BFM3323-02 Rectifier (AC-DC) (1)

8/11/2019 BFM3323-02 Rectifier (AC-DC) (1)

1/51

1

Chapter 2

AC 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


2/51

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


3/51

3

Single-phase, half-wave, R-load

mm

mRMSo

mm

mavgo

VV

tdtVV

VV

tdtVVV

5.02

)sin(2

1,

(rms),tageOutput vol

318.0)sin(2

average),or(DCtageOutput vol

2

0

0

1

+

vs_

+

vo_

voio

vs

t

A K


4/51

4

1 2

V RL

5

Given the supply voltage v = 120 Vrms, 60 Hz. Determine:

a) The average load voltage and currentb) The load voltage in rms

c) The average power absorbed by the load, RL

d) The power factor of the circuit

Answer

a) 54 V & 10.8 Ab) 84.9 V

c) 1440 W

d) 0.707

Example


5/51

5

Half-wave with R-L load

tan

)(

:where

)sin()(

:isresponseforceddiagram,From

response,naturalresponse;forced:

)()()(

:Solutioneqn.aldifferentiorderFirst

)()()sin(

:KVL

1

22

R

L

LRZ

tZ

Vti

ii

tititi

td

tdiLRtitV

vvv

mf

nf

nf

m

LRs

+

vs_

+

vo

_

+

vR_

+

vL_

i


6/51

6

R-L load

tm

mm

m

tmnf

tn

et

Z

Vti

Z

V

Z

VA

AeZ

Vi

A

AetZ

Vtititi

RLAeti

td

tdiLRti

)sin()sin()(

as,giveniscurrenttheTherefore

)sin()sin(

)0sin()0(

:i.e,conductingstartsdiodethebeforezerois

currentinductorrealisingbysolvedbecan

)sin()()()(

Hence

;)(

:inresultswhich

0)(

)(

0,sourcewhenisresponseNatural

0


7/51

7

R-L waveform

:i.e,decreasingiscurrentthebecausenegativeis

:Note

dt

diLv

v

L

L

t

vo

vs,

io

vR

vL


8/51

8

Extinction angle

otherwise

0

0for

)sin()sin(

)(

load,L-Rithrectfier wthesummariseTo

and0betweenconductsdiodetheTherefore,

y.numericallsolvedbeonlycan

0)sin()sin(

:toreduceswhich

0)sin()sin()(

.angle,tiontheextincasknownispointThis

OFF.turnswhendiodeiszeroreachescurrent

point whenheduration)Tthatduringnegative

issourcethe(althoughradiansnlonger tha

biasedforwardinremainsdiodethat theNote

t

etZV

ti

e

eZ

Vi

tm

m

=L/R


9/51

9

RMS current, Power

RMSRMSs

RMSRMSs

RMS

RMS

o

IV

Ppf

IVS

S

P

S

Ppf

RI

tdtitdtiI

tdtitdtiI

.

.

i.esource,

by thesuppliedpowerapparenttheis

load.by theabsorbedpowerthetoequalwhich

source,by thesuppliedpowerrealtheiswhere

:definitionfromcomputedisFactorPower

P

:isloadby theabsorbedPower

NCALCULATIOPOWER

)(2

1

)(2

1

:iscurrentRMSThe

)(2

1)(

2

1

:iscurrent(DC)averageThe

,

,

2

o

0

22

0

2

0

2

0


10/51

10

1 2

V= vmsinwt

L

0.1 H

RL

100

Given Vm= 100 V , = 377 rads-1. Determine:

a) An expression for current i & extension angleb) The average current

c) The rms current

d) The power absorbed by RLe) The power factor of the circuit

Answer

a) 0.936sin(t-0.361)+0.936e-t/0.377, 201o(3.5 rad)b) 0.308 A

c) 0.474 A

d) 22.4 W

e) 0.67

Example


11/51

11

Half wave rectifier, R-C Load

sin

OFFisdiodewhen

ONisdiodehenw)sin(

/

m

RCt

mo

Vv

eV

tVv

+

vs_

+

vo_

iD

Vm

Vmax

vs

vo

Vmin

iD

Vo

Time constant = t = RC


12/51

12

Operation

Let C initially uncharged. Circuit isenergised at t=0

Diode becomes forward biased as the

source become positive

When diode is ON the output is the sameas source voltage. C charges until Vm

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

The source becomes less than the outputvoltage

Diode reverse biased; isolating the loadfrom source.

The output voltage decays exponentially

(with time constant RC)


13/51

13

Estimation of

mm

m

m

RCmm

RCtm

RCtm

mm

VV

RC

RCRC

RC

RCV

V

eRC

VV

t

eRC

V

td

eVd

tVtd

tVd

sinand

Thereforewave.sinetheofpeakthetocloseveryis

22-tan

:thenlarge,iscircuits,practicalFor

tantan

1

tan

1

1

sin

cos

1sincos

equal,areslopesthe,At

1sin

)(

sin

and

cos)(

sin

:arefunctionstheofslopeThe

11

/

/

/


14/51

14

Estimation of

forynumericallsolvedbemustequationThis0)(sinsin(

or

)sin()2sin(

,2tAt

)2(

)2(

RC

RCmm

e

eVV


15/51

15

Ripple Voltage

fRC

V

RCVV

RCe

eVeVVV

eVeVv

t

VV

VVVV

VVV

tV

mmo

RC

RCm

RCmmo

RCm

RCmo

m

mmmm

o

2

21:expansoinSeriesUsing

1

:asedapproximatisvoltagerippleThe

)2(

:is2atevaluatedtageoutput volThe

2.thenconstant,istageoutput volDC

such thatlargeisCand2,andIf

sin)2sin(

2atoccurstageoutput volMin.istageoutput volMax

2

22

2222

minmax

max

Note: Sin(A+B)=sinAcosB+cosAsinB


16/51

16

Capacitor Current

)2(t)(i.e

OFF,isdiodewhen

sin

)2(t)(2i.e

ON,isdiodewhen

)cos(

),(ngsubstitutiThen,

OFFisdiodewhensin

ONisdiode when)sin()(

But)(

)(

:,ofIn terms

)(

)(:asexpressedbecancapacitorin thecurrentThe

/

/

RCtm

m

c

o

RCtm

m

o

oc

oc

eR

V

tCV

ti

tv

eV

tVtv

td

tdvCti

t

td

tdvCti


17/51

17

Peak Diode Current

R

VCVi

R

V

R

Vi

CVCVI

iiii

mmpeakD

mmR

mmpeakc

CRDs

sincos

:iscurrentpeakdiodeThe

sin)(2sin)(2

.

:obtainedbecan)(2atcurrentResistor

cos)(2cos

Hence..)(2atoccurscurrentdiodepeakThe

:thatNote

,

,

Note: cos(A+B)= cosAcosB - sinAsinB


18/51

18

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

load is =500 Ohm, C=100uF. Assume and are calculatedas 48 and 93 degrees respectively. Determine (a) Expressionfor 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

m

o

m

o

o

m

e

t

eV

ttVtv

VradV

rad

rad

VV

Vm

Vmax

vs

vo

Vmin

iD

Vo


19/51

19

Example (cont)

A

radradu

R

VCVi

e

t

eR

V

tCV

ti

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

:currentdiodePeak(d)

(OFF)A339.0

(ON)A)cos(4.6

(OFF))sin(

(ON))cos(

:currentCapacitor(c)

7.5610050060

7.1692

:ionApproximatUsing

43sin)2sin(

:Using

:(b)Ripple

,

)85.18/(62.1

)/(

minmax


20/51

20

Controlled half-wave

+

vo_

+

vs_

igia

2

2sin1

2]2cos(1[

4

sin2

1

voltageRMS

cos12

sin2

1

:voltageAverage

2

2

,

mm

mRMSo

mmo

Vtdt

V

tdtVV

VtdtVV

t

v

vo

ig

t

t

v s

+ VSCR -

Note power sinus: sin x= (1-cos2x)2


21/51

21

Controlled h/w, R-L load

eZ

VA

AeZ

Vi

i

AetZ

Vtititi

m

m

t

mnf

sin

sin0

,0:conditionInitial

sin)()()(

+

vs_

i

+

vo

_

+

vR_

+

vL_

t

vs

vo

io

+ VSCR -


22/51

22

Thyristor waveform


23/51

23

Controlled R-L load

RIP

:loadby theabsorbedpowerThe

dtti2

1I

:currentRMS

dti2

1I

:currentAverage

coscos2

V

dtsinV2

1

V

:voltageAverage

angel.conductionthecalledisAngle

esinsinZ

V0i

ynumericallsolvedbemustangleExtinction

otherwise0

tforesintsinZ

V

ti

g,simplifyinandAforngSubstituti

2

RMSo

2

RMS

o

m

mo

)(

m

t)(

m


24/51

24

Thyristor Triggering


25/51

25

Examples

1. A half wave controlled rectifier has a source of 120V

RMS at 60Hz. R = 20 ohm, L= 0.04 H, and the delay

angle is 45 degrees. Determine: (a) the expression for

current 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.


26/51

26

Freewheeling diode (FWD)

Note that for single-phase, half wave rectifierwith 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

(a)

(b) (c)


27/51

27

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 waverectifier, 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.


28/51

28

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

currentiD2

vo

t


29/51

29

Why single-phase full-wave ?

To produce purely DC (less ripple) voltage

or current

Suitable for high power application

Average current in the AC source is zero,thus avoiding problem associated with

non-zero average source current especially

in transformer


30/51

30

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 CTis half.

Diodes ratings

for CT is twice

than FB m

mmo

m

mo

VV

tdtVV

ttV

ttVv

637.02

sin1

:voltage(DC)Average

2sin

0sin

circuits,bothFor

0

+

vs_

is

iD1

+

vo_

io

Full Bridge

D1

D2

D4

D3


31/51

31

Bridge waveforms

+

vs_

is

iD1

+

vo_

io

Full Bridge

D1

D2D4

D3

Vm

Vm

-Vm

-

Vm

vs

v

o

vD1

vD2

vD3 vD4

io

iD1 iD2

iD3 iD4

i

s


32/51

32

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


33/51

33

Full wave bridge, R-L load

+

vs_

is

iD1

+

vo_

io

+

vR_

+

vL_

vo

vs

io

iD1 , iD2

iD3 ,iD4

is

t

2

1

4

3


34/51

34

Approximation with large L

,for,2

:i.e.terms,harmonicthe

alldroptopossibleisitenough,largeisIf

.increasingryrapidly vedecreasesThus

decreases.harmonicincreases,As

:currentsharmonicThe

curentDCThe

1

1

1

12

termsharmonicstheand

2

termDCthewhere

)cos()(

Series,FourierUsing

...4,2

RLR

V

R

VIti

L

nI

Vn

LjnR

V

Z

VI

R

V

I

nn

VV

VV

tnVVtv

moo

n

n

n

n

nn

o

o

mn

mo

nnoo

(average value)

(Zn increases)


35/51

35

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

:loadthetodeliveredPower

,2

currenteApproximat


36/51

36

Examples

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

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

i) determine the average current in the load

ii) determine the first two higher order harmonics of

the load current

iii) determine the power absorbed by the load


37/51

37

Controlled full wave, R load

R

VP

V

tdtVV

VtdtVV

RMSo

m

mRMSo

mmo

2

2

,

:isloadRby theabsorbedpowerThe

42sin

221

sin1

VoltageRMS

cos1sin1

:voltage(DC)Average

+

vs_

is

iD1

+

vo_

ioT1

T4 T2

T3


38/51

38

+

vs_

is

iD1

+

vo_

io

+

vR_

+

vL_

Controlled, R-L load

vo

Discontinuous mode

io

vo

Continuous mode

+

io

T1, T2ON

T3, T4

ON

2

1

4

3


39/51

39

Discontinuous mode

zero.angreater thbemust)(tatcurrentoperationcontinousFor

).(isexpressioncurrentoutputtheinwhenismodecurrentusdiscontino

andcontinousbetweenboundaryThe

0)(

:conditiony withnumericallsolvedbemustandangleextinctiontheisthatNote

)(

:ensuretoneedmode,usdiscontinoFor

;tanand

)(

for

)sin()sin()(

:loadL-Rwithwavehalfcontrolledsimilar toAnalysis

1

22

)(

o

tm

i

R

L

R

LLRZ

t

etZ

Vti


40/51

40

Continuous mode

cos

2VdtsinV

1V

:asgivenistageoutput vol(DC)Average

R

Ltan

mode,currentcontinuousforThus

RLtan

forSolving

0,e1)sin(

),sin()sin(

:identityryTrigonometUsing

0)esin()sin(

0)i(

m

mo

1

1

)(

)(

t


41/51

41

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 D3reverses (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


42/51

42

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

D1

D6

D1

D3

D3

D2D3

D4

D5

D4

ia

ib

ic


43/51

43

Three-phase waveforms

Top group: diode with its anode at thehighest 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 thebottom 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 .


44/51

44

Three-phase, average voltage

phase.-singleaofnhigher thamuchisrectifierphase-threea

ofcomponentvoltageDCoutputthat theNote

955.03

)cos(3

)sin(3

1

:voltageAverage

radians.3ordegrees60overaverageitsObtainsegments.sixtheofoneonlyConsiders

,,

323

,

32

3

,

LLmLLm

LLm

LLmo

VV

tV

tdtVV

vo

Vm, L-L

vo


45/51

45

3 Phase Current Waveform

Output current is assumed to be DC


46/51

46

Controlled, three-phase

Vm

van vbn vcn

T1

+

vo_

vpn

vnn

ioT3

T2

T6

+ vcn -

n

+ vbn -

+ van -

T5

T4

vo


47/51

47


waveforms


48/51

48


waveforms


49/51

49

Output voltage of controlled

three phase rectifier

cos3

)sin(3

1

:ascomputedbecanvoltageAverage

SCR.theofangledelaythebeletFigure,previoustheFrom

,

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.


50/51

50

Q1. Consider an uncontrolled full wave diode rectifier

with series R and L load. The supply voltage is given as

340 sin (314t) V.

a) By assuming the load inductance is large enough,

sketch the waveform of

i) Output voltage, Vo

ii) Diode current iD1, iD2, iD3 and iD4iii) Input supply current, i

s

b) If R-L load are 10 and 100 mH respectively, and theinstantaneous voltage across the load is given as

Where

i) Derive the average output voltage, Voii) Calculate the average load current, Ioiii) Calculate the amplitude of harmonic voltage, Vn for n =

2, 4iv) Calculate the amplitude of harmonic current, Vn for n =

2, 4

v) Calculate the load current in rms

4,2

)cos()(n

nwtVnVotvo

1

1

1

12

nn

VmVovn


51/51

Q2. a) Draw the circuit diagram of a single-phase

controlled full-wave rectifier using four power devices

with resistive load connected to it. Assuming the input

voltage of the rectifier is v (t) = Vmsin(t) and the delayangle is ,

i) Sketch the load voltage and current waveforms

ii) Derive the equations for the average load voltage and

current.

b) Consider using the rectifier as DC supply to a highly

inductive load (R = 2.4 and L = 0.6 H) which keeps theload current smooth and ripple-free. The rectifier is

supplied from 1414 Vrms, 50 Hz supply.

i) Sketch the load voltage waveform and show that therelationship between the average load voltage and the

delay angle is (2Vm/) cos .

ii) If the load requires a current of 450 A, calculate the

delay angle that will give this value.

iii) Prove that the load current is continuous.

iv) Determine the power absorbed by the load.

v) Explain how the power absorbed by the load is

increased by including a freewheeling diode i

Documents

BFM3323-02 Rectifier (AC-DC) (1)