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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