AC to DC Conversion Rectifire

Power Electronics: Dr. Zainal Salam,

May 2002

1

AC to DC ConversionRectifier

• Basics of rectifier circuits– Half-wave, full-wave, three phase with

different loads• Rectifier performance• “Commutation” effect.• Line current issues

– Line current distortion– Neutral current


May 2002

2

Rectifier

• DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches.

• Basic block diagram

• Applications: DC welder, DC motor drive, Battery charger,DC power supply, HVDC

AC input DC output


May 2002

3

Ideal Rectifiers:Single-phase, R-load

ωtvo io

vs

ωt

+vs

_

+vo

_

( ) mmrmso

mm

mDCavgo

VtdtVV

VV

tdtVVVV

5.0sin21

tage,output vol RMS318.0

)sin(21

tageoutput vol DC""

0

2,

0

==

==

===

∫

∫

π

π

ωωπ

π

ωωπ


May 2002

4

Controlled half-wave

ωtvs

voia

α

ig

α

ωt

+vs

_

ig

ia

( ) [ ]

( )[ ]

( )πα

πα

ωωπ

ωωπ

απ

ωωπ

π

α

π

α

π

α

22sin1

2]2cos(1[

4

sin21

volatgeRMS

cos12

sin21

: voltageAverage

2

2

+−=−=

=

+==

∫

∫

∫

mm

mRMS

mmo

VtdtV

tdtVV

VtdtVV

+vo

_


May 2002

5

Half wave, R-L load

+vTNB

_

+

vo

_

+vR

_

+vL

_

i

tan and )(

:where

)sin()(

:is response forced diagram, From ly.respective response, natural""

and forced"" asknown are , :where

)()()(

:of form in the isSolution equation. aldifferentiorder first a is This

)()()sin(

122

=+=

−⋅

=

+=

+=

+=

−RLLRZ

tZ

Vti

ii

tititi

dttdiLRtitV

vvv

mf

nf

nf

m

LRs

ωθω

θω

ω


May 2002

6

R-L load

[ ]ωτω

τ

τ

τ

θθωω

θθ

θ

θω

τ

tm

mm

m

tmnf

tn

etZ

Vti

ZV

ZVA

AeZ

Vi

A

AetZ

Vtititi

RLAeti

dttdiLRti

−

−

−

+−⋅

=

⋅

=−⋅

=⇒

+−⋅

=

+−⋅

=+=

==

=+

=

)sin()sin()(

as, written becan current theTherefore

)sin()sin(

)0sin()0(

:i.e ,conducting starts diode thebefore zero iscurrent inductor realisingby solved becan

)sin()()()(

Hence

; )(:in resultswhich

0)()(0,source when is response Natural

0


May 2002

7

Extinction angle

[ ]

[ ]

≤≤

+−⋅

=

=+−

=+−⋅

=

−

−

−

otherwise0

0for

)sin()sin()(

load, L-Rith rectfier w thesummarise To

and 0between conducts diode theTherefore,

y.numericall solved beonly can

0)sin()sin(

: toreduceswhich

0)sin()sin()(

. angle, extinction theasknown ispoint This OFF. turns diode when is zero reachescurrent point when The

negative) is source the(although radian than longer biased forwardin remains diode that theNote

βω

θθωω

β

β

θθβ

θθββ

β

π

ωτω

ωτβ

ωτβ

t

etZ

V

ti

e

eZ

Vi

tm

m


May 2002

8

Waveforms

ωt

vo

vs,

π 2π

io

β

vR

vL

3π 4π0

:i.e ,decreasing

iscurrent thebecause negative is :Note

dtdiLv

v

L

L

=


May 2002

9

R-L-DC source load

+vs

_

+

vdc

_

i

R L

vs

io

β

vdc

α

ωτα

ωτω

θα

βωαθωω

ω

αα

eR

VZ

VA

tAeR

VtZ

Vti

Vdt

tdiLtiRtV

VVVV

dcm

tdcm

dcm

m

dcdcm

+−=

<<

+−−⋅

=

++=

=⇒=

−

−

)sin(

where

for otherwise 0

)sin()(

Solving,

)()(sinloop, thearound KVL Writing

sin sin

enconduct wh start to willdiode The1


May 2002

10

Half-wave, R-C load

+vs

_

+vo

_

iD

π 2π 3π 4π

Vm

Vmax

vs

vo

Vmin

π /2

iD

3π /2

α θ

∆Vo

( )

θ

ω

θ

ωθωθsin

OFF is diodewhen ON is diodehen w)sin(

/

m

RCtm

o

VVeV

tVv

== −−


May 2002

11

Ripple

fRCVVVVV

fRCV

RCVV

RCe

eVeVVV

eVeVv

tVV

VVVVVVV

tV

mm

omo

mmo

RC

RCm

RCmmo

RCm

RCmo

m

mmmmo

22

:asgiven is voltageload average The

2

211

: thatNote

1

:as edapproximat is voltageripple The

) 2(

:is 2at evaluated tageoutput vol The2. thensuch that large is C and 2, and If

sin)2sin(

:is ripple thediagram, toReffering.2at occurs tageoutput volMin . is tageoutput volMax

2

22

2222

minmax

max

−=∆

−=∆

=

=∆⇒

−=−

−=−≈∆

==+

+=≈==

−=+−=−=∆

+=

−

−

−

−

−+

−

ωπωπ

απ

απωπαπθ

ααπ

απω

ωπ

ωπ

ωπ

ωπ

ωπππ

θ


May 2002

12

Controlled half-wave, R-L load

+vs

_

i

+vo

_

+vR

+vL

ωt

vo

vs

π 2π

ωtα

i

π β 2π

( )( )

( ) ( )

( ) ωτα

ωτα

ωτω

θα

θαα

α

θωωωω

−

−

−

−⋅

−=

+−⋅

==

=

+−⋅

=+=

eZ

VA

AeZ

Vi

i

AetZ

Vtititi

m

m

tm

nf

sin

sin0

,0 :condition Initial

sin)()()(

( )

( ) ( )

≤≤

−−−⋅

=

−−

otherwise 0tfor

sinsin

g,simplifyin and for ngSubstituti)(

βωα

θαθω

ω

ωτωα t

m etZ

V

ti

A


May 2002

13

Center-tappedis

+vs

_

− vo +

iD1

iD2

io

+vs1

_

+vs2

_

π 2π 3π 4π

Vm

Vm

-2Vm

-2Vm

vs

vo

vD1

vD2

io

iD1

iD2

is


May 2002

14

Full bridge, R-L load

+vs

_

is

i D1

+

vo

_

io

+vR

_+vL

_

D1

D2

D3

D4

π 2π 3π 4π

vo

vs

io

iD1 , iD2

iD3 ,iD4

output

supplyis


May 2002

15

Approximation for large L

( )

below.shown is L largeion with approximat The

,for ,2

:i.e. terms,harmonic theall drop topossible isit enough, large isIf

RLRV

RVIti

L

moo >>==≈ ωω

ω

π 2π 3π 4π

vo

is

io

approx.

2Vm/R

iD1 , iD2

iD3 ,iD4

output

supply

exact


May 2002

16

Approximation with large LIo

+vs

_

is

D1 D3 +

vo

_

D4 D2

is

Iovooutput

2Vm/R

supply


May 2002

17

Controlled full-wave

+vs

_

is

i D1

+vo

_

i oT1

T2

T3

T4

( ) [ ]

( )[ ]

( )

RVP

V

tdtVV

VtdtVV

RMSo

m

mRMS

mmo

2

2

:is load R by the absorbedpower The

42sin

221

sin1

cos1sin1

=

+−=

=

+==

∫

∫

πα

πα

ωωπ

απ

ωωπ

π

α

π

α


May 2002

18

Controlled, R-L load

+vs

_

is

i D1

+

vo

_

+vR

_+vL

_T4

T3T1

T2

π 2π

vo

Discontinuous mode

βα π+α

io

( )

απ

ωωπ

ωα

πα

α

cos2

sin1

:asgiven is voltageoutput (DC) Average

tan

mode,current continuousFor

1

m

mo

V

tdtVV

RL

=

=

≤

∫+

−

π 2π

vo

Continuous mode

π+α

α β

io


May 2002

19

Three-phase output voltagevo

0

π/3 2π/3

Vm, L-L

vo

π/3

[ ]

phase.-single a ofn higher thamuch isrectifier phase- threea

ofcomponent voltageDCoutput that theNote

955.03

)cos(3

)sin(3

1

: voltageAverage

radians.3or degrees 60over average itsObtain segments.six theof oneonly Considers

,,

323

,

32

3,

LLmLLm

LLm

LLmo

VV

tV

tdtVV

−−

−

−

==

=

= ∫

π

ωπ

ωωπ

π

ππ

π

π


May 2002

20

Controlled three-phase

+vo

_

vp

n

vn

n

ioD3

D2D6

+ vcn

-

n+ vbn

-

+ van

-

D5

D4

Vm

van vbn vcnα

vo

απ

ωωπ

απ

απ

cos3

)sin(3

1

:as computed becan voltageAverage

,32

3, ⋅

== −

+

+−∫ LLm

LLmoV

tdtVV


May 2002

21

Performance Parameters

ss

dcdc

dc

ac

dc

rms

dcrmsac

ac

dc

rmsrmsrmsrms

dcdcdcdcdc

IVP

VAPTUF

VVFF

VVFF

VVV

PP

IVIV

IVPIV

×==

=

=

−=

=

×

×=

rating

:ration utilisatior Transforme

:content ripple theof measure theis which factor, Ripple

:tageoutput vol theofshape theof measure thei.e factor, form The

:tageoutput vol theofcomponent effective The

valueac 2) valueDc 1):components twoof composedbeing as considered becan tageoutput vol The

:ratio)tion (rectifica Efficiency

:power acOutput :currentoutput of valueRMS :tageoutput vol of valueRMS

:power dcOutput :currentoutput Average :tageoutput vol Average

2

22

η


May 2002

22

Performance Parameters

1,0,1,0,0%,100

:have shouldRectifier Ideal

device. of ratingscurrent peak especify th toquotedoften isIt value.RMS its tocompared

current input peak of measure a is CF factor,Crest

coscos

:as defined isfactor power input The

1

current,input theof distortion harmonic Total

cos

factor,nt displaceme the voltage,andcurrent input theofcomponet lfundamenta ebetween th angle theis If

,

11

2

2,1

,2

,1

2,1

2,

=======

=

=××

=

−

=

−=

=

DPFPFTHDTUFRFV

II

CF

II

IVIVPF

I

I

I

IITHD

DF

ac

s

peaks

s

s

ss

ss

rmss

rmss

rmss

rmssrmss

η

φφ

φ

φ


May 2002

23

ExamplesThe half-wave rectifier has a purely R-load. It is fed by a 1:1 transformer. Determine (a) efficiency (b)from factor, (c) ripple factor, (d) transformer utilisation factor, (e) the CF of the input current. Repeat fro full-wave with R-load.

( )( )( )( )

( )

25.0

:FactorCrest

voltage.DC pure a frompower deliver toused being isit n when larger tha times3.496 be

muster transfrom that thesignifies 496.3286.0/1 i.e. /1:

286.05.0707.0

318.0707.02 ;5.0

er, transform theofsecondary On the

%121157.11:Factor Ripple

%157318.0

5.0:Factor Form

%5.405.0

318.0:Efficiency

5.0318.0

5.02

;5.02

318.0 ;318.0wave,-halfFor

,

2

22

2

2

2

2

===

=

=×

=×

=

=====

=−=−=

===

===

=×==×=

====

====

RVRV

II

CF

TUFNote

IVPTUF

VVVRVII

FFRFV

VVVFF

RVRV

PP

RVIVPRVIVP

RVVIVVVR

VR

VIVVV

m

m

s

peaks

ss

ac

mmsmrmss

m

m

dc

rmsm

m

ac

dc

mrmsrmsac

mdcdcdc

mmrmsm

mrms

mmdcm

mdc

η

ππ


May 2002

24

Commutation currents in half wave with FWD


May 2002

25

Commutation interval, u

• Prior to ωt=0, Vs is (-ve); current Id is circulating through D2 with Vd=0 and is=0.

• At ωt=0, VD1 is (+ve), thus D1 is forward-biased.

• Since D2 is also conducting, it provides a short circuit path for is to build up: iD2=Id-is

• Current is builds up to a value Id during the commutation interval ωt=α. Note that id2 is positive and D2 conducts.

• D2 stops conducting at ωt=u.

• Note that is cannot excessd Id


May 2002

26

Commutation

( )

( )

−=

=−

−=

−==

==

=

<<==

−

∫

∫∫

s

ds

dss

s

uss

u

u

u

ds

I

sss

ud

s

ssss

sssL

VILu

ILuV

uV

tVtdtVA

A

ILdiLtdtV

Ii

dtdiLdtdiL

utdtdiLtVv

d

21cos

)cos1(2Hence,

)cos1(2

)cos(2sin2

,Area"" theisequation theof sideleft The

sin2

interval,n commutatio during to0 from goes that realising and sidesboth gIntegratin

But,

0 ; )sin(2

1

00

00

ω

ω

ωωω

ωωωω

ω

ωω


May 2002

27

Commutation interval (2)

• If Ls=0, then cosu=1, or u=0; i.e current commutation will be instantaneous.

• For a given frequency w, the commutation interval – increases with Ls and Id

– Decreases with increasing Vs.

( )

( )

( ) ( )

ds

s

s

u

s

su

d

sssdo

ILV

tdtVtdtV

tdtVV

VVtdtVV

πω

ωωπ

ωωπ

ωωπ

πωω

π

π

π

π

245.0

sin221sin2

21

sin221

Ls, finiteWith

45.02

22sin221

:intervaln commutatio with tageOutput vol

00

0

−=

−=

=

===

∫∫

∫

∫


May 2002

28

Commutation in full-wave


May 2002

29

Commutation currents• Prior to ωt=0, D3 and D4 are conducting and is=-Id.• Subsequent to ωt=0, vs is (+ve) D1 and D2 become

forward biased because of the short circuit path provided by D3 and D4.

• If all diodes are identical, all diodes conduct during commutation interval, and therefore vd=0.

. to- from changes Lsroughcurrent th theD2, and D1 toD4 and D3 fromcurrent

ofn commutatio thisDuring Id. and u,tat hereforeinterval.Tn commutatio theof end the

at tobeginning at the zero from up builds where

2Hence

2

21

43

21

dd

sdDD

du

uds

udDD

uDD

II

iIii

Ii

iIi

iIii

iii

====

+−=

−==

==

ω


May 2002

30

Commutation angle

( )

( )

πω

π

ωω

ωωωω

dss

udod

ds

sdssu

ds

I

Isss

u

u

ILVAVV

IVLu

ILuVA

ILdiLttdVAd

d

29.0n,commutatio todue tageoutput vol average The

221cos

2cos12

2sin2

1

0

−=−=

−=

=−=⇒

===

−

−∫∫


May 2002

31

Line voltage distortion

+vs

_

is

i D1

+

vo

_

io

+vR

_

+vL

_

∑

∑

≠

≠

−=

−=

−

−=

−=

11

11

11

11

1

1

1

:is harmonicscurrent toduecomponent distortion voltage theHence

,components harmonics andlfundamenta its of in terms Expressing

sinusoidal be toassumed is where

h

shsPCC

isssPCC

h

shs

isssPCC

ss

isssPCC

dtdiLv

dtdLvv

dtdiL

dtdLvv

vv

dtdLvv


May 2002

32

Neutral currents in three-phase, four wire system

+vb

n

+vb

+va

ia

ib

ic

ia

Single phase rectifier system

∑

∑

∑

∑

∞

+=

∞

+=

∞

+=

∞

+=

−−+−−=

−−+−−=

=

−+−=

+=

12

0111

12

0111

12111

121

)240sin(2)240sin(2

)120sin(2)120sin(2

supply,utility phase threebalanced a Assuming...3,2,1 where

)sin(2)sin(2

on rectifer diode phase-single ofpact Im

khhhsh

osc

khhhsh

osb

khhhshs

khahaa

n

tItIi

tItIi

k

tItI

iii

i

φωφω

φωφω

φωφω


May 2002

33

Neutral current

conductor. line theascurrentmuch asleast at carry toable abe toconductor

neutral requires now which , codes wiringelectricalin changes toled hasn realisatio This

large. quite becan current wireneutral thecurrent, lfundamenta of percentage

tsignifican quite iscurrent harmonic third theSince

conductor! line in the current rms harmonic third the timesely threeapproximat

iscurrent neutral theof valuerms that themeanswhich

3only, harmonic third thegConsiderin

23

,rmsIn

)sin(23

zero, toup addcomponent frequency lfundamenta theand harmonics triplen none all that Realising

3

)12(3

2

)12(3

sn

khshn

khhhshn

cban

Ii

Ii

tIi

iiii

=

=

−=

++=

∑

∑

∞

+=

∞

+=φω

AC to DC Conversion Rectifire

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Transcript of AC to DC Conversion Rectifire