20142015 lecture2_diode rectifier1_puan ida

8/16/2019 20142015 Lecture2_diode Rectifier1_puan Ida

1/71

1


2/71

Ideal vs real supplies

• An Ideal supply looks likeo An absolutely fat DC level, oro A pure sinusoid

• Real supplies are usually distorted by 1.the nature o the load, pulling the current

out o phase with the voltage or phasedistortion (in A.C. supplies )

2.unwanted signals harmonic distortion!.the "uality o the supply supplying the

source

2


3/71

3

Resistive circuit-all circuit power is dissipated by R-Voltage and current are in phase with eachother.

Inductive circuit

-No circuit power is dissipated by theload(s).-Rather, power is alternately absorbed fromand returned to the ! source.-Voltage and current are "# o out of phasewith each other.

!ircuit with resistance and reactance-there will be more power dissipated by theload(s) than absorbed$returned,-Voltage and current in such a circuit will beout of phase by a value somewhere between# o and "# o.

1. Phase Distortion in A.C. supplies


4/71

4


5/71

5


6/71

2. Harmonic distortion

• Disadvantage o power electronics devices #during conversion process (recti ying orinverting or chopping$, har%onic co%ponentswill be introduced into the syste%.

• &ar%onic co%ponent %ay consist o voltageand/or current distortion .

• 'he a%ount o distortion can be %easure bycalculating thei. 'otal &ar%onic Distortion ('&D$,ii. Displace%ent actor (D $ andiii. )nput *ower actor (* $.

6


7/71

i) Displacement factor

+ is the displacement angle, between fundamental input current(i s1)and input voltage (Vs)

!isplacement factor (!")

φ= cos!"

V s, I s and I s1 are #$% values


8/71

ii) THD

-

&'! of the input current alsonown as 'armonic "actor ('")

*nput +ower "actor (+")


φ=φ= cos

*

*cos

*V

*V +"

s

s1

ss

s1s

21

221

2

22

1I

I

I

I-ITHD

s1

s

s1

s1s

−

=

=

iii) Power factor


9/71

3. Filters to fght distortion

Two common solutions to the harmonic problem:%.Produce less distortion! Design better supplies that minimize switching harmonics & use specialised components'. Filter the rest: Eradicate unwanted harmonics and noise using filters

ain function of filters: to either smoothen out the rectified voltage orcurrent.

The simplest filter types:i. ! filters - inductors capacitorsii.R! filters & resistors capacitors (!apacitors resist voltage change)

iii.R filters & resistors inductors (Inductors resist current change)*ore sophisticated filters use combinations of these.


10/71

• 'o reduce har%onic co%ponents, lters %ust beintroduced and also good grounding shouldshield these devices ro% e/ternal inter erences.

1-

Before rectif ier– Remove input

harmonics

After rectifier– Remove output

harmonics


11/71

i) LC filterLC filter to limit the voltage ripple

11


12/71

12

ii) RC filterarallel Capacitance to limit the voltage ripple

+o filter out unwanted high-fre uency signals , choose R and ! such thatthe cutoff fre uency is comfortably below the lowest harmonic fre uency.

ut, the lower the cutoff fre uency, the larger R and C have to be/ larger R causes larger power ( R" losses in the filter

/ larger C is e#pensi$e

+he best solution is usually a tradeoff betweeno !ost ando effectiveness.


13/71

1!

iii) RL filter"eries inductance to limit the current ripple

Inductors are rather e0pensive and bul1y compared tocapacitors, so are not used nearly as often as R! filters.


14/71

Application:

14


15/71

15

PERFORMA !E PARAMETER"OF RE!TIFIER"

%. 2o,dc'. 2 o,ac3. 4fficiency5. 6orm factor 7. Ripple factor 8. +ransformer

utili9ation factor ./ crest factor

Regulation measures how close a real power suppl% is tothe ideal&


16/71

dc#a$era%e $al&e

10

0verage (or dc) value denoted b ++ # 0% with

the subscript ( dc ) or ( ave/ )

( )tωdt)ωsin(XT1X2

1

t

tmdc/ave ∫ =

dtx(t)T1

X2

1

t

tdc/ave ∫ =


17/71

1.

[ ][ ]

π

ππ

ωπ

ωωπ

ω

π

π

m!5

m!5

-m

!5

-m!5

!5aveo

m

266

)cos(cos(-)66

t)cos(2

266

td t)sin(621

26

666t),sin(66(t)7iven

=

−=

−=

=

===

∫

2 pulses per c cle


18/71

RM" $al&evaluation of the rms components depends on

the shape of the waveform/ "or the # &*"* # we normall have an instantaneous sinusoidal waveform as the input

1-


19/71

1

A! !omponent (in composite waveform)&he effective (rms) value of the ac component of output voltage is


20/71

1' utput dc po!er

w(erePdc, O&tp&t dc power Vdc, A$era%e $al&e of the output (load)

voltage,Idc, A$era%e $al&e of the output (load)current

2

dcdcdc I V P =


21/71

' O&tp&t ac power

wherePac, 8utput ac powerV rms, root9mean9s:uare of the output voltage

Irms, root9mean9s:uare of the output current,

21

rmsrms phac I V P =)1( rmsrms phac I V P 3)3( =


22/71

*' Efficienc+

22

Efficiency(Practical definition)Notes: include the power loss.

or

Efficiency(Text Book definition)Notes: measuring the quality of the

output dc in relation to the output rmsalue.

rms srms s

dcodco

acin

dco

I V

I V

P

P =≡η

lossdco

dco

acin

dco

P P

P

P

P +

=≡η

rmsormso

dcodco

aco

dco

I V

I V

P

P =≡η


23/71

' !rest factor

2!

rest "actor ( ") is a measure of the pea input current against therms input current/ sed to specif pea current ratings


s

s(pea )

**

" =


24/71

-' Form factor

23

"orm factor is a measure of the shape of theoutput voltage

dcV rms

V "" =

#ipple factor is a measure of ripple content

.' Ripple factor


25/71


26/71

UNITEN Generates Professionals

!HAPTER *!HAPTER *

Diode RectifiersDiode Rectifiers


27/71

hapter 3 =earning 8utcomes

At the end o" this Chapter# $ou should %e a%le&o understand the characteristics and operation of the

uncontrolled diode rectifiers&o determine performance parameters of diode rectifiers&o learn the techni:ues to anal ;e and design diode

rectifiers

2


28/71

2>


29/71

Ass&mptions!iodes used are ?ideal@ A&rr (reverse recover time) B - and V ! B-

2


30/71


31/71

# %*%&*V load 9C urrent and voltage output waveform should be in phase/&he word 'alf means the circuit is conducted

for half of the c cle (&) i/e/ during positiveregion 8D=E

!1

a) Description


32/71

!2

%implest t pe not used in industr

8nl one diode #esistive load # *nput is single phase 0


33/71

4) Vo

!!

[ ]

[ ]

mm

o

mo

-m

o

-mo

!5aveo

ss

-/31>V V

V

)cos(cos(-)2V

V

t)cos(2V

V

td t)sin(V 2

1V

V V V t),sin(V (t) v 7iven

==

−=

−=

=

===

∫

π

ππ

ωπ

ωωπ

ω

π

π


34/71

c) E5ample *'1 6Te5t 7oo8)

!3

&he rectifier in figure above has a purelresistive load of #/ !etermine F the efficienc ,

"", #", & ", +*V of the diode !1, " of theinput current and +"/


35/71

!4


36/71

36

d) E5ample

onsider the half9wave rectifier circuit with aresistive load of 25 Ω and a 6-'; ac source of11-Vrms/

alculate the a$era%e $al&es of Vo and *o/ Gustifthe significant value of Vo and *o/

alculate the rms $al&es of Vo and *o/

alculate the average and effective power (+ o,dc,+o,rms ) and efficienc delivered to the load/

alculate the converter efficienc


37/71

3.

(i) &he average values of Voand *o are given b

1/ >0 25

4 /52#V *

and,

4 /52V

(11-)2

V V

oo

m

o

===

===

(ii) &he rms value of the ofVo and *o

3/110 25

../.>

#

V *

and,

../.>V 2(11-)2

2V

2&&VmV

rms,orms,o

m-rms

===

====

(iii) + o,dc, +o,rms andefficienc

4-I61--242

>61--

++

)definitionboo(teJt

>K 1/ >


38/71

3>

e) E5ample *'*9 Half wa$e rectifier wit( R load

"or a half9wave rectifier, the source is asinusoid of 12-Vrms at a fre:uenc of6-';/ &he load resistor is 5 Ω/ !etermine

(i) the average load current,(ii) the effective power absorbed b the load (iii) the power factor of the circuit/


39/71

3

%olution(i) &he average load current

1-/>0 (5)

16 /.

#

V *

and,

16 /.V (12-)2V

s

o

m

===

==

144-K 5

>4/#

V +

and >4/ V 2

(12-)22

V V

22rms

mrms

===

===

(ii) &he effective power, + o,rms absorbed b the load

-/.-.

2(5)(12-)2

(12-)

144-

2#V

V

+*V

+%

++"

mrmss,

rmss,rmss,

rmso,

i

rmso,

=

=

===(iii) +ower "actor


40/71

' "in%le P(ase Half wa$erectifier wit( R3 3oad


41/71

#= load 9C urrent and voltage output waveform are not in phase/ &he arelagging each other/

&he word 'alf means the circuit isconducted for half of the c cle (&) or insimple word during positive region 8D=E

31

a) Description


42/71

*f Vs B V m sin ( 5 t),Vs is positive when - L 5 t L pi, andVs is negative when pi L 5 t L2pi/

Khen Vs starts becoming positive , the diode starts to conduct and the source eeps the diode in conduction till Vs C V # / MeforeNt reaches pi radians, Vs B V # at some instant/ 0t this instant,there is still current through the load circuit Nt reaches piradians and there is some energ stored in the inductor/%ince the energ stored in a s stem cannot changeinstantaneousl , the current through the inductor does not fallto ;ero immediatel / =et O be the instant at which the Vs B V # /&hen, for the period defined b O L wt L pi radians, the load

inductor acts as a source and the inductor supplies the potentialdifference between voltage across the load resistor and thesource voltage/

42


43/71

b) Kaveforms

dt di

LV L =

ntil certain time (L π), V sCV # (hence V =BV s9V # is positive), the current builds upand inductor stored energ increases/

0t maJimum ( " )of V # , V sBV # hence, V = B-V/ Me ond this point, V = becomes negative

(means releasing stored energ ), andcurrent begins to decrease/

0fter &B π, the input, V s becomes negativebut current still positive and diode is stillconducts due to inductor stored energ /&he load current is present at certainperiod, but never for the entire period,regardless of the inductor si;e/

&his will results on reducing the averageoutput voltage due to the negativesegment/ &he larger the *nductance, thelarger negative segment 43


44/71

4433

:

&he diode remains 8Das long as there is a Pve

current through it/&he diode turns 8"" atthe instant which itscurrent reaches ;ero/(NtBQP )

c) Determination of t(e


45/71

c) Determination of t(et&rn OFF an%le 6 ;


46/71

30

t(e sol&tion of t(is DE (as t(e followin%

form9

to determine t(e constant A , consider t(e initial condition9

T(e loadan%le


47/71

3

9the current e:uation becomes

9&o determine the turn off angle substitute the condition Rat NtB PQ, is B-S

&he above nonlinear e:uation can be solved iterativel for (Q)/T(eload

an%le


48/71

3-

*' "in%le P(ase Half wa$e

rectifier wit( R3 3oad =Freew(eelin% Diode

4>


49/71

3

'alf wave rectifier with #= load has ver poor :ualit 9

mainl due to the negative voltage that appears across theload/

"reewheeling diode can be applied to eliminate thenegative load voltage, and hence improve the :ualit /


50/71

5-

Dote that both ! 1and ! 2 cannot be turned on at thesame time/

"or a positive c cle voltage source,! is on , ! m is off &he voltage across the #= load is thesame as the source voltage/

"or a negative c cle voltage source, ! is off ,! m is on&he voltage across the

#= load is ;ero /'owever, the inductorcontains energ frompositive c cle / &he loadcurrent still circulates through the #= path/

a& Description


51/71

51

4) 2a$eforms

dco,dco,dco,

dco,dco,

mdco,


52/71

42

' "in%le P(ase Half wa$erectifier wit( R3E 3oad

52


53/71

53


54/71

a) Description

54

#e:uirement for diode to conduct the diode

must be forward biased/

*f Vs C (source), the i o will be able to flow

*f C Vs (instantaneous values), the diode will bereversed biased hence diode will not conduct,therefore output i o B - 0/


55/71


56/71

56

E

vo

i o

Turn !N angle

Turn !"" angle

π


57/71

(t)i(t)i(t)i

Edt(t)id

#(t)i$ tsin%

&'Diode )%E

(sin Esin%

nf

m

m

1-m

+=

++=ω

=α=α

(t)i f is determined sin s *er*osition for t+e t,o so rce ( E tsin%m ω )

τ=

θω=t-

n

mf

e(t)i$

E-)-tsin(

.

%(t)i

/ 0)(i

)e$ E

)-sin(.

%(- 0)(i

ot+er,ise 0

t e$ E

-)-tsin(.

%

t)(i

m

t-m

=β⇒=β+θα=⇒=α

β≤ω≤α+θω=ω

ωτα

ωτω


58/71

5>

O&tp&t

Volta%eFor RE 3oad

'(ample) Hal"*!ave +ectifer ,ith +-

4) 0$+= 3/4


59/71

( p ) ,-oad

4

22 74 &8

D1

1 9h%s

2 %&

i si

o

22 74 &8

D1

1 9h%s

2 %&

i si

o

!etermine the performance parameters of the circuit shown

Solution

When the diode current is positive, i.e.0


60/71

6-

( )

( )

π ω

θ

ω

φ ω

ω

ω

π

θ ω

ω

σ π ω

t

s

s

t

s

t

m s

m s

et i

A A

it at Aet i

Z

Ae Z

t V i

t t V v

t for

tan

22

tan

1)132sin(324

1)1320sin(3240

00)132sin(324

1325111100602010

sin

31sin2220sin

0

−

−

−

+−=

=⇒+−===

+−=

°∠=+=

+−=

==+


61/71

61

*oad $oltage and currentwa$eforms


62/71

62

Determination of performance parameters-+he DC output $oltage

[ ] V V dc 281)2212cos(12 2220 =−= π -+he R - of the output $oltage

( ) 1 5 22%2sin211

= +− σ σ π π rmsV

+he dc current

( ) t d et Z

V I

t m

dc ω

π

σθ

ω

∫ + −

+−=0

tansinsin

2

1

alternati$el%, and much simpler, since the inductor $oltage a$er age is zero

A RV

I R I V V dcdcdcdc Rdc 1 28 ==⇒==

( )

3 7

0

2m

3 7

0

2

mrms

t)sin2(

2

ωt

2π

%

tdtsin(%2π

1

% ω

−=ωω= ∫


63/71

63

+o determine rms , assume

( )

( )

( ) ( )

A I I I

At d e I

At d t I

e Z

V it

Z

V i

e Z V t Z V iii

rmsrmsrms

t

rms

rms

t mm

t

mm s

4313

03sin5111

222021

2813)sin(5111622220

sinsin

sinsin

22

21

73

0

102

2

2212

0

2

2

2

1

tan21

tan21

=+=

= =

=−=

=−=

+−=+=

∫

∫

−

−

−

ωπ

ωπ

θ

θ

π

ω

θω

θ

ω

and

:where


64/71

64

7251281

1 5 ===dc

rms

V V

FF

27470== s s

dc

I V P

TUF

35 20==ac

dc

P P η

1112 =−= FF RF

+he 6orm 6actor

+he ripple 6actor

+he transformer utili9ation factor

+he conversion efficiency

+he results showed circuit with R* load the half wa$e rectifierhas $er% poor .ualit%mainl% due to the negati$e $oltage appears across the load&

/

Freewheeling diode can be applied to eliminate the negati$e load $oltage, andhence impro$e the .ualit%& +o $erif% this determine the performance

parameters of the freewheeling diode circuit&


65/71


66/71

66

15Freewheeling Diode


67/71

6.

0 30 60 90 120 150 180 210 240 270 300 330 360

0

5

10

0 30 60 90 120 150 180 210 240 270 300 330 3600

10

20

30I supply

0 30 60 90 120 150 180 210 240 270 300 330 3600

20

40Output Current

0

200

400Output Voltage


68/71

6>


69/71

6

+o determine s,rms

( )

( ) ( )

A I I I

At d e I

At d t I

e

Z

V t

Z

V iii

rmsrmsrms s

t

rms

rms

t mm

s

1513

0 43sin5111

2220

2

1

1713)sin(5111622220

sinsin

22

21

0

102

2

150

0

22

2

1

tan21

=+=

=

=

=−=

+−=+=

∫

∫ −

−

ππ

ω

θω

ωπ

ωπ

θ

+he rms of the load current is gi$en b%

( )

A I

At d e I

I I I

rmso

t

rms D

rms Drms srmso

55131 231513

1 231121

22

2 102

22

=+=

==+=

∫ −−π

π

π

π ω

ω π


70/71

.-


71/71

20142015 lecture2_diode rectifier1_puan ida

Documents