Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 1 Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier –Uncontrolled:

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

1

Chapter 2AC to DC CONVERSION

(RECTIFIER)• Single-phase, half wave rectifier

– Uncontrolled: R load, R-L load, R-C load– Controlled– Free wheeling diode

• Single-phase, full wave rectifier– Uncontrolled: R load, R-L load, – Controlled – Continuous and discontinuous current mode

• Three-phase rectifier – uncontrolled – controlled


2

Rectifiers

• 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

• Input can be single or multi-phase (e.g. 3-phase).

• Output can be made fixed or variable

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

AC input DC output


3

Single-phase, half-wave, R-load

mm

mRMSo

mm

mavgo

VV

tdtVV

VV

tdtVVV

5.02

)sin(21

,

(rms), tageOutput vol

318.0)sin(2

average),or (DC tageOutput vol

2

0

0

1

+vs

_

+vo

_

vo io

vs

t


4

Half-wave with R-L load

tan

)(

:where

)sin()(

:is response forced diagram, From

response, natural response; forced:

)()()(

:Solution eqn. aldifferentiorder First

)()()sin(

:KVL

1

22

RL

LRZ

tZ

Vti

ii

tititi

tdtdi

LRtitV

vvv

mf

nf

nf

m

LRs

+vs

_

+

vo

_

+vR

_

+vL

_

i


5

R-L load

tm

mm

m

tmnf

tn

etZ

Vti

Z

V

Z

VA

AeZ

Vi

A

AetZ

Vtititi

RLAeti

tdtdi

LRti

)sin()sin()(

as,given iscurrent theTherefore

)sin()sin(

)0sin()0(

:i.e ,conducting starts diode thebefore zero is

current inductor realisingby solved becan

)sin()()()(

Hence

; )(

:in resultswhich

0)(

)(

0,source when is response Natural

0


6

R-L waveform

:i.e ,decreasing iscurrent thebecause negative is

:Note

dtdi

Lv

v

L

L

t

vo

vs,

io

vR

vL


7

Extinction angle

otherwise

0

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,tion theextincasknown ispoint This

OFF. turns whendiodeis zero reachescurrent

point when heduration)T that during negative

is source the(although radians n longer tha

biased forwardin remains diode that theNote

t

etZ

V

ti

e

eZ

Vi

tm

m


8

RMS current, Power

RMSRMSs

RMSRMSs

RMS

RMS

o

IVP

pf

IVS

S

PSP

pf

RI

tdtitdtiI

tdtitdtiI

.

.

i.e source,

by the suppliedpower apparent theis

load. by the absorbedpower the toequalwhich

source, by the suppliedpower real theis where

:definition from computed isFactor Power

P

:is load by the absorbedPower

NCALCULATIO POWER

)(21

)(21

:iscurrent RMS The

)(21

)(21

:iscurrent (DC) average The

,

,

2o

0

22

0

2

0

2

0


9

Half wave rectifier, R-C Load

sin

OFF is diodewhen

ON is diodehen w)sin(

/

m

RCt

mo

Vv

eV

tVv

+vs

_

+vo

_

iD

Vm

Vmax

vs

vo

Vmin

iD

Vo


10

Operation

• Let C initially uncharged. Circuit is energised at t=0

• Diode becomes forward biased as the source become positive

• When diode is ON the output is the same as source voltage. C charges until Vm

• After t=/2, C discharges into load (R).

• The source becomes less than the output voltage

• Diode reverse biased; isolating the load from source.

• The output voltage decays exponentially.


11

Estimation of

mm

m

m

RCmm

RCtm

RCtm

mm

VV

RC

RCRC

RC

RCV

V

eRC

VV

t

eRC

V

td

eVd

tVtd

tVd

sin and

Therefore wave.sine theofpeak the toclose very is 22

-tan

: thenlarge, is circuits, practicalFor

tantan

1tan

1

1sin

cos

1sincos

equal, are slopes the,At

1sin

)(

sin

and

cos)(

sin

:are functions theof slope The

11

/

/

/


12

Estimation of

for y numericall solved bemust equation This

0)(sinsin(

or

)sin()2sin(

, 2tAt

)2(

)2(

RC

RCmm

e

eVV


13

Ripple Voltage

fRC

V

RCVV

RCe

eVeVVV

eVeVv

t

VV

VVVV

VVV

t

V

mmo

RC

RCm

RCmmo

RCm

RCmo

m

mmmm

o

2

21 :expansoin Series Using

1

:as edapproximat is voltageripple The

) 2(

:is 2at evaluated tageoutput vol The

2. then constant, is tageoutput vol DC

such that large is C and 2, and If

sin)2sin(

2at occurs tageoutput volMin

. is tageoutput volMax

2

22

2222

minmax

max


14

Capacitor Current

)2(t )( i.e

OFF, is diodewhen

sin

)2(t )(2 i.e

ON, is diodewhen

)cos(

),( ngsubstituti Then,

OFF is diodewhen sin

ON is diode when )sin()(

But

)(

)(

:, of In terms

)(

)(

:as expressed becan capacitor in thecurrent The

/

/

RCtm

m

c

o

RCtm

mo

oc

oc

eR

V

tCV

ti

tv

eV

tVtv

td

tdvCti

t

td

tdvCti


15

Peak Diode Current

R

VCVi

R

V

R

Vi

CVCVI

iiii

mmpeakD

mmR

mmpeakc

CRDs

sincos

:iscurrent peak diode The

sin)(2sin)(2

.

:obtained becan )(2at current Resistor

cos)(2cos

Hence. .)(2at occurscurrent diodepeak The

: thatNote

,

,


16

Example A half-wave rectifier has a 120V rms source at 60Hz. The

load is =500 Ohm, C=100uF. Assume and are calculated as 48 and 93 degrees respectively. Determine (a) Expression for output voltage (b) peak-to peak ripple (c) capacitor current (d) peak diode current.

(OFF) 5.169

(ON) )sin(7.169

(OFF) sin

(ON) )sin(7.169 )sin()(

:tageOutput vol (a)

;5.169)62.1sin(7.169sin

843.048

;62.193

;7.1692120

)85.18/(62.1

/

t

RCtm

mo

m

o

o

m

e

t

eV

ttVtv

VradV

rad

rad

VV

Vm

Vmax

vs

vo

Vmin

iD

Vo


17

Example (cont’)

A

radradu

R

VCVi

e

t

eR

V

tCVti

VufRC

V

RCVV

VVVVVV

VVV

mmpeakD

t

RCtm

m

c

mmo

mmmmo

o

50.4)34.026.4(500

)62.1sin(7.169)843.0cos(7.169)100)(602(

sincos

:current diodePeak (d)

(OFF) A 339.0

(ON) A )cos(4.6

(OFF) )sin(

(ON) )cos(

:currentCapacitor (c)

7.5610050060

7.1692

:ionApproximat Using

43sin)2sin(

:Using

:(b)Ripple

,

)85.18/(62.1

)/(

minmax


18

Controlled half-wave

+vo

_

+vs

_

ig

ia

22sin

12

]2cos(1[4

sin2

1

voltageRMS

cos12

sin21

: voltageAverage

2

2

,

mm

mRMSo

mmo

Vtdt

V

tdtVV

VtdtVV

t

v

vo

ig

t

t

v s


19

Controlled h/w, R-L load

eZ

VA

AeZ

Vi

i

AetZ

Vtititi

m

m

tm

nf

sin

sin0

,0 :condition Initial

sin)()()(

+vs

_

i

+vo

_

+vR

_

+vL

_

t

vs

vo

io


20

Controlled R-L load

RIP

dtiI

dtiI

VtdtVV

eZ

Vi

etZ

V

ti

A

RMSo

RMS

o

mmo

m

tm

2

2

)(

)(

:load by the absorbedpower The

2

1

:current RMS

2

1

:current Average

coscos2

sin2

1

: voltageAverage

.angel conduction thecalled is Angle

sinsin0

y numericall solved bemust angle Extinction

otherwise 0

tfor sinsin

g,simplifyin and for ngSubstituti


21

Examples

1. A half wave rectifier has a source of 120V RMS at 60Hz. R=20 ohm, L=0.04H, and the delay angle is 45 degrees. Determine: (a) the expression for i(t), (b) average current, (c) the power absorbed by the load.

2. Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply. Determine the power factor absorbed by the resistance.


22

Freewheeling diode (FWD)• Note that for single-phase, half wave rectifier

with R-L load, the load (output) current is NOT continuos.

• A FWD (sometimes known as commutation diode) can be placed as shown below to make it continuos

+vs

_

io

+vo

_

+vR

_

+vL

_

+vs

_

+vo

_

D1 is on, D2 is off

io

vo= vs

io

+vo

_

io

D2 is on, D1 is off

vo= 0


23

Operation of FWD

• Note that both D1 and D2 cannot be turned on at the same time.

• For a positive cycle voltage source,– D1 is on, D2 is off

– The equivalent circuit is shown in Figure (b)

– The voltage across the R-L load is the same as the source voltage.

• For a negative cycle voltage source,– D1 is off, D2 is on

– The equivalent circuit is shown in Figure (c)

– The voltage across the R-L load is zero.

– However, the inductor contains energy from positive cycle. The load current still circulates through the R-L path.

– But in contrast with the normal half wave rectifier, the circuit in Figure (c) does not consist of supply voltage in its loop.

– Hence the “negative part” of vo as shown in the normal half-wave disappear.


24

• The inclusion of FWD results in continuos load current, as shown below.

• Note also the output voltage has no negative part.

FWD- Continuous load current

iD1

io

output

Diode current

iD2

vo

t


25

Full wave rectifier

Center-tapped

D1

is

+vs

_

vo +

iD1

iD2

io

+vs1

_

+vs2

_

D2

+ vD1

+ vD2

• Center-tapped (CT) rectifier requires center-tap transformer. Full Bridge (FB) does not.

• CT: 2 diodes

• FB: 4 diodes. Hence, CT experienced only one diode volt-drop per half-cycle

• Conduction losses for CT is half.

• Diodes ratings for CT is twice than FB m

mmo

m

mo

VV

tdtVV

ttV

ttVv

637.02

sin1

: voltage(DC) Average

2 sin

0 sin

circuits,both For

0

+vs

_

is

i D1

+vo

_

i o

Full Bridge

D1

D2

D4

D3


26

Bridge waveforms

+vs

_

is

i D1

+vo

_

i o

Full Bridge

D1

D2

D4

D3

Vm

Vm

-Vm

-Vm

vs

vo

vD1 vD2

vD3 vD4

io

iD1 iD2

iD3 iD4

is


27

Center-tapped waveforms

Center-tapped

D1is

+vs

_

vo +

iD1

iD2

io

+vs1

_

+vs2

_

D2

+ vD1

+ vD2

Vm

Vm

-2Vm

-2Vm

vs

vo

vD1

vD2

io

iD1

iD2

is


28

Full wave bridge, R-L load

+vs

_

is

i D1

+

vo

_

io

+vR

_+vL

_

vo

vs

io

iD1 , iD2

iD3 ,iD4

is

t


29

Approximation with large L

,for ,2

:i.e. terms,harmonic the

all drop topossible isit enough, large is If

. increasingry rapidly ve decreases Thus

decreases. harmonic increases, As

:currents harmonic The

curent DC The

11

112

termsharmonics theand

2

termDC thewhere

)cos()(

Series,Fourier Using

...4,2

RLR

V

R

VIti

L

nI

Vn

LjnR

V

Z

VI

R

VI

nn

VV

VV

tnVVtv

moo

n

n

n

n

nn

oo

mn

mo

nnoo


30

R-L load approximation

vo

vs

io

iD1 , iD2

iD3 ,iD4

is

t

RIP

IIII

R

V

R

VI

RMSo

oRMSnoRMS

moo

2

2,

2

:load the todeliveredPower

,2

current eApproximat


31

Examples

Given a bridge rectifier has an AC source Vm=100V at

50Hz, and R-L load with R=100ohm, L=10mH

a) determine the average current in the load

b) determine the first two higher order harmonics of the load current

c) determine the power absorbed by the load


32

Controlled full wave, R load

R

VP

V

tdtVV

VtdtVV

RMSo

m

mRMSo

mmo

2

2

,

:is load R by the absorbedpower The4

2sin

22

1

sin1

Voltage RMS

cos1sin1

: voltage(DC) Average

+vs

_

is

i D1

+vo

_

i oT1

T4T2

T3


33

+vs

_

is

i D1

+

vo

_

io

+vR

_

+vL

_

Controlled, R-L load

vo

Discontinuous mode

io

vo

Continuous mode

+

io


34

Discontinuous mode

zero.an greater th bemust )(tat current operation continousFor

).( is expressioncurrent output thein when is modecurrent usdiscontino

and continousbetween boundary The

0)(

:conditiony with numericall solved bemust and angle extinction theis that Note

)(

:ensure toneed mode, usdiscontinoFor

; tan and

)( for

)sin()sin()(

:load L-R with wavehalf controlled similar to Analysis

1

22

)(

o

tm

i

RL

RL

LRZt

etZ

Vti


35

Continuous mode

cos2

sin1

:asgiven is tageoutput vol (DC) Average

tan

mode,current continuousfor Thus

tan

for Solving

,01)sin(

),sin()sin(

:identityry Trigonomet Using

0)sin()sin(0)(

1

1

)(

)(

mmo

VtdtVV

RL

RL

e

ei


36

Single-phase diode groups

• In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential. Therefore, the diode with its anode (+) at the highest potential will conduct (carry) id.

• For example, when vs is ( +), D1 conducts id and D3 reverses (by taking loop around vs, D1 and D3). When vs is (-), D3 conducts, D1 reverses.

• In the bottom group, the anodes of the two diodes are at common potential. Therefore the diode with its cathode at the lowest potential conducts id.

• For example, when vs (+), D2 carry id. D4 reverses. When vs is (-), D4 carry id. D2 reverses.

+vs

_+vo

_

vp

vn

io

D1

D3

D4

D2

vo =vp vn


37

Three-phase rectifiersD1

vo =vp vn

+vo

_

vpn

vnn

ioD3

D2

D6

+ vcn -

n+ vbn -

+ van -

D5

D4

Vm

Vm

van vbn vcn

vn

vp

vo =vp - vn


38

Three-phase waveforms• Top group: diode with its anode at the

highest potential will conduct. The other two will be reversed.

• Bottom group: diode with the its cathode at the lowest potential will conduct. The other two will be reversed.

• For example, if D1 (of the top group) conducts, vp is connected to van.. If D6 (of the bottom group) conducts, vn connects to vbn . All other diodes are off.

• The resulting output waveform is given as: vo=vp-vn

• For peak of the output voltage is equal to the peak of the line to line voltage vab .


39

Three-phase, average voltage

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

vo

Vm, L-L

vo


40

Controlled, three-phase

Vm

vanvbn vcn

T1

+vo

_

vpn

vnn

ioT3

T2

T6

+ vcn -

n+ vbn -

+ van -

T5

T4

vo


41

Output voltage of controlled three phase rectifier

cos3

)sin(3

1

:as computed becan voltageAverage

SCR. theof angledelay thebe let Figure, previous theFrom

,

32

3,

LLm

LLmo

V

tdtVV

• EXAMPLE: A three-phase controlled rectifier has an input voltage of 415V RMS at 50Hz. The load R=10 ohm. Determine the delay angle required to

produce current of 50A.

Documents

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 1 Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier –Uncontrolled: