (RL-load)

L

R

L

R

LLL

LL

eR

EeItiI

ERidt

diL

1

0

011

11

tL

R

SL

SL

SLL

eZ

V

R

EI

R

Et

Z

VI

tVERidt

diL

sin2

sin2

sin2

12

22

Mode 1

t0

Mode 2

tR

L 1tan 22LRZ




Single-Phase Semiconverter (RL-load)

RMS Current

for Thyristor

tdiI LR

2

22

1

RMS Current

for Thyristor

tdiI LA 2

2

1

RMS Output

Current

tditdiI LLrms

2

20

2

12

1

2

1

AVG Output

Current

tditdiIdc 2

01

2

1

2

1




The single-phase semiconverter has an RL load of

L = 6.5mH, R = 2.5 Ohm, and E = 10 V. The input

voltage is VS = 120 V(rms) at 60 Hz. Determine

(a) the load current IL0 at , and the load

current IL1 at , (b) the average thyristor current IA

(c) the rms thyristor current IR (d) the rms output current Irms

and (e) the average output current Idc

0t 60t




Single-Phase

Full Converter

Rectification

Mode

Inversion

Mode




Single-Phase Full Converter

2

sin2

2

cos2

sin2

2

22 mmrms

mmdc

VtdtVV

VtdtVV




Single-Phase Full Converter (RL-load)

tL

R

SL

SL e

Z

V

R

EI

R

Et

Z

VI

sin2

sin2

0

Mode 1 = Mode 2 R

L 1tan 22LRZ




Single-Phase Full Converter (RL-load)

RMS Current

for Thyristor

tdiI LR

2

2

1

RMS Current

for Thyristor

tdiI LA

2

1

RMS Output

Current RRRrms IIII 222

AVG Output

Current AAAdc IIII 2




Dual Converter




Single-Phase Dual Converter

High-Power Variable-Speed Drives

21

22

11

cos2

cos2

dcdc

mdc

mdc

VV

VV

VV




Three-Phase Semiconverter




3 Phase Controlled Rectifiers

• Operate from 3 phase ac supply voltage.

• They provide higher dc output voltage.

• Higher dc output power.

• Higher output voltage ripple frequency.

• Filtering requirements are simplified for

smoothing out load voltage and load current.




• Extensively used in high power variable speed industrial dc drives.

• Three single phase half-wave converters

can be connected together to form a three phase half-wave converter.




3-Phase

Half Wave Converter (3-Pulse Converter)

with RL Load

Continuous & Constant Load Current Operation




Vector Diagram of 3 Phase Supply Voltages

VAN

VCN

VBN

1200

1200

1200 RN AN

YN BN

BN CN

v v

v v

v v




3 Phase Supply Voltage

Equations

We deifine three line to neutral voltages

(3 phase voltages) as follows




0

0

0

sin ;

Max. Phase Voltage

2sin

3

sin 120

2sin

3

sin 120

sin 240

RN an m

m

YN bn m

m

BN cn m

m

m

v v V t

V

v v V t

V t

v v V t

V t

V t




van vbn vcn van




io=Ia

Constant Load

Current

Ia

Ia

Each thyristor conducts for 2/3 (1200)




To Derive an

Expression for the

Average Output Voltage of a

3-Phase Half Wave Converter

with RL Load

for Continuous Load Current




0

1

0

2

0

3

0

306

5 150

6

7 270

6

2Each thytistor conducts for 120 or radians

3

T is triggered at t

T is triggered at t

T is triggered at t




5

6

6

5

6

6

3sin .

2

3cos

2

3 5cos cos

2 6 6

mdc

mdc

mdc

VV t d t

VV t

VV




0 0

0

Note from the trigonometric relationship

cos cos .cos sin .sin

5 5cos cos sin sin

6 63

2co

cos 150 cos sin 150 sin3

2 cos 30

s .cos sin sin6 6

.cos

mdc

mdc

A

VV

B A B A B

VV

0sin 30 sin




0 0

0 0 0 0

0 0

0 0

0

0

0

0

0 0

Note: cos 1

cos 180 30 cos sin 180 30 sin3

2 cos 30 .cos sin 30 sin

cos 30 cos sin 30 sin3

2 cos 30 .cos sin 30 s

80 30 cos 30

sin 180 30 sin 30

in

mdc

mdc

VV

VV




032cos 30 cos

2

3 32 cos

2 2

3 3 33 cos cos

2 2

3cos

2

Where 3 Max. line to line supply voltage

mdc

mdc

m mdc

Lmdc

Lm m

VV

VV

V VV

VV

V V




max

The maximum average or dc output voltage is

obtained at a delay angle 0 and is given by

3 3

2

Where is the peak phase voltage.

And the normalized average output voltage is

mdmdc

m

ddcn n

VV V

V

VV V

cosc

dmV




15 26

2 2

6

1

2

The rms value of output voltage is found by

using the equation

3sin .

2

and we obtain

1 33 cos 2

6 8

mO RMS

mO RMS

V V t d t

V V




3 Phase Half Wave

Controlled Rectifier Output

Voltage Waveforms For RL

Load

at

Different Trigger Angles




0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

Van

V0

V0

Van

=300

=600

Vbn

Vbn

Vcn

Vcn

t

t

=300

=600




030

060

090

0120

0150

0180

0210

0240

0270

0300

0330

0360

0390

0420

0

V0

Van

=900

Vbn Vcn

t

=900




3 Phase Half Wave Controlled Rectifier With

R Load and

RL Load with FWD




a a

b b

c c

R

V0

L

R V0

+

T1

T2

T3

n n

T1

T2

T3




3 Phase Half Wave

Controlled Rectifier Output

Voltage Waveforms For R Load

or RL Load with FWD

at

Different Trigger Angles




0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

Vs

V0

Van

=0

=150

Vbn Vcn

t

VanVbn Vcn

t

=00

=150




0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

V0

=300

VanVbn Vcn

t

V0

=600

VanVbn Vcn

t

=300

=600




To Derive An

Expression For The Average Or Dc Output Voltage Of A

3 Phase Half Wave Converter With Resistive Load

Or RL Load With FWD




0

1

0 0

1

0

2

0 0

2

0

306

30 180 ;

sin

5 150

6

150 300 ;

sin 120

O an m

O bn m

T is triggered at t

T conducts from to

v v V t

T is triggered at t

T conducts from to

v v V t




0

3

0 0

3

0

0

7 270

6

270 420 ;

sin 240

sin 120

O cn m

m

T is triggered at t

T conducts from to

v v V t

V t




0

0

0

0

0

0

180

30

0 0

180

30

180

30

3.

2

sin ; for 30 to 180

3sin .

2

3sin .

2

dc O

O an m

dc m

mdc

V v d t

v v V t t

V V t d t

VV t d t




0

0

180

30

0 0

0

0

3cos

2

3cos180 cos 30

2

cos180 1, we get

31 cos 30

2

mdc

mdc

mdc

VV t

VV

VV




Three Phase Semiconverters

• 3 Phase semiconverters are used in Industrial dc drive applications upto 120kW

power output.

• Single quadrant operation is possible.

• Power factor decreases as the delay angle

increases.

• Power factor is better than that of 3 phase

half wave converter.




3 Phase

Half Controlled Bridge Converter

(Semi Converter)

with Highly Inductive Load &

Continuous Ripple free Load

Current




Wave forms of 3 Phase

Semiconverter for

> 600




0 0

1

3 phase semiconverter output ripple frequency of

output voltage is 3

The delay angle can be varied from 0 to

During the period

30 210

7, thyristor T is forward biased

6 6

Sf

t

t




1

1 1

If thyristor is triggered at ,6

& conduct together and the line to line voltage

appears across the load.

7At , becomes negative & FWD conducts.

6

The load current contin

ac

ac m

T t

T D

v

t v D

1 1

ues to flow through FWD ;

and are turned off.

mD

T D




1

2

1 2

If FWD is not used the would continue to

conduct until the thyristor is triggered at

5, and Free wheeling action would

6

be accomplished through & .

If the delay angle , e3

mD T

T

t

T D

ach thyristor conducts

2for and the FWD does not conduct.

3mD




0

0

0

We deifine three line neutral voltages


sin ; Max. Phase Voltage

2sin sin 120

3

2sin sin 120

3

sin 240

RN an m m

YN bn m m

BN cn m m

m

v v V t V

v v V t V t

v v V t V t

V t

V

is the peak phase voltage of a wye-connected source.m




3 sin6

53 sin

6

3 sin2

3 sin6

RB ac an cn m

YR ba bn an m

BY cb cn bn m

RY ab an bn m

v v v v V t

v v v v V t

v v v v V t

v v v v V t




Wave forms of 3 Phase

Semiconverter for

600




To derive an

Expression for the


of 3 Phase Semiconverter

for > / 3 and Discontinuous Output Voltage




76

6

76

6

For and discontinuous output voltage:3

the Average output voltage is found from

3.

2

33 sin

2 6

dc ac

dc m

V v d t

V V t d t




max

3 31 cos

2

31 cos

2

3 Max. value of line-to-line supply voltage

The maximum average output voltage that occurs at

a delay angle of 0 is

3 3

mdc

mLdc

mL m

mdmdc

VV

VV

V V

VV V




17 2

62

6

The normalized average output voltage is

0.5 1 cos

The rms output voltage is found from

3.

2

dcn

dm

acO rms

VV

V

V v d t




Three Phase Dual Converters

• For four quadrant operation in many industrial

variable speed dc drives , 3 phase dual

converters are used.

• Used for applications up to 2 mega watt output

power level.

• Dual converter consists of two 3 phase full

converters which are connected in parallel & in

opposite directions across a common load.




Outputs of Converters 1 & 2

• During the interval (/6 + 1) to (/2 + 1),

the line to line voltage vab appears across

the output of converter 1 and vbc appears

across the output of converter 2




0

0

0

We deifine three line neutral voltages


sin ;

Max. Phase Voltage

2sin sin 120

3

2sin sin 120

3

sin 240

RN an m

m

YN bn m m

BN cn m m

m

v v V t

V

v v V t V t

v v V t V t

V t




To obtain an Expression for the Circulating

Current

If vO1 and vO2 are the output voltages of

converters 1 and 2 respectively, the

instantaneous voltage across the current

limiting inductor during the interval

(/6 + 1) t (/2 + 1) is given by




1 2

3 sin sin6 2

3 cos6

The circulating current can be calculated by

using the equation

r O O ab bc

r m

r m

v v v v v

v V t t

v V t




1

1

6

6

1

max

1.

13 cos .

6

3sin sin

6

3

t

r r

r

t

r m

r

mr

r

m

r

r

i t v d tL

i t V t d tL

Vi t t

L

Vi

L




Four Quadrant Operation




• There are two different modes of

operation.

Circulating current free

(non circulating) mode of operation

Circulating current mode of operation




Non Circulating Current Mode Of Operation

• In this mode of operation only one converter is

switched on at a time

• When the converter 1 is switched on,

For 1 < 900 the converter 1 operates in the

Rectification mode

Vdc is positive, Idc is positive and hence the

average load power Pdc is positive.

• Power flows from ac source to the load




• When the converter 1 is on,

For 1 > 900 the converter 1 operates in the Inversion mode

Vdc is negative, Idc is positive and the average load power Pdc is negative.

• Power flows from load circuit to ac source.





For 2 < 900 the converter 2 operates in

the Rectification mode

Vdc is negative, Idc is negative and the

average load power Pdc is positive.

• The output load voltage & load current

reverse when converter 2 is on.

• Power flows from ac source to the load





For 2 > 900 the converter 2 operates in the

Inversion mode

Vdc is positive, Idc is negative and the average

load power Pdc is negative.

• Power flows from load to the ac source.

• Energy is supplied from the load circuit to the ac

supply.




• Both the converters are switched on at the same

time.

• One converter operates in the rectification mode

while the other operates in the inversion mode.

• Trigger angles 1 & 2 are adjusted such that

(1 + 2) = 1800

Circulating Current

Mode Of Operation




When 1 < 900, converter 1 operates

as a controlled rectifier. 2 is made greater than 900 and converter 2 operates as an Inverter.

• Vdc is positive & Idc is positive and Pdc is positive.




• When 2 < 900, converter 2 operates

as a controlled rectifier. 1 is made greater than 900 and converter 1 operates as an Inverter.

• Vdc is negative & Idc is negative and Pdc is positive.



UNIT III

DC Choppers






Introduction • Chopper is a static device.

• A variable dc voltage is obtained from a constant dc voltage source.

• Also known as dc-to-dc converter.

• Widely used for motor control.

• Also used in regenerative braking.

• Thyristor converter offers greater efficiency, faster response, lower maintenance, smaller size and smooth control.






Choppers are of Two Types

• Step-down choppers.

• Step-up choppers.

• In step down chopper output voltage is less than input voltage.

• In step up chopper output voltage is more than input voltage.






Principle Of

Step-down Chopper

V

i0

V0

Chopper

R

+

−






• A step-down chopper with resistive load.

• The thyristor in the circuit acts as a switch.

• When thyristor is ON, supply voltage appears across the load

• When thyristor is OFF, the voltage across the load will be zero.






Vdc

v0

V

V/R

i0

Idc

t

t

tON

T

tOFF






verage value of output or load voltage.

verage value of output or load current.

Time interval for which SCR conducts.

Time interval for which SCR is OFF.

Period of switching

dc

dc

ON

OFF

ON OFF

V A

I A

t

t

T t t

=

=

=

=

= + = or chopping period.

1 Freq. of chopper switching or chopping freq.f

T= =







.

duty cycle

ONdc

ON OFF

ONdc

ON

tV V

t t

tV V V d

T

tbut d

t

=

+

= =

= =






2

0

Average Output Current

RMS value of output voltage

1 ON

dcdc

ONdc

t

O o

VI

R

tV VI d

R T R

V v dtT

=

= =

= ∫






2

0

2

But during ,

Therefore RMS output voltage

1

.

.

ON

ON o

t

O

ONO ON

O

t v V

V V dtT

tVV t V

T T

V d V

=

=

= =

=

∫






2

2

Output power

But

Output power

O O O

OO

OO

O

P V I

VI

R

VP

R

dVP

R

=

=

∴

=

=






Effective input resistance of chopper

The output voltage can be varied by

varying the duty cycle.

i

dc

i

VR

I

RR

d

=

=






Methods Of Control

• The output dc voltage can be varied by the following methods.

– Pulse width modulation control or constant frequency operation.

– Variable frequency control.






Pulse Width Modulation

• tON is varied keeping chopping frequency ‘f’ & chopping period ‘T’ constant.

• Output voltage is varied by varying the ON time tON






V0

V

V

V0

t

ttON

tON tOFF

tOFF

T






Variable Frequency Control

• Chopping frequency ‘f’ is varied keeping either tON or tOFF constant.

• To obtain full output voltage range, frequency has to be varied over a wide range.

• This method produces harmonics in the output and for large tOFF load current may become discontinuous






v0

V

V

v0

t

t

tON

tON

T

T

tOFF

tOFF






Step-down Chopper

With R-L Load

V

i0

V0

Chopper

R

LFWD

E

+

−






• When chopper is ON, supply is connected across load.

• Current flows from supply to load.

• When chopper is OFF, load current continues to flow in the same direction through FWD due to energy stored in inductor ‘L’.






• Load current can be continuous or discontinuous depending on the values of ‘L’ and duty cycle ‘d’

• For a continuous current operation, load current varies between two limits Imax and Imin

• When current becomes equal to Imax the chopper is turned-off and it is turned-on when current reduces to Imin.






Outputvoltage

Outputcurrent

v0

V

i0

Imax

Imin

t

t

tON

T

tOFF

Continuouscurrent

Outputcurrent

t

Discontinuouscurrent

i0





Expressions For

Load Current

iO For Continuous Current

Operation When

Chopper Is ON (0 ≤ t ≤ tON)





V

i0

V0

R

L

E

+

-






( ) ( ) ( )

( )

( )

min

min

Taking Laplace Transform

. 0

At 0, initial current 0

OO

O O O

O

O

diV i R L E

dt

V ERI S L S I S i

S S

t i I

IV EI S

RRSLS S

LL

−

−

= + +

= + − +

= =

−= +

++






( ) min

Taking Inverse Laplace Transform

1

This expression is valid for 0 ,

i.e., during the period chopper is ON.

At the instant the chopper is turned off,

load c

R Rt t

L L

O

ON

V Ei t e I e

R

t t

− −

−= − +

≤ ≤

( ) maxurrent is O ONi t I=






When Chopper is OFF

i0

R

L

E






( )

( ) ( ) ( )

( ) max

When Chopper is OFF 0

0

Talking Laplace transform

0 0

Redefining time origin we have at 0,

initial current 0

OFF

OO

O O O

O

t t

diRi L E

dt

ERI S L SI S i

S

t

i I

−

−

≤ ≤

= + +

= + − +

=

=






( )

( )

max

max

Taking Inverse Laplace Transform

1

O

R Rt t

L LO

I EI S

R RS LS S

L L

Ei t I e e

R

− −

∴ = −

+ +

= − −






( ) min

The expression is valid for 0 ,

i.e., during the period chopper is OFF

At the instant the chopper is turned ON or at

the end of the off period, the load current is

OFF

O OFF

t t

i t I

≤ ≤

=






( )

( )

min

max

max

max min

min

From equation

1

At ,

To Find &

1

R Rt t

L L

O

ON O

dRT dRT

L L

V Ei t e I e

R

t t dT i t I

V EI e I e

I I

R

− −

− −

−= − +

= = =

−∴ = − +






( )

( )

( )

max

min

From equation

1

At ,

1

R Rt t

L LO

OFF ON O

OFF

Ei t I e e

R

t t T t i t I

t t d T

− − = − −

= = − =

= = −






( ) ( )1 1

min max

min

max min

max

1

Substituting for in equation

1

we get,

1

1

d RT d RT

L L

dRT dRT

L L

dRT

L

RT

L

EI I e e

R

I

V EI e I e

R

V e EI

R Re

− −− −

− −

−

−

∴ = − −

−= − +

− = −

−






( ) ( )

( )

max

1 1

min max

min

max min

Substituting for in equation

1

we get,

1

1

is known as the steady state ripple.

d RT d RT

L L

dRT

L

RT

L

I

EI I e e

R

V e EI

R Re

I I

− −− −

= − −

− = −

−

−






( )

max min

max min

Therefore peak-to-peak ripple current

Average output voltage

.

Average output current

2

dc

dc approx

I I I

V d V

I II

∆ = −

=

+=






( )( )

min max

min

max minmin

Assuming load current varies linearly

from to instantaneous

load current is given by

. 0

O ON

O

I I

I ti I for t t dT

dT

I Ii I t

dT

∆= + ≤ ≤

− = +






( )

( )

( )

( )

( )

2

0

0

2

max min

min

0

2

min max min2 2max minmin

0

RMS value of load current

1

1

21

dT

O RMS

dT

O RMS

dT

O RMS

I i dtdT

I I tI I dt

dT dT

I I I tI II I t dt

dT dT dT

=

− = +

−− = + +

∫

∫

∫






( )

( )( )

12 2

max min2

min min max min

2

0

0

2

max minmin

0

RMS value of output current

3

RMS chopper current

1

1

O RMS

dT

CH

dT

CH

I II I I I I

I i dtT

I II I t dt

T dT

−= + + −

=

− = +

∫

∫






( )( )

( )

12 2

max min2

min min max min3

Effective input resistance is

CH

CH O RMS

i

S

I II d I I I I

I d I

VR

I

−= + + −

=

=






Where

Average source currentS

S dc

i

dc

I

I dI

VR

dI

=

=

∴ =






Principle Of Step-up Chopper

+

−

VOV

Chopper

CLOAD

DLI

+ −






• Step-up chopper is used to obtain a load voltage higher than the input voltage V.

• The values of L and C are chosen depending upon the requirement of output voltage and current.

• When the chopper is ON, the inductor L is connected across the supply.

• The inductor current ‘I’ rises and the inductor stores energy during the ON time of the chopper, tON.






• When the chopper is off, the inductor current I is forced to flow through the diode D and load for a period, tOFF.

• The current tends to decrease resulting in reversing the polarity of induced EMF in L.

• Therefore voltage across load is given by

. ., O O

dIV V L i e V V

dt= + >






• A large capacitor ‘C’ connected across the load, will provide a continuous output voltage .

• Diode D prevents any current flow from capacitor to the source.

• Step up choppers are used for regenerative braking of dc motors.






Expression For Output Voltage Assume the average inductor current to be

during ON and OFF time of Chopper.

Voltage across inductor

Therefore energy stored in inductor

= . .

Where

When Chopper

period of chopper.

is ON

ON

ON

I

L V

V I t

t ON

=

=






( )

(energy is supplied by inductor to load)

Voltage across

Energy supplied by inductor

where period of Chopper.

Neg

When Chopper

lecting losses, energy stored in inductor

is OFF

O

O OFF

OFF

L V V

L V V It

t OFF

L

= −

= −

=

= energy supplied by inductor L






( )

[ ]

Where

T = Chopping period or period

of switching.

ON O OFF

ON OFF

O

OFF

O

ON

VIt V V It

V t tV

t

TV V

T t

∴ = −

+=

=

−






1

1

1

1

Where duty cyle

ON OFF

OON

O

ON

T t t

V Vt

T

V Vd

td

T

= +

= −

∴ =

−

= =






For variation of duty cycle ' ' in the

range of 0 1 the output voltage

will vary in the range

O

O

d

d V

V V

< <

< < ∞






Performance Parameters • The thyristor requires a certain minimum time to

turn ON and turn OFF.

• Duty cycle d can be varied only between a min. & max. value, limiting the min. and max. value of the output voltage.

• Ripple in the load current depends inversely on the chopping frequency, f.

• To reduce the load ripple current, frequency should be as high as possible.






Problem

• A Chopper circuit is operating on TRC at a frequency of 2 kHz on a 460 V supply. If the load voltage is 350 volts, calculate the conduction period of the thyristor in each cycle.






3

460 V, = 350 V, f = 2 kHz

1Chopping period

10.5 sec

2 10

Output voltage

dc

ONdc

V V

Tf

T m

tV V

T

−

=

=

= =×

=






3

Conduction period of thyristor

0.5 10 350

460

0.38 msec

dcON

ON

ON

T Vt

V

t

t

−

×=

× ×=

=






Problem

• Input to the step up chopper is 200 V. The output required is 600 V. If the conducting

time of thyristor is 200 µsec. Compute

– Chopping frequency,

– If the pulse width is halved for constant frequency of operation, find the new output voltage.






6

200 , 200 , 600

600 200200 10

Solving for

300

ON dc

dc

ON

V V t s V V

TV V

T t

T

T

T

T s

µ

µ

−

= = =

=

−

=

− ×

=






6

6

Chopping frequency

1

13.33

300 10

Pulse width is halved

200 10100

2ON

fT

f KHz

t sµ

−

−

=

= =×

×∴ = =






( )

6

6

Frequency is constant

3.33

1300

Output voltage =

300 10200 300 Volts

300 100 10

ON

f KHz

T sf

TV

T t

µ

−

−

∴ =

= =

∴

−

×= = −






Problem

• A dc chopper has a resistive load of 20Ω and input voltage VS = 220V. When chopper is ON, its voltage drop is 1.5 volts and chopping frequency is 10 kHz. If the duty cycle is 80%, determine the average output voltage and the chopper on time.






( )

( )

220 , 20 , 10

0.80

= Voltage drop across chopper = 1.5 volts


0.80 220 1.5 174.8 Volts

S

ON

ch

ONdc S ch

dc

V V R f kHz

td

T

V

tV V V

T

V

= = Ω =

= =

= −

= − =






3

3

3

3

Chopper ON time,

1Chopping period,

10.1 10 secs 100 secs

10 10

Chopper ON time,

0.80 0.1 10

0.08 10 80 secs

ON

ON

ON

ON

t dT

Tf

T

t dT

t

t

−

−

−

=

=

= = × =×

=

= × ×

= × =






Problem

• In a dc chopper, the average load current is 30 Amps, chopping frequency is 250 Hz, supply voltage is 110 volts. Calculate the ON and OFF periods of the chopper if the load resistance is 2 ohms.






3

30 , 250 , 110 , 2

1 1Chopping period, 4 10 4 msecs

250

&

30 20.545

110

dc

dcdc dc

dc

dc

I Amps f Hz V V R

Tf

VI V dV

R

dVI

R

I Rd

V

−

= = = = Ω

= = = × =

= =

∴ =

×= = =






3

3 3

3

Chopper ON period,

0.545 4 10 2.18 msecs

Chopper OFF period,

4 10 2.18 10

1.82 10 1.82 msec

ON

OFF ON

OFF

OFF

t dT

t T t

t

t

−

− −

−

= = × × =

= −

= × − ×

= × =






• A dc chopper in figure has a resistive load

of R = 10Ω and input voltage of V = 200 V. When chopper is ON, its voltage drop is 2 V and the chopping frequency is 1 kHz. If the duty cycle is 60%, determine

– Average output voltage

– RMS value of output voltage

– Effective input resistance of chopper

– Chopper efficiency.






V

i0

Chopper

+

−

R v0

200 , 10 , 2

0.60, 1 .

chV V R Chopper voltage drop V V

d f kHz

= = Ω =

= =






( )

[ ]

( )

( )


0.60 200 2 118.8 Volts


0.6 200 2 153.37 Volts

dc ch

dc

O ch

O

V d V V

V

V d V V

V

= −

= − =

= −

= − =






( )22

0

0 0

Effective input resistance of chopper is

118.811.88 Amps

10

20016.83

11.88

Output power is

1 1

i

S dc

dcdc

i

S dc

dT dT

ch

O

V VR

I I

VI

R

V VR

I I

V VvP dt dt

T R T R

= =

= = =

= = = = Ω

−= =∫ ∫






( )

[ ]

( )

2

2

0

0

0.6 200 22352.24 watts

10

Input power,

1

1

ch

O

O

dT

i O

dT

ch

O

d V VP

R

P

P Vi dtT

V V VP dt

T R

−=

−= =

=

−=

∫

∫






( )

[ ]0.6 200 200 22376 watts

10

Chopper efficiency,

100

2352.24100 99%

2376

ch

O

O

O

i

dV V VP

R

P

P

Pη

η

−=

× −= =

= ×

= × =






Problem • A chopper is supplying an inductive load with a

free-wheeling diode. The load inductance is 5 H

and resistance is 10Ω.. The input voltage to the chopper is 200 volts and the chopper is operating

at a frequency of 1000 Hz. If the ON/OFF time

ratio is 2:3. Calculate

– Maximum and minimum values of load current

in one cycle of chopper operation.

– Average load current






5 , 10 , 1000 ,

200 , : 2 : 3

Chopping period,

1 11 msecs

1000

2

3

2

3

ON OFF

ON

OFF

ON OFF

L H R f Hz

V V t t

Tf

t

t

t t

= = Ω =

= =

= = =

=

=






3

2

3

5

3

3

5

31 10 0.6 msec

5

ON OFF

OFF OFF

OFF

OFF

T t t

T t t

T t

t T

T−

= +

= +

=

=

= × × =






( ) 3

3

3

max

1 0.6 10 0.4 msec

Duty cycle,

0.4 100.4

1 10

Maximum value of load current is given by

1

1

ON OFF

ON

ON

dRT

L

RT

L

t T t

t

td

T

V e EI

R Re

−

−

−

−

−

= −

= − × =

×= = =

×

− = −

−






3

3

max

0.4 10 1 10

5

max 10 1 10

5

Since there is no voltage source in

the load circuit, E = 0

1

1

200 1

101

dRT

L

RT

L

V eI

Re

eI

e

−

−

−

−

× × ×−

× ×−

− ∴ =

−

−

= −






3

3

0.8 10

max 2 10

max

min

120

1

8.0047A

Minimum value of load current with E = 0

is given by

1

1

dRT

L

RT

L

eI

e

I

V eI

Re

−

−

− ×

− ×

−=

−

=

− =

−






3

3

0.4 10 1 10

5

min 10 1 10

5

max min

200 17.995 A

101

Average load current

2

8.0047 7.9958 A

2

dc

dc

eI

e

I II

I

−

−

× × ×

× ×

−

= = −

+=

+= ≈






Problem

• A chopper feeding on RL load is shown in figure,

with V = 200 V, R = 5Ω, L = 5 mH, f = 1

kHz, d = 0.5 and E = 0 V. Calculate

– Maximum and minimum values of load

current.

– Average value of load current.

– RMS load current.

– Effective input resistance as seen by source.

– RMS chopper current.






3

3

V = 200 V, R = 5 , L = 5 mH,

f = 1kHz, d = 0.5, E = 0

Chopping period is

1 11 10 secs

1 10T

f

−

Ω

= = = ××

i0

v0

Chopper

R

LFWD

E

+

−






3

3

3

3

max

0.5 5 1 10

5 10

max 5 1 10

5 10

0.5

max 1

Maximum value of load current is given by

1

1

200 10

51

140 24.9 A

1

dRT

L

RT

L

V e EI

R Re

eI

e

eI

e

−

−

−

−

−

× × ×−

×

× ×−

×

−

−

− = −

−

−

= − −

−= =

−






3

3

3

3

min

0.5 5 1 10

5 10

min 5 1 10

5 10

0.5

min 1

Minimum value of load current is given by

1

1

200 10

51

140 15.1 A

1

dRT

L

RT

L

V e EI

R Re

eI

e

eI

e

−

−

−

−

× × ×

×

× ×

×

− = −

−

−

= − −

−= =

−






( )

( )( )

1 2

12 2

max min2

min min max min

Average value of load current is

2

for linear variation of currents

24.9 15.120 A

2

RMS load current is given by

3

dc

dc

O RMS

I II

I

I II I I I I

+=

+∴ = =

−= + + −






( )

( )( )

( )

( )

12 2

2

1

2

24.9 15.115.1 15.1 24.9 15.1

3

96.04228.01 147.98 20.2 A

3

RMS chopper current is given by

0.5 20.2 14.28 A

O RMS

O RMS

ch O RMS

I

I

I d I

−= + + −

= + + =

= = × =






Effective input resistance is

= Average source current

0.5 20 10 A

Therefore effective input resistance is

20020

10

i

S

S

S dc

S

i

S

VR

I

I

I dI

I

VR

I

=

=

= × =

= = = Ω






Classification Of Choppers

• Choppers are classified as

– Class A Chopper

– Class B Chopper

– Class C Chopper

– Class D Chopper

– Class E Chopper






Class A Chopper

V

Chopper

FWD

+

−

v0

v0

i0

i0

LOAD

V






• When chopper is ON, supply voltage V is connected across the load.

• When chopper is OFF, vO = 0 and the load current continues to flow in the same direction through the FWD.

• The average values of output voltage and current are always positive.

• Class A Chopper is a first quadrant chopper .






• Class A Chopper is a step-down chopper in which power always flows form source to load.

• It is used to control the speed of dc motor.

• The output current equations obtained in step down chopper with R-L load can be used to study the performance of Class A Chopper.






Output current

Thyristorgate pulse

Output voltage

ig

i0

v0

t

t

ttON

T

CH ON

FWD Conducts






Class B Chopper

V

Chopper

+

−

v0

v0

−i0

i0

L

E

R

D






• When chopper is ON, E drives a current through L and R in a direction opposite to that shown in figure.

• During the ON period of the chopper, the inductance L stores energy.

• When Chopper is OFF, diode D conducts, and part of the energy stored in inductor L is returned to the supply.






• Average output voltage is positive.

• Average output current is negative.

• Therefore Class B Chopper operates in second quadrant.

• In this chopper, power flows from load to source.

• Class B Chopper is used for regenerative braking of dc motor.

• Class B Chopper is a step-up chopper.






Output current

D conducts Chopper

conducts

Thyristorgate pulse

Output voltage

ig

i0

v0

t

t

t

Imin

Imax

T

tONtOFF





Expression for Output Current






( ) min

For the initial condition i.e.,

During the interval diode 'D' conduc

at 0

The solution of the ab

ts

voltage equation

ove equation is obtained

along similar lines as in s

is given by

OO

O

LdiV Ri E

dt

i t I t

= + +

= =

tep-down chopper

with R-L load






( )

( ) ( )

min

max

max min

During the interval chopper is ON voltage

equation is g

1 0

At

1

0

iven by

OFF OFF

R Rt t

L LO OFF

OFF O

R Rt t

L L

OO

V Ei t e I e t t

R

t t i t I

V EI e I e

R

LdiRi E

dt

− −

− −

−∴ = − + < <

= =

−= − +

= + +






( )

( )

( )

max

max

min

min max

Redefining the time origin, at 0

The solution for the stated initial condition is

1 0

At

1ON ON

O

R Rt t

L LO ON

ON O

R Rt t

L L

t i t I

Ei t I e e t t

R

t t i t I

EI I e e

R

− −

− −

= =

= − − < <

= =

∴ = − −






Class C Chopper

V

Chopper

+

−

v0

D1

D2

CH2

CH1

v0i0

i0

L

E

R






• Class C Chopper is a combination of Class A and Class B Choppers.

• For first quadrant operation, CH1 is ON or D2 conducts.

• For second quadrant operation, CH2 is ON or D1 conducts.

• When CH1 is ON, the load current is positive.

• The output voltage is equal to ‘V’ & the load receives power from the source.

• When CH1 is turned OFF, energy stored in inductance L forces current to flow through the diode D2 and the output voltage is zero.






• Current continues to flow in positive direction.

• When CH2 is triggered, the voltage E forces current to flow in opposite direction through L and CH2 .

• The output voltage is zero.

• On turning OFF CH2 , the energy stored in the inductance drives current through diode D1 and the supply

• Output voltage is V, the input current becomes negative and power flows from load to source.






• Average output voltage is positive

• Average output current can take both positive and negative values.

• Choppers CH1 & CH2 should not be turned ON simultaneously as it would result in short circuiting the supply.

• Class C Chopper can be used both for dc motor control and regenerative braking of dc motor.

• Class C Chopper can be used as a step-up or step-down chopper.






Gate pulseof CH2

Gate pulseof CH1

Output current

Output voltage

ig1

ig2

i0

V0

t

t

t

t

D1 D1D2 D2CH1 CH2 CH1 CH2

ON ON ON ON






Class D Chopper

V+ −v0

D2

D1 CH2

CH1

v0

i0

L ER i0






• Class D is a two quadrant chopper.

• When both CH1 and CH2 are triggered simultaneously, the output voltage vO = V and output current flows through the load.

• When CH1 and CH2 are turned OFF, the load current continues to flow in the same direction through load, D1 and D2 , due to the energy stored in the inductor L.

• Output voltage vO = - V .






• Average load voltage is positive if chopper ON time is more than the OFF time

• Average output voltage becomes negative if tON < tOFF .

• Hence the direction of load current is always positive but load voltage can be positive or negative.






Gate pulseof CH2

Gate pulseof CH1

Output current

Output voltage

Average v0

ig1

ig2

i0

v0

V

t

t

t

t

CH ,CH

ON1 2 D1,D2 Conducting






Gate pulseof CH2

Gate pulseof CH1

Output current

Output voltage

Average v0

ig1

ig2

i0

v0

V

t

t

t

t

CH

CH1

2

D , D1 2






Class E Chopper

V

v0

i0L ER

CH2 CH4D2 D4

D1 D3CH1 CH3

+ −






Four Quadrant Operation v0

i0

CH - CH ON

CH - D Conducts1 4

4 2

D D2 3 - Conducts

CH - D Conducts4 2

CH - CH ON

CH - D Conducts3 2

2 4

CH - D Conducts

D - D Conducts2 4

1 4






• Class E is a four quadrant chopper

• When CH1 and CH4 are triggered, output current iO flows in positive direction through CH1 and CH4, and with output voltage vO = V.

• This gives the first quadrant operation.

• When both CH1 and CH4 are OFF, the energy stored in the inductor L drives iO through D2 and D3 in the same direction, but output voltage vO = -V.






• Therefore the chopper operates in the

fourth quadrant.

• When CH2 and CH3 are triggered, the load current iO flows in opposite direction & output voltage vO = -V.

• Since both iO and vO are negative, the chopper operates in third quadrant.






• When both CH2 and CH3 are OFF, the load current iO continues to flow in the same direction D1 and D4 and the output voltage vO = V.

• Therefore the chopper operates in second quadrant as vO is positive but iO is negative.






Problem • For the first quadrant chopper shown in figure,

express the following variables as functions of V,

R and duty cycle ‘d’ in case load is resistive.

– Average output voltage and current

– Output current at the instant of commutation

– Average and RMS free wheeling diode current.

– RMS value of output voltage

– RMS and average thyristor currents.






V

i0

v0

Chopper

FWD

+

−

LOAD






Average output voltage,

Average output current,

The thyristor is commutated at the instant

output current at the instant of commutation is

since V is the output v

ONdc

dcdc

ON

tV V dV

T

V dVI

R R

t t

V

R

= =

= =

=

∴

oltage at that instant.






( )2

0

0

Free wheeling diode (FWD) will never

conduct in a resistive load.

Average & RMS free wheeling diode

currents are zero.

1

But during

ONt

O RMS

O ON

V v dtT

v V t

∴

=

=

∫






( )

( )

( )

2

0

2

1

Where duty cycle,

ONt

O RMS

ON

O RMS

O RMS

ON

V V dtT

tV V

T

V dV

td

T

=

=

=

=

∫






( )

RMS value of thyristor current

= RMS value of load current

Average value of thyristor current

= Average value of load current

O RMSV

R

dV

R

dV

R

=

=

=






Boost Converter or

Step Up converter

Buck-Boost

Converter

Buck Converter or

Step Down Converter

Simple DC-DC Converter Topologies






SMPS benefits

– Very wide input voltage range.

• For example: most personal computer power supplies are SMPSs - accepting AC input 90V to 250V.

– Lower Quiescent Current than linear regulators

– Less heat than an equivalent linear regulator.

• Much Lower Green House Gas emissions

– Overall Smaller geometry components are used

– Lighter Weight

– Lower running cost - Lower total cost of ownership (TCO).

– Battery operated devices - longer lifetime.






SMPS disadvantages

– Significant Output Ripple

• May need a post filter to decrease ripple

• May need a secondary linear low drop out

regulator to ensure damaging voltage transients

keep away from voltage sensitive elements -

electronics.

• An SMPS May add too much cost.

– How much is too much?





UNIT-IV

INVERTERS






Single-Phase Inverters

Half-Bridge Inverter

One of the simplest types of inverter. Produces a square wave output.







(cont’d) Full Bridge (H-bridge) Inverter

Two half-bridge inverters combined.

Allows for four quadrant operation.







(cont’d) Quadrant 1: Positive step-down converter

(forward motoring)

Q1-On; Q2 - Chopping; D3,Q1 freewheeling







(cont’d)

Quadrant 2: Positive step-up converter

(forward regeneration)

Q4 - Chopping; D2,D1 freewheeling







(cont’d) Quadrant 3: Negative step-down converter

(reverse motoring)

Q3-On; Q4 - Chopping; D1,Q3 freewheeling







(cont’d) Quadrant 4: Negative step-up converter

(reverse regeneration)

Q2 - Chopping; D3,D4 freewheeling







(cont’d)

Phase-Shift Voltage Control - the output of

the H-bridge inverter can be controlled by

phase shifting the control of the

component half-bridges. See waveforms

on next slide.







(cont’d)







(cont’d)

The waveform of the output voltage vab is a quasi-

square wave of pulse width φ. The Fourier series of vab

is given by:

The value of the fundamental, a1=

The harmonic components as a function of phase

angle are shown in the next slide.

( )1,3,5...

4sin cos

2

dab

n

V nv n t

n

φω

π=

=

∑

( )4

sin / 2dVφ

π







(cont’d)






Three-Phase Bridge

Inverters

Three-phase bridge inverters are widely

used for ac motor drives. Two modes of

operation - square wave and six-step. The

topology is basically three half-bridge

inverters, each phase-shifted by 2π/3, driving each of the phase windings.






Three-Phase Bridge Inverters

(cont’d)







(cont’d)

The three square-wave phase voltages can

be expressed in terms of the dc supply

voltage, Vd, by Fourier series as:

1

0

1,3,5...

2( 1) cos( )nd

a

n

Vv n tω

π+

=

= −∑

1

0

1,3,5...

2 2( 1) cos( )

3

ndb

n

Vv n t

πω

π+

=

= − −∑

1

0

1,3,5...

2 2( 1) cos( )

3

ndc

n

Vv n t

πω

π+

=

= − +∑







(cont’d)

The line voltages can then be expressed as:

0 0

1,3,5...

2 3cos( / 2) cos( 2)d

bc b c

n

Vv v v t n tω π ω π

π =

= − = − − −∑

0 0

1,3,5...

2 3cos( 5 / 6) cos( 5 6)d

ca c a

n


π =

= − = + − −∑

0 0

1,3,5...

2 3cos( / 6) cos( 6)d

ab a b

n


π =

= − = + − +∑







(cont’d)

The line voltages are six-step waveforms and

have characteristic harmonics of 6n±±±±1, where n is an integer. This type of inverter is

referred to as a six-step inverter.

The three-phase fundamental and harmonics

are balanced with a mutual phase shift of

2π/3.







(cont’d)

If the three-phase load neutral n is isolated from the the

center tap of the dc voltage supply (as is normally the

case in an ac machine) the equivalent circuit is shown

below.







(cont’d)

In this case the isolated neutral-phase

voltages are also six-step waveforms with

the fundamental component phase-shifted

by π/6 from that of the respective line voltage. Also, in this case, the triplen

harmonics are suppressed.







(cont’d)

For a linear and balanced 3Φ load, the line currents

are also balanced. The individual line current

components can be obtained from the Fourier series

of the line voltage. The total current can be obtained

by addition of the individual currents. A typical line

current wave with inductive load is shown below.







(cont’d)

The inverter can operate in the usual inverting or

motoring mode. If the phase current wave, ia, is

assumed to be perfectly filtered and lags the phase

voltage by π/3 the voltage and current waveforms are

as shown below:






Three-Phase Bridge Inverters The inverter can also operate in rectification or regeneration

mode in which power is pushed back to the dc side from the ac side. The waveforms corresponding to this mode of operation with phase angle = 2π/3 are shown below:







(cont’d)

The phase-shift voltage control principle

described earlier for the single-phase

inverter can be extended to control the

output voltage of a three-phase inverter.







(cont’d)







(cont’d)

The three waveforms va0,vb0, and vc0 are of

amplitude 0.5Vd and are mutually phase-

shifted by 2π/3.

The three waveforms ve0,vf0, and vg0 are of

similar but phase shifted by φ.







(cont’d)

The transformer’s secondary phase voltages,

vA0, vB0, and vc0 may be expressed as follows:

where m is the transformer turns ratio

(= Ns/Np). Note that each of these waves is a

function of φ angle.

0 0 0( )A ad a d

v mv m v v= = −

0 0 0( )B be b e

v mv m v v= = −

0 0 0( )C cf c f

v mv m v v= = −







(cont’d)

The output line voltages are given by:

While the component voltage waves va0, vd0, vA0 … etc. all

contain triplen harmonics, they are eliminated from the

line voltages because they are co-phasal. Thus the line

voltages are six-step waveforms with order of harmonics

= 6n±1 at a phase angle φ.

0 0AB A Bv v v= −

0 0BC B Cv v v= −

0 0CA C Av v v= −







(cont’d)

The Fourier series for vA0 and vB0 are given

by:

( )0

1,3,5...

4sin cos

2

dA

n

mV nv n t

n

φω

π=

=

∑

( )0

1,3,5...

4sin cos 2 / 3

2

dB

n

mV nv n t

n

φω π

π=

= −

∑







(cont’d)

The Fourier series for vAB is given by:

Note that the triplen harmonics are removed

in vAB although they are present in vA0 and

vB0.

( )1,5,7,11...

4 2sin cos cos

2 3

d

n

mV nn t n t

n

φ πω ω

π=

= − −

∑

0 0AB A Bv v v= −






PWM Technique

While the 3Φ 6-step inverter offers simple control and low switching loss, lower order

harmonics are relatively high leading to high

distortion of the current wave (unless

significant filtering is performed).

PWM inverter offers better harmonic control

of the output than 6-step inverter.






PWM Principle

The dc input to the inverter is “chopped” by

switching devices in the inverter. The

amplitude and harmonic content of the ac

waveform is controlled by the duty cycle of

the switches. The fundamental voltage v1

has max. amplitude = 4Vd/π for a square wave output but by creating notches, the

amplitude of v1 is reduced (see next slide).






PWM Principle (cont’d)






PWM Techniques

Various PWM techniques, include:

• Sinusoidal PWM (most common)

• Selected Harmonic Elimination (SHE)

PWM

• Space-Vector PWM

• Instantaneous current control PWM

• Hysteresis band current control PWM

• Sigma-delta modulation






Sinusoidal PWM

The most common PWM approach is

sinusoidal PWM. In this method a

triangular wave is compared to a

sinusoidal wave of the desired

frequency and the relative levels of the

two waves is used to control the

switching of devices in each phase leg

of the inverter.






Sinusoidal PWM

(cont’d)

Single-Phase (Half-Bridge) Inverter

Implementation






Sinusoidal PWM (cont’d)

when va0> vT T+ on; T- off; va0 = ½Vd

va0 < vT T- on; T+ off; va0 = -½Vd






Sinusoidal PWM

(cont’d)






Sinusoidal PWM (cont’d)

Definition of terms:

Triangle waveform switching freq. = fc (also called

carrier freq.)

Control signal freq. = f (also called modulation

freq.)

Amplitude modulation ratio, m = Vp

VT

Frequency modulation ratio,

mf (P)= fc / f

Peak amplitude

of control signal

Peak amplitude

of triangle wave






Multiple Pulse-Width Modulation

• In multiple-pulse modulation, all pulses are the same width

• Vary the pulse width according to the amplitude of a sine wave evaluated at the center of the same pulse






Generate the gating signal

2 Reference Signals, vr, -vr






Comparing the carrier and reference signals

• Generate g1 signal by comparison with vr

• Generate g4 signal by comparison with -vr






Comparing the carrier and reference signals






Potential problem if Q1 and Q4 try to turn ON at the same time!






If we prevent the problem

Output voltage is low when g1 and g4 are

both high






This composite signal is difficult to generate






Generate the same gate pulses with one sine wave






Alternate scheme






rms output voltage

• Depends on the modulation index, M

2

1

pm

o S Sm

pV V V

δ δ

π π=

∑= →

Where δm is the width of the mth pulse






Fourier coefficients of the output voltage

( ) ( )2

1

4 3sin sin sin

4 4 4

1, 3, 5, ..

pS m m m

n m mm

V nB n n

n

n

δ δ δα π α

π=

∑ = + − + +

=






Harmonic Profile






Compare with multiple-pulse case for p=5

Distortion Factor is considerably less






Series-Resonant Inverter






Operation

T1 fired, resonant pulse of current flows through the load. The current falls to zero at t = t1m and T1 is “self – commutated”.

T2 fired, reverse resonant current flows through the load and T2 is also “self-commutated”.

The series resonant circuit must be underdamped,

R2 < (4L/C)






Operation in Mode 1 – Fire T1

11 1

1

1(0)

(0) 0

(0)

C S

C C

diL Ri i dt v V

dt C

i

v V

+ + + =

=

= −

∫






21 1

12 2

2

11

0

1

( ) sin

1

4

( ) sin

2

Rt

Lr

r

s c

t r

ts cr

r

i t A e t

R

LC L

V VdiA

dt L

V Vi t e t

L

R

L

α

ω

ω

ω

ωω

α

−

=

−

=

= −

+= =

+=

=






To find the time when the current is

maximum, set the first derivative = 0

( )

1

1

1

0

sin cos 0

.....

tan

tan

1tan

2

t ts cr r r

r

rr m

r mr m

rm

r

di

dt

V Ve t e t

L

t

tt

t

α αα ω ω ωω

ωω

α

ωω

α

ω

ω

− −

−

−

=

+− + =

=

=

=






To find the capacitor voltage, integrate the

current

( )

( )

1

1

1

1

1

0

0

1

1 1

1( ) ( )

1( ) sin

...

( ) ( ) ( sin cos ) /

0 ( )

( ) r

t

C c

t

ts cC r C

r

t

C s C r r r r s

m

r

C m C s C s

v t i t dt VC

V Vv t e t dt V

C L

v t V V e t t V

t t

v t V V V e V

α

α

απ

ω

ωω

α ω ω ω ω

π

ω

−

−

−

= −

+= −

= − + + +

≤ ≤

= = + +

∫

∫

The current i1 becomes = 0 @ t=t1m






Operation in Mode 2 – T1, T2 Both OFF






t2m






Operation in Mode 3 – Fire T2

3

3 2 1

33 3

3

1(0) 0

(0) 0

(0)

C

C C C

diL Ri i dt v

dt C

i

v V V

+ + + =

=

= − = −

∫






1

3 1

1

3

3

3

0

3

( ) sin

1( )

( sin cos )( )

0 ( )m

C t

r

r

t

C C

t

C r r r

C

r

r

Vi t e t

L

v t i dt VC

V e t tv t

t t

α

α

ωω

α ω ω ω

ω

π

ω

−

−

=

= −

− +=

≤ ≤

∫






3 3 1

1 1

1

1

3

1

( )

( ) ( )

.

.

1

1

1

r

m

r

m

C C C C

C C S C S

C S z

z

C S z

C S C

v t V V V e

v t V V V e V

V Ve

eV V

e

V V V

πα

ω

πα

ω

−

−

= = =

= = + +

=−

=−

+ =






• Space Vector Diagram

α1V

0V

3V

2V

4V

5V

6V

βj

POO

PPOOPO

OPP

OOP POP

refV

θ

OOOPPP

SECTOR ISECTOR III

SECTOR IV SECTOR VI

SECTOR V

SECTOR

II

ω• Active vectors: to (stationary, not rotating)

• Zero vector:

1V

6V

0V

• Six sectors: I to VI






• Space Vectors

• Three-phase voltages

0)()()( =++ tvtvtvCOBOAO

• Two-phase voltages

=

)(

)(

)(

3

4sin

3

2sin0sin

3

4cos

3

2cos0cos

3

2

)(

)(

tv

tv

tv

tv

tv

CO

BO

AO

ππ

ππ

β

α

• Space vector representation

)()()( tvjtvtV βα +=

(2) →→→→ (3)

[ ]3/43/20)()()(

3

2)(

ππ j

CO

j

BO

j

AOetvetvetvtV ++=

where xjxejx sincos +=

(3)

(1)

(2)

(4)






• Space Vectors (Example)

Switching state [POO] →→→→ S1, S6 and S2 ON

dBOdAOVtvVtv

3

1)(,

3

2)( −==

dCOVtv

3

1)( −=and

(5) →→→→ (4)

(7)

(5)

(6) 0

13

2 j

deVV =

Similarly,

3)1(

3

2 π−

=kj

dkeVV

.6...,,2,1=k

α1V

0V

3V

2V

4V

5V

6V

βj

POO

PPOOPO

OPP

OOP POP

refV

θ

OOOPPP

SECTOR ISECTOR III

SECTOR IV SECTOR VI

SECTOR V

SECTOR

II

ω






• Active and Zero Vectors

Space Vector Switching State (Three Phases)

On-state Switch Vector

Definition

[PPP] 531 ,, SSS Zero

Vector 0V

[OOO] 264 ,, SSS

00 =V

1V

[POO] 261 ,, SSS 01

3

2 jd eVV =

2V

[PPO] 231 ,, SSS 32

3

2π

=j

d eVV

3V

[OPO] 234 ,, SSS 3

2

33

2π

=j

d eVV

4V

[OPP] 534 ,, SSS 3

3

43

2π

=j

d eVV

5V

[OOP] 564 ,, SSS 3

4

53

2π

=j

d eVV

Active

Vector

6V

[POP] 561 ,, SSS 3

5

63

2π

=j

d eVV

• Active Vector: 6

• Zero Vector: 1

• Redundant switching

states: [PPP] and [OOO]

1S

2S

3S 5S

4S 6S

B

C

P

N

dV

A






(8)

• Reference Vector Vref

• Definition

α1V

0V

3V

2V

4V

5V

6V

βj

POO

PPOOPO

OPP

OOP POP

refV

θ

OOOPPP

SECTOR ISECTOR III

SECTOR IV SECTOR VI

SECTOR V

SECTOR

II

ω

• Angular displacement

∫=t

dtt0

)( ωθ (9)

θjrefref eVV =

• Rotating in space at ω

fπω 2=






• Relationship Between Vref and VAB

• Vref is approximated by two active

and a zero vectors

• Vref rotates one revolution,

VAB completes one cycle

• Length of Vref corresponds to magnitude of VAB

1V

2V

refV

θ

1VT

T

s

a

2VT

T

s

b






• Dwell Time Calculation

• Volt-Second Balancing

++=

++=

0

0021

TTTT

TVTVTVTV

bas

basref

(10)

• Ta, Tb and T0 – dwell times for and ,21

VV

0V

• Ts – sampling period

• Space vectors

d

j

refrefVVeVV

3

2,

1==

θ 3

23

2 πj

deVV =

0

0=V

, and

(11) →→→→ (10)

=

+=

bdsref

bdadsref

TVTV

TVTVTV

3

1)(sin

3

1

3

2)(cos

θ

θ

:Im

:Re

(11)

(12)

1V

2V

refV

θ

1VT

T

s

a

2VT

T

s

b






• Dwell Times

Solve (12)

−−=

=

−=

bas

d

refs

b

d

refs

a

TTTT

V

VTT

V

VTT

0

sin3

)3

(sin3

θ

θπ

3/0 πθ <≤ (13)






• Vref Location versus Dwell Times

refV

Location 0=θ 6

0π

θ << 6

πθ =

36

πθ

π<<

3

πθ =

Dwell Times 0

0

=

>

b

a

T

T

ba TT > ba TT = ba TT < 0

0

>

=

b

a

T

T

1V

2V

refV

θ

1VT

T

s

a

2VT

T

s

b

SECTOR I

Q






• Modulation Index

−−=

=

−=

cbs

asb

asa

TTTT

mTT

mTT

0

sin

)3

(sin

θ

θπ

(15)

d

ref

aV

Vm

3= (16)






• Modulation Range

• Vref,max

32

3

3

2max,

d

dref

VVV =×= (17)

α1V

0V

3V

2V

4V

5V

6V

βj

POO

PPOOPO

OPP

OOP POP

refV

θ

OOOPPP

SECTOR ISECTOR III

SECTOR IV SECTOR VI

SECTOR V

SECTOR

II

ω

(17) →→→→ (16)

• ma,max = 1 →→→→

• Modulation range: 0 ≤≤≤≤ ma ≤≤≤≤ 1 (18)






• Switching Sequence Design

• Basic Requirement:

Minimize the number of switchings per

sampling period Ts

• Implementation:

Transition from one switching state to

the next involves only two switches in

the same inverter leg.






• Seven-segment Switching Sequence

dV

2

0T

2

aT

2

bT

2

aT

BNv

ANv

CNv

0

1V

1V

2V

0V

2V

POOOOO PPO PPP PPO POO OOO

dV

dV

4

0T

4

0T

2

bT

sT

0

0

0V

0V

• Total number of switchings: 6

• Selected vectors: V0, V1 and V2

• Dwell times: Ts = T0 + Ta + Tb






• Undesirable Switching Sequence

• Vectors V1 and V2 swapped

dV

2

0T

2

aT

2

bT

2

aT

BNv

ANv

CNv

0

1V

1V

2V

2V

POOOOO PPO PPP PPOPOO OOO

dV

dV

4

0T

4

0T

2

bT

sT

0

0

0V

0V

0V

• Total number of switchings: 10






• Switching Sequence Summary (7–segments)

Sector Switching Sequence

0V

1V

2V

0V

2V

1V

0V

I

OOO POO PPO PPP PPO POO OOO

0V

3V

2V

0V

2V

3V

0V

II

OOO OPO PPO PPP PPO OPO OOO

0V

3V

4V

0V

4V

3V

0V

III

OOO OPO OPP PPP OPP OPO OOO

0V

5V

4V

0V

4V

5V

0V

IV

OOO OOP OPP PPP OPP OOP OOO

0V

5V

6V

0V

6V

5V

0V

V

OOO OOP POP PPP POP OOP OOO

0V

1V

6V

0V

6V

1V

0V

VI

OOO POO POP PPP POP POO OOO

Note: The switching sequences for the odd and ever sectors are different.






• Simulated Waveforms

ABv

AOv

0

0

0

Ai

dV

3/2 dV

π π2 π3

π π2 π3

VIVI

Sector

III

IIIIV

V

III

IIIIV

V

f1 = 60Hz, fsw = 900Hz, ma = 0.696, Ts = 1.1ms






• Waveforms and FFT

n

ABv

AOv

Ai

dV

3/2 dV

π2

dVVAB 566.01 =

dn VVAB /

π2π π3






• Waveforms and FFT (Measured)






dnAB VV /

am

1=n

2=n

dnAB VV /

am

1=n

• Waveforms and FFT (Measured)

Hz601 =f sec720/1=sT ( and )






• Even-Order Harmonic Elimination

BNv

ANv

CNv

5V

4V

0V

0V

0V

ABvdV−

4V

5V

dV

dV

dV

Type-A sequence (starts and ends with [OOO])

BNv

ANv

CNv

ABvdV−

5V

4V

dV

dV

dV

4V

5V

0V

0V

0V

Type-B sequence (starts and ends with [PPP])







1V

3V

2V

4V

5V

6V

°30

°30

Space vector Diagram







AOv

ABv

d

n

V

VAB

• Measured waveforms and FFT







am

dnAB VV /

1=n

Hz601 =f sec720/1=sT ( and )






• Five-segment SVM

dV

2

aTbT

2

aT

BNv

ANv

CNv

0

1V

1V

2V

0V

POOOOO PPO POO OOO

dV

sT

0

0

0V

2

0T

2

0T

dV

aT

1V

2V

0V

PPP PPO POO PPP

dV

sT

0V

2

0T

2

0T

(a) Sequence A

2V

PPO

dV

2

bT

2

bT

(b) Sequence B






• Switching Sequence ( 5-segment)

Sector Switching Sequence (A)

0V

1V

2V

1V

0V

I

OOO POO PPO POO OOO 0=

CNv

0V

3V

2V

3V

0V

II

OOO OPO PPO OPO OOO 0=

CNv

0V

3V

4V

3V

0V

III

OOO OPO OPP OPO OOO 0=

ANv

0V

5V

4V

5V

0V

IV

OOO OOP OPP OOP OOO 0=

ANv

0V

5V

6V

5V

0V

V

OOO OOP POP OOP OOO 0=

BNv

0V

1V

6V

1V

0V

VI

OOO POO POP POO OOO 0=

BNv






• Simulated Waveforms ( 5-segment)

dV

π2 π4

1gv

3gv

5gv

A Bv

0

0

Ai

3/2π

π2 π4

π2 π4

• No switching for a 120° period per cycle.

• Low switching frequency but high harmonic distortion

• f1 = 60Hz, fsw = 600Hz, ma = 0.696, Ts = 1.1ms





UNIT V

AC voltage controller and

cycloconverter






4.1.1 Single-phase AC voltage controller

R u 1 u o

i o

VT 1

VT 2

u

O

u 1

u o

i o

VT

ω t

O ω t

O ω t

O ω t

The phase shift range

(operation range of phase

delay angle): 0 ≤≤≤≤αααα≤≤≤≤ ππππ






• Resistive load, quantitative analysis


RMS value of output current

RMS value of thyristor current

Power factor of the circuit

( ) ( )π

απα

πωω

π

π

α

−+== ∫ 2sin

2

1dsin2

11

2

1o UttUU (4-1)

(4-2) R

UI o

o =

(4-3) ( ) )2

2sin1(

2

1sin2

2

1 1

2

1

π

α

π

αω

ω

π

π

α+−=

= ∫ R

Utd

R

tUIT

(4-4) π

απα

πλ

−+==== 2sin

2

1

1

o

o1

oo

U

U

IU

IU

S

P






Inductive (Inductor- resistor) load , operation principle

R

L

u 1 u o

i o

VT 1

VT 2

O

u 1

u o

i o

u VT

O

O

ω t O

u G1

u G2

O

O

ω t

ω t

ω t

ω t

ω t

The phase shift range:

φ ≤≤≤≤α ≤≤≤≤ ππππ






4.2 Other AC controllers

4.2.1 Integral cycle control—AC power controller

Circuit topologies are the same as AC voltage controllers.

Only the control method is different.

Load voltage and current are both sinusoidal when thyristors are conducting.

R u 1 u o

i o

VT 1

VT 2

π

M

Line period

Control period = M *Line period = 2 π

4 π

M O

Conduction

angle =

2 π N

M

3 π

M

2 π

M

u o

u 1 u o , i o

ω t

U 1 2






4.3 Thyristor cycloconverters

4.3.1 Single- phase thyristor-cycloconverter

• Circuit configuration and operation principle

P N

Z

ωt ο

uo ap= 2

π Output

voltage

ap=0

Average

output voltage

ap=

2

π






• Single- phase thyristor-cycloconverter

Modes of operation

t

t

t

t

t

O

O

O

O

O

u o , i o

u o i o

t 1 t

2 t 3

t 4 t 5

u o u P

u N

u o

i P

i N

u P u N u o

i o i N i P

blocking P

N

Rectifi

cation

Inver

sion

blocking Rectifi

cation

Inver

sion






Typical waveforms

1

O

O

2

3 4

5

6

u o

i o

ω t

ω t






• Modulation methods for firing delay angle

Calculation method

– For the rectifier circuit

ωt

ωt

αcosd0o Uu =

tUu oomo sinω=

ttU

Uoo

d0

om sinsincos ωγωα ==

)sin(cos o

1tωγα −=

(4-15)

–For the cycloconverter

output

(4-16)

–Equating (4- 15) and (4-16)

–therefore (4-17)

(4-18)

Principle of cosine

wave-crossing method

u2 u3 u4 u5 u6 u1

ap3 ap4

uo

us2 us3 us4 us5 us6 us1






Output voltage ratio

(Modulation factor)

)10(0

≤≤= γγd

om

U

U

π

2 2 π π ωο t 3 π

2

0

30

60

90

120

150

Output voltage phase angle

α / ( º) γ=0

γ=0.1 0.2

0.3

0.8 0.9 1.0

0.1 0.2 0.3

0.8

0.9

1.0






4.3.2 Three- phase thyristor-cyclo converter

• The configuration with common input line






• The configuration with star-connected output






Typical waveforms

200 t / ms

Output voltage

Input current with

Single-phase output

Input current with

3-phase output

200 t / ms

200 t / ms






• Input and output characteristics

The maximum output frequency and the harmonics in the output voltage are the same as in single-phase circuit. Input power factor is a little higher than single-phase circuit. Harmonics in the input current is a little lower thanthe single- phase circuit due to the cancellation of some harmonics among the 3 phases.

To improve the input power factor:

–Use DC bias or 3k order component bias on each of the 3 output phase voltages

• Features and applications

Features:

–Direct frequency conversion—high efficiency

–Bidirectional energy flow, easy to realize 4- quadrant operation

–Very complicated—too many power semiconductor devices

–Low output frequency

–Low input power factor and bad input current waveform

Applications:

–High power low speed AC motor drive






4.4 Matrix converter

• Circuit configuration

input

output

a) b)

a b c

u

v

w

S 1

1

S 1

2

S 1

3

S 2

1

S 2

2

S 2

3

S 3

1

S 3

2

S 3

3

S ij






• Usable input voltage

a) b) c)

a) Single-phase input

voltage

b) Use 3 phase voltages

to construct output

voltage

c) Use 3 line-line voltages

to construct output

voltage

Um

U1m

Um 1

2

√3

2 U1m






• Features

Direct frequency conversion—high efficiency can realize good input and output waveforms, low harmonics, and nearly unity displacement factor

Bidirectional energy flow, easy to realize 4- quadrant operation

Output frequency is not limited by input frequency

No need for bulk capacitor (as compared to indirect frequency converter)

Very complicated—too many power semiconductor devices

Output voltage magnitude is a little lower as compared to indirect frequency converter.





UNIT I Power Semiconductor Devices -...

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Transcript of UNIT I Power Semiconductor Devices -...