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'*.+ dsr-

ChapterAC to DC CONVERSION

(RECTIFIER). Single-phase, alf wave ectifier

Uncontrolled: R load, R-L load, R-C loadControlledFree wheeling diode

. Single-phase, ull wave ectifierUncontrolled: load, R-L load,ControlledContinuous nd discontinuous urrent mode

. Three-phase ectifier

uneontrolledcontrolled

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Rectifiers

DEFINITION: Converting AC (frommains or other AC source) o DC power byusing power diodes or by controlling hefiring angles of thyristors/controllableswitches.

o Basic block diagram

AC input DC output

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

o Output can be made ixed or variable

' Apptications: DC welder, DC motor drive,Battery charger,DC ower supply, HVDC

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Single-phase, alf-wave,R-load

Output voltage DC or average),1t

Vo=Vour = -7_

[V*sin(rrr)dar -

.Q/0 o

Output voltage rms),

= 0.3I\Vm

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vm

; lf-sin(rcr)dcu)/Yo'RMS - - 0.5vm

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Half-wave with R-L load

KVLI vs =vR *vL

V* sin(or) i(ar)R + Ldi{{fi)dar

First order differential eqn. Solution :i(ffi)=if(ca)*inGrt)

iy :fotced response; n natural esponse,

From diagram, orced response s :(v\

iyGn)

l+ l'sin(ar

0)" \ L)

where:

vo

+yR

+vL

+4

Z* + (aL)z2

' [ * )

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0 = tan-

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R-L load

Natural esponse s when source 0,

i(ar)R+LqP Qdcn

which results n :

ir(cu)= trr-cnf 'n ;T= LlRHence

i(ffi)= f (or) inGa) (ryJ .sin(ar q +Ae-ffi)\z )A can be solved by realising nductor currentis zero before he diode starts onducting, .e:

lrt \t(0) [ry f sin(O0)* Ae-olm\z )+ A=(ry1,i,,1-) k\ sin(d)

\Z) \Z)\

Therefore he current s given as,

(v \ |o\ r oi ,( A\--culrnli(c,x)l + l.[sin(rrr0)+ sin(d)e r\ Z )

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R-L waveform

Note:

v7 is negative ecause he current s decreasing, .e

-diVt - L-'r

dtPower Electronics and

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Extinction angleNote that the diode remains n forward biasedlonger than n radiarts although he source snegative during that duration)The point whencuffent reaches ero s whendiode urns OFF.

This point is known as heextinction angle, .(v \ | o,---1

i(P)=l + | lri"fB 0)+sin(d)r-F'*]- e\z,/which reduces o :

sin(p - 0)+ sin(d)e-F an - 0f "*ronly

be solved numerically.Therefore, he diode conducts between 0 and p

To summarise he ectfier withR

-L load,

i(ax1

otherwisePower Electronics nd

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k). [tr,1* q+ sin(de-unl

\Z )L '

for 0

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RMS cuffent, PowerThe average DC) current s :

,2n lPIo=* !t1o4a* ! li7,Dac,xLte o zodThe RMS current s :

, TT nusl* Ii,

@)drni* !,r

r*>dat

POWER CALCULATION

Power absorbed y the oad s :A

/ \ ,Po = (Iru,fS ' .RPower Factor s computed rom definition

Ppf =;J

where P is the real power supplied y the source,

whieh equal o the power absorbed y the oad.S s the apparent ower supplied y thesource, .e

/ \s - (v".RMSIU*s)

+pf=+{(%,nr,rs lI nms

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vo

Half wave ectifier, R-C Load

when diode s ON

when diode s OFF

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fv*sin(cu) lrrr-(cu-e)tc'xc

=V*sin d

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Operation

Let C initially uncharged. Circuit isenergised t cI=0

Diode becomes orward biased as hesource become positive

When diode s ON the output is the sameas source oltage. C charges ntil V-After er=17/2, discharges nto load (R).

The source becomes ess han he outputvoltage

Diode reverse iased; solating he oad

from source.

. The output voltage decays exponentially.

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Estimation f 0The slope of the functions area(v*sinc,r)

d'(ux)=V* cos t)t

and

d.V*sin ,-(*-eY r'hc \Id(rfi)

( t \ t ,=v-"sind - I l.r-(*-e\ c'xc"' \ nRC)At efi = O,the slopes are equal,

( 1r

V*cosl=Vmsin

d |- -:^

l. t-te-q)t(IR'c

' \ nRC)

_.V*cosd _ 1----tV*sin9. oRC

11= _

tan? - aRC1 , r 1 / \

0 = tan-t (- z;PtC)= tan-'(tnC) + 7tFor practical circuits, rtR.C s large, hen :

0 = -tan(*)+ r -++ r =g22

d is very close o the peak of the sine wave.Therefore

and V*sin? =VmPower Electronics nd 11

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Estimation f cr

At Gx.=2n a,

v* sin(Zn+ a) = (v* sin0)e-(2n+a-0) @RC

or

sin(a (sin01t-rzn+u-o)!fuRC 0This equation nust be solved numerically or a

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Ripple Voltage

Max output voltage s Vrrr*.Min output voltage occurs t rtr - 27r+ aLVo =Vm^ -V*in

=V* -Vmsin(2n+ a)

=V* -vmsinaIf Vg=V* and e - Irl2, and C s large uch hat

DC output voltage s constant, hen e = ft12.The output voltage valuated t ef, = 2n + s rs

=v*e#*) *,-(#)The ripple voltage s approximated s-( 9 ( -r+l)

LVo=V* -V*e \nlRC -V*l l- e \(DRC |

t)-( zo\ AUsingSeries xpansoin e [anc -I., +(rlP-C

+ LVo=v*( g-\-u*'\r,xc) fRC

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Capacitor CurrentThe current n the capacitor can be expressed s

i-ft)= QdvoU)- / d(t)In terms f ffi,:

i"(a)= qcdvo{c*)d(c,x)But

lv*sin(rrr)vo\ffi)lr*sin d r-(*-elt

r*cwhen diodeis ON

when diodeis OFF

Then, substituting vo(cfi),a{Vm cos(at)when diode s ON,i.e (2n + a) < (rx3 (2n + 0)

i"{eu)

V*sin0 .u-(*-e)tc,xcR

when diode s OFF,

i.e (0) < (rr" (2n + a)Power Electronics nd

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Peak Diode Current

Note that :

ir=iD=ip+ig

The peak diode current occurs at (Zrc+ a),Hence.

I ,,prrk - caCVmcos(Zn a) = uCVmcos

Resistor uffent at (2n + a) can be obtained:

ip(zn a) -v*sinQn+ a)

-v*sinaRRThe diode peak current s :

iD,prak otVmcos a +V*slnR

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ExampleA half-wave ectifier has a 120V nns source t 60H2. Theload s =500 Ohm, C=100uF. Assume o and 0 are calculatedas 48 and 93 degrees espectively. etermine a) Expressionfor output voltage b) peak-to peak ipple (c) capacitorcurrent d) peak diode current.

vm 120"12 t6g.7v:

0 =93o =I.62rad;

a - 48o -0.843rad

V* sin0 - 169.7 in(1 62rad) = 169.5V

(a) Output voltage:

lv* sinla) = |69 l sin(6)v"(0I) = Io *-'

lv*sing -(m-eY 'xc(oN)

(oFF)

.169.7sin(ra)

=ft ol.s"-kt-,ozxts'ss)(oN)

(oFF)

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Example cont')

(b)Ripple:

Using:LVo V** -V-in

LVo=V* -vmsin(Zn + d) -vm -vmsin a = 43VUsing Approximation

( 2n\ r l 169.7LV^=V,| Ll ,

l -- '*=-=56.7V

"'\(DRC) JRC 60x500x1002

(c) Capacitor urrent:

frrcV*cos(rrr) (ON)i,(cx)=

1_v^sr!{0). -(a_e)Kanq (oFF)l.n

(d) Peak diode current:

iD,peak caCV*cosa ,#"

* (2x nx60)(100u)r69.7 os(0. 43rad) 169 7 sn(l 62rad)s00

= (4.26 0.34) 4.5AA

[6.4cos( fi) A (oN)=t- 0.339.-(cn-toz)r(ts.ss) (oFF)

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Controlled half-wave

Average oltage:

r4vo=*lr*sin(cufieft=-ro

a

RMS voltage

u

ht*cosal27t '

t /Yo.RMS -1I t

;!V*sin(r,l)l atnd

v=. / -m [tt- cos(Zatildcu !-ry

\4oi-

\ r 2t-9+

1T

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Controlled h/w, R-L load

+vo

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+yR

+vL

i(or) if (ca)+in(m) (?).sin(rrr 0)*er#

Initialconditio : (a)* 0,

i(a)= =(y). sina o\ e,#

+A=-l(Y" 'r-a

L\ z-|sin(a- t))'

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Controlled R-L loadSubstituting or A and simplifying,

fru r T 'o-*'1llyul .l rin(ra e)- sin(a 0)r-* lfora < a

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Examples

1. A half wave ectifier has a source f 120V RMS at 60H2.R=20 ohm, L=0.04H0 nd he delay angle s 45 degrees.Determine: a) the expression or i( cil), (b) averagecurrent, c) the power absorbed y the oad.

2. Design a circuit to produce an average oltage of 40Vacross 100 ohm oad from a 120V RMS, 60Hz supply.Determine he power actor absorbed y the resistance.

Power Electronics and 2l