EE432_1

1

EE432 POWER ELECTRONICS

Dr. Ali M. Eltamaly, King Saud University

Dr. Ali M. EltamalyKing Saud University


Chapter 1

Introduction

1.1 .Definition of Power Electronics

Power electronics refers to control and conversion of electrical power by power semiconductor devices wherein these devices operate as switches

Electronic power converter•Rectifier converting an AC voltage to a DC voltage, •Inverter converting a DC voltage to an AC voltage, •Chopper or a switch-mode power supply that converts a DC voltage to another DC voltage, and •Cycloconverter and cycloinverter converting an AC voltage to another AC voltage.


1.2 Rectificationuncontrolled and controlled rectifiers

DC-To-AC Conversion

•Emergency lighting systems, •AC variable speed drives, •Uninterrupted power supplies, and, •Frequency converters.

DC-to-DC Conversion

•Step-down switch-mode power supply, •Step-up chopper, •Fly-back converter, and ,•Resonant converter.

typical applications•DC drive, •Battery charger, and,•DC power supply.


1.5 AC-TO-AC Conversion

cycloconverter or a Matrix converter converts

Adjustable Speed Drives (ASD)


Diode Circuits or Uncontrolled Rectifier

Rectification: The process of converting the alternating voltages

and currents to direct currents


The main disadvantages of half wave rectifier are :

• High ripple factor,• Low rectification efficiency,• Low transformer utilization factor,

and,• DC saturation of transformer

secondary winding.


Performance Parameters

22dcrmsac VVV

acdc PP / rectification effeciency

dcrms VVFF /

11 22

222

FFVV

VVV

VV

RFdc

rms

dc

dcrms

dc

ac

form factor

ripple factor


121

2

21

21

2

S

S

S

SSi

II

IIITHD

121

2

21

21

2

S

S

S

SSv

VV

VVV

THD

FaactorntDisplacemeFactorDistortionII

IVIV

IVPPF

S

S

SS

SS

SS

*

coscos1

111


Single-phase half-wave diode rectifier with resistive load.


0

sin21 m

mdcV

tdtVV

2sin

21

0

22 mmrms

VtdtVV

RV

RVI mdc

dc

RV

RV

I mrmsrms 2

the load and diode currents RV

II mDS 2


Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1.

mm

mdcVV

tdtVV ))0cos(cos(2

)sin(21

0 RV

RV

I mdcdc

2)sin(

21

0

2 mmrms

VtVV

R

VI m

rms 2

%53.40

2*

2

*

**

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc

57.12

2

m

m

dc

rmsV

V

VVFF

211.1157.11 22 FFVV

RFdc

ac

)d (It is clear from Fig2.2 that the PIV is mV


Half Wave Diode Rectifier With R-L Load

Fig.2.3 Half Wave Diode Rectifier With R-L Load


ttViRdtdiL m 0),(sin*

Divide the above equation by L we get:

ttL

ViLR

dtdi m 0),(sin*

Adtt

LVeeti mdt

LRdt

LR

sin*)(

Adtt

LVeeti mt

LRt

LR

sin*)(

t

LR

m AetLtRLwR

Vti

cossin)( 222

wLjRZ


2222 LwRZ

cosZR sinZL RL tan

R

wLZ

t

LR

m AettZ

Vti

cossinsincos)(

t

LR

m AetZ

Vti

sin)(

tansinsin)(t

mt

LR

m AetZ

VAetZ

Vti

00 i sinZ

VA m

tansinsin)(t

m etZ

Vti


tansinsin)(t

m etZ

Vti

0sinsin)( tan

eZ

Vi m

)cos1(*2

sin*2 0

mmdc

Vtdt

VV

)2sin(1(5.0*2

)sin(*21

0

2

VmdwttVV mrms


)cos1(*2

sin*2 0

mmdc

Vtdt

VV

)2sin(1(5.0*2

)sin(*21

0

2

VmdwttVV mrms


Half wave diode rectifier with free wheeling diode


tansinsin)(t

m etZ

Vti t0

20* tforiRtdidL

tan)(

t

eBti Bi )(

BeZ

Vi m

)sin(sin)( tan

tantansinsin)(

tm eeZ

Vti 2 t


Example 2 A diode circuit shown in Fig.2.3 with R=10 , L=20mH, and VS=220 2 sin314t.

(a) Determine the expression for the current though the load in the period 20 t and determine the conduction angle .

(b) If we connect free wheeling diode through the load as shown in Fig.2.5 Determine the expression for the current though the load in the period of 30 t .

Solution: (a) For the period of t0 , the expression of the load current can be obtained from (2.24) as following:

.561.010

10*20*314tantan3

11 radRL

and 628343.0tan

8084.11)10*20*314(10)( 23222 LRZ

t

tm

et

etZ

Vti

5915.1

tan

*532.0561.0sin8084.11

2220

sinsin)(

tetti 5915.1*0171.14561.0sin3479.26)(


The value of can be obtained from the above equation by substituting for 0)( i . Then, 5915.1*0171.14561.0sin3479.260 e

By using the numerical analysis we can get the value of . The simplest method is by using the simple iteration technique by assuming

5915.1*0171.14561.0sin3479.26 e and substitute different values for in the region 2 till we get the minimum value of then the corresponding value of is the required value. The narrow intervals mean an accurate values of . The following table shows the relation between and :

1.1 6.49518

1.12 4.872781.14 3.231861.16 1.578851.18 -0.0798081.2 -1.73761

It is clear from the above table that 18.1 rad. The current in 2 wt will be zero due to the diode will block the negative current to flow.


(b) In case of free-wheeling diode as shown in Fig.2.5, we have to divide the operation of this circuit into three parts. The first one when t0 (D1 “ON”, D2 “OFF”), the second case when 2 t (D1 “OFF” and D2 “ON”) and the last one when

32 t (D1 “ON”, D2 “OFF”). In the first part ( t0 ) the expression for the load current can be obtained as

In case (a). Then: wtetwti 5915.1*0171.14561.0sin3479.26)( for t0

the current at t is starting value for the current in the next part. Then Aei 1124.14*0171.14561.0sin3479.26)( 5915.1

In the second part 2 t , the expression for the load current can be obtained from (2.30) as following:

tan)(

t

eBti where AiB 1124.14)(

Then teti 5915.11124.14)( for ( 2 t ) The current at 2t is starting value for the current in the next part. Then

Ai 095103.0)2( In the last part ( 32 t ) the expression for the load current can be obtained

from (2.36):

tan2

sin22sin)(

tmm eZ

VitZ

Vti

25915.1532.0*3479.26095103.08442.6sin3479.26)( tetti

25915.11131.148442.6sin3479.26)( tetti for ( 32 t )


Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier


m

mdcVtdtVV 2sin1

0

RVI m

dc 2

2

sin1

0

2 mmrms

VtdtVV

R

VI mrms 2

PIV of each diode = mV2

RVII m

DS 2

Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current.


%05.81

2*

2

2*2

**

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc

11.1222

2

m

m

dc

rmsV

V

VVFF

483.0111.11 22 FFVV

RFdc

ac

The PIV is mV2


Center-Tap Diode Rectifier With R-L Load


)sin(* tViRtdidL m

tansinsin)(t

m etZ

Vti t0

tansin)(

t

m AetZ

Vti

i()=i(2)=i(3)=……..=Io

tansinsin)( eZ

ViI mo

tansin)(

AeZ

VIi mo

sinZ

VIA mo


tansinsin)(

tm

om e

ZVIt

ZVti

tantansinsin)(

t

o

tm eIetZ

Vti

tan2

tan2

sin2sin)(

t

o

tm eIewtZ

Vti 32 t


Single-Phase Full Bridge Diode Rectifier With Resistive Load


Example 4 single-phase diode bridge rectfier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.

VV

tdtVV mmdc 956.190

2sin1

0

A

RV

I mdc 7324.12

2

VVtdtVV mmrms 132.212

2sin1

2/1

0

2

%06.81rmsrms

dcdc

ac

dc

IVIV

PP

11.1dc

rms

VV

FF

482.011 22

222

FFVV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac The PIV=300V

Input power factor = 1cosRe

SS

SS

IVIV

PowerApperantPoweral


.............,5,3,14

cos0cos2

cos2

sin*20

0

nforn

In

nI

tnn

ItdtnIb

oo

oon

)..........9sin917sin

715sin

513sin

31(sin*4)( tttttIti o

%46151

131

111

91

71

51

31))((

2222222

tITHD s

24

1o

SI

I

%34.4814

21

24

1))((2

2

2

1

o

o

S

Ss I

III

tITHD


Example 5 solve Example 4 if the load is 30 A pure DCFrom example 4 Vdc= 190.986 V, Vrms=212.132 V AIdc 30 and rmsI = 30 A

%90rmsrms

dcdc

ac

dcIVIV

PP 11.1

dc

rmsVVFF

482.011 22

222

FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

The PIV=Vm=300V AII o

S 01.272

30*42

41

Input Power factor= PowerApperant

PoweralRe

LagI

IIV

IV

S

S

SS

SS 9.01*30

01.27cos*cos* 11

Full Bridge Single-phase Diode Rectifier with DC Load Current

.............,5,3,14

cos0cos2

cos2

sin*20

0

nforn

In

nI

tnn

ItdtnIb

oo

oon

)..........9sin917sin

715sin

513sin

31(sin*4)( tttttIti o

%46151

131

111

91

71

51

31))((

2222222

tITHD s

24

1o

SI

I

%34.4814

21

24

1))((2

2

2

1

o

o

S

Ss I

III

tITHD

Example 5 solve Example 4 if the load is 30 A pure DCFrom example 4 Vdc= 190.986 V, Vrms=212.132 V AIdc 30 and rmsI = 30 A

%90rmsrms

dcdc

ac

dcIVIV

PP 11.1

dc

rmsVVFF

482.011 22

222

FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

The PIV=Vm=300V AII o

S 01.272

30*42

41

Input Power factor= PowerApperant

PoweralRe

LagI

IIV

IV

S

S

SS

SS 9.01*30

01.27cos*cos* 11

Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier .


0dtdiLV S

sS

Multiply the above equation by td then, SsS diLtdV

o

o

I

ISs

u

m diLtdtV

sin

Then; osm ILuV 2coscos Then; osm ILuV 2cos1

Then; m

osV

ILu 21cos

Then;

m

osV

ILu 21cos 1

m

osV

ILut

21cos1 1


dtdiLv S

srd

oS

I

ISS

u

rd ILdiLtdvo

o

2

the total reduction per period is:

oS

u

rd ILtdv

42

oSoS

rd ILfILV 42

4

u

rd tdv

is the reduction area in one commutation period t

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 42tan


the rms value of the supply current

2u

oI

oI 2u

uIo2

uIo2

sI

2u

2u

22 u

2

2u

2u

2u

2u

22

u

usI

2


]2

[2 2/

0

2

2/

22

u

uo

os tdItdt

uI

I

322

228342 23

2

2 uIuuu

II oo

s

2u

oI

oI 2u

uIo2

uIo2

sI

2u

2u

22

u

2

2u

2u

2u

2u

22 u

usI

2


2u

oI

oI2u

uIo2

uIo2

sI

2u

2u

22 u

2

2u

2u

2u

2u

22 u

usI

sJ 2u

2u

2u

2u

sI 0 0 0 0

sI uIo2

uIo2

uIo2

uIo2

sJ 2u

2u

2u

2u

sI 0 0 0 0

sI uIo2

uIo2

uIo2

uIo2

It is an odd function, then 0 no aa

m

sss

m

sssn tnJ

ntnJ

nb

11sin1cos1

2sin

2sin

2sin

2sin2*11 unununun

uI

nnb o

n

2sin*8

2nu

unIb o

n

2sin*8

1u

uIb o

2sin*

28

1u

uII o

S


32

sin2

32

2cos

2sin4

2cos

322

2sin*

28

2cos*

21

uu

uuu

uu

u

uI

uu

Iu

IIpf

o

o

S

S


2sin*

28

1u

uII o

S

322 2 uII o

s


m

osV

ILu 21cos 1 oSoS

rd ILfIL

V 42

4

osm


322 2 uII o

s

2sin*

28

1u

uI

I oS

32

sin2uu

upf

Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance Ls=5mH supply to feed 200 A pure DC load, find: (i) Average DC output voltage, (ii) Power factor. And (iii) Determine the THD of the utility line current.

VVm 155562*11000

VV actualdc 9703200*005.0*50*415556*2

.285.015556

200*005.0*50**2*21cos21cos 11 radV

ILum

os

917.0

3285.

2285.0

285.0sin*2

32

sin*22

cos*1

uu

uuIIpf

S

S

AuII o

S 85.1933285.0

2200*2

322 22

Auu

II oS 46.179

2285.0sin*

285.0*2200*8

2sin*

28

1

%84.40146.17985.1931

22

1

S

Si I

ITHD

Three-Phase Half Wave Rectifier

mm

mdc VVtdtVV 827.0233sin

23 6/5

6/

RV

RVI mm

dc*827.0

**233

mmmrms VVtdtVV 8407.08

3*321sin

23 6/5

6/

2

RVI m

rms8407.0

RV

RVII mm

Sr 4854.03

08407

ThePIV of the diodes is mLL VV 32

Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side and the load resistance is R=20. If the source inductance is negligible, determine (a) Rectification efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode.

VVVV mS 59.3752*58.265,58.2653

460

mm

dc VVV 827.0233

R

VRVI mm

dc0827

233

mrms VV 8407.0R

VI mrms

8407.0

%767.96rmsrms

dcdc

ac

dcIVIV

PP

%657.101dc

rmsVVFF

%28.1811 22

222

FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

The PIV= 3 Vm=650.54V

Three-Phase Half Wave Rectifier With DC Load Current and zero source induct

New

axis

321 3/

3/0

oo

ItdIa

harmonicstrepleanallfor

nfornI

nfornI

tnnIdwttnIa

o

o

oon

0

17,16,11,10,5,43*

,....14,13,8,7,2,13*

sincos*1 3/3/

3/

3/

...8cos

817cos

715cos

514cos

412cos

21cos3

3)( ttttttIItI OO

s

%24.1090924.119

*21

23

3/1))((

2

2

2

1

O

o

S

Ss I

III

tITHD

Example 8 Solve example 7 if the load current is 100 A pure DC

VVVV mm

dc 613.310827.0233

AIdc 100

VVV mrms 759.3158407.0

%37.98100*759.315100*613.310

rmsrms

dcdc

ac

dcIVIV

PP

%657.101dc

rmsVVFF

%28.1811 22

222

FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac



Three-Phase Half Wave Rectifier With Source Inductance


m

oVLIu

321cos 1

m

oVLIut

321cos1 1

oo

rd ILfLIV 32

3

om

Actualdc ILfVV 3233


Example 9 Three-phase half-wave diode rectifier connected to 66 kV, 50 Hz , 5mH supply to feed a DC load with 500 A DC, fined the average DC output voltage.

Solution: Vvm 538892*3

66000

(i) oceinduc

sourcewithout

dcActualdc ILfVV 3tan

VILfVV om

Actualdc 44190500*005.0*50*32

53889*3*33233

Three-Phase Full Wave Rectifier With Resistive Load

1 3 5

4 6 2

b

c

IL

V LIs

Ip

a

LLmLLm

mdc VVVVtdtVV 3505.1654.12333sin33 3/2

3/

RV

RV

RV

RVI LLLLmm

dc3505.123654.133

LLmmmrms VVVtdtVV 3516.16554.14

3*923sin33 3/2

3/

2

RVI m

rms6554.1

RV

RVI mm

r 9667.03

6554.1

RVI m

S 29667.0

Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20ohms. If the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode .

VVVV mm

dc 226.621654.133

A

RV

RVI mm

dc 0613.31654.133

VVVV mmrms 752.6216554.14

3*923

AR

VI mrms 0876.316554.1

%83.99rmsrms

dcdc

ac

dcIVIV

PP

%08.100dc

rmsVVFF

%411 22

222

FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac


Three-Phase Full Wave Rectifier With DC Load Current

....,.........15,14,12,10,9,8,6,4,3,2,0

.....),........3(132

),3(112

),3(7

2),3(

52

,32

cos2

sin*2

1311751

6/56/

6/5

6/

nforb

Ib

Ib

Ib

Ib

Ib

tnn

ItdtnIb

n

ooooo

oon

ttttv�tItI o

s

13sin13111sin

1117sin

715sin

51sin32)(

%31251

231

191

171

131

111

71

51))((

22222222

tITHD s

oS II32 oS II

3*2

1

%01.311/3*23/21))(( 2

2

1

S

Ss I

ItITHD

Power Factor =S

S

S

SII

II 11 )0cos(*

LL

oSV

ILu 21cos 1o

ord fLILIV 6

26

oLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan

632 2 uII o

S

2sin

621

uuI

I oS

63

sin*32

cos

632

2sin62

2cos

21

uu

uuuI

uu

Iu

IIpf

o

o

S

S

Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find;Commutation time and commutation angle.DC output voltage.Power factor.Total harmonic distortion of line current.

oradu 61.14.2549.0

LL

oSV

ILu 21cos 1

dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan

VVdcactual 43830300*008.*50*633000*35.1 9644.0

62549.0

3*2549.0

2549.0sin3

63

sin*3

uu

upf

AuII d

s 929.2396

2549.03

*300*263

2 22

Auu

II o

S 28.2332

2549.0sin*2*2549.0*

300*3432*2

sin2

341

9644.02

2549.0cos*929.23928.233

2cos*1

uIIpf

s

S

%05.24128.233

929.239122

1

S

Si I

ITHD

For three phase uncontrolled rectifier with pure DC current load without source inductance shown in Fig.1, draw the following: (a) The output voltage waveform in waveform (2) of Fig.2. (b) The currents waveforms of switches 1 to 6 in the waveforms (3) to (8) of Fig.2, (c) Secondary current of phase b in waveform (9) of Fig.2, (d) Primary current of phase B in waveform (10) of Fig.2, Then,

(e) Derive an equation of secondary current tib waveform by using Fourier transforms. (f) Derive an equation of Primary current tiB waveform by using Fourier transforms. (g) Find its input power factor at the supply (primary) terminals .

A

B

C

a

b

c

IA

IB

IC

Ia

Ib

Ic

IAB

IBC

ICA n

2:1D1 D3 D5

D4 D6 D2

30A

=I'a

=I'b

=I'C

EE432_1

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