File 0003

AI{D F'ItiTEKS KECTIT'IERS -

3.1 lntroduction ' --r ^r^*ant

In ^f Anv tvoe of an electronic circuit'

4:]I";J**'i"o's'deck'rV "".ii";ffi5:T:',ii;:*xrri:11[nHil;ff o1partrv:1fi?:ii-::*;*X#1.?";J,il:t:;"*: ru'v fully or Parrrv"" operate of the d'c' power whichoperate c R.which vV.C.ir. T;'; 5;;;; ;;il"; functioning

il"$::lifi :tt;"nff *;;*u';:l^:i'"{i$i*"#..xi'Ftrii;:fgT;'5 ques'iion rhen'ihe Hz so i"::I:r::1v' :i "::liliid; ? ff il"l*:y*:T:,iffi;' get a d'c' Powersupply and?evices circuits ti"tt'o"it the various is, how

A typical d'c' power "'ppll^-t1l.1sts

of various stages'The

tlvrssti^tt of qa typicar,l;. diagram dragram ur 3'1' ,llr1;:#+":il?11{;";;;il ii also shown in the Fig'

Fig'3'1 shows the block

of cilcuitsrhenature

voltages at various P

M Primary

Transfo" Unregulatecl

A.C. malns 230V,50Hz

Fig' 3'1 Block diagramof

a typicald'c' powersupply

v' 50Hz)is'"T:":',1,:t::*'nX",iii::i:f;f::T:ltil: (230 rhea.c'volta5e circuit

;::*X';Ji'"',kli.lli:?t:ii;1Y::*:::::;n:*t::1 ,,"Jf,ff '"to"du'y.,oltage'Therectifier g"ia"rl."Ju"' *t" a r",r" .,"*r Thus,with suita,. A pulsatin*d'c' voltagemeans I

I I I I I

$

into a oli:lt^".t iI';;i;g"' *tt"itipple in it' Thefilter convertsthis a'c'voltage *7ulyi"e,t*r*"t furg" the.pulsating voltageiontaining ru=.1-.ii'r,l* unidirectionat .Jd.,.u, the ripple contentin ripple' This .i?."if.whicir urt"r some used circuitis stilt iien the filter outpuicontains d.c.and tries to -u;1;;;o,*,.r. (133)

E.D.c.-f 134 -E

fJ,L;,fi,i"$:il:r,rated d.c.voltage. A circuit used arl thed.c.u6,tug" keeps <eeps

Rectifiers and Filters

filter is a regulator smoorh,., -,Ii-tle. the the d." ^,,n,...tl-Tlk"t dc ou,,"i uoiii inputa.. uortog" It keeps ",],1;l;H,?ffiffi:i::l;:*::ff:ij:ilHr:i"; i?ll.Tl': .r"r,"", rhe ;:T:41 ourput iT;",r?;:',',";::ttoush i'ri;',"r'"";::X,lj',; Ii::_:.,,* under ,,".,-,-],*1r" :'J;';#l'.".ru,.1 "i,n"1 oru ""rr"r" 'J*-,,'ru'ioijiffiir':T",:;;T,",:,Tfil",1:1:fjlii::i: u al:ilJ'I#lf-1^':g:1:'o'i".uir"i;:::;;TITIT*'-"-1'll?"LE,ai""J € days,complete.;d;;;.1iil o1r" -'"'" turtqrtrons/ complete

*iei*ili;il1'r#il;h.litft ::;t"::l:lllifi:in,jti:ii: whichther- canbe rJgulatorcircuits ."nn".i.d. ,vo, l;l'#l\:r? *rrrrule'- in '::-:"" ilrho,^*^^_^?ul t;,",',i:il.tilrT.i6H[: mtegrated In Inthis "." ttrischapter, circuit chaprer, (rc) we we for

3.2 Recfifiar. Rectifiers _

shalr sha' -- *'**/ sr'rrr,,^^,.^-:tt'6i"it study,n" 'te v?rlous rectifier circuits and filter "r-l^.1circuits.

A rectifier is a device which

..r,,,-_ a'c' converts voltase ropursating JtT:Jnt'n d.c,vortage, -' usins *"rr'l5urtt one p-n junction

"'i::;:;ilffi

tlu brased while practi ",^

diode conductsonrrzi- ^^^ r.

j:i+:}fi j:."ii":::r*4X"i#*:{:;*## fi,:$,j,Txil; ;"'',:iT i:""'.T;:'iytllt'f;t:#[_!u*;i; nlll;ruilfit":#iTfi j:J

:'*:;l*r" *.i'jtf ,::'1T.."""*:,1 T"':::l;il "'i: * :1",1 "':H::li

ou unidirectional I'e' d'c' Thus p-n l]31",.ttbj:cted ro an a.c. uortug"u-Jtr"r"' putsating;..:;;;;" iunction s a recrifier ..;;;;; rng alternating voltage to a

3'2'1Thermportant characferisticsof a Rectifiercircuit . Theimportant

points to be studie, _^rr" analysingthe *" various va a) Waveform of rectifiercircuirs the load curren, are, , ;" ;;;^:,: As rectifier important to converts-a.c.to pursating d.c.,it is determines,r?nurrr"-,t* ffi;;

b) Regur arionJ ;:'""fi:,

lHi"'

throu ghr*'i'ii'i.'uu'i*u,",,

'""iL-." ::Til, : changes.Practicalty ff the load curr, {sr' r;"J regu

c) d)

* li tionistostudy *, u*".i'o?d.i ^' rJ' ."i#' li:rr.,:XTT;' ::*;T' :; " Rectifier effic.

:Itsisni rie'-nli,l power,^.o..i |.',1 "i#:T;: :;H ::';:*:iT Peak value of

".

' rnuput,:*1.1, circuir ft: maximum ,"fl*;i,T:,'j:,1111li'1,ll;..ffi'l':rcircuii'Thisdecides tnu

e) peakvarueortt" "r"ir"ri,"i;#;Tr::tiifier (PIv) : when ,Illtl.g". acrossthe

.uiir-,g or

rhe.revers edirection acros s,n" .,.il "##x#:i1ili; :,!T."#J1 -'- or voltase i.".iri ru,ing ofa "".f,'";i;r;;';:".:;:tlXt""rt"T diode.

?If,j[:

\

:w

135

E.D.C.-l t

Rectifiersand Filters

Ripple factor : The output of the rectifier is of pulsating d.c. type. The amount of a.c. content in the output can be mathematically expressed by a factor called rip-rple factor.

Using one or more diodes following rectifier circuits can be designed. 1. Half wave rectifier 2. Full wave rectifier 3. Bridge rectifier Let us discussthe various rectifier circuits in detail.

3.3 Half Wave Rectifier ,' In h.a[wa.r4erpctifier, rectifying element c-ondrictsonly during positiv_ehalf cycle of iniut a.c._supply-The negative half cylles of a.c. supply are eliminited from ihe output. (This rectifier circuit consists of resistive load, rectifying element, i.e. p-n junction diode, and the sorlrce of a.c. voltage, all connected in series); The circuit diagram is shown in Fi9.3.2. Usually, the rectifier circuits are operated from ac mains supp$ To obtain the desired d.c.voltage across the load, Fig. 3.2 Halfwave rectifier the a.c. voltage is applied to rectifier circuit using suitable step-up or step-down transformer, mostly a step-down one, with necessaryturns ratio.

\

The input voltage to the half-wave rectifier circuit shown in Fig. 3.2 is a sinusoidal a.c.voltage, having a frequency which is the supply frequency, 50 Hz. ,,Thetransformer decides the peak value of the secondary voltage. If the N, are primary number of turns and N, are secondary number of turns and E",., is the peak value of the primary voltage theri; i'",

!r

=

tr,,,

Eo-

where E.- is the peak value of the secondary a.c. voltage. As the nature of E,. is sinusoidal the instantaneous value will be, ' €9 = E'- sin ot I

I

I 1

t

tw=Znf f = supply frequency

E.D.C.-t

136

Rectifiersand Filters

Let R1represents the forward resistance of the diode. Assume that, under relerse biased condition, the diode acts almost as open circuit, conducting no current.

E.D. \

3.3.1Operation of the Circuit */

During the positive half cycle of secondary a.c voltage, terminal (A) becomes positive with respect to terminal (B). The diode is forward biased and the current florr's in the circuit in the clockwise direction, as shown in Fig. 3.2. The current will flors for almost full positive half cycle. This current is also flowing through load resistance R, hence denoted as ir, the load current. During negative half cycle when terminal (A) is negative with respect to ternunal (B), diode becomesreverse biased. Hence no current flows in the circuit. Thus the circuit current, which is also the load current, is in the form of half sinusoidal pulses.

As nc o:t=2

The load voltage, being the product of load current and load resistance,rtill al-cobe in the form of half sinusoidal pulses. The different waveforms are illustrated in Fig. 3 3. e" = E.rsin ot

z

\

l"u = loc 'olt

Appl

wl shoul

9.3.3

Iti

Subst Fig.3.3 Loadcurrentand loadvoltagewaveformsfor halfwaverectifier The d.c. output waveform is expected to be a straight line but the half n'ave rectitier gives output in the form of positive sinusoidal pulses. Hence the output r-scalled pulsating d.c. It is discontinuous in nature. Hence it is necessaryto calculate the average value of load current and average value of output voltage.

Th comP

3.3.2AverageDC Load Current( Inc ) The average or dc vafue of alternating current is obtained by integration. For finding out the average value of an alternating waveform, we have to determine the area under the curve over one complete cycle i.e. from 0 to 2nand then dividing it by the basei.e.2r

Bu to1.ft

E.D.C.-l

137

Mathematically, current waveform

Rectifiers and Filters

can be described as, = I_sin alt

i,

ir=0 where

I.

for 0 s 0* < ,r for n< c* < 2n

= peak value of load current

rnc =

*'f o''o

,rd(art)=

*'[r^sin(rrrt)d(ort) tfto

As no current flows negative half cycle of ac input voltage, i.e. between cot= n to .d":i"g cot= 2 4 we change the limits oI integrati,on.

Ioc =

*f t," sin(rrrt)d(cot) tfr.

=

I

=-]t f [-.o'(rot)]j

*fO-cos(0)l

= -*t-t-11=+ -I loc

= -rL = average value

Applying Kirchhoff's voltage law we can write, I_ = -- l-r*__ R 1+ R 1 + R ,

windins ortransrormer. rr'\ is notsivenit

,n"ilfiT*di::"iiff.:ffi;#ary

3.3;3AverageDC Load Voltage(Eoc) It is the product of average D.C. load current and the road resistance R,. Enc = IncRl Substituting value of lpq,

..-l*":t:qg

," ,.t#::$:l

Eoc= t^ RL=

E;+i

RL RJ,T

resistanceR. and forward diode resistance Rl are practically very small

#*'#,|t

.o'"oaredto R1,(Rr+ R )/RLis neglisibly smaucompared Eoc =

E lsm G

E.D.C.-l

138

Rectifiers and

3.3.4R.M.S.Value of Load Current (Inrus) The R.M.S mear

andthenfindingsquare root.

R.M.s. uur,,.of il:ilHlll;_;j,lf,,li"T::: Invs =

d(rot) *jU^sinr,r92 -,.0

r

r,,td(,0) /+l(t-'sin2 -

t

lt f [-cos(2rot)]d(art)

'm.il-

Utnb T -m

-rmI

-I* =. '2

z

i f rot sin(Z,rt))n _-J,

2"17- 4

F7A

! 2"1,)

as sin(2n) = sin (0) = Q

-I

5"r=* lnl

.

L

Note : studentsmust remember that this R-M.s.value'is 'vr for .i half wave rectifiedwavefo henceit is I,n/2. For full sinewave lr Iriit O.

3.3.5D.C.PowerOutput(pnc) Thed.c.porveroutput canbe

obtainedas, Poc = Eocloc=l3cR.

D.C.Poweroutput = I6cR. = E*lto, - -r*2

LTrJ

Poc =

12

*nt fr-

where

I-

.rr nC

-

E'o' Rr+R.+R. E"- 2R, n'[Rr +R, +R, ]2

i_ol

E.D.C.-l

139

Rqcllifiersand Filters

3.3.6A.C. power Input (pac) input taken from the secondary of transform, ,"_^1n,"O.ower

:$:":::ilH:"U*"*:';'J';: " R,. the,r*t*"1 J"'i"i#T,1"i# "'i:,:x: Pac = tflr r lR,,+I{ r +R.]

but

,r R M s = I * ; r2

P n c=

for half wave,

?[R.+R,+R,J

3.3.7RectifierEfficiency(4) The rectifier efficiency is defined as the ratio of output d.c. power to input a.c. !or\'er. D.C. f ttplrt powet,

_

p,.,.

A.C.inputpower a* -:_--

-

ur

t l

= t:

---''-r-'L

*

h -

1* R, '' n2

[Rir-R1

+R. l

(l / n2\R, (\ RI, + R - ' L, ' + " sR l -)

0.406

r*i&*,\ ) (Rr.

) me'tionedearlier,we getthemaximum rheorerical hrUlrf":?"-.*lrr."1*,rs efficiency of t ;

o/ot1^u^= 0.406x100= 4A.6,,h, fhus in half wave rectifier, maximum 40.6oh a..,

inihe,""1^,ljl"e{ri5i1ncr oirectirier is,40,/n

,il"'J#!:ii,:",,,:ltf:":ff"r#; power'It is presentintermsof ripples i" ti"'n.,tput which is fluctuating component output'Tliusmorethe rectifie.ufiiciercy, iess are tn" ,;ppru tTtil;il'n" contentsin J.3.8RippteFacror(y) It is seen that the output of half wave rectifier

is not pure d.c. but a pulsating Theoutput contain.pulsatingt:flp:Tll" d.c. called ripp.les. r.i""ilyir-,".lltrouta not be a'y ripples in the rectifier outpu"t.The measureof such ,rp_ples pr"i",.,, ir.'ti.,"ortp.rt is with the help of a factor calledripple factor a-.,ot",rii r,.ti t"tt, no* ,*oott. is the output. smaller the ripple factor .tore, is the.outp;i;; ;;".e d.c. The ripple ractor expresses how much successfulthe circr.ritis in obtuii;; ili d.c.from a.c. input. Mathematicallyripple factor is defined as the ratio of R.Iv{.s. varue of the a.c. componentto the averageor d.c.component.

-_\...

E.D.C.-l

140 Ripple factor

y -

Rectifiers and Filters

E-I).c

R.M.S.valueofa.c.component Averageor d.c.component

Now the output current is composed of a.c. component as well as d.c. component. Iu. = r.m.s.value of a. c. component present in output

Let

T ri-n I

Ioc = d.c. component present in output Inus = R.M.S. value of total output current

Inus =

frr;E

Ripplefactor = ,lut

rePr

as per definition

loc

The

:rF i

Iu. = Now

6c3

i-ri .{-

33

drn rde voI

This is the general expression circuit.

can be used for anv rectifier

Now for a half wave circuit, I""s = I ^

2

while

I

rat

Ir D c -_- - m

y = 1.271 This indicates that the ripple contents in the output are 1..2'1.1. times the d.c. component i.e. 12L.1% of d.c. component. The ripple factor for half wave is ven'high which indicates that the half wave circuit is a poor converter of a.c. to d.c. The ripple factor is minimised using filter circuits along with rectifiers.

3.3.9Load Current The load current i, which is composed of a.c. and d.c. components can be erpressed using Fourier seriesas,

3-

;ffi

and Filters

E.D.C.-l

141

-l 1sir ,r t- 3cos2r ot- 2 cos4r ot I-" ' L .. . E* n 2 1Sn 3n I

t-

ponent, sent

Rectifiersand Filters

This expression shows that the current may be considered to be the sum of an infinite number of current components, according to Fourier series. The first term of the series is the average or d.c. value of the load current. The second term iq a varying componont having frequercy Jglle as that of a.c. r@

IsEu"a f.rndu*"ntul-.otr.;;t";toftt.e?F;*t*uing-Ga

gfr;;t.n6;;fu.-

The thitd t"tt" ir a v@e,i[ilcy t*ice *te rteq"etrc/or "gain supply voltage.This is calledsecondhut^o"i. .o*po."""t€imilarly ' all- the other terms repieient the a.c.componenlsand areffi Thus ripple in the output is due to the fundamental component alongwith the various harmonic components.And the averagevalue of the total pulsating d.c. is the d.c.value of the load current,given by the constantterm in the series,I./ n

1,3.t0PeakInverseVoltage(PIV) The Peak Inverse Voltage is the peak voltage acrgss the diode in the reverse direction i.e. when lhg {iq{elr_s leyerse bia.sed.In half wave rectifier, the load current is ideally zero when the diode is reverse biased and hence the maximum value of the voltage that can exist acrossthe diode is nothing but 8.n,. PIV of diode = Er- = Maximum value of secondary voltage

:ectifier

=- , ! Lr F D-C^ Il I D C = 0

This is called PIV rating of a diode. So diode must be selected based on this PIV rating and the circuit specifications.

3.3.11TransformerUtilizationFactor (T.U.F.) -

The factor which indicates how much is the utilization of the transformer in the circuit is called Transformer Utilization Factor (T.U.F.) The T.U.F. is defined as the ratio of d.c. power delivered to the load to the a.c power rating of the transformer. While calculating the a.c. power rating, it is necessary to consider r.m.s. value of a.c. voltage and current. l.c. - ;.th

T

'''5^.

e

iPle

The T.U.F. for half wave rectifier can be obtained as, A.C. power rating of transformer = Epy5lpyg -

Errn.I* - Er* Irn

Jt p::SSed

2

2J,

Remember that the secondary voltage is purely sinusoidal hence its r.m.s. value is 1/J2 times maximum while the current is half sinusoidal hence its r.m.s. value is 1,/2 of the maximum, as derived earlier. D.C. power delivered to the load = Iil

Rt

...E-

E.D.C.-l

Rectifiers and Filters --

/ |^m ',2 i -l r\l r: |

I

\n/ D.C. Powerdeliveredto the load

T.U.F. =

A.C. Powerratingof the transformer lr

\2

l:g, ln, -

\n/ /_

, \ Enir" I I l, 2"'Z )

Neglecting the drop acrossR, and R, we can write,

E.* = I-R, i

;

tl ! .L.r

I_ 2 R, .zJl n2

I*'R,

= ?j4 n2 = 0.287 The vaiue of T'U.F' is lorv which sholvs ihat in half wave circuit, the transformer is not fuily utiUzed.

3.3.12Disadvantages of Half WaveRectifierCircuit .-,L. The ripp_lefactor of half r,va'e rectifier circuit is 1.21,w.hich is quite h*lh. The ouiput containslot of varying components. 2.

The maximum theoretical rectification efficiency is found to be 40%. The practical vaiue will be less than tiris. Tiris indicates that iralf wave rectifier circuit is quite inefficient.

a

The circuit has low transforrner utilization factor, showing that the transformer - -- is not fully utilized.

A a.

The dc current is flowing through the secondary rvinding of the transformer which may cause dc sattiration of the core of the transformer. To minimize the saturatton, transformer size have to be increased accordingly. This increasesthe cost.

Because'of all these disadvantages,the half-wave rectific.rcircuit is normally not used as a power rectifier circuit. Ex. 31 :

A lmlf wauerectificrcir*it is suppliedfrarn a 230 V, 50 Hz supply with a stepdou,n rntio af 3 :7 to a resistiueload of 10 ld>.Thc ,l.iode forward resistanceis 75 tt tuhile 'l() trsnsfortnersecandaryresistance is et.CnlculatcmarifilLntr,auerlge, KMS aalueso! clffrent, D.C. output 'toltage,efficienq af rectificationnnd riltple factor.

ismntlFtfT'*

E.D.C.-l

143

Rectifiersand Filters

Sol. : The circuit is shown in the Fig. 3.4

es

Output

N n :N 2i s 3 : 1

Fig.3.4 The given values are, R r = 7 5 O , R l = 1 . 0k O , R , = 1 0 C ) The given supply voltages are always r.m.s. values. 3.

Ep(RMS)= z3av, *= N2 N2

Nt

-l.e.i_=-

1

1Nr3

E"(RMS)

-

Fn(RMS)

\

1 _ ES(RMS) 3 230 ES(RMS) = 76.667V This is r.m.s.value of the transformer secondary voltage. tr

t'-

=Jlx76667 = lSri;r.lYt

I..

=

tr "sm

_T

R. +R'. + R, 108.423

Im=

10+i5+10><1d 10.75mA r Iuu =

ln 7tr -Lv'tJ

T

-'m rDC_ lL

)L

3.422m4 rI I^ _R M S

-

^*

for half wave

2 r0.75 = 5.375mA

E.D.C.-l

144

Rectifiersand Filters

EDc = d.c output voltage= Ioc Rr_

= 3.422x10-3x 10x 103

PDC

= 34.22Y = d.c.output power = EocIoc

ED.

5oL trf a terII

= 34.22x3.422x 10-3 = 0.1171W This also can be obtained as, r2

Poc = 4R, 'L

- (ro'zs"roj)'

x 10x 103

n2

= 0.11,71W Pec = a.c.input power = I S u s [ R+, R r + R L ]

= (s.azs,ro4)2[ro+75+1ox103]

3.4

= 4.2913W o'1171 o/o11= t* ,1oo too = " Poc 0.2913 = 40.19"/o The ripple factor is constant for half wave rectifier and is 1.21. 1f - 1.21,

Ex.3.2|

a) Assumingidealdiode,calculatethed.c.outputaoltage the netutorkshownin for Fig.3.5. part (a)if theidealdiodeis replaced U Repeat bya silicondiode,haainga cut-in aoltage of 0.7 V. Neglectdiodeforward-resistance.

rfu VRc

T R1=1kO

F trans

3.{.

c

and t Fig.3.5

Filters

Rectifiersand Filtere

145

E.D.C.-l

Sol. : In the circuit of Fig. 3.5, the diode will be forward biased during negative half cycle of a.c. input voltage, and d.c. output voltage will be negative w.r.t. common ground terminal, as shown. a) For an ideal diode, cut-in voltage Vy = 0, Rr = 0 - Maximum value of a.c. input voltage

D.C.output voltage =

'! =

-15

= -4.77V Negativesign indicatesthat voltageis negativew.r.t. ground. b) For a silicondiode,Y,r= 0.7V, R1 is assumedto be zero. - [ MaximumA'c' voltage- v' ] D.c. output voltage = It

=

- n 5- 0 . 7 1 fr

= -4.55V

3.4FullWaveRectifier D1

@ |

|

|

|

+

Fr.

+rdl

rl 3._t_

@

l. l<

t

;

ir

ii

t

trDc l.

le.

r*'':- '.tl

+ Fl

+lac

N-r--

D2

t- . :":.-c

Fig. 3.6 Full wave rectifier

RL

The full wave rectifier conductsduring both positive and negative half cycles of input a.c. supply. In order to rectify both the half cycles of a.c.input, two diodesare used in this circuit. The diodes feed a common load R, with the a center tap help of transformer. The a.c. voltage is applied through a suitable power transformer with turns ratio. proper

The full wave rectifier circuit is shown in the Fig.3.6. For the proper operation of the circuit, a center-tap on the secondary winding of the transformer is essential.

3.4.1Operation of the Circuit Consider the positive half cycle of ac input voltage in which terminal (A) is positive and terminal (B) negative. The diode D, will be forward biased and hence will conducl

E.D.C.-l

Rectifiersand Filters

146

EJ).C

while diode D, will be reverse biased and will act as open circuit and will not conduct. This is illustrated inFig.3.7.

i-t

Dl

@ l+ I

l-r.

------->

I Arr gll AC.suppry JI

ld.l

V iL

>l 2t

|

RL

-

?ll

l___l Fig.3.7Currentflow duringpositivehalfcycle The diode D, supplies the load current, i.e. i, = i6' .This current is flowing through upper half of secondary winding while the lower half of secondary winding of the transformer carries no current since diode D, is reverse biased and acts as open circuit. In the next half cycle of ac voltage, polarity reverses and terminal (A) becomes negative and (B) positive. The diode D, conducts, being forward biased, while D, does not, being reverse biased. This is shown in Fig. 3.8.

Th

orcult :eed,r--l

3.d2 l

Fig.3.8Currentflow during negativehalf cycle The diode D, supplies the load current, i.e. i, = iaz Now the lower half of the secondary winding carries the current but the upper half does not. It is noted that the load current flows in both half cycles of ac voltage and in the same direction through the load resistance. Hence we get rectified output across the load. The load current is sum of individual diode currents flowing in corresponding half cycles. It is also noted that the two diodes do not conduct simultaneously but in alternate half cycles.The individual diode currents and the load current are shown in Fig.3.9

rnd Filters

E.D.C.-l

147

Rectifiersand Filters

t :onduct. Secondary F -sm voltage

; \:

(one half)

i/

0 ior

: tirough r: of the ;::cuit. be'comes : D. does

-I rav - rDC

I

ort Load voltage e, Em E"u = EDc

ort

Fig.3.9 Loadcurrentand voltagewaveformsfull waverectifier Thus the full wave rectifier circuit essentially consists of two half-wave rectifier circuits working independently (working in alternate half cycles of a c) of each other but feeding a common load. The output load current is still pulsating d.c. and not pure d.c.

3.4.2Maximum Load Current Let

l.: : the

Rf = forward resistanceof diodes R.

winding resistanceof each half of secondary

RL

load resistance

es

instantaneous a.c. voltage acrosseach half of secondary

d in the .--

::

rL^ tl tg

i- : half : -.i in : '..nin

e^ (D

E.-

Ern..sin ot 2nf maximum value of a.c.input voltage acrosseach half of secondary winding

looking Hence we can write the expression for the maximum value of load current, at equivalent circuit shown in Fig 3.10.

Dl

Rf

Upperhalf of secondary Rs E"t

Lowerhalf of secondary E"t

RL

'-/

(b)

Fis.3.l0

{a)

where

Er.

'r'

=

I.

= maximum value of load current iL

[laPrl[,

3.4.3AverageDC Load Current (Ips) Consider one cycle of load current it from 0 to 2n to obtain the average value which is d.c. value of load current. il=I-sinrot

O(alt(n

But for nto 2n, the current i1 is again positive while sin ort term is negative during nto2n. Hence in the region rto2n the positive iL can be rePresented as negative of In..sin (olt).

Fig.3.11

n< al < 2n

-l-sinrot

tL=

I

t- t+.

-

Ioc=i Iira1rtl zn'o

./l

=

*[i',

+J-,. ,*'' d('t)"J sinotd(cut)

*[J,*'t

-J'i''''' o('')] d(
- (-cos'01' ] l? [t-.*'t)fi

and Filters

Rectifiersandllllglg

149

E.D.C.-l

:nt, looking

= but

[*.otn+ coso+ cos2n-cosr]

f

cosTr -

-1

= l*t
*-,

for full waverectifier

+

of two half wave For half wave it is I*/n and full wave rectifier is the combination obviously the d'c' value for circuits acting urt"rn"1"i'y in two half cyclesof input. Hence full wave circuit is 2l^/rt 3.4.4 Average DC Load Voltage (Eoc) The d.c.load voltageis,

E D c= I o c R r = + Substitutingvalue of I,,,, EDc =

2 Esm RL

n[Rp +Rr+Rr]

=;['+gl 2 Er,n

I current it eragevalue :1'.t.

t : is again Li negative F,".n ;r to 2 rc e-nted as

*ta* *' But as Rl and & << Rr' Rhence L " Ep6

t =

2E.t

-;

3.4.5RMS Load Current ( Inus) The R.M.S.value of current, IsMs,is obtalnectas follovrs

:

l,2n

I n v s= , l * l i 2 t a l r o q Y-ru0

Cl,:)

since two half wave rectifier are similar in operation we rIRMS -

-

*i

can write,

o," sinortl2d(cot)

=r,ffiassin2
Rectifiersand Filters

150

E.D.C.-l

E.D.

L - cos 2
3.4.

= Im

assin(2n)=sin(0)=g

Im Invs =

"t,

?A.6DCPowerOutput(Poc)

But

D.C.PoweroutPut = Encloc= I3c Rl /.r

\2

poc

=['?J *' I3.R,

Poc

4t1*, n-

Substituting value of I* we get,

ti*

P o c= 7

3.4

,r"*,

*t (R, + R, + Rr)z

Note : Instead of remembering this formula students can use the expression EoaIoa or t2o.n, to calculate Po6 while solving the problems'

'

3"4.7LC Power Input (Pac) Thea.c.powerinput is givenbY, Pec = I3-r(R, +R, +Rr)

= ^ ^pA.U

/t

[tJ

\2

(Rr+R'+R')

_ tl(nr+R,+nr) 2

Substituting value of I- we get, Pnc =

E3* (Rr +R. +Rr)2 "1*(Rr+R,+Rr)

:\P

ilters

Rectifiersand Filters

151

E.D.C.-l

-2 tr rtt!-

D

rAC

2(Rr +R,+Rt)

3.4.8Rectifier EfficiencY(r)

Poa output Pos inPut

n=

2 , 1 n+, n rk :s_\:L--:l-

+R --L'', r)

2 8 Rr-

n=- n t ( R r + R r + R t ) it from denominator But if Rp+ R. (( R1, neglecting 8Rr _8 n2(RL) Tt'

o/or1,,'u*=

100= 812% ix

efficiency of full wave rectifier' This is the maximum theoretical

3.4.9RiPPleFactor (Y)

is given by a general wave rectifier the ripple factor As clerived earlier in caseof half

expression, Ripple factot' =

For full wave

Iws

-- \ lJ2

and Ioc=21^ln

so'

substitutingin the aboveequation' Ripple factor

RiPPlefactor= T=0'48 Thisindicatesthattheripplecontentsintheoutputare43%ofthed.c.component wave circuit' ,tf',i.frlt t""ch lessthan that for half

Rectifiersand Filters

152

E.D.C.-l

3.4.10Load Current (i1) the series for The fourier series for the load current is obtained by taking the sum of half cycles, i'e' there the individual rectifier current. The two diodes conduct in alternate Hence, is a phase difference of nradians between two diode currents' tdt

and

- ,,

[

+l sinrot

fr*tatt'..] ficos?tnt-

io, = ia, with <,rt replaced bY (
io,

+ - 'o' a@t+r)"'] * *)- 3.os2(cot + sin(rot "1 fr ] [i - t, - rn.t - cos(zax +aT + zn)- cos(+ot )"'] fr fr ] [| = t. - lsinot- L cos?t'tt-t*nt'"'] fr fl

- t,

Then the fourier seriesfor the load current is, io, * ia, L

=

- .ora't"'] zot t^n- +-"os fi

the The first term in the above series representsthe average or dc value' while twice i.e' 2f, is ripple the of remaining terms "ripple". It is seenthat the lowest frequency load current of the the supplj, fruquur,Cyof ac supply. The lowest ripple frequencyin the full-wiveconnection, is double than that in ihe half-wave connection' . As seen from Fig. 31 and,Fig 3.8 the individual diode currents are flowing in Hence the net opposite directions ihrougt the haro-halvesof the secondarywinding' s,econdarycurrent will be differenceof individual diode currents. ir". = iot - io, Thus, series of The fourier series of ir.. is obtained by the difference between the write, individual diode currents.Using aboverelationswe can i""t = I*sinrot No d'c' Hence under ideal conditions,the secondarycurrent is purely sinusoidal' This saturation' component flows through the secondaryhence there is no danger of circuit' Thus the reduces the transfor*"i lorrus and overall size and cost of the transformergetsutilised effectively. 3.4.11 Peak Inverse Voltage (PIV) biased It can be observedfrom the circuit diagram that when the diode is reversed across drop The it' then full transformer secondary voltage gets impressed across to voltage which conducting diode is assumedzero. ThuJthe peak val-u9of the inverse is voltage acrossboth the parts of the transformer secondary' diode gets"subjected

t

I I

Filters

E.D.e.-l

153

Rectifiersand Filterc

pIV of diode = 2 Er* n e si o r | :rere

=_

(

r, rF! *D^Cl l t D C _ 6

where Esm=maXimum value of a.c.voltage acrosshalf the transformer secondary.

3.4.12TransformerUtilizationFactor(T.U.F.) In full wave rectifier, the secondarycurrent flows through each half separatelyin every half cycle. Wtrite the primary of transformer carries current continuously. Hence T.U.F is calculatedfor primary and secondarywindings separatelyand then the average T.U.F.is determined.

llr

SecondarvT.U.F =

DC power to load

AC power rating of secondary

= IfuR, = ERMS I,,,,,

Neglecting forward resistance \ of diode, Er,',- [^R .

ile the , :''\'ice c: the

SecondaryT.U.F. =

3"^'o'- , [-'R,

n2

2 = 0.812

L

i

lrt

l'.e r.et

les of

The primary of the transformer is feeding two half-wave rectifiers separately.These two half-wave rectifiers work independently of eachother but feed a common load. We havealreadyderived the T.U.F.for half wave circuit to be equalto 0.287. Hence T.U.F.for primary winding = 2x T.U.F.of half wave circuit = 2x0.287 -- 0.574. The averageT.U.Ffor fullwave circuit will be AverageT.Ll.F.for

- -:.ls

=

T.U.F of primary+ T.U.F of secondary

2

full wave rectifier circuit

t-* :.".g

0.574+0.812 2 l:a-d A::-TSS ir'::.h

= 0.693 .'.AverageT.U.F.for full-wave rectifier= 0.693 Thus in full-wave circuit transformer gets utilized more than the half wave rectifier circuit.

Rectifiers and Filters

154

E.D.G.I

3.4.13Voltage Regulation

with- respect to the load current' The The secondary voltage should not change the chutge in d'c' output voltage as voltage regulation i, inl iu.tor which tells us"about toaa Inanfes from no load to full load condition' If (Vot)r'rr'= D'C' voltageon no load (V6.)rr- = D'C' voltageon full load then voltage regulation is defined as

(va')NL - (va')rr'

Voltageregulation =

...(3.1)

1V6.)nr_

circuit' ,:l.essthe value of voltageregulation,betteris the performanceof rectifier For a full wave circuit,

...( 3.2)

= (V6.)Nr+ and

...(3.3)

(V6r)rr- = Inc Rt-

as' The regulationcanbe exPressed 2E'- -IocRrtr xL00 %R = IocRr-

Now

Er^

Im=

R, +Rt +R,

Er,,. = I - ( R r + R t + R . ) and

)l

TIDC

; )l -^*

%R=

n

- nrl-?n' [Rr+ R ,+

x 100

fi

R6+R1 +R, -Rt

x 100

RL

=

R tj n t RL

,.100

canbe expressedas' Neglectingwinding resistanceR' the regulation

ntrroo %R =

where

Rf = forward resistanceof the diode'

ry-

ters

Rectifiersand Filters

155

E.D.C.-l

3.4.14Comparisonof Full Wave and Half f{vl l:e E. :i

Circuit

For comparison, we assume that the full-wave and half-wave circuits use identical diodes, identical load resistancesand the voltage across half the secondary winding of transformer used in full-wave circuit is the same as the voltage across the secondary winding of the transformer used in half-wave circuit' 1.

2.

The d.c. load current in case of full wave circuit is twice to that in half wave circuit; similarly the D.C. load voltage in full wave circuit is twice that in half wave circuit. The lowest ripple frequency in full wave circuit is twice that in half wave circuit. Now to remove ripple the additional circuits called filter circuits are used along with rectifier circuits. But as the frequency is more in full-wave, the capacitor values required in capacitance filter are much less hence smaller elements are sufficient in filter circuits used with full wave circuit to reduce ripple.

3.

Becausethere is no net d.c. current through windings of the transformer used in full wave circuit, the lossesare less as compared to lossesin transformer used in half wave circuit.

The full wave connection gives dc Power output four times as large, when compared with half wave connection. 5 . The efficiency of rectification in a full wave connection is twice that for half wave connection.

4.

6. The ripple-factoris lessfor full-wave, i.e. rectificationis more nearly complete for full wave as comparedto half-wave' secondary haainga center-tapped Afull-waaerectifiercircuitisfedfroma transformer lf to centertapis 30V. thediode winding.Thermsaoltage from eitherendof secondary is 8 {2,for a loadof is 2 Clandthato! thehalfsecondnry forward resistance '1.ldl, calculate to load, s) Powerdeliaered atfull load, b) % Regulatiort of rectification, c) EfficiencY of secondarY. TUF il S o 1 . :G i v e n : E . : 3 0 V , R - r = 2 0 , R , : 8 f ) , R L : l k Q

Ex.3.3:

\r

Es = ERl,ts= 30 V

Er,,, = E, ,[2 : 30J1 volt=42'426Y E rn,

T rm

Rf+RL+Rs

=

42mA

za ft-

4 00+8 -_ 2 + "1" 0

..-i --.--

E.D.C.-l

156 Ioc =

-) n

Rectifiersand Filters

I,o=26.74mA

a) Power delivered to load

= I r o " R , : ( 2 6 . - ' tx4 l o - 3 ) 2 ( t t < o ) = 0.71.5W b)

V p 6 , n o l o a d= 3 r . .

=?r30Ji

= 27V Vo., full load = Ioc Rr, = (26.74mA) (1 16)) = 26.74V % Regulation =

VNl, - Vpt-

vrl

"

1gg

_ 27 26.74x 100 26.74 = 0.97% c)

D.C.output Efficiencyof rectification _ A.C.input 8 n2

,-Rf

*R.

t,-

RL

- 8x ln2

r * (2+8) l , -

1000

= 0.802i.e. 80.2% d) Transformersecondaryrating

= Enus Inus

= r3ou E='Al LJZI

= 0.89W D'c' PoweroutPut T.U.F. = A.C. rating

=

07L5 0.89

= 0.802 Ex.3.4\

For thefull-waaerectifiercircuitshownin theFig.3.1.2V is a sinusoidalaoltage.lf the maximumallowableaaerage d.c,currentin eachdiodeis 1.A, calculatethemaximum aalueof V. Assumelwo diodesto beidmtical,andneglectdiode allowablepeak-to-peak resistance in forwarddirection.

]tH;Ew!'6t!r'*

s

Rectifiersand Filters

157

E.D.C.-l

Fig.3.12 = Sol.: Given: F.W. rectifier with R1 100O, A.C. input voltage is V Let V- is maximum value or amplitude of sinusoidal voltage, across each half of the secondary winding. Maximum lp"for each diode

=

1A

A full-wave rectifier essentially consists of two independent half-wave rectifiers feeding a common load. The averageIo" per H.W. rectifier diode

rv

=l

I-

=-1 --q fiKl

u,'hglgltn

_vm RL

vm

1A=l fr

RL

Vm = nRr=nx100 = 314.16 V "

Vpeak ro pear

=

Vru,

t

2

= 628.32V

3.5BridgeRectifier The bridge rectifier circuits are mainly used as, a) a power rectifier circuit for converting ac Power to dc power, and b) a rectifying systemin rectifier type ac meters,.suchas ac voltmeter, in which the ic voltage under measurementis first converted into dc and measured with conventional meter. In this system, the rectifying elements are either coPPer 9|ide tYPeor seleniumtYPe. The basicbridge rectifiercircuit is shown in Fig. 3.13. [Lr

E.D.C.-l

158

Rectifiersand Filters

Jrr

ACSupp'v

E

-DC

l-,[" Fig.3.13Bridgerectifiercircuit The bridge rectifier circuit is essentially a full-wave rectifier circuit, using four diodes, forming the four arms of an electrical bridge. To one diagonal of the bridge, the ac voltage is applied through a transformer if necessary,and the rectified dc voltage is taken frop the other diagonal of the bridge. The main advantage of this circuit is that it does not require a center tap on the secondary winding of the transformer. Hence wherever possible, ac voltage can be directly applied to the bridge-

3.5.1Operation of the Circuit Considerthe positivehalf of ac input voltage.The point A of secondary becomes positive. The diodes D, and D, will be forward biased, while D, and Dn reverse biased. The two diodes D, and Dr conduct in series with the load and the current flows as shown in Fig. 3.14.

Fig. 3.14Currentflow during positivehalf cycle In the next half cycle, when the polarity of ac voltage reverses hence point B becomes positive diodes D, and Dn are forward biased, while D, and D, reverse biased. Now the diodes D, and D. conduct is series r,r'ith the load and the current flows as shown in Fig.3.1.4. lt is seen that in both cvcles of ac, the load current is flowing in the same direction hence, we get a full-wave rectified output"

Rectifiersand Fj!!9q9

159

E.D.C.-l

exactly same as shown before for The rt'aveforms of load current and voltage remain full- wave rectifier.

;-l

Fig. 3.15Currentflow during negativehalf cycle

for VariousParameters 3.5.2Expressions

full-wave rectifier circuit; all the The bridge rectifier circuit, being basically a circuit using two diodes, are the characteristicdiscussedpreviously for a full-wave of a bridge rectifiercircuit' characteristic load current and Ioa ' Ip"5 remains The relation between I* the maximum value of circuit' same as derived earlier for the full wave rectifier

3Ipc =

ur't4 torr,=

3

This will be clear from the equivalent The exPression for I- will change slightly. circuit shown in the F i 9 . 3 . 1 6 .

I r

= F-sm

Rs i

II

x X,.

Fig.3.16

,".

E.D.C.l

160

Rectifiersand Filters

each half cycle two diodes conduct simultaneously.Hence maximum value of . l" load currentis, E't

=

I-

R, +2R, + R, So the only modification is that instead of R6, which is forward resistance of each diode, the term 2 Rl appears in the denominator. The remaining expressions are identical to those derived for two diode full wave rectifier and reproduced for the convenience of the reader. Eoc = I*Ra

itr =-"Sm

t4

Poc = I3c Rr.= *r31 nIn'

Pac = Ii',r(R. +2R, +Rr)

_ t l ( z n ,+ R ,+ R r )

nTorl-u,.

=

y=

2 8Rr n ' ( R , + 2 R ' "+ R r ) 81.2% 0.48

The E"* is the maximum value of a.c. voltage across full secondary winding of the transformer used. As the current flows through the entire secondary of the transformer for all the time, the transformer utilization factor is 0.8L2. This is more than the T.U.F for full wave rectifier circuit. The basic voltage regulation expression remains same as,

%R =

(va.)Nr.-(va.)er ,too (Va.)rI.

Approximately it can be expressedas' .ro

%R -

"'f x100 RL

3.5.3Advantages of BridgeRectilierCircuit ,-t)

The current in both the pri and for the entire cycle and bente_fal small size arJdlesscostmay be used.

of the

ut,

transformer flows r tra

E.D.C.-| ",

i6i

RectillerrandFi]len

2) -l!q-slt"r-tgp is-lequired in the lransfgrmersqqqldary.Hence,wherever possible,aqtqqllqg_e cql dilectly beappliedto thebridge.

"3)

pe-@ of the triuufqrmeriq-rneppositedireqrionin two half cycles.Hencer€! e!.c..gomponertlfefgllgj! -- ----------s- rglg *hi.b_eedrees_&"=brq",

and danger of saturath

,A)

Due to pure a-lternating cr{r91t_1qr-9ec9ry!gy of transformer,. the transls,lmer

gets utilised effectively and henceth_ecirqu.itii zuiEble lble fgl fq!_qpp!&egqnswhere large powers are reqlired.

5) As two diodesconductin seriesin eachhalf cyg!g,1nv_e1s_e_vq!lage appearing gqloqr {tg4e€eIgr11ea. ff;nce ttre .ir.,rit-'."r, b9 _uted_forhigh voltage :ppltsglbsqtuch a peakreJ/er:eyoltageappearingacrossdiode is c1Ue4peit i$rersevoItage;a gng_(B[!) gf_diode. 3.5.4 Disadvantagesof Bridge Rectifier The only disadvantage of bridge rectifieris the useof four diodesgEgomparedto two diodes in normal fgll_lgave-Eetrfier. ", t"9i."ted bu r"r^ 2 E t y_9!taee. Ex.3.5:

A 5 Kl loadisfedfromabridgerectifierconnected across a tranformer secondary whose primaryis connected to 460V, 50Hz supply.Theratioofnumberofprimaryturnsto secondary turnsis 2:1. Calculate d'c.Ioad current, d.c.Ioad ooltage, rippleaoltage andP.LV.ratingofdioite. S o l :. R , = 5 1 6 . ! 5 = x 1 0 3C ) ,N r : N 2i s 2 : 1 Ep = 460Vn.u.svalue E " N r 1- . 3 = =

EPNI

E. =

2 1

;tEp

=230V

Er,n = {f r E, =230x J1=j25.269y. now

IDc =

+

wheret^=

R1. eneglecting

2Et,,. -2x325'269 I.,r- = trRt- nx 5 x l0l3 = 41.41mA D.C.loadvoltageEDc =' Ipc * Rr = 41.41x10-3x$x 103

= 207.072Y Ripplevoltage = Ripplefactorr Vnc

Qs.Ew