rectifier 2002

Power Electronics and Drives (Version 2),

Dr. Zainal Salam, 2002

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– R load– R-L load, – Controlled R, R-L load– 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 with R-load

+vs

_

+vo

_

ωtvo io

vs

ωt

mm

mavgo

VV

tdtVVV

318.0

)sin(

(average), tageOutput vol

0

==

== ∫

π

ωωπ



4

RMS voltage

( )2,

0

Output voltage (RMS)

1 sin( )2

Output current (DC),0.318

mo RMS m

o mo

VV V t d t

V VIR R

π

ω ωπ

= =

= =

∫

• DC voltage is fixed at 0.318 or 31.8% of the peak value

• RMS voltage is reduced from 0.707 (normal sinusoidal RMS) to 0.5 or 50% of peak value.

• Half wave is not practical because of high distortion supply current. The supply current contains DC component that may saturate the input transformer



5

Half-wave with 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



6

R-L load

[ ]

[ ]ωτω

τ

τ

τ

τ

θθωω

θθω

θθ

θ

θω

τ

tm

tm

mm

m

tmnf

tn

etZ

Vti

etZ

Vti

ZV

ZVA

AeZ

Vi

A

AetZ

Vtititi

RLAeti

dttdiLRti

−

−

−

−

+−⋅

=

+−⋅

=

⋅

=−⋅

=⇒

+−⋅

=

+−⋅

=+=

==

=+

=

)sin()sin()(

or

)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



7

R-L waveform

ωt

vo

vs,

π 2π

io

β

vR

vL

3π 4π0

:i.e ,decreasing

iscurrent thebecause negative is :Note

dtdiLv

v

L

L

=



8

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. turnsdiode

whenis zero reachescurrent point when The

duration) that during negative is source the(although radians n longer tha

biased forwardin remains diode that theNote

βω

θθωω

ββ

θθβ

θθββ

β

π

ωτω

ωτβ

ωτβ

t

etZ

V

ti

e

eZ

Vi

tm

m



9

RMS current, Power Factor

( )

( )( )

( )( )RMSRMSs

RMSRMSs

RMS

RMS

o

IVPpf

IVS

S

PSPpf

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

)(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



10

Half wave rectifier, 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

ωθωθ

sinOFF is diodewhen

ON is diodehen w)sin( /

m

RCtm

o

VveV

tVv

=

=−−



11

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.



12

Estimation of θ

( )

( )( )( )

( )

( ) ( ) πωωθ

ωθ

ωθθ

ωθθ

θω

ωθ

ωθ

ωω

ω

ωθθ

ωθω

ωθω

+−=−=

−=

−=⋅

⇒

⋅

−⋅=

=

⋅

−⋅=

⋅

=

−−

−−

−−

−−

RCRC

RC

RCVV

eRC

VV

t

eRC

Vtd

eVd

tVtd

tVd

m

m

RCmm

RCtm

RCtm

mm

11

/

/

/

tantan

1tan

1

1sincos

1sincos

equal, are slopes the,At

1sin)(

sinand

cos)(

sin

:are functions theof slope The



13

Estimation of α

( )

α

θα

θαπ

απω

θ

ππ

ππθ

ω

ωθαπ

ωθαπ

for y numericall solved bemust equation This

0)(sinsin(

or)sin()2sin(

, 2tAt

sin and

22-tan

: thenlarge, is circuits, practicalFor

)2(

)2(

=−

=+

+=

=

=+−=+∞=

−+−

−+−

RC

RCmm

mm

e

eVV

VV

RC



14

Ripple Voltage

−=−≈∆

==+

+=

≈==

−=+−=

−=∆

+=

−

−

−

−+

−

RCm

RCmmo

RCm

RCmo

m

mmmm

o

eVeVVV

eVeVv

t

VV

VVVV

VVV

t

V

ωπ

ωπ

ωπ

ωπππ

θ

απ

απω

παπθ

ααπ

απω

22

2222

minmax

max

1

:as edapproximat is voltageripple The

) 2(

:is 2at evaluated tageoutput vol The

2. then constant, is tageoutput vol DCsuch that large is C and 2, and If

sin)2sin(

:is ripple thediagram, toReffering

2at occurs tageoutput volMin

. is tageoutput volMax



15

Voltage ripple-cont’d

2

Approximation of exponent term yields:

21Substituting,

2

The output voltage ripple is reduced by increasing C. As C is increased, the conduction interval

for diode d

RC

mo m

eRC

VV VRC fRC

π ω πω

πω

− ≈ −

∆ ≈ =

•

•ecreases.

Therefore, reduction in output voltageripple results in larger peak diode current.•

• EXAMPLE: The half wave rectifier has 120V RMS source at 60Hz. R=500 Ohm and C=100uF. Determine (a) the expression for output voltage, (b) voltage ripple.



16

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

_



17

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

_

ωt

vo

vs

π 2π

ωtα

io

π β 2π



18

Extinction angle

( )

( ) ( )

( ) ( ) ( )

( )degrees. for conducts diode thei.e

.angel conduction thecalled is Angle

y.numericall solved beonly can which

sinsin0

,0 when defined is angle, Extinction

otherwise 0tfor

sinsin

g,simplifyin and for ngSubstituti

(

)(

γθβ

θβθββ

β

βωα

θαθω

ω

ωτβα

ωτωα

−

−−−

==

=

≤≤

−−−⋅

=

−

−−

eZ

Vi

i

etZ

V

ti

A

m

tm



19

RMS voltage and current

( ) [ ]

( ) ( )

RIP

dtiIdtiI

VtdtVV

RMSo

RMSo

mmo

⋅=

==

−==

∫∫

∫

2

2

:is load by the absorbedpower The

21

21

current RMS current Average

coscos2

sin21

: voltageAverage

ωωπ

ωωπ

βαπ

ωωπ

β

α

β

α

β

α

• EXAMPLES• 1. 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.

• 2. 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.



20

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

_

io

+vo

_

io

D2 is on, D1 is off

vo= 0+vs

_

+vo

_

D1 is on, D2 is off

io

vo= vs



21

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.



22

FWD- Continuous load current

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

• Note also the output voltage has no negative part.

π 2π 3π 4π

iD1

io

output

Diode current

iD2

vo

0

ω t



23

Full wave rectifier with R load

+vs

_

is

i D1

+vo

_

i o

Bridge circuit

is

+vs

_

− vo +

iD1

iD2

io

+vs1

_

+vs2

_

Center-tapped circuit



24

Notes on full-wave• Center-tapped rectifier requires center-tap

transformer. Bridge does not.

• Center tap requires only two diodes, compared to four for bridge. Hence, per half-cycle only one diode volt-drop is experienced. Conduction losses is half of bridge.

• However, the diodes ratings for center-tapped is twice than bridge.

{

( ) mm

mo

mm

o

VVtdtVV

ttVttVv

637.02sin1: voltageDC

2 sin 0 sin

circuits,both For

0===

≤≤−≤≤=

∫ πωω

π

πωπωπωω

π



25

Bridge waveforms

π 2π 3π 4π

Vm

Vm

-Vm

-Vm

vs

vo

vD1

vD2

vD3 vD4

io

iD1 iD2

iD3

iD4

is



26

Center-tapped waveforms

π 2π 3π 4π

Vm

Vm

-2Vm

-2Vm

vs

vo

vD1

vD2

io

iD1

iD2

is



27

Full wave bridge, R-L load

+vs

_

is

i D1

+

vo

_

io

+vR

_

+vL

_

π 2π 3π 4π

vo

vs

io

iD1 , iD2

iD3 ,iD4

output

supplyis



28

R-L load analysis: approximation with large L

output. eapproximat to sufficient termsfew aOnly . increasing

for rapidly decreases makes This

decreases. harmonicth for the amplitude voltageincreases, as that Note

:are currents harmonic and DC The

11

112

2where

)cos()(

:as described is tageoutput vol Series,Fourier Using

...4,2

nI

nn

LjnRV

ZVI

RVI

nnVV

VV

tnVVtv

n

n

n

nn

oo

mn

mo

nnoo

ω

π

π

πωω

+===

+−

−=

=

++= ∑∞

=



29

R-L load analysis

( )

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



30

Examples

( )RIP

IIII

RMSo

oRMSnoRMS

2

2,

2

:load the todeliveredPower =

=+= ∑

• EXAMPLE: Given a bridge rectifier has an AC source Vm=100V at 50Hz, and R-L load with R=10ohm, 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



31

Controlled full wave, R load

+vs

_

is

i D1

+vo

_

i o

( ) [ ]

( )[ ]

( )

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



32

+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



33

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

τω

θ

ωβωα

θαθωω ωταω

+=

+

=

+<

=

=

+=≤≤

−−−⋅

=

−

−−

o

tm

i

RL

RL

LRZt

etZ

Vti



34

Continuous mode

[ ]

( ) απ

ωωπ

ωα

ωα

α

αθ

αθθαπ

θαπθαπαπ

πα

α

ωτπ

ωτααπ

cos2sin1

: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

==

≤

=

≥−−

−=−+

≥−+−−+≥+

∫+

−

−

−

−+−



35

Single-phase diode groups

+vs

_+vo

_

vp

vn

io

D1

D3

D4

D2

vo =vp −vn

• 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.



36

Three-phase rectifiersD1

+vo

_

vpn

vnn

ioD3

D2

D6

+ vcn -

n+ vbn -

+ van -

D5

D4

vo =vp −vn

Vm

Vm

van vbn vcn

vn

vo =vp - vn

vp

3π 4ππ 2π0



37

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 .



38

Three-phase, average 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



39

Controlled, three-phase

+vo

_

vpn

vnn

ioD3

D2

D6

+ vcn -

n+ vbn -

+ van -

D5

D4

D1

Vm

van vbn vcnα

vo



40

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

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.

rectifier 2002

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