power electronicspe.gastli.info/chapter3/pe_ch3.pdf · 2006. 11. 11. · dr. adel gastli rectifiers...

POWER ELECTRONICSPOWER ELECTRONICS

RECTIFIERSRECTIFIERS((AC to DC Converters)AC to DC Converters)

Dr. Adel GastliEmail: [email protected]

Dr. Adel Gastli Rectifiers (DC-DC Converters) 2

Learning ObjectivesLearning ObjectivesTo understand the operation and characteristics of rectifiers.

To learn the types of rectifiers.

To understand the performance parameters of rectifiers.

To learn the techniques for analysis and design of rectifiers using Matlab/Simulink, and PSIM.

To study the effects of load inductance on the load current.


CONTENTSCONTENTSCONTENTS1. Introduction2. Applications3. Topologies4. Performance Parameters5. 1-Phase Half-Wave Rectifier6. 1-Phase Full-Wave Rectifier7. Line Quality Issues8. 3-phase Rectifiers9. Comparison of Rectifiers10. Summary


INTRODUCTIONINTRODUCTION

AC to DC converters without control are known as diode rectifiers. They are designed using diodes.

Their designs are not expensive and are very popular in the industrial applications.

AC to DC controlled converters are designed using thyristors.

The rectifiers are required to supply ripple-free dc voltage or dc current to the load.


The rectifiers usually draw highly non-sinusoidal current from the electric utility supply, giving rise to poor power factor and thus poor efficiency.

Improving power factor is a very important design objective. (techniques will not be discussed in this chapter)

Another design concern is the reduction of high frequency distortions in the line current, which are caused by switching loads or switch-mode converters as loads.


In this chapter we will study the followings:

Basic rectifier concepts

Some typical rectifier topologies

Distortion issues


APPLICATIONSAPPLICATIONS

1. DC power supplies for computers and electronic equipments:

• Low power level (<500W). • Source is single-phase• The dc voltage can be processed by a

switch-mode dc-to-dc converter to provide multiple output voltages with minimum ripple content.

• Draw highly non-sinusoidal current, hence, the line quality issues are very critical.


2. Battery charging systems• With more portable instrumentation such as

in wireless mobile communications and computers the demand for inexpensive battery charging systems has exploded.

3. High voltage dc (HVDC) transmission converters.

• Power transmission over high voltage dc lines is becoming more popular than on the ac lines. It is more economical.

• There is no problem of differences in frequencies (50Hz or 60Hz) and frequencies transients.


TOPOLOGIESTOPOLOGIES

ac-dc Converter (switch network)

ac source

dc load

is io

vovs

1-phase2-phases3-phases

4May involve a step-down or -up transformer4Center-tap transformer4Δ or Y connection transformers


vo

vp

Ls

S1

1-phase

vo

vp

Ls

S2

vp

Ls

S1

A B

2-phase

vo

vp

Ls

S3

vp

Ls

S2

B C

vp

Ls

S1

A

3-phase

HalfHalf--Wave RectifiersWave Rectifiers

Switches are usually diodes that are self-controlled or thyristors (SCR) that are controlled, unidirectional and unipolar.

They can turn on when voltage across them is positive (forward biased). When on, the voltage becomes zero. They turn off when voltage across them becomes negative or current becomes zero

and tends to reverse (reverse biased).


FullFull--Wave RectifiersWave Rectifiers

voLs

vp

S2

S4

S1

S3

1-phase full-wave2-pulse

vo

Ls

vp

A

Ls

vp

B

Ls

vp

C

S1

S4

S3

S6

S5

S2

3-phase Y-full-wave6-pulse

voA

vLL

B

Ls

C

S1

S4

S3

S6

S5

S2

3-phase Δ-full-wave6-pulse

Number of sinusoidal peaks in a period


RemarksRemarksHalf-wave topology has less semiconductor switches but requires higher component stresses.

Full-wave topology has more switches but is capable of handling high power with minimum component stresses.


Types of LoadsTypes of Loads• Resistive Load: (L=0, E=0)• Inductive-Resistive: (medium L, low R and

E=0) magnetic lift and relays.• Inductive-Voltage Sink: (medium L, low R and

E) DC motors, HVDC bus and battery charging circuit.

• Current Sink: (high L, low R and E) DC motors, heavy magnetic pick ups and relays.

• Capacitive-Resistive: (L=0, R and E replaced by capacitor)

• Voltage Sink: (L=0, low R and high C) DC power supplies


PERFORMANCE PARAMETERSPERFORMANCE PARAMETERS

• Although output voltage is dc, it is discontinuous and contains harmonics.

• A rectifier is a power processor that should give:– Minimum amount of harmonic contents.– Sinusoidal input current (as possible).– Close to unity power factor.

• Knowledge about harmonic contents of input and output voltage and current waveforms is required.

• Thus, Fourier series expansions are to be used.


Fourier series expansion:

∑∞

=

++=,...2,1

000 )sin(n

nndc nIIi φθ

Average value. Peak amplitude of nth harmonic.

Phase angle of nth

harmonic with respect to source voltage.

Output current:

Input current:

∑∞

=

+=,...2,1

)sin(n

snsns nIi φθ


dcV : Average value of the output voltage

dcI : Average value of the output current

rmsV : rms value of the output voltage

rmsI : rms value of the output current

The output voltage of a rectifier circuit is composed of 2 components:

one dc component, Vdc and one ac, Vac (Fourier Series).

22dcrmsac VVV −= : effective (rms) value of ac component

R

L

sv

D

si

Lvpv


dcdcdc IVP = : Output dc power

rmsrmsac IVP = : Output ac power

ac

dc

P

P=η : efficiency or rectification ratio

There are different types of rectifier circuits and their performances are evaluated in terms of the following parameters:

dc

rms

V

VFF = : form factor (shape of output)

dc

ac

V

VRF = : ripple factor (ripple of output) 12 −= FFRF

ss

dc

IV

PTUF = : Transformer utilization factor


φcos=DFDisplacement factor :

12

121

21

2

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

−==

s

s

s

ss

I

I

I

IITHDHF

Harmonic factor or Total harmonic distortion factor of the input current:

φφ coscos 11

s

s

ss

ss

I

I

IV

IVPF ==Input power factor.

φ : angle between the fundamental components of voltage and current. It’s called displacement angle

Is1 is the rms value of fundamental component of the input current. Is rms value of input current.


s

peacks

I

ICF )(= Crest factor of input current.

NotesNotes:1. HF is a measure of the distortion of a waveform and also

known as total harmonic distortion (THD).2. If is is purely sinusoidal, Is1=Is and PF=DF. DF becomes the

impedance angle θ=tan-1(ωL/R) for an RL load.3. DF is also known also as displacement power factor (DPF).4. An ideal rectifier should have η=100%, Vac=0, RF=0,

TUF=1, HF=THD=0, and DPF=1.


11--PHASE HALFPHASE HALF--WAVE DIODE WAVE DIODE RECTIFIERRECTIFIER

Resistive Load:

R

D

vs

iD

+ vD -

Vo

+

_

0 0.005 0.01 0.015 0.02-400

-300

-200

-100

0

100

200

300

400

io

vo

vs

R

vii

v

vvv

Vv

D

s

ss

ms

00

0 0 if 0

0 if

)sin(

==

⎩⎨⎧

≤>

=

= θ


Example 1Example 1

Determine:

a) Efficiency

b) FF: form factor

c) RF: ripple factor

d) TUF: transformer utilization factor

e) PIV: peak inverse voltage of Diode.

f) CF: Crest factor

g) PF: power factor

R

D

sv

tωπ π20

( )tVv ms ωsin= 0v

mV

pv

+

_

+

_

+

_

0v

tωπ π20

mV


Solution:

⎟⎠⎞

⎜⎝⎛ −−=== ∫∫ 1

2cos)sin(

1)(

1 2/

00 0

T

T

VdttV

Tdttv

TV m

T

m

T

dc

ωω

mm

dc VV

V 318.0==π

The average output voltage Vdc is defined as

The rms value of a periodic waveform is defined as

fTf

1,2 == πω

R

V

R

VI mdc

dc 318.0==

m

T

m

T

Lrms VdttVT

dttvT

V 5.0)(sin1

)(1 2/

0

22

0

2 === ∫∫ ω

R

V

R

VI mrms

rms 5.0==


a)( )

( )%5.40

5.0

318.02

2

====rmsrms

dcdc

ac

dc

IV

IV

P

Pη

b) %157or 57.1318.0

5.0===

dc

rms

V

VFF

c) 121%or 21.1157.11 22 =−=−= FFRF

d) Transformer secondary voltage rms:

mm

s VV

V 707.02

==

This rectifier has a high ripple factor

This rectifier has a low efficiency


Transformer secondary current is same as that of the load:

R

VI m

s 5.0=

VA rating of the transformer is:

R

VIV m

ss

2

5.0707.0 ×=

( )286.0

5.0707.0

318.0 2

=×

==ss

dc

IV

PTUF

e) Peak inverse (or reverse) blocking voltage: mVPIV =


f) R

VI m

peaks =)(R

VI m

s 5.0=

Crest factor of input current: 2)( ==s

peaks

I

ICF

g) Input power factor for a resistive load can be found as:

707.05.0707.0

5.0 2

=×

==VA

PPF ac

Note:1/TUF=1/0.286=3.496 signifies that the transformer must be 3.5 times larger than when it is used to deliver power from a pure ac voltage. In addition it has to carry a dc current, whichresults in a dc saturation problem of the transformer core.

(Lagging)


R-L Load:

)2( 0

)0( )0(]sin)[sin()(

0

/0

/

220

πθσπ

σπθααθω

ωθωθ

<<+=

+<<++−+

= −−

i

eIeLR

Vi LRLRs

L

D

vs

i0

+ vd -

vo

+

_

R

( )

( ) σπθθθ

ω

ω

+<<=+

=+

0 sin

,sin

00

00

m

m

Vd

diLRi

tVdt

diLRi

⎟⎠⎞

⎜⎝⎛= −

R

Lωα 1tan

0 0.005 0.01 0.015 0.02-400

-300

-200

-100

0

100

200

300

400

vo

io

vdσ


Skip proof

• It is assumed that the load current flows during the period

( )tVRidt

diL m ωsin0

0 =+D conducts:

0 σπω +<< t

The homogeneous equation is defined by:

0)()(

or 0 00

00 =+=+ θ

θθω Ri

d

diLRi

dt

diL

Proof:


The solution to this homogeneous equation is called the complementary integral:

( ) L

R

Aei ωθ

θ−

=0

The particular solution is the steady-state response.

( ) ( )αωθ −= tZ

Vi m sin0

( )

( ) ( )2222

221

sin,cos

and tan where

LR

L

LR

R

LRZR

L

ω

ωαω

α

ωωα

+=

+=

+=⎟⎠⎞

⎜⎝⎛= −


The total solution is the sum of both the complimentary and the particular solution and it is shown as:

( ) ( )αθθ ωθ

−+=⎟⎠⎞

⎜⎝⎛ −

sin0 Z

VAei mL

R

( ) ( )αsin000 Z

VAi m=⇒=

( ) ( ) ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡−+=

⎟⎠⎞

⎜⎝⎛ −

αθαθ ωθ

sinsin0L

R

m eZ

Vi

As ωt increases, the current would keep decreasing.

For some value of ωt, say π+σ , the current would be zero.

If ωt > π+σ , the current would evaluate to a negative value.


Since the diode blocks current in the reverse direction, the diode stops conducting when ωt reaches π+σ.

Then an expression for the average output voltage can be obtained.

[ ]

R

VI

Vd

VV

dcdc

smmdc

=

+−== ∫+

)cos(12

.sin2 0

σππ

θθπ

σπ

)2( 0 ),(0 ,0 00 πθσπσπθ <<+=+<<== vvvV sL

Average voltage drop across L

Note that Vdc

is maximum for σ=0.


Simulink SimulationSimulink Simulation

Try to run “psbdiode.mdl” and change the value of L to notice its effect on waveforms.


R-L Load with freewheeling diode:

L

D1

vs

id

+ vd -

vo

+

_

R

Dm

L

D1

vs

id

+ vd -

vo

+

_

R

Dm

Mode 1

)0( )0(]sin)[sin()(

/0

/

220 πθααθω

ωθωθ <≤++−+

= −− LRLRs eIeLR

Vi

LRLRs eIeLR

VI ωπωπα

ωπ /

0/

220 )0(]1)[sin()(

)( −− +++

=

( ) )0 ( sin00 πθθ

θω <<=+ mV

d

diLRi


( ))2( )()( 00 πθππθ ω

πθ

<<=−

−L

R

eIi

)( 0 00

0 iid

diLRi L ==+

θω

L

D1

vs

id

+ vd -

vo

+

_

R

Dm

Mode 2

L-R

L-Rs

e

eα

LR

VII ωπ

ωπ

ωπ

/

/

2200 1sin

)()0()2(

−+==

)0( )0(]sin)[sin()(

)( /0

/

220 πθααθω

θ ωθωθ <≤++−+

= −− LRLRs eIeLR

Vi

Mode 1

Mode 2

Prove this equation

Diode Dm short-circuits the load: v0=0


⎥⎦⎤

⎢⎣⎡ +== ∫∫∫ θθ

πθ

ππ

π

ππdididiIdc

2

00 0

2

0 0 2

1

2

1

R)L(L

V

LR

LV

IIII

s

s

>>=

+=

−=−

ωω

ωω

π

)(

)0()(

22

00min0max0

DC component

Ripple component

The load current has 2 components:

Prove this equation


0 0.005 0.01 0.015 0.02-400

-300

-200

-100

0

100

200

300

400

vo

io

vD

Negative part of the voltage is

removed


Voltage Sink Load: (i.e. Battery charger)

tω

R

EVm −

0i

sv

tωπ π20

mVE

Rv

tω0EVm −

α βαπβα −== − ,sin 1

mV

E

R

EtV

R

Evi ms −

=−

=ωsin

0

βωα << t

i0

E

R

D

sv 0vpv+

_

+

_

+_

Rv


ExampleExampleBattery voltage E=12V and its capacity is 100Wh.The average charging current should be Idc=5A. The primary voltage is Vp=120V, 60Hz, and the transformer turn ratio is n=2:1. Calculate:

a) Conduction angle δ of the diode.b) The current-limiting resistance, Rc) The power rating PR of Rd) The charging time h0 in hourse) The rectifier efficiency ηf) The PIV of the diode

i0

E

R

D

sv 0vpv+

_

+

_

+_

Rv


Solution:Solution:

a) Conduction angle:

V85.842

V,602/120/,V120,V12

==

=====

sm

psp

VV

nVVVE

o

o

74.163

87.17113.8180

=−=

=−=

αβδ

β

o13.885.84

12sinsin 11 === −−

mV

Eα

αβδ −=

b) Average charging current:

( ) ( ))2(cos22

1sin

2

1 πααπ

ωωπ

β

α−+=

−= ∫ EV

Rtd

R

EtVI m

mdc


( ) Ω=−+= 26.4)2(cos22

1 πααπ

EVI

R mdc

c) The rms battery current Irms is:

( )

( )

W4.28626.42.8A2.8

4.67

sin42sin2

222

1

sin

2

1

2

22

2

2

=×=⇒=

=

−+−⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎟⎠⎞

⎜⎝⎛ −

= ∫

R

mmm

mrms

P

EVV

EV

R

tdR

EtVI

αααππ

ωωπ

β

α

RIP rmsR2=


d) The power delivered Pdc to the battery is:

W60512 =×== dcdc EIP

%32.171004.28660

60

powerinput total

battery todeliveredpower

=×+

=

+==

Rdc

dc

PP

Pη

h667.160

100100Wh 100 00 ===⇒=

dcdc P

hPh

e) Efficiency η is:

f) Peak inverse voltage PIV is:

V85.961285.84PIV =+=+= EVm


Fourier series expansion of output voltage:Fourier series expansion of output voltage:

( ) ( ) ( )( )∑∞

=

++=K,2,1

0 cossinn

nndc tnbtnaVtv ωω

( )π

ωωπ

ωπ

ππm

mdc

VtdtVtdvV === ∫∫ )(sin

2

1)(

2

10

2

0 0

( ) ( ) ( )

K2,3,4,5,6,nfor 0

1nfor 2

)(sinsin2

1)(sin

2

10

2

0 0

==

==

== ∫∫m

mn

V

tdtntVtdtnva ωωωπ

ωωπ

ππ


( ) ( ) ( )

( )

K

K

1,3,5,nfor 0

,6,4,2nfor 1

1-1

)(cossin2

1)(cos

2

1

2

0

2

0 0

==

=−

+=

== ∫∫

n

V

tdtntVtdtnvb

nm

mn

π

ωωωπ

ωωπ

ππ

( )

K−

−−−+=

tV

tV

tV

tVV

tv

m

mmmm

ωπ

ωπ

ωπ

ωπ

6cos35

2

4cos15

22cos

3

2sin

20


11--PHASE HALFPHASE HALF--WAVE WAVE CONTROLLED RECTIFIER CONTROLLED RECTIFIER

+

_

v0

i0T1

+ _vT1

vs=Vmsin(ωt)

+

__

+

vp

(a) Circuit

i0Idc

v0Vdc

0

(b) Quadrant

v0/R

ωt

ωt

ωt

2π

2π

2ππ

π

πα

α

α

i0

v0Vm

V1

VT1

V1

Vm

vs

0

0

0

02π

α π

(c) waveforms

ωt

-Vm

( )απ

θθπ

π

α

cos12

sin2

1)(0

+=

= ∫m

mdc

V

dVV

R

VI dc

dc)(0

)(0 =

⎟⎠⎞

⎜⎝⎛ +−=

= ∫

2

2sin1

2

sin2

1 22)(0

ααππ

θθπ

π

α

m

mrms

V

dVV


11--PHASE FULLPHASE FULL--WAVE WAVE DIODE RECTIFIERDIODE RECTIFIER

Center-tapped transformer:

0 π 2π

voVm

ωt

Diode D1

+

_

Diode D2

+

_viac supply

Load resistance R

vs

vo+ –

tVv ms ωsin=

-2Vm

ωtvD

tVv ms ωsin=

0 π 2π

Vm

ωt

vs


Bridge rectifier:tVv ms ωsin=

+

_

+

_

vp vs

0 π 2π

0 π 2π

vo

Vm

Vm

ωt

ωt

-Vm

ωt

vD

vs

D1

vo

+

_

D3

D4 D2

R

( )R

VP

R

V

R

VI

VV

V

mdc

mdcdc

mm

dc

26366.0

6366.0

6366.02

=

==

==π


R

VP

R

V

R

VI

VV

Vm

ac

mrmsrms

mm

rms

2707.0

707.02

2

=

⎪⎪⎭

⎪⎪⎬

⎫

==

==

a) Efficiency η:

%81100707.0

6366.02

2

=×==ac

dc

P

Pη

b) Form Factor:

11.16366.0

707.0FF ===

dc

rms

V

V


c) Ripple Factor:

%2.48or 482.0111.11RF 22 =−=−= FF

d) TUF:

mm

s VV

V 07072

==

rms of transformer secondary voltage:

rms of transformer secondary current is same as that of output current:

R

VI m

s

0707=

ss

dc

IV

PTUF =


%81or 81.0707.0

6366.02

2

===ss

dc

IV

PTUF

e) Peak reverse blocking voltage, PIV=Vm.

2707.0

1

/707.0

/)( ====RV

RV

I

ICF

m

m

s

peaks

f) Crest factor:

g) Input PF for a resistive load can be found from:

1707.0

707.02

2

===VA

PPF ac

Much higher then that of a center-taped transformer (PF=0.707 see textbook p.79))


Current Sink Load (highly inductive load):

Io

is

D1 L

vs

vo

+_

D4

D2

D3

Mode 1

Mode 2

Mode 1: 0<θ<π

Mode 2: π<θ<2π

πθθ

ππ

mmdc

VdVV

2 sin

10

=⋅= ∫

Io

vsis

vo

( )θπ

nn

Ii

ns sin

4

.5,3,1

0∑∞

=

=L

Odd harmonics only

Mode 1 Mode 2 Mode 1


[ ] [ ]( )

[ ] [ ]( ))cos()2cos()cos(1

)cos()cos(

)sin()sin(1

)sin(1

0

20

0

0

2

00

2

0

ππππ

θθπ

θθθθπ

θθπ

ππ

π

π π

π

π

nnnn

I

nnn

I

dnIdnIdnib sn

−+−=

−−−=

⎟⎠⎞⎜

⎝⎛ −== ∫ ∫∫

[ ] [ ]( )ππ n

I

n

Ib

b

00,5,3,1

,6,4,2

4)1(111

0

=−−++=

=

L

L

[ ] [ ]( ) 0)sin()sin(

)cos()cos(1

)cos(1

20

0

0

2

00

2

0

=−=

⎟⎠⎞⎜

⎝⎛ −== ∫ ∫∫

ππ

π

π π

π

π

θθπ

θθθθπ

θθπ

nnn

I

dnIdnIdnia sn

PROOFPROOF


4834.018

14

21

222

1

=−=−⎟⎟⎠

⎞⎜⎜⎝

⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ππ

s

s

I

ITHD

π2

4 01

IIs =

00

20

1IIIs == ∫

π

π( )θ

πn

n

Ii

ns sin

4

.5,3,1

0∑∞

=

=L

(Study example 3(Study example 3--6 in your textbook)6 in your textbook)


Voltage Sink Load (Continuous Conduction Mode):

io

is

D1 LR

vs E+_D4

D2

D3

Mode 1: Mode 1: D1 & D3

conduct0≥sv

Mode 2: Mode 2: D2 & D4

conduct0≤sv

R

EVIV

ERivEvvv

L

RL

−=⇒=

−−=−−=

00

000

0

v0

θsin0 mVv =

svv =0

svv −=0

θsin00

0 mVERidt

diLv =++= i0 can be found by solving

this differential equation.


% nc_sp_bg_VSm.mVm =max(vs);w =2*pi*50;O =w*tout;O1 =O*180/pi;Im =max(i0);plot(O1,vs/Vm,O1,v0/Vm,O1,E/Vm,O1,is/Im,O1,i0/Im);axis([0 max(O1) -1.2 1.2]);legend('vs/Vsm','vo/Vsm','E/Vsm','is/Im','io/Im',3);


Clock

Single phase bridge rectifierwith current sink loadload

+

-v

+

-v

+ -vis

tout

vs

v0

E

i0

Press toPlot Results

R

L

+i-

+i-

ak m

D4

ak m

D3

ak m

D2

ak m

D1

220V50Hz

Vm = 220VE = 180VR = 0.1ΩL = 3mH


0 100 200 300 400 500 600 700

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

vs/Vm

vo/Vm

E/Vm

is/Iom

io/Iom

π 2π 3π 4π

Mode 1 Mode 2 Mode 1 Mode 2


Voltage Sink Load (Discontinuous CM):

io

is

D1 LR

vs E+_D4

D2

D3

Mode 1: Mode 1: ii00(0)=0(0)=0

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

mV

E1sinα

D1 & D3 start conducting

πβθα ≤<<

Evs ≥

Evs =)(α

D1 & D3 stop conducting

Mode 2: at Mode 2: at ββ

00 =i

Ev =0

svv =0

απθβ +<<


Evs =+− )( απ D2 & D4 become forward biased and start conducting

Mode 3: at Mode 3: at π+απ+α

sii =0 Evs ≥− βπθαπ +<<+

Mode 4: at Mode 4: at π+βπ+β

D2 & D4 stop conducting because negative current cannot flow in them.

00 =i

Ev =0 απθβπ +<<+ 2


% nc_sp_bg_VSm.mVm =max(vs);w =2*pi*50;O =w*tout;O1 =O*180/pi;Im =max(i0);plot(O1,vs/Vm,O1,v0/Vm,O1,E/Vm,O1,is/Im,O1,i0/Im);axis([0 max(O1) -1.2 1.2]);legend('vs/Vsm','vo/Vsm','E/Vsm','is/Im','io/Im',3);


Clock

Single phase bridge rectifierwith current sink loadload

+

-v

+

-v

+ -vis

tout

vs

v0

E

i0


R

L

+i-

+i-

ak m

D4

ak m

D3

ak m

D2

ak m

D1

220V50Hz

Vm = 220VE = 180VR = 0.1ΩL = 0.3mH


0 100 200 300 400 500 600 700

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

vs/Vm

vo/Vm

E/Vm

is/Im

io/Im

α β π+α π+β 2π+α


( )

( ) 0coscos

0sin

=−−−⇒

=−=⋅ ∫∫αββα

θθθβ

α

β

α

m

mL

V

E

dEVdvNon linear equation. Can be solved for ββ

graphically or by numerical methods using

a computer program.

( ) ( )αθω

θαω

θω

θθ

α−−−== ∫ L

E

L

Vd

L

vi mL coscos)(0

θπ

β

αdiI ∫= 00

1 ( )παω

αω

−+= 2cos20 L

E

L

VI m

peak

Peak current),0,0(for

0

0 πβαω

===== EL

V

I

ICF mpeak

Crest Factor

απθ −=

(Study example 3(Study example 3--7 in your textbook, Use 7 in your textbook, Use PSIMPSIM))


FullFull--Wave Bridge Controlled Rectifier Wave Bridge Controlled Rectifier

Highly inductive load: Ia=constant

R

VI

V

dVV

dcdc

m

mdc

)(0)(0

)(0

cos2

sin2

2

=

=

= ∫+

απ

θθπ

απ

α

arms

sm

mrms

II

VV

dVV

=

==

= ∫+

)(0

22)(0

2

sin2

2 απ

αθθ

π


Bridge Rectifier (RL load)

( )

( )

( / )1

22 1

2 sin 0

2sin

tan ( / )

oo s o

R L to

diL Ri E V t for i

dt

V Ei t A e

Z R

Z R L L R

ω

ω θ

ω θ ω

−

−

+ + = ≥

= − + −

= + =

• Case 1: Continuous conduction, io(0) >0• Case 2: Discontinuous conduction, io(0) =0


Bridge Rectifier (RL load)

( )

( )

( ) ( )

( / )( / )0

( / )( / )

1 ( / )( / )

0

2sin

2sin 0

sin sin2

10

so

R L tsLO

R Ls

LO L R L

V Ei t

Z R

VEI e for i

R Z

eV EI I

Z e Rfor i

π ω

π ω

π ω

ω θ

α θ

α θ α θ

− −

−

−

= − −

⎡ ⎤+ + − − ≥⎢ ⎥

⎣ ⎦− − − −

= = −−

≥

• Case 1:Case 1: Continuous conduction, io(0) =IL0>0

Beginning of mode 1End of mode 1


( ) 0

1

1sin

210 =−

−

+−−==

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

R

E

e

e

Z

VII

L

R

L

R

sLL

ωπ

ωπ

θα

( )R

Z

V

E

e

e

sL

R

L

R

21

1sin −=

−

+−⇒

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

ωπ

ωπ

θα

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×

+

−−=⇒

−

−

−

θθα

θπ

θπ

cos1

1sin

tan

tan1 x

e

ec

Bridge Rectifier (RL load)• Case 2:Case 2: Discontinuous conduction, io(0)=0

Critical value of α

00 =⇒≥ Lc Iαα

sV

Ex

R

Z

2 ,

cos

1==

θ


Example 10.2

( )1,3,5,..

1

4( ) sin

40.90

2

as

n

as a

s a

Ii t n n t n

n

II I

I I

ω απ

π

∞

=

= −

= =

=

∑

1

cos( )

2 2cos( ) cos( )S

S

DF

IPF

I

α

α απ

= −

= − = −ωt

2ππα π+α

T3,T4 T1,T2 T3,T4On

Vm

0

Ia

Ia

i0

is

0

2ππα π+α

v=Vmsin(ωt)

-Ia


LINE QUALITY ISSUESLINE QUALITY ISSUES

The current flowing in one load has an effect on the voltage applied to other loads.Current harmonics generated by one load affect the power quality of the power system, thus, the performances of other loads connected to the same system.

G

Load 1

Load 2

Load 3Vg ve

vs

iline

i3

i2

i1

ZL


33--Phase Bridge RectifiersPhase Bridge Rectifiers

( )( )o

mcn

ombn

man

tVv

tVv

tVv

240sin

120sin

sin

−=

−=

=

ω

ω

ω

Single-phase: High output voltage ripple

Low ripple frequency (2fs) Limitations

Limitations can be overcome or minimized using multiphase (3φ) input sources.

( )( )( )o

mca

ombc

omab

tVv

tVv

tVv

210sin3

90sin3

30sin3

−=

−=

+=

ω

ω

ω

io

vcn

v0

+_

D1

D4

D3

D6

D5

D2

a

c

b

ia

ib

ic

van

vbn

n


Constant Output Current

io

Ls

v0

+_

D1

D4

D3

D6

D5

D2

A

B

C

iA

iB

iC

Ls

Ls

Mode 1:Mode 1:

CABCAB vvv & >

D1 & D6 conduct

0>ABv

io

Ls

v0

+_

D1

D4

D3

D6

D5

D2

A

B

C

iA

iB

iC

Ls

Ls

Mode 2:Mode 2:

BCABCA vvv & >

D1 & D2 conduct

0>ACv


io

Ls

v0

+_

D1

D4

D3

D6

D5

D2

A

B

C

iA

iB

iC

Ls

Ls

Mode 3:Mode 3:

CAABBC vvv & >

D3 & D2 conduct

0>BCv

io

Ls

v0

+_

D1

D4

D3

D6

D5

D2

A

B

C

iA

iB

iC

Ls

Ls

Mode 4:Mode 4:

CABCAB vvv & >

D3 & D4 conduct

0>BAv


io

Ls

v0

+_

D1

D4

D3

D6

D5

D2

A

B

C

iA

iB

iC

Ls

Ls

Mode 5:Mode 5:

BCABCA vvv & >

D5 & D4 conduct

0>CAv

io

Ls

v0

+_

D1

D4

D3

D6

D5

D2

A

B

C

iA

iB

iC

Ls

Ls

Mode 6:Mode 6:

CAABBC vvv & >

D5 & D6 conduct

0>CBv


% nc_3p_bg_CSm.mIo =max(i0); Vm =max((vA-vB)); V0 =mean(v0); O1 =2*50*t*180;subplot(211)plot(O1,(vA-vB)/Vm,O1,(vB-vC)/Vm,O1,(vC-vA)/Vm,O1,v0/Vm);axis([0 max(O1) -1.5 1.5]); xlabel('Angle (^o)'); ylabel('Voltages'); grid;subplot(212)plot(O1,i0/Io,O1,iA/Io,O1,iB/Io,O1,iC/Io);axis([0 max(O1) -1.5 1.5]); grid; xlabel('Angle (^o)'); ylabel('Currents');


Vm = 220VI0 = 10A

current sink loadThree phase bridge rectifier with

+ -v

+ -v

+ -v

+

-v

vC

vB

vA

iC

iB

iA

vC

vB v0

vA

i0


+i-

+i-

+i-

+i-

10

I0

ak m

D6

ak m

D5

ak m

D4

ak m

D3

ak m

D2

ak m

D1

sign

al


0 100 200 300 400 500 600 700-1.5

-1

-0.5

0

0.5

1

1.5

Angle (o)

Vol

tage

s

0 100 200 300 400 500 600 700-1.5

-1

-0.5

0

0.5

1

1.5

Angle (o)

Cur

rent

s

vo

vAB vBCvCA

iA iB iCi0


( ) )2/6/( , 6/sin30 πθππθ <≤+−=−= mab Vvv

The output voltage v0 is periodical with a period of 60o.

The average output voltage can be calculated over one period from π/6 to π/2 (mode 4).

( )

mm

m

VV

dVV

654.133

6/sin33/

1 2/

6/0

==

+−= ∫

π

θπθπ

π

π0

6/5

6/

20 3

21IdIIL == ∫

π

π

θπ

rms value of the line current

( )

mm

mrms

VV

dVV

6554.14

39

2

3

6/sin33/

1 2/

6/

22

=×+=

+= ∫

π

θπθπ

π

π


Purely resistive load: Purely resistive load: Run “nc_3p_bg_R.mdl”

Three phase bridge rectifier with Purely Resistive load

+ -v

+ -v

+ -v

+

- v

vC

vB

vA

id

iC

iB

iA

vC

vB v0

vA

vd

i0


R

+ i-

+ i-

+ i-

+i-

Demux

ak m

D6

ak m

D5

ak m

D4

ak m

D3

ak m

D2

ak m

D1


0 100 200 300 400 500 600 700 800

-1

0

1

Angle (o )

Vol

tage

s

0 100 200 300 400 500 600 700 800

-1

0

1

Angle (o )

Dio

de C

urre

nt

0 100 200 300 400 500 600 700 800

-1

0

1

Angle (o )

Line

Cur

rent

vo

vAB vBCvCA

iD1

ia

π/6 π/2


R

VI m

m

3=Peak current through each diode:

Line rms current (secondary transformer current):

( )

mm

ms

II

dR

VI

7804.03

sin2

1

6

2

3/sin3

2

8 2/

3/

2

2

=⎟⎠⎞

⎜⎝⎛ +=

−⎟⎟⎠

⎞⎜⎜⎝

⎛= ∫

πππ

θπθπ

π

π


rms current through each diode:

mss

r III

I 5518.022

2

===

Study example 3.10 p. 94 in your textbook Study example 3.10 p. 94 in your textbook


RL load: RL load: Run “nc_3p_bg_RL.mdl”

Three phase bridge rectifier with RL load

+ -v

+ -v

+ -v

+

-v

vC

vB

vA

id

iC

iB

iA

vC

vB v0

vA

vd

i0


RL

+i-

+i-

+i-

+i-

Demux

ak m

D6a

k m

D5

ak m

D4

ak m

D3

ak m

D2

ak m

D1


R=10Ω, L=5mH

0 100 200 300 400 500 600 700 800

-1

0

1

Angle (o)

Vo

ltag

es

0 100 200 300 400 500 600 700 800

-1

0

1

Angle (o)

Load

Cur

rent

0 100 200 300 400 500 600 700 800

-1

0

1

Angle (o)

Cur

rent

s

Current waveform is smoother then that of R Load


RL and voltage sink load: RL and voltage sink load: Run “nc_3p_bg_RLE.mdl”

Three phase bridge rectifier feeding a dc motor

+

-v

+

-v

+

-v

+ -v

+ -v

+ -v

+

-v

vC

vB

vA

vL

iC

iB

iA

vC

vB

v0

vA

vd

E

i0


R

L

+i-

+i-

+i-

+i-

ak m

D6

ak m

D5

ak m

D4

ak m

D3

ak m

D2

ak m

D1

R=9.2Ω, L=11.7mH, E=60V


0 100 200 300 400 500 600 700 800-300

-200

-100

0

100

200

300

vsvoEvd

0 100 200 300 400 500 600 700 800-30

-20

-10

0

10

20

30

isio


)sin(300 tVERi

dt

diL m ω=++

ABvvt =≤≤ 0,3

2

3

πωπ

( )

R

EeAt

Z

Vi tLRm −+−= − /

10 )sin(3 θω

( )22 LRZ ω+= ⎟⎠⎞

⎜⎝⎛= −

R

Lωθ 1tan

For simplicity, refer to Fig. 3.14 in your textbook, where a phase shift is considered.


For ωt=π/3, i0=I0

( )( )ωπθπ 3//01 3

sin3 LRe

ZR

EIA ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−+=

( )( )

R

Ee

Z

V

R

EI

tZ

Vi

tLRm

m

−⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−+

+−=

−ωπθπ

θω

3//0

0

3sin

3

)sin(3


At steady state:

000 )3/2()3/( Iii == ππUsing the previous slide equation of i0 and the above equation together we can obtain:

( )( )

( )( ) R

E

e

e

Z

VI

LR

LRm −

−−−−

= −

−

ωπ

ωπθπθπ3//

3//

0 1

)3/sin()3/2sin(3

for 00 ≥I

I0 can be substituted in the previous slide equation of i0.


[

( )( )( )( )

R

Ee

e

tZ

Vi

tLRtLR

m

−⎥⎦⎤

−−−−

+

−=

−−−

1

)3/sin()3/2sin(

)sin(3

3//3//

0

ωπωπ

θπθπ

θω K

00 ≥i3

2

3

πωπ≤≤ tfor and

Very complex expression that can be manipulated using numerical methods (i.e. integral calculation for determining the average and rms values of the current).

(Study Example 3.11 in your textbookStudy Example 3.11 in your textbook)


ThreeThree--Phase FullyPhase Fully--Controlled Controlled Rectifier Rectifier

Figure 10.5


33--Phase 6Phase 6--Pulse SCRPulse SCR--Bridge RectifierBridge Rectifier

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls

pulse number conducting current angle voltages(mode) thyristor transition

1 T6 T5 to T1 π/3+α (va>vc, vb<0)

2 T1 T6 to T2 2π/3+α (va>0, vc<vb<0)3 T2 T1 to T3 π+α (vb>va, vc<0)4 T3 T2 to T4 4π/3+α (vb>0, va<vc<0)5 T4 T3 to T5 5π/3+α (vc>vb, va<0)6 T5 T4 to T6 2π+α (vc>0, vb<vc<0)


Constant Output Current (Ls=0)

Mode 1:Mode 1:

CABCAB vvv & >

T1 & T6 conduct

0>ABv

Mode 2:Mode 2:

BCABCA vvv & >

T1 & T2 conduct

0>ACv

ABvv =0

ACvv =0

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls


Mode 3:Mode 3:

CAABBC vvv & >

T3 & T6 conduct

0>BCv

Mode 4:Mode 4:

CABCAB vvv & >

T3 & T4 conduct

0>BAv

BCvv =0

BAvv =0

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls


Mode 5:Mode 5:

BCABCA vvv & >

T5 & T4 conduct

0>CAv

Mode 6:Mode 6:

CAABBC vvv & >

T5 & T6 conduct

0>CBv

CAvv =0

CBvv =0

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls


0 100 200 300 400 500 600 700

-1

0

1

v LL

0 100 200 300 400 500 600 700

-1

0

1

i 0

0 100 200 300 400 500 600 700

-1

0

1

Angle ( o)

i s

v0 vAB vAC vBC vBA vCA vCB


Waveforms and Conduction Times

/ 2

( ) / 6

/ 2

/ 6

3

33 sin

6

3 3cos

o dc ab

m

m

V v d

V d

V

π α

π α

π α

π α

θπ

πθ θπ

απ

+

+

+

+

= =

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

∫

∫

/ 2 2 2( ) / 6

33 sin

6

1 3 33 cos 2

2 4

o rms m

m

V V d

V

π α

π α

πθ θπ

απ

+

+

⎛ ⎞= +⎜ ⎟⎝ ⎠

= +

∫

α=π/3


3-Phase Bridge Rectifier (RL Load)

' '

'

2 sin( ) ( ) ( )6 6 2

22 sin ( ) ( )

3 3

2 sin 0

ab ab

ab

LL ab L

v V t for t

V t for t

diL Ri E V t for i

dt

π π πω α ω α

π πω α ω α

ω

= + + ≤ ≤ +

= + ≤ ≤ +

+ + = ≥


ThreeThree--Phase FullPhase Full--Converter: Current EquationConverter: Current Equation

( )

( ) ( )[ ]

( ) ( ) ( )( )

( )( ) 01

3/sin3/2sin2

03

sin2

'sin2

3/

3//

1

'/3//1

≥−−

−+−−+=

>⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+−++

−−=

−

−

−+

R

E

e

e

Z

VI

eZ

V

R

EI

R

Et

Z

Vi

RL

LRab

L

tLRabL

abL

ωπ

ωπ

ωαπ

θαπθαπ

θαπ

θω

• Case 1Case 1: Continuous conduction, io(0)>0

( ) ( ) ( ) 1122 3/3/2,tan, LLL Iii

R

LLRZ =+=+⎟

⎠⎞

⎜⎝⎛=+= − απαπωθω

Considering:


ThreeThree--Phase FullPhase Full--Converter: Current EquationConverter: Current Equation

( ) ( ) ( )( )

( )( ) 01

3/sin3/2sin23/

3//

1 =−−

−+−−+= −

−

R

E

e

e

Z

VI

RL

LRab

L ωπ

ωπθαπθαπ

• Case 2:Case 2: Discontinuous conduction, io(0)=IL1=0

Can be solved for the critical α=αc for known value of x and θ.

( ) ( ) ( )( )

( )( ) θθαπθαπθπ

θπ

cos1

3/sin3/2sin

2 tan3/

tan3//

⎥⎦

⎤⎢⎣

⎡−

−+−−+== −

−

RL

LRcc

ab e

e

V

Ex


A new mode is created whenever current is steered between thyristors.For example, a new mode 1x is created when the current is steered away from T5 to T1.

Prior to mode 1x, T5 and T6 are conducting.

T1 is triggered at π/3+α.

Ls keeps T5 conducting until the current in T5 decreases to zero while the current in T1 increases to I0 (duration u).

u: is called the commutation angle

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls

Constant Output Current Line Inductance Constant Output Current Line Inductance LLss>0>0


120 140 160 180 200 220 240 260 280 300 320

0

0.2

0.4

0.6

0.8

1

1.2

Angle (o)

Vol

tage

s

vAB vBC vCA

vCA vCB

u

Note that during u, vAC is shorted by T1 and T5 through two line inductances

Causes line notching in power lines and degrades the quality of the power from the utility.

io

Ls

v0

+_

T1

T4

T3

T6

T5

T2

A

B

C

iA

iB

iC

Ls

Ls


Effect of line inductanceEffect of line inductance

The output voltage average value decreases which is equivalent to a voltage drop in a resistance.

The input line voltage becomes non-sinusoidal and presents notches which affect the quality of the power supply, hence, the performance of other loads.


MATLAB SIMULATIONMATLAB SIMULATION

Synchronization Voltages

DC motor equivalent circuit

Three-phase Bridge Thyristor Rectifier

208 V rms L-L3-phase Source

+ i-

iB

+ i-

iA

+- v

Vd

+- v

Vca

Vc

+- v

Vbc

Vb

+- v

Vab

Va

V & I

vBC

vCA

vAB

vd

iBiA

i

A

B

C

pulses

+

-

Thyristor Converter

alpha_degABBCCABlock

pulses

Synchronized6-Pulse Generator


Mux

+ i-

Id

30

0

iA & iB

Vd

(wc_3p_bg_RLE.mdl)


1100 1200 1300 1400 1500 1600 1700 1800

-1

0

1

)

Vol

tage

s

vABvBCvCAvo

1100 1200 1300 1400 1500 1600 1700 1800

-1

0

1

)

Load

Cur

rent

1100 1200 1300 1400 1500 1600 1700 1800

-1

0

1

Angle (o)

i A,i B


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-300

-200

-100

0

100

200

300vAB

vAB1

Notching: vab=0


COMPARISSON OF RECTIFIERSCOMPARISSON OF RECTIFIERS

The goal of a rectifier is to yield a dc output voltage at given dc output power.Therefore, it is more convenient to express the performance parameters in terms of Vdc and Pdc. For example, the rating turns ratio of the transformer in a rectifier circuit can be easily determined if the rms input voltage to the rectifier is in terms of the required output voltage Vdc.


Due to their relative merits, the single-phase and three-phase bridge rectifiersare commonly used in ac-dc conversion.

See Table 3.2 p. 102 in the textbook for an example of comparison of some of the performance parameters of diode rectifiers with a resistive load.


CHAPTER SUMMARYCHAPTER SUMMARY

• This chapter has described the techniques of conversion and control of ac-dc converters.

• The important design information gained in this chapter may be summarized as follows:

– DC voltage from an ac-dc converter can be controlled by the control of firing angle α.

– A multiphase ac-dc converter gives high ripple frequency and thus the filter requirements in the output circuit are less constraining.


CHAPTER SUMMARYCHAPTER SUMMARY

– Single phase ac-dc converters are used in low-medium power applications. High power applications use three-phase converters

– The presence of line (source side) inductance introduces the commutation angle constraints and gives rise to an equivalent output resistance.

• This resistance is responsible for output voltage drop at a higher load current.

• On the input side it also causes line notching.

power electronicspe.gastli.info/chapter3/pe_ch3.pdf · 2006. 11. 11. · dr. adel gastli rectifiers...

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