Chapter 2 AC to DC CONVERSION (RECTIFIER) · Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 1 Chapter 2 AC to DC CONVERSION (RECTIFIER) • Single-phase, half

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

===

====

�

�

π

π

ωωπ

πωω

π

+vs

_

+vo

_

vo io

vs

ω tπ 2π


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

��

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�=

+=

−⋅��

��

�=

+=

+=

+=

−RL

LRZ

tZ

Vti

ii

tititi

tdtdi

LRtitV

vvv

mf

nf

nf

m

LRs

ωθ

ω

θωω

ωωω

ωωωω

+vs

_

+

vo

_

+vR

_

+vL

_

i


5

R-L load

[ ]ωτω

ωτ

ωτω

ωτω

θθωω

θθ

θ

θωωωω

τω

ωωω

tm

mm

m

tmnf

tn

etZ

Vti

ZV

ZV

A

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Vi

A

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tdtdi

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−

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+−⋅��

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⋅��

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)sin()sin()(

as,given iscurrent 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


6

R-L waveform

:i.e ,decreasing iscurrent thebecause negative is :Note

dtdi

Lv

v

L

L

=

ωt

vo

vs,

π 2π

io

β

vR

vL

3π 4π0


7

Extinction angle

[ ]

[ ]

�

�

≤≤

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=

=+−

=+−⋅��

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biased forwardin remains diode that theNote

βω

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ω

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β

π

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t

etZ

V

ti

e

eZ

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m


8

RMS current, Power

( )

( )( )

( )( )RMSRMSs

RMSRMSs

RMS

RMS

o

IVP

pf

IVS

S

PSP

pf

RI

tdtitdtiI

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.

. i.e source,

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: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

=�

=

=

⋅=

==

==

��

��

βπ

βπ

ωωπ

ωωπ

ωωπ

ωωπ


9

Half wave rectifier, R-C Load

( )

θ

ω

θ

ωθωθsin

OFF is diodewhen

ON is diodehen w)sin( /

m

RCtm

o

Vv

eV

tVv

=

��

= −−

+vs

_

+vo

_

iD

α θ

π 2π 3π 4π

Vm

Vmax

vs

vo

Vmin

π /2

iD

3π /2

∆Vo


10

Operation

• Let C initially uncharged. Circuit is energised at ωt=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 ωt=π/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 θ

( )

( )( )( )

( )

( ) ( )

( )

mm

m

m

RCmm

RCtm

RCtm

mm

VV

RC

RCRC

RC

RCVV

eRC

VV

t

eRC

V

tdeVd

tVtd

tVd

=

=+−=+∞=

+−=−=−

=

−=⋅

�

⋅��

��

�−⋅=

=

⋅��

��

�−⋅=

⋅

=

−−

−−

−−

−−

θθ

ππππθ

ωπωωθ

ωθ

ωθθ

ωθθ

θωω

θ

ωθ

ωω

ω

ωθθ

ωθω

ωθω

sin andTherefore wave.sine theofpeak the toclose very is

22-tan

: thenlarge, is circuits, practicalFor tantan

1tan

1

1sincos

1sincos

equal, are slopes the,At

1sin

)(sin

and

cos)(

sin:are functions theof slope The

11

/

/

/


12

Estimation of α

αθα

θαπ

απω

ωθαπ

ωθαπ

for y numericall solved bemust equation This0)(sinsin(

or

)sin()2sin(

, 2tAt

)2(

)2(

=−

=+

+=

−+−

−+−

RC

RCmm

e

eVV


13

Ripple Voltage

fRCV

RCVV

RCe

eVeVVV

eVeVv

t

VV

VVVV

VVV

t

V

mmo

RC

RCm

RCmmo

RCm

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m

mmmm

o

=��

��

�=∆�

−=

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�

�

��

�

�

−=−≈∆

==+

+=≈

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−=∆+=

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��

�−

��

��

�−��

��

�−

��

��

�−��

��

� −+−

ωπ

ωπ

απ

απωπα

πθααπ

απω

ωπ

ωπ

ωπ

ωπ

ωπππ

θ

2

21 :expansoin Series Using

1

:as edapproximat is voltageripple The

) 2(

:is 2at evaluated tageoutput vol The2. then constant, is tageoutput vol DC

such that large is C and 2, and If

sin)2sin(

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. is tageoutput volMax

2

22

2222

minmax

max


14

Capacitor Current

( )

( )

( )

( )( )

�

�

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⋅−

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=

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sin

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)cos(

),( ngsubstituti Then,

OFF is diodewhen sin

ON is diode when )sin()(

But)()(

:, of In terms)()(

:as expressed becan capacitor in thecurrent The

/

/

απωθ

θ

θπωαπ

ωω

ω

ω

θ

ωω

ωωωω

ω

ωθω

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RCtm

m

c

o

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mo

oc

oc

eR

V

tCV

ti

tv

eV

tVtv

tdtdv

Cti

ttdtdv

Cti


15

Peak Diode Current

RV

CVi

RV

RV

i

CVCVI

iiii

mmpeakD

mmR

mmpeakc

CRDs

ααω

ααπαπ

απ

αωαπωαπ

sincos

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sin)(2sin)(2

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cos)(2cosHence. .)(2at occurscurrent diodepeak The

: thatNote

,

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+=

=+=+

+

=+=+

+==


16

ExampleA half-wave rectifier has a 120V rms source at 60Hz. The load is =500 Ohm, C=100uF. Assume α and θ 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(7.169sin843.048

;62.193

;7.1692120

)85.18/(62.1

/

t

RCtm

mo

m

o

om

e

t

eV

ttVtv

VradV

rad

rad

VV

ω

ωθω

ω

θ

ωωω

θα

θ

π 2π 3π 4π

Vm

Vmax

vs

vo

Vmin

π /2

iD

3π /2

α θ

∆Vo


17

Example (cont’)

( ) ( )

( )

A

radradu

RV

CVi

e

t

eR

V

tCVti

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VRC

VV

VVVVVV

VVV

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t

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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

=+=

+××=

+=

��

⋅−=

�

�

⋅−=

=××

==��

��

�=∆

=−=+−=∆−=∆

−−

−−

π

ααω

ω

θωω

ω

ωπ

ααπ

ω

ωθω


18

Controlled half-wave

+vo

_

+vs

_

ig

ia

( ) [ ]

( )[ ]

( )π

απαωω

π

ωωπ

απ

ωωπ

π

α

π

α

π

α

22sin

12

]2cos(1[4

sin21

voltageRMS

cos12

sin21

: voltageAverage

2

2

,

+−=−=

=

+==

�

�

�

mm

mRMSo

mmo

Vtdt

V

tdtVV

VtdtVV

ωtv

vo

α

ig

ωt

ωt

v s


19

Controlled h/w, R-L load

( )

( )

( ) ( )

( ) ωτα

ωτα

ωτω

θα

θαα

α

θωωωω

−

−

−

�

��

� −⋅��

��

�−=�

+−⋅��

��

�==

=

+−⋅��

��

�=+=

eZ

VA

AeZ

Vi

i

AetZ

Vtititi

m

m

tm

nf

sin

sin0

,0 :condition Initial

sin)()()(

+vs

_

i

+vo

_

+vR

_

+vL

_

ω t

vs

π 2π

vo

α

io

β


20

Controlled R-L load

( ) ( ) ( )

( ) ( ) ( )

( )

( ) [ ]

( )

( )

RIP

dtiI

dtiI

VtdtVV

eZ

Vi

etZ

Vti

A

RMSo

RMS

o

mmo

m

tm

⋅=

=

=

−==

−=

�

�

��

�

�−−−�

�

��

�==

�

�≤≤

�

�

��

�

�−−−⋅�

�

��

�=

�

�

�

−−

−−

2

2

)(

)(

:load by the absorbedpower The

21

:current RMS

21

:current Average

coscos2

sin21

: voltageAverage

.angel conduction thecalled is Angle

sinsin0

y numericall solved bemust angle Extinction

otherwise 0

tfor sinsin

g,simplifyin and for ngSubstituti

ωωπ

ωωπ

βαπ

ωωπ

θβγ

θβθββ

β

βωαθαθωω

β

α

β

α

β

α

ωτβα

ωτωα


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(ωt), (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

π 2π 3π 4π

iD1

io

output

Diode current

iD2

vo

0

ω t


25

Full wave rectifier

Center-tapped

D1is

+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===

��

≤≤−≤≤

=

� πωω

π

πωπωπωω

π

+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

π 2π 3π 4π

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 −

π 2π 3π 4π

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

ω tπ 2π


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 The1

11

12

termsharmonics theand

2 termDC thewhere

)cos()(

Series,Fourier Using

...4,2

RLR

VR

VIti

L

nI

Vn

LjnRV

ZV

I

RV

I

nnV

V

VV

tnVVtv

moo

n

n

n

n

nn

oo

mn

mo

nnoo

>>==≈

+==

=

��

��

�+

−−

=

=

++= �∞

=

ωπ

ω

ω

ω

π

π

πωω


30

R-L load approximation

vo

vs

io

iD1 , iD2

iD3 ,iD4

is

ω tπ 2π

( )

RIP

IIII

RV

RV

I

RMSo

oRMSnoRMS

moo

2

2,

2

:load the todeliveredPower

,2

current eApproximat

=

=+=

==

�

π


31

Examples

Given a bridge rectifier has an AC source Vm=100V at 50Hz, and R-L load with R=100ohm, L=10mHa) determine the average current in the loadb) determine the first two higher order harmonics of the

load currentc) determine the power absorbed by the load


32

Controlled full wave, R load

( ) [ ]

( )[ ]

( )

RV

P

V

tdtVV

VtdtVV

RMSo

m

mRMSo

mmo

2

2

,

:is load R by the absorbedpower The4

2sin22

1

sin1

Voltage RMS

cos1sin1

: voltage(DC) Average

=

+−=

=

+==

�

�

πα

πα

ωωπ

απ

ωωπ

π

α

π

α

+vs

_

is

i D1

+vo

_

i oT1

T4T2

T3


33

+vs

_

is

i D1

+

vo

_

io

+vR

_

+vL

_

Controlled, R-L load

π 2π

vo

Discontinuous mode

βα π+α

io

π 2π

vo

Continuous mode

π+α

α β

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

)(

απω

απβ

β

β

παβ

τωθωβωα

θαθωω ωταω

+=

+

=

+<

=��

��

�=

+=≤≤

−−−⋅��

��

�=

−

−−

o

tm

i

RL

RL

LRZt

etZ

Vti


35

Continuous mode

[ ]

( ) απ

ωωπ

ωα

ωα

α

αθ

αθθαπ

θαπθαπαπ

πα

α

ωτπ

ωτααπ

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

π 2π0 4π

Vm

Vm

van vbn vcn

vn

vp

vo =vp - vn

3π


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

−−

−

−

==

=

= �

π

ωπ

ωωπ

π

ππ

π

π

vo

0

π/3 2π/3

Vm, L-L

vo

π/3


40

Controlled, three-phase

Vm

vanvbn vcn

T1

+vo

_

vpn

vnn

ioT3

T2

T6

+ vcn -

n+ vbn -

+ van -

T5

T4

α

vo


41

Output voltage of controlled three phase rectifier

απ

ωωπ

α

απ

απ

cos3

)sin(3

1

:as computed becan voltageAverage

SCR. theof angledelay thebe let Figure, previous theFrom

,

32

3,

⋅��

��

�=

=

−

+

+−�

LLm

LLmo

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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

Chapter 2 AC to DC CONVERSION (RECTIFIER) · Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003 1 Chapter 2 AC to DC CONVERSION (RECTIFIER) • Single-phase, half