'.

• I.i'

.';"'~'

·1 • " -, (' , .-' ....... _ J.

-<:>t : .v-

· .A_ cj,Jte~t::. with !1lc..~ou.r.Cg.. diQd~ and RC 19a'd is :sho.w.nin Fig .. 3.2: (a). Wb-el} -~w!i,t<;hS is

dl@s~d·at i-·g, KVJ., gIV~S "" .

or

1f -, .tu ; C uu-.»,Itl;l,Laplace tr~sfarm is R I(s) + ~ [liS) +w]= Vs '.

; C S S S

As the initi~ volt~ge ac~oss C js- zero, q(o) = O.With this, Eq. (3.1) becomes

I(S)[R +..!_J= VSCS S

1(8)=: CVa = Vs. 1

"R~S+ R~) R ..~+~~

...(3.1) .

"

_iRC

, ",

(0) (b)

,Fig. 3.2. Diode circuit with RC load (a) circuit diagram and (b) waveforms ..V .

i(t) = _! .e-tiRCR'Its Laplace inverse is

The voltage across capacitor is'. t V t

(t) =.! r idt = _8 r -tIRCVc C Jo " RC Jo e r • ,

= V8 (1 - i t;/w) ". ! \ '.•..'.(3:&,1) ,

=Vo (1- ((~) ....(3.3b)·:

where ~ = RC is the time constant far RC circuit. "Fro~ Eq. (3~3a), initial rate of c'ba.ng~ ef

'capacit~J."valtage is given by 'fd ~l:)" " ~.'[v.e~tlRC. _,"1J = Vs' " .:, .

i " lcl:t. /=0' s, " 'RC t='o RC., .,:, ...(3.4.).: source voltage, VB

Titme constant, p,C = Cd v ; at) _ ", ,c t-O

, In: ;Fig. 3~.2(~), cUrtent a~ui'voltage variations are as shown ..

,','

I'

3.1.3. ar, Load . "., . When,siWltea'S,:jsclosed at,t = () in. the RL .and 'di0~e ciscudt of Fig, 3,3 (~), KVL ~yes_, .

, Ldi Vtu» 'dt= S

, , For RL circuit, ~ = 't is the time constant. The waveforms of current and voltage aCl10SS

inductance are sketched in Fig. 3.3 (b).

3.1.4. JLCLoadA diode circuit with de source voltage Vs' switch S and LC load is shown in Fig. 3.4:'(a),

Whel'l switch S is closed at t = 0, the voltage equation governing its performance is given by

L di 1J'd V" '" '," -3, ,t +, C ~, t = S'~'\ ,I ;.',' I I, t,. ·~.t\· .;/ j tv

, Its Laplace transform is L [s l(s) ~ i(O)] +1.[I(S) +~] = VB ', C s s s, -

,I As, the circuit is initially relaxed, i(O) = 0 and Vc (0),=0 or q(O) = C . vc(O) = 0

I(S)[SL'; l]= Vs·sO . s

(0)

Fig. 3.3. Diode circuit ~th RL.load (a) circuit diagram and (b) waveforms.

With initial current in the inducto~ as zero, the solution of Eq. (3.5) gives. V _!it:W = J! (1- e L )

Initial rate of rise of current is '

The voltage across L is

VB=y

, ..,'

t

,..(3.6)

... (3.7)

... (3.8)

"

01'

Let (

Its

v.

-mvc(tO)

E

F

·8)

3.7)

S8

l.y,

, [Art. 3.1]

0r'" V

l(s) =o~. 1L '2 I,

S'+-"-LC

. I' , ' V ' '(I)" .; Jc COoLet (1)0=0 1LC ' This gives l(s) =0 -L S , -2-

0- = Vs . '\I'[ . -2--2

,,' " , ' , . (1)0 S +(1)0 S +(1)0

5, Ili:)

+----+ t

t, ,I)L

t

·~I I-vs t

(b)

(0)

Fig. 3.4: Diode circuit with LC load (a) circUit diagram and (b) waveforms.

Its Laplace inverse is i(t) = Vs . \[f sin uJot ,..(3.9)

Here 000 = :..jIe is calledzesoncnz frequency of the circuit. Capacitor voltage is given by

1 ft 1J' _/cvc(t) = C 0 i(t) . dt =C 0 Vs' '\II sin 000 t

= Vs (1- cos OOot) ... (3,IOa)Voltage across inductance is given by

L diet) ,vL(t) = dt = Vcos 000 t ... (3.10b)

WHen OOoto=n or when to='nlooo, from Eq. (3.9), i(to) = 0 and from Eq. (3.le,a),vc(tO) = 2 Vs and uL(tO) = - Va .

Here to = nl 000 ~ conduction time of diode = n ...JLC .

From Eq. (3~!!}),circuit or dioel:e current at to/2 = ~ attains a peak value of, • '. 2000 ,

I;.; Vs . Y'CIL as shown in Fig. 3.4 (b). V01tage across diode, soon 'a'fte~diode stops conduction 'at ~ois given by ,

ill) == _ UL _ Vc +- Vs = (;)"_2 Vs 1- V¥ ~ _ VS'

Wa:v:efOJ.ims M i:~t~"IJ.€,vr, and u» are sk.~t4i!h~d iiJ;l Fig. _3.4 (b). it is, seen tliit 'at.. ,

:-;tb.f2==--..1L dio,de cureent reaches p~ak value,' vc=Vs and vL,=;,(i). Also .~t,~ . ,2 Wo' ':_ " i,. , ',.- • ' :', , c , " , _, , ~ _'. '

, ~d'= ~/.o)Q = 1t C'-lILC), ,mode"cUl'tent decays ta zero and capacitor-is charged to voltage 2V~. SlOonafter to, voltage aCFQSSL is zero and diode voltage ,vD = ~ Vs.

JBl~mJ!llte 3.Jl.. For the circuit shown in F~g. 3.a (a), the capacitor is initially charged, to a,.l)_Ql,t:ag~:Vo with upper plate pa'sitiue. Switch S is closed at t = O. Derive expressions for the,e/Jirrent in the, circuit and voltage across capacitor. C. What is the pea,k value of diode currennFind aliso the energy dissipated in the circuit: -

§lI}n1\ll~ioRll. When switch S is closed, KVL gives

, 1f ,Is i; C idt=O

Its Laplace transform, including the iriitlal voltage across capacitor, is

R 1'[~' CVo] ,l(s)+- , _-, =0C s s

l(s) [R + ~s] = :0vIts solution, as per Art. 3.~1.2,is i(t) = ;. e-tiRC

or

iko

~~{~-"':, (b)

Fig. 3.5. Pertaining to Example 3.1 (a) circuit diagram and (b) waveforms.

:. ,Peak. diode current = ~o

(a)

uC(i) = ~ f: idt - Va

1Jt v ' '-, . _Q e- tlRC o1t ''i'T-CoR" "a-va

., =.-Voe;-tIRC,; :p1:1fpen:~~,~) and voltage vc(t) ar~ sketched in Fig, 3.5 (b).

, -.. l' . ',',.'E:aeFgy dlssipated in the circuit, =iCV~Jo~les

>EXl

ehal'ge(fop cun

:Finvoltagethe uol;

~o:

Its

vc(t) = (V. - Va) '\[f J; sin (1)0 t.+ Va

= (V. - Va) (1-:cos (l)ot)-{-Vavc(t) = Vo" 'vc(t) ='Vs and at (1)0 t =n, vc(t) = 2 (Vs - Vo) + Vo = 2V - V

8 o·~~~~~~~~~~

~dtaufor the:""ent?

'Il' a/I; a..t30 '" a,ts· SOOl).

Of, ....

[Art. 3.1]. 37

Emmpne 3.2. In the diode and LC network shown in Fig. 3.6 (a), the capacitor is, initiallycharged to voltage Vo with upper plate positiue. Switch S is closed at t= O. Derive' expressionsfor current through and uoltage across C,

Find the conduction time of diode, peak current through the diode and final steady-stateuoltage across C in case V. = 400 V; Vo = 100 V; L = 100 IJH and C = 30).IF. Determine alsothe uoltage across diode after it stops conduction:

§ou~Jion : When switch S is closed, KVL for Fig. 3.6 (a) gives

Tdi If'd V'b'-+- £ t= 'dt C . s

Its Laplace transform gives

L [I . ! 1 [IrS) C· Vo] V.S (s)-£(O)]+- ~+-- =-

C S S, I S

As initially i(O) = 0, the above equation becomes

[1J v, - Vo/(s) sL+- =--se s

• • • to!..o-Tl/2wo~ I

I :

Vc ! : .'.~Vs --------- J (2Vs-Vo)

'IVS : •

I.t

(a)

t- (Vs - Va )--.6.===

(b)

Fig. 3.6. Pertaining to Example 3.2 (a) circuit diagram and (b) waveforms.

This equation in s-domain can be solved as in Art 3.1.4. Its solution is '

i(t) = '(VB - Va) . '\[f sin (1)0 t

At (1)0 t = 0,At (1)0 t = 1t!2,

" '

")'" ,

""I, ,

J8 [Aiwt. 3.t]

Di~de conduction time, . to = ..2"_ = 7i; -.fLC = w/30 x 100 x 10- 6 == 54.77 usmo '

Peak current through diode, Ip = (VS - Vo) -{f ,= 300 - f30 = 164.32 A-'J1oQ

Steady state voltage across C occurs when moto = 1t,

.. V C = 2 (VS - Vo) + Vo = 2Vs - Vo = 2 x 400 - 100 = 700 V

Voltage across diode, after it stops conducting, is given byVD = - "i:- Vc + Vs = 0 - (2Vs - Vo) + Vs = - Vs + V1j = - 400 + 100

=- 300 V.Example 3.S. In the circuit shown in Fig. 3.7 (aJ, the capacitor has initial uoltage of Vo

with upper plate positive. The circuit is switched at t = O. Derive expressions for current andvoltage across capacitor. Find the conduction time for diode and steady-state capacitor voltage.

Sohrtion. The voltage equation for the circuit of Fig. 3.7 (a), after switch S is closed att = 0, is

"IIf

L di 1 f 'd 0dt+C z t=

Its Laplace transform, including initial voltage across capacitor, is

l(s) , sL +1. [!.J§)_ _ CVo] = 0C s s

I(S)[SL+ s~]= :0Here minus sign is put before Vo, because for the direction of positive current flow, polarity

of Vo is opposite.

Solution of above s-domain equation, from Art. 3.1.4, is

t

t"

III

-~ !IDOh IOF to =!.. -'- -:>It

Wo -V]b _o

(a) .. (b)'Fig. 3.7. Pertaining to Example 3.3 (a) circuit diagram and (b) wavp.fnl'n1"

" .. G",_,_ .;;>",;';', J.:J.~.~. 't:lh:~f.~·l~'~:A'~.!"':o~~.Jt ......_"",'~a:·j·v.,_.......:t._·_:_,~ '•._\ ...i..1~ : .l.'''' if .........L.r." •• - OJ __ - •

Diode C--VoH

Dio!

StelVol1Wa,

3.1.

Ad·switch;

Wit

, or

equatio

or

where

is caller

is calle!

Vond

at

'Dip-de'·€il'€uits, and Rectifiers [Art. 3.1] ._ 39·

t

~ . ~Fig. 3.8: Diode Circuit with RLC load Cci) circuit diagram and tbywaveforms.

With zero initial conditions, the Laplace transform of above equation is

[ 1] v,l(s) R+sL+- =-sC s··V 1,

.. or l(s) = LS

' 2 R 1s +Ls+LC'

Here S2 + R s + _!_ = 0 is the characteristic equation in s-domain. The roots of thisL LC W'l

ty

Voltage across C is vc(t) = v, '\[[ S; sin. wot - Vo = - Vo cos wot

to '" .2£. = 1t .,fLCWo

Diode conduction time, ;

Steady-state capacitor voltage = - Vo cos 1t = + VoVoltage across diode; " vD == - Vo.

Waveforms for i, Vc and vD are sketched in Fig. 3.1 (b). . \",

3.1.5. RLC LoadA diode in series with RLC circuit is shown in Fig. 3.8 (a). KVL for this circuit, when

switch S is closed at t = 0, is given by

'R' L di 1f idt V :. ,~z+ ¢t+C Z ='s

s R Underdamped _Cdtically damped

~._. <overdamped

" ' " ." ......

"<, -'-'~-...... _

r1-..:...', i.,,_, ' ~__.J'TJ",

+----VR - •

. L

equation are.' ,

or

where

is caned the damping-factor.

-is called resonant frequency in rad/sec ., \ ' \,

.{

" :' ,

...~3.11)

.~.(3.12)

...(3.13)

and "'," ~id:ij·~.vO)~ - ~2 = ringing frequency in rad /sec .

• 1 2 -2Alaa 0)0 = '1O)r + ~Depending upon the values of~ and 0)0' the solution fl!>l' the current can have three

saJu.tians.CBlSe1. In case;~ < 0)0, it is seen from Eq. (3.11) that theroots are complex and the

is saidto be underdamped. The two roots areS1 ;" - ~ +j to, and' 82 =- ~- j O)r

and' the current is given by, . '.' . V .... ,

i(t) =_8_. e-E,t sin (.0 rt. c.o,.LC!Ii§e 2. In; > (.00' the two roots are real an~'the'circ~it is said to be ouerdamped, The

roots are

. . VB '_l... _~-::-:2:"'let) = _IY'F. 2 . Sw..u '1<;~ - (.00 tL '11;- - (.00

Csse 3. In case 's = (.00' the roots. are equal and the circuit is said to be critically damped.The roots are S1 ,,; 82 = - s and the solution for the cu~ent is .

Vi(t) =.2. t 'e-E,I. L

Waveforms of current for the three different Ievels of damping are sketched in Fig.(b) .

.JE~mll!llljp)Ke 3.41. For trWcircuit of Fig. 3.8 (a), th~'data is as under ;'R=10n, L:;;J mH, C=5 ~ Vs=230V

The circuit is initially relaxed. With switch closed at t = 0, determine (a) current i(t) ~)conduction time of diode (c) rate of change of current at t = O.

~olrutq;noJlll: (a) From Eq. (3.12),

lOx 1000~ ;:: 2 x 1 = 5000 rad / sec

1 1 105 . '.:~.

From Eq. (3.13), roo= ~LC ~ [1 x 1?- 3 x 5 x 10- 6t2 = To = 14142.136 ra4! s

From Eq. (3.14), Ol, =[ !~~..- (5000)' J1I2 ~ 13228.76 rad I s

H~te as ~ < roo,the circuit is underdamped. The current is, therefore, given. by Eq. (3:16)

230 x 1000 - - 5000t •i(t) = 13228:76xl' e ., sin 0.3228,7S)t

_ 1Ii' ~M4",e - 5000! . sin (132:28_.7_6 t)

.fiiod~ eiFel!

(b) Diol

., Gen.1

(c) Fr~

3.2. FJRJE:II.l Fil

S is nowof arc atcalled freunderstatwo mod!

ModVo,S,D,

Fr

.JVJIa'vo]

fOl'wBcurrech.1~uiin Fig;i"Vel1

·6)

7)

8

; III'~ - IIWiode·Ciwcui~ mod Redifiel'S. - -

[Art. 3.2]- -

fb) Diode stops 'conducting when (Ort1 = 1t

r. :. Conduction time of diode, .

.'

(c) From Eq. (3.15),

1t 1tt1 = (.()r = 13228.76 = 237.48211s

di Vs [- ~t • - - (I') - lit]""- = -_ e . (0 cos (O~- sm.(O~· - '0 edt (.()rL r

di I = Vs = 230 x 1000 = 230000 Als.dt L 1 "'t=o -

~il

t5)

3.2. 1FRlEE'WJB[ElElLThTGDIODESIn Fig. 3.3 (a), steady state current, after switch S is closed, is equal to V/R. If switch

S ,is now opened, the energy stored [ = ~ L <VsIR)2] in inductance L will appear in the form

of arc at the opening contacts of switch S. In order to avoid such an occurrence, a diode FD,called freewheeling, or flywheel, diode, is connected across RL as shown in Fig. 3.9. (a), Forunderstanding how FD comes into play, the working of circuit of Fig. 3.9 (a) is divided intotwo modes.

iWoiLlle I : When switch S is closed in Fig. 3.9 (a) at t = 0, current flows throughVs' S, D, R and L as shown in Fig. 3.9 (b). In this circuit, current i is given by

'NO

D if+

R :Rt R R

VS FP ,~, Vs FD+t tl<1!LVL L L dl

if +.

(0) (b) (c)

Fig. 3.9. Circuit of Fig. 3.3 with freewheeling diode., '

Final value of current,

V -~ti = R (l-e L)

1= VsR

.,M~«lleJill: ':Yhen switch S is opened at t.= 0, current in the circuit ~ends to decay and soa - ·It' L di : . d d'" L hi h .ve : ·age· , dt' I'S In, uce . In W IC CUrrent

forward biases freewheeling diode. Thecurrent is, therefore, transferred to thecircuit consisting «r». R and L as shown

.)~ Fig. 3',9 (c). In this' circuit; current isgwen. by . .

...(3.1~)

Fi~.3.9 Cd). Current variation in tho ";_..,,,,;+- n~ 1:': _ "n

...(3.18)

.4Jl

time

;.' I,

42 [Alit. 3.4]

Consider the circuit of Fig. 3.10 (a) where de source feeds L through diode D. ~freewheeling diode FD is connected across L. When switch S is closed at t = 0, KVL gives

diVs =L dt

. Vs=t:or , ... (3.20)

+

'kdWhenf

\orward bisoositivetet

lEnergyThe en

her.eit is\',.5. SiNG:

Rectifi(c) (b) It A. . di d b f 0 age,Fig, 3.10. Diodecircuitwith FD amdL load (c) circuit iagram an (l wave orms. utput vol

This shows that current i rises linearly with time t. In case switch S is opened at tv load In thi:

V . 'D As here i . t . th . it f . d b deal as blcurrent ~ tl begins to flow through F. t ere ISno reSISance m e circur orme yL 3.5.1.

Land FD, current continues to flow at its constant value of i t1. Energy stored ~ the ~his ..apphcatio

1 (Vs)2 1 '~ . .. the princiinductance is - - tl .L = - . -L tl Joules. Current waveforms are shown m Fig. 3.10 (b). \ I2L 2 nas:ofoutput,

3.4. RECOVERY OF TRAPPED ENERGY '.'. The I.In the ideal circuit of Fig. 3.10 (a), the energy stored in the inductor is trapped. This are now (

tra?ped energy is not dissipa~c~de~en whe~ FD conducts be~ause circui~ does not contain (a)- ]aresistance, The 'best way of utilization of this trapped energy IS to return It to the source. In 3.12 (a).this manner, not energy taken from the source is reduced and the system efficiency improves.rot= bO tc

One way of return:ing this trapped energy back to the source is to add a second winding current iclosely coupled with the inductor winding as shown in Fig. 3.11. A diode Dis also placed in negativeseries with the second winding: The inductor now behaves like a transformer. 'I'he twO state. 0windings are so arranged that their polarity markings are opposite to each other. cot = 2n:,

Whe:nswitch S is closed, current i begins to flow and energy is stored in the inductanceof primary winding with N} turns. The polarity of the secondary winding voltage V2 is asshown. The diode D is reverse biased by voltage (V2 + Vs).

IoI

:'" tl---~'- 'I

I

'1~'ld I

'~ftl

t

IArt. 3,$j 43

d

.20)

(a) '. .'. " (b)Fig. 3.11. Energy-recovery circuit (a) switch S closed and (b) switch S opened.

. When switch S is opened, polarities of voltages VI and V2 get reversed" the diode is nowforward biased by voltage (\12 - Vs), As a result, diode begins to conduct a current il into thepositive terminal of source voltage Vs and so the trapped energy is fed back to the source.

Energy fedback to de source = Vs x current il dependent upon (V2 - Vs)'

The energy stored in t. of NI turns is transferred to secondary winding of N2 turns fromwhere it is fed'back into the de source, " .

i" 3.S. §lINGlJE-PHA§E DJIODE RlEICTWIER§Rectification is the process of conversion of alternating input voltage to direct output

voltage. As stated before, a rectifier converts ac power to de. In diode-based rectifiers, theoutput voltage cannot be controlled, '

In this section, uncontrolled single-phase rectifiers are studied, The diode is assumedideal as before. , .

.3.5.1. Smgle-Phase Half-wave RectifierThis is the simplest type of uncontrolled rectifier. It is never used in industrial

applications because of its' poor performance, Its study is, however, useful in understandingthe principle of rectifier operation.

In a single-phase half-wave rectifier, for one cycle of supply voltage, there is one half-cycleof output, or load, voltage, As such, it is also called single-phase one-pulse rectifier.

The load on thy output side of rectifier may be R, RL or RL with a flywheel diode. Theseare now q~~S'E!O briefly. . .-. '(a)) ~d : The circuit diagram of a single-phase' half-wave rectifier is shown in Fig.

3.12 (a). During the positive half cycle, diode is forward biased, it therefore conducts fromwt = 09 to rot= n. During the positive half cycle, output voltage Vo = source voltage v~ and loadcurrent io= VoiR. At rot = n, Vo = 0 and for R load, io is.also zero. A.f, soon as Vs tends t~!be~(i)menegative after rot:: n, dioUeD is reverse biased, it is therefore turned .off and goes into blockingstate ', OU~'p'J;ltvoltage, as weI~ as output current; are zero 'Hom (I) t:: 'It to '(I) t = 2n. After(j) t =;=.2n, dlC!l.cle,IS again forward biased and conduction begins. .

,F~r a'l'esi~tiille~0a~,.0u.tJ>litcurrent io has the same,waveflDFmas that oitke~ output voltageL10· thode voltage v» IS zero. when diode conducts .. Diode is reverse biased. from (J) t = 1t to

------'----'--=-.<:. .•-,.._:'~~,....:~::..~·i-·~j.___:Z~~--"-·'·~-·--: __ J1X\ '+;'!r

.O'il__ [_A_~_·a_. _3_.5...:J ~»~.f~)'l~~e~.r:_E!£11I:tl"~~d"" C::. r""Dlo e I.

((){ t:: 2n: ae SllOW Th -..::\_~ ~ n e wavefo- f . .voltuge'" lIDS 0 V., Vo, ~o and un are sketched in Fig. :1.12 (1;). Retr. p~. IS sUlusoidal i e V - V· KVL .c .... . . - .Q'J"

" s - m sin rot. lor the circuit of FJg. 3.12 (a.) glV(;H U "" r) " Ifj rj rli,

\ Inpt

Pow<

(b) 1en B

+ +

b;SinwtJ R ruo0_

tllI '

IL\lio

I ,

I I

Vo I

o~--~;-~~~-~3~nwi

At. (a) (b) or

Fig. 3.12. Single-phase half-wave diode rectifier with R load (a) circuit diagram and (b) waveform,

A~e:rage value of output (OT load) voltage,

vo=ffr VmSinrotd(rot)]~~LoVm v;

::: 211: I-cosrot I~=n

'7 __[l t -r ,Q • 2 s . d ,)1/2J~m.s value of output voltage, t 07' - 2n Jo v~"sm em (ro.)

'(T r ..ro: ,. }1/20:: ;i2;'c lJo 1-:~s 2ro~ . d(rot)

V~;c:--

Su

S((b) ..

A

2 ... (3,2

Here the subscript 'r' is used to denote rms value.Average value of load current,-,./ V. V

I -_Q-~0- R - Kit.

1 d I __Vor __v,Rms value of oa current, or l!- 2R

r.{e81~rvalue of load, or diode, currentVm

'=71

... (3,i

... (3J

... (3,1

Peak inverse voltage, PIV, is an important parameter in the design of rectifier CirC~:'1PlV is the maximum voltage thai appears across the device (here diode) during its blocktstate. In F'ig. 3.12, l~IV= Vm =..,f2 . VB = V2 (rms value of transformer secondary voliag:Y;:.:-,

ki~·t.,:~.\.,~~~/~.,:<:.\...,;;;:::,,\.o,:,{)\~.~»~:.' .>.:~\~\l<'.....! ~

orDrIllS.

3.22)

.28)

~III

I"

!~~~~ ?~o~eCi~cuUsand Rectifiers [Art. 3.5]

~t" Power delivered to resistive load = (rms load voltage) (rms load current)~ .

10'1- ~ ~~~ = V .I = Vm . Vm = __E!. = ~ =]2 R .or or 2 2R 4R 2R or

Power delivered to loadInput VA

Vor . lor v; 'l[2V.= -- = - = -- = 0.707 lag.

Vs ' lor V. 2Vs(b) lL Ioad : Single-phase half-wave diode rectifier with L load is shown in Fig. 3.13 (a).

When switch S is closed at rot = 0, diode starts conducting. KVL for this circuit gives

diov =v =L-=V sinrots 0 dt m

...(3.26)

~'.f".,/-, ,

Input power factor

. VmJto =L sin ro t .dt

Vm= - roL cos rot +A

VmO=-roL+A

A=VrnlroLSubstitution of the value of A in Eq. (3.27 rei) gives

or ~)

...(3.27 a)

At rot = 0, io = 0, :.

Output voltage,

. Vmto = roL (1- cos (ot)

di; v'nvo:;:L dt =L roL [sin rot] (0= Vm sin rot = Vs

...(3.27 b)

Source voltage Vs and botn output voltage v(j 9,Udoutput current io are plotted in Fig. 3.13(b).

Average value of output voltage, Vo= 0

wt

. ". (a) '. .... . (b).FIg. 3.13. Single-phase one-pulse rectdier With ~ load (a) circuit diagram and (b) waveforms:

, :~"0I.iW;):I.wil (J),

l.!(;ll,lli; <{11.},'W tt}' {lfj1,',lliJ)),I; l~I(I~nf;(}I?,T!i3 at (J)t ;= 11:

V: ' 2V'-l,~",,;= r;: (1 + 1)= .: 'I'm

" "', (}.)'-' WLJ

,1'f27tVm " ,.AYf.)J:t~tj(j v!ilw) OJ' c!)o'un;f;, To;: -2' -:r(1-'~0S rot) d(rot)

11: 0 O.Jl.J

Vm 1::: 0)£ :=:2'1=

Rtl'W 'V'f).1.rwof ftmd;xmcn1;al CT~r.r.en.t,111'i~ given by'

1,,= [i.(~ir.(00' rot)'d(rot)j'"Vm : Vs 10

:=:~12·OJL OJL:=: T2Rms value (if ;r:cr;'MfJed current :=:[fa + Iq1l2

[&]112

= fa + ; = 1.22510

Go

CODsta

shoWl]

Voltage aC1:(!HHdiode, Un:=:O.

R

T1

voltag

.. ,(3.3( F(

N

.k... (3.3i

(c) 0 Load : In. J!'jU. :~.14 (a), when switch S is closed at OJt = 0, the equation governinthe behaviour of the circuit is

, do; d .~o= edt = c at (Vs,sm (1) t)

R

Ewt 50H;

the p,

I SI vmlII I

rr 2n 3rr t.rr wt Rt •

II

wCvm:I.

3n/2 . Iwt F

I

! ,I

2n' I

IT 3n 7n/2 ,wt

]

(rt) . .' (b) ,I

S. r 1 _·nhM!.: half-wave diode rectifier ......qth C load (a): circuit diagram and (b) wavefl)1ili'

F' a 3 14. Ing e r " - ,.,,:.',.. . .1,;;," , . _ . " ~:::-,J:;t~:::'!..~i.b.,tt

= roC Vm <}0S ro~,..~. . ..

...(£.32)

t:'\ m t 1+ 1 f'd V··· ._.vl!l>~.p1!I! - v.0. 'Gage, Vo = ~ t t = sm rot = v = vc.:. . ,C' . m s

'Cap~~it0r'is charged to vbltage-'Vm at rot=:i and 'subseque~tiy this voltage remains

constant at Vm• ~hi~ is showri is Vo:;: Vc in FIg. 3.14 (b)

'Capacit0Feurrent er-lead current is maximum at rot = O. Its value at rot = 0 is ro CV 17' asshswn.

The diede conducts f0~27t seconds onlyfrom rot = 0 to rot = ~ . During this interval, diode,co 2voltage is, therefore, zero. After rot = 7t/2 , diode voltage vn is given by

v1)= -'uo+ v,.:;_ Vm'+Vm sin rot= V (sin rot - 1) ...(3.33)

For Eq. (3.33), the time originis redefined at rot = 7t/2.

After rot = 7t/2, diod~ voltage is plotted as shown in Fig. 3.14 (l». At rot = 327t,vo = - 2 Vm·

Average value of voltage across diode,"1f27t .VD = -2 ' Vm (sin, rot - 1) d(rot)

. 7t 0

= 11 = '-I2vm s

. i

...(3.34 a)

Rms value of fundamental component of voltage across diode,

. [1 f27t 2 :. 2' ]112 v,, V lr == 27t 0 V;" sin rot d(rot) ,=,~

Rms value of voltage across diode, = -vr-V'71'b-+-V-'tr-=1:225 Vm

...(3~34b)

...(3;35)

.' Example 3.5 •.A single-phase.230 V, 1 kW heater is connected across single-phase 230 V,50 Hz supply through a diode. Calculate the power delivered to the heater element. Find alsothe peak diode current and input power factor.

2302

Solution. Heater resistance, R = 1000 Q

Rms value of output voltage, from Eq. (3.22), is( ,. I •. ' ,. V2 X 230

v; 2,'I

Power absorbed by heater element

_ ~r-R

2 X 2302 .i000 _, '.x--' = 50'0W

4 2302

Peak value ~t~ioclecwrent, frem ~q. (3.25), isgtven by .. V2 x 230 _', .

'2302 X 1000 = 6.1478 A

(a) (b)Fig. 3.15. Single-phase half-wave diode rectifier with RE load.

would not conduct at rot = 0 because diode is reverse biased until source voltage vs equalsE. When Vm sin 91=E, diode D starts conducting and the turn-on angle 91 is given by

91= sin-1 (:m) , ...(3.36)

The diode now conducts from rot = 91 to rot = (n - (1)' i.e. conduction angle for diode is(n - 2(1) as shown in Fig. 3.15 (b). During the conductionperiod of diode, the voltage equationfoil' the circuit is

!'ll1ilutP€lwe~facter = ~r = J2x230 x _1_ _ .'.Vs 2 230 - 0.707fag.

(d) RlE Load • Sin 1 hemf E is shown . •F' g e-p ase half-wave rectifier with load resistance·R and load cOUnte

:'. m Ig. ~.15 (c), If the switch S is closed at rot = 0° or when Vs = 0, then died!

\.\

Oil'

Vm sin rot =E +ioRVm sin rot -·E

io R

AV9r:age value of this current is given by

10= ~ [f.:-" rv, sin rot - E) d(rot)]1= 21tR [2 Vm cos 91-E (n - 281)]

Rms value of the load current of Eq. (3.37) is

t;",[~(,-e, (Vm SinBrot - E} d(rot)r

= [ 2~2 J:,-" tV! sin' q,t+E' - 2 V~ E sin rot) d(OOt)r

wI

(c

,poV

wi

It j

perioddiode i

ExFig.3 ..

(a)(b)(c)(d)(e)

So

wI

Av

... (3.37 (b

'•.. (3.381PI

(c

[Art. 3.5] 49

[. ]1/2

=. 2~2, {(V; + E2) (1t- 291)+ v; sin~'91 - 4 Vm E cos e1}

. Power delivered to load,

P=EIO+I!rR wattsSupply pf Power delivered to load

(Source voltage) (rms value of source current)E 10 +l!rR

Vs' lor

..,(3.39)

...(3.40)

...(3.41)

It is seen from Fig. 3.15 (a) that at (JJf; = 0°, vo = ., E and at wt = 91, vo = O. During theperiod diode conducts, vo = O. When wt = 31t/2, Vs = - 'IIm and VD = - (Vm +E). Thus PIV fordiode is (Vm + E).

_Example 3.6. A de battery of constant emf E is charged through a resistor as shown inFig, 3.15 (a). For source uoltage of 235 V, 50 Hz and for R = 80., E = 150 V,

(a) find the value of average charging current,(b) find the power supplied to battery and that dissipated in the resistor,(c) calculate the supp'ly pf, ,(d) find: the charging time in case battery capacity' is 1000 Wh and(e) find rectifier efficiency and PIV of the diode.Solution: (a) The diode will start conducting at an angle 91>where

9 . -1 150 274660,1 = SID T2 X 230 = .

Average value of charging current, from Eq. (3.38), is

10 = 21t~ 8 [2 . .,f2 x 230 cos 27.4660 -150 (1t_ 2 x 2~.:~6 x 1t)}

= 4.9676 A(q) Power delivered to battery

= E 10 = 150 x x 4.9676 = 745.14 W

Rms value of charging current, from Eq. (3:39),'is

lor = [21t ~ 64{(2302 + 1502) (1t - 2 x 27.466 x' 1~0) +'230

2sin 27.466

- 4 -{2 X 230 150 cos 27.4660}] = 9.2955 A .

Power dissipated in resistor

= Z;r R = (9.2955)2 X 8 = 691.25 W

(c) From Eq. (3.41), the supply

f= 745.14 + 691.25 = 0 6721p 230 x 9.2955" ag

,(d) (Power delivered to battery) (charging time in hours)

= Battery' capacity'in Wh.

~,

~:·1

50 [A<r-t. ·3.5J

.', Charging time 1000:=: 74m::::: 1.34:2h.

::: ~w()): deli~:ed to battery'f()tal inpllt power745.14

:::745. i4 -I- 691..25 x 100::: fiil.876%::: V;n +1£:::{2 x 230 + 15@:::475.22V.

Theis and tb

(e) Rectifier efficiency

"\\Vh:e

. (e) RL Load : A single-phase one-pulse diode rectifier feeding RL load is shown:in3.16 (a). Current io ccmtinues to flow-even after Source voltage Vs has become negative;is because of the presence of inductance L in the load circuit. Voltage V R ::: io R has thewaveshape as that of io· Inductor voltage VL::: VB _ vn is also shown. The current io :Rowsthe two areas A and B are equal. Area A represents the energy stored by L an~ area Benergy released by L. It must be noted that average value of voltage lJL across inductor Lzero.

where cp

The

(a)

Its S!

Tota:wt

wherewt -Cons

At (1]

'wt

wt "

S'ubs(b) "

Fig. 3.16. Single-phase half-wave diode rectifier with RL load '(a) circuit diagram and (b) waveforms.

When io = ° at rot= P; "t.= 0, vR = ° and voltage Us appears as re~erse bias acrossD as shown. At p, voltage uD across diode jumps from zero to Vm sin p where p > n. "''''II~O:r.lCU1tl.el

~'= y is also the conduction angle of the diode.Average value of output voltage,

v: :::_!_ i~V sin rot· d(rot)o 21t 0 m

Vm A,~ 21t (1- c~s 1-')

, al e of load or-output currentAverage v u, , Vo', Vm - A

10 = II::: 21tR .(1,:- C0S 1-')

where <I> = tan-1 i and X = oiL. Here <I> is the angle by which rms current Is lags Vs'

·r L h The transient component it can be obtained from force-free equation

di,Rit+L dt = 0

. _IitIts solution gives It = A e L

Total solution for current io is, therefore, given byV R

io = is + it = -i sin (rot- 4» +A e-r;t

where Z=...JR~+X2

Constant A can be obtained from the boundary condition at rot = O.At rot= 0, or at t = 0, io = 0. Thus, from Eq. (3.44)

VmO=-Zsin<l>+A

VmA=Zsin<P

Diode Circuits and Rectifiers [Art. 3.5]

A general expression for output current io for 0, < rot< ~ can be obtained as under :When diode is conducting, KVL for the circuit of Fig. 3.16 (a) gives

~ dio .RLo +Lat = Vni sin rot

The load, or output, current io consists of two components, one steady state componentis and the other transient component it. Here is is given by .

.vis = . r:;:;<Tm2 sin (rot - <1»

'1Ri'fX

...(3.44)

Substitution of A in Eq. (3.44) gives

. Vm [. ( th) . '" _Iit]to =Z sin rot - 't' + sin 't' . e L

for 0 s rots ~diadl It is also seen from the waveform of io in Fig. 3.16 (b) that when rot=~, io= O. With thisBer' .condition, Eq. (3.45) gives

...(3.45)

sin (P - $) + sin $ . exp [ - !~]= 0

The solution of this transcendental equation can give the val~e of extinction angle ~.(j) RL load with freewheeling diode : Performance of single-phase one-pulse diode

rectifier with RL load can be improved by connecting a freewheelmg diode across the load asshown in Fig. 3,.17 (a). Output voltage is Vo = Vs for 0 $; rot$1t. At rot= n, source voltage Vs iszero, but output 'current io is not zero because of L in the load circuit. Just after rot = 11:, asUs tends te reverse, 'negative polarity of Vs reaches cathode of FD through conducting diollleD, whereas positive polarity of'», reaches anode 0fFD direct. Freewheeling (or flywheel) >diocie'

I . .

- (b)Fig. 3.17. Single-phase one-pulse diode rectifier with RL load and freewheeling diode

(a) circuit diagram and (b) waveforms. . diode D

After rot = n, current freewheels through circuit RL and FD. The energy stored in L currentnow dissipated in R. When energy stored in L = energy dissipated in R, current falls to Z! to 'b'; cat rot = f3 < 2n. Depending upon the valu~ of Rand L, the current may not fall to zere el experiewhen rot = 2n, this is called continuous conduction. But in Fig. 3.17 (b), load current dee Dl andto zero before rot = 2n; load current is therefore discontinuous.

The effects of using freewheeling diode are as under :(i) It prevents the output (or load) voltage from becoming negative.

(ii) As the energy stored in L is transferred to load R through FD, the system ef,ficieris improved.

(iii) The load current waveform is more smooth, the load performance is thereiimproved.

The waveforms for VS' vo' io, VD' is'and ifd are drawnin Fig. 3.17 {b).The expression for the load current io can be obtained from Art. 6.1.2 if.reqwre,d.!

seen frQIll Fig. 3.~7 {b) that

F'b> . . \, therefore . " ,from. D .; gets f0rward-hias . ' Power Electrolli.' to FD as v tena ed. As a result, loacl C " . • .reverse VQltao:e . s s to reverse. After (J)t _ . urrent ~oIS Immediately transfetr

. 0 WIth PIV equal to Vm

. - n, dIOdecurrent ill = 0 and it is sUbject~d~,01

singhto prjcondt

is LoUo

+ +

UR1 27f 311 I wII

m - I I I

+ :~:lJL t I I I -ild IT 2rr 371' wI

I wI

iS~

I

(0) I

l~I .

I I I wIo----l FO :- r- 0 _,,_J FD :-III I I

271 3'71 wI

average ~utput voltage, In vVo = 21 Vm sin rot d( rot) = .2!:.

1t o· 1t

of sourcrectifiier

Ave]

AveJ

Rms

.•• (3, Rms

PewEIn.pu'

-wi

diode

[Art. 3.S~ 53

.' (i) Single-phase full-wave mid-point diode r~ctifi~r : Fig. 3.113(a) illustrates a:lp.~l~~fl:nas~fun-wave mi~-peint ~ecti~~r using diodes. The turns ratio from each secondaryo prunary IS take~ as unity for simplicity, When 'a' is positive with respect to 'b' ; diode D1

conducts for TC radians. In the next half cycle, 'b' is positive with respect to 'a' and therefore

~ ~Fig. 3.18.Single-phase full-wave mid-point diode rectifier

(a) circuit diagram and (b) waveforms. .

diode D2 conducts. The output voltage is shown as Vo in Fig. 3.18 (b). The waveform for outputcurrent io (not shown in the figure) is.similar to Vo waveform. When la' is positive with respectto 'b', diode 02 is subjected to a reverse voltage of 2vs' In the next" half cycle, diode D1experiences a reverse voltage of 2us' This is shown in Fig. 3.18 (b). Thus, for diodesDl andD2, peak inverse voltage is 2Vm• Waveforms of Fig. 3.18 (b) show that for one cycleof source voltage, there are two pulses of output voltage. So single-phase full-wave dioderectifier can also be called single-phase two-p~.JliQ_clcueclliier.

. 1f" 2VAverage output voltage; Vo = 1t 0 Vm sin (I) t d( (I) t) = 1tm

Vo10=-R

1/2

Rms value of output voltage, Vor =[~ f: ~ sin2 (J) t d«(I)t)]

Vm=12=Vs

Average output current,

Rms value of load current,

Pewer delivered to leadInput voltamperes .

:. Input

I w~

iI;

4rrwt

...(3.48)

...(3.49)

54) [Art. 3.5]

Perce

(ii) §mgle-ph ful Power Electronirectifier emplo . as~ I-wave diode 10rid e . . I:! Diode eidiodes D1 D'2 Ylllg

ddiodes is shown in Fig 3 gI9 TC"( e)ctWhlfler : A single-phase full-wave brido

, can uct to th '. a en 'a' is iti ith .~is bi • ge er so that 0 t ' POSl rve W1 respect to 'b'su ~ected to a reverse volta f u put voltagn is uub' Each of the diodes D3 and n'

respect to 'a',diodes D3 D4 gde0 Us as shown in Fig. 3.19 (b), When 'b'is positive "'"t~

D1 ' con uct togeth d .., 'I Wit!and D2 experience are er an output voltage is ubu' Each of the two diode. verse voltage of Us as shown. 1

(a)f(3.50), is

Perce

wIVs Vo ~~w (b) F(

21( 3rr 41T

\7 .wI-Vm

(a) (b)Fig. 3.19. Single-phase full-wave diode brdidge rectifier (a) circuit diagram and (b) waveforms. , It isA comparison of Figs. 3.18 (b) and 3.19 (b) reveals that a diode in mid-point full-wa operation

rectifier is subjected to PIV of 2Vm whereas a diode in full-wave bridge rectifier has PIV JEX8!ll!

Vm only. source. Ti

Three-phase bridge rectifiers using diodes are discussed in Chapter 5. Example 3.7 i constant.formulated to illustrate the effect of reverse recovery time on the average output voltage. (a) av

(b) au) lExaIDplie 3.1. In a single-phase full-wave diode bridge rectifier, the diodes have a rever;recovery time of 40 us, For an ac input voltage of 230 V, determine the effect of reverse recave/. (c) rmtime on the average output voltage for a supply frequency of (a) 50 Hz and (b) 2.5 kHz. §oli1lll.1

shown in(a) A\

§oltntion. Single-phase full-wave diodebridge rectifier is shown in Fig. 3.19 (a) andoutput voltage Vo is shown in Fig. 3.19 (b). Ifreverse recovery time is taken intoconsideration, the diodes D1 and D2 will not beoff at rot == 11: in Fig. 3.19 (b), but will continue

to conduct until t == :'+ t.; as depicted in Fig.

3 20 The reduction in output voltage is givenb' the cross-hatched area. Average value of

Y. d ti n in oumut voltage IS grven bythis re ucno =r1Itrr

Vr = - Vm sin rotd (rot)11: 0

u Ao

Avera

(b) Av

Fig. 3.20. Effect of reverse recovery timeon output voltage.

Rms\

v

[Art. 3.5}

Vm= -.- (1 - cos rot )1t rr ...(3.50)

With zen') reverse recovery time, average output voltage, from Eq. (3.48), isVo = 2 -.[2 x 230 ~ 207.04 V

. . 1t(a) For f= 50 Hz and t.; = 40 JlS, the reduction in the average output 'voltage, from Eq.

(3.50), is .

VmVr =~ (1 - cos 21tf trr)

-.[2 x 230 (I 180)= 1- cos 21tx'50 x 40 x 10-6 x--1t I 1t=8.174mV

Percentage reduction in average output voltage

8.174x 10-3 '-3207.04 x 100 = 3.948 x 10 %

(b) For I= 2500Hz, the reduction in the average output voltage, from Eq. (3.50), is

-.[2 x 230 ( 180)Vr = --- 1 - cos 21tx 2500 x 40 x 10- 6 X --1t . 1t= 19.77V

Percentage reduction in average output voltage = 21~7~;4 x 100 = 9.594%.

. It is seen from above that the effect of reverse recovery time is negligible for diodeoperation at 50 Hz, but for high-frequency operation of diodes, the effect is noticeable.

Example 3.8. A single-phase full bridge diode rectifier is supplied from 230 V, 50 Hzsource. The load consists of R = 10 .Q and a large inductance so as to render the load currentconstant. Determine

(a) average values of output voltage and output current,(b) average and rms values of diode currents,(c) rms values of output' and input currents, and supply pf.§GHutJion. The circuit diagram and relevant waveforms for this uncontrolled rectifier are

shown in Fig. 3.21.(a) Average value of output voltage,

Vo = 2Vm = 2 -Y2 x 230 = 207.04 V1t 1t,

Average value of output current,

. 1~Vo = 207.04 = 20 704Ao R 10 .

(b) Average value of diode current, .1 I' .

[DAV= ~~1t.=; = 20.~04 = 10.352 A

, - ...rr;n [0 20 704Rms value of di0de current, [Dr = -" 21t =:::r2 = ~ = 14.642 A

55

j. J\..r 7! 2rr

ioll---_:_! _---.-~I__ -+__ -. flo

I I I, \ iilo -fro Il-lo 1 wI

.j wI

[ ...t.rr wI

R

Jr' 21risL

)

(a) (b)Fig. 3.21. Pertaining to Example 3.8 (a) circuit diagram and (b) waveforms.

As load, or output, current is ripple free" rms value of output current= average value of output current = 10 = 20.704 A

Rms value of source current, Is = ~ =10 = 20.704 A

Load power = Volo = 207.04 x 20.704 WInput power = Vsls cos <I>

.. 230 x 20.704 x cos <I> = 207.04 x 20.704

S I f '" 207.04 0 90 I ..: upp y p = cos 'I' = 230 = v, aggmg.

EX81mplle 3.~o A diode whose internal resistance is 20 n is to supply pouier to a 10~load from a 230 V (rms) source of supply. Calculate (a) the peak load current (b) the de 10

current (c) the de diode voltage (d) the percentage regulation from no load to given load.(lAS., 1-~.

,§el"uJJ.11;iollll. A voltage of 230 V supplying power to 1000 n, through a single diode, is sboiN 'Fig. 3.22 (a). Waveforms for the source voltage, load current io and diode voltage VDI

shown in Fi~. 3.22 (b).(a) It is seen from the waveform of io that peak load current lorn is given by

Vm ..[2 x'230lorn == R +Rf) = i020 == O,318~ A

wI

piode Cir~

(b) DC

(c) DC

(d) At

At givi

:. Volt

3.6. ZlEN]B

Zenervoltage.

Circuitbehaves asvoltage acbreakdoWlivoltage ~af1current. Tla Source w

,For th.a voltage E(c) is nece~

- .wt

+11/2 IT 27f . 3rr

VD~'OX~D- -' ~

wt

R=1000n -rrw21[ 37f(a) (b)

Fig. 3.22. Pertaining to Example 3.9 (a) circuit diagram and (b) waveforms.

wt......t

(b) DC load current, 1 f"10= 27t 0 lom·sin rotd(Olt)

10m=-=0.10151A7t

(c) DC diode voltage, 1 f"VD =10RD - - 230-{2sin rotd(rot)27t 0

v, 230-{2=loRD--=O.10151x 20 ---=-101.5 V

7t 7t,

v: ;, Vm = ..[2 x 2aO= 103.521Von 7t 7t

v: '= 230,[2 , 1000= 101 491V01 '7t X 1020 .

Von - VOl 100 = 103.521- 101.491 = 1 96101 _"Von X 103.52.1 . -10.

(d)' At no load, load voltage,

At given load, load voltage,

:. Voltage regulation

3.6. ZENER DIODESZener diodes are specially constructed to have accurate and stable reverse breakdown

voltage. ',if

Circuit symbol for Zener diode is shown in Fig. 3.23 (a). When it is forward biased, itIDeh~vesas a normal diode, When reverse biased, a small leakage current flows. If the reversevoltage across' Zener diode is increased, a value of voltage is reached at which reverse'oreakdeW1l occurs. 'Fhls is indicated by a sudden increase of Zener current, Fig. 3.23 (b). Thevoltage _aOOterreverse breakdown remains practically constant over a wide range of Zener0mTent. 'Fflis'm-iikes it s:uitaIDlefar use as a voltage regulator to furnish constant ,,{!01ta!gefroma source whose voltage 'may vary noticeably. '

. Far the operation 0fZener diode asa voltage regulator, (i) it must be reverse biased witha Vi(i)ltage'greater than its breakdown, er Zener" voltage and (ii) a series resistor B.s' Fig. 3.23(c) is necessary to 1imi,t the reverse current through the diode below its rated value.

+1 , 'fl;l.e·

Breakdown Rs 1,5 l,,' -.-,~'volta~e + -:

~~.v,

_. 'E;g;aJI 1z

where G-v +v v. Vz Load resista;mvz -, of the d,;"

§@ll.l

-I(b) (c)

Anode

(a)

Fig. 3.23. Zener diode (a) circuit symbol (b) V-I characteristics(c)use as a voltage regulator.

If Vz ::: voltage across Zener diode, then it is seen from Fig. 3.23 (c) that sourceIs is

Load, or output, current, 10:::iwhere R::: load resistance. Current through Zener

lz =ls -10Power rating of a Zener diode is Vz' lz. These are available in a voltage range f!om

volts to about 280 V. .

)

Example 3.10. Design a Zener voltage regulator, shown in fig. 3.24, to me~t thespecifications : .

Load ~oltage = 6.8 V; Source voltage Vs is 20 V ± 20% and load current is 30 mA ±. The Zener requires a minimum current of 1mA to breakdown. The diode D has a

voltage drop of 0.6 V.Solution. When source voltage is maximum and load current is minimum, then

resistance should be maximum.

".~. Vs. max == VL + (IL. miT).+ Iz) Rs. max

.. R:. min.::: (20 X 0.8) -: 6.8 :::1075 Q. .' [30 X 0.5 + 1] X 1O-::!,Similarly, " ..

Vs. min.::: VL + (lL"max +j~)n, ., ~ 'm&n

,:'. R '.::: (~Q ,X 0.8) - 6.S . ,s v mm [30 X 1.5 + ll~X 10- 3 ::: 200 Q

Mruximum, load resistance. .~, ,

+

Fig. 3.24. Pertaining to E*am,~le3.10'.

Ri.,. max:::: _,._VL~ ~ 6.8 .~. . 1'L. min. . 3.6X 0:5 X 1:0-3 ::: 41ii3.3 Q .

4\../f' • 1 d . t . R' VL:l.V.l'nn!ll'l'l!lm ea reSIS anee, [.;.min. ::::y-=--:::: _ _ 6..S. _. .L'm~' 3I!>XUixl(')-3-11ii1.§U

, "

or

or

As :t

AlS(

3.1.I:or bothSketch t

.Wh~

.' .

[Art. 3..6] 59

The voltage rating OF the Zener diode is6.8 - 0.6 = 6.2 V.

"c ~Example~.n.The complete circuit shown in. Fig. :J.25 (a) represents a 25 Vdc uoltmeter- where G is a PMMP galvanometer having full-scale deflection current Ifsd = 200 micro-A GJ,!-dresistance RG == 500 ohms, and D is a 20- V Zener diode. Find Rl and R2• What is the functionof the diode D in this circuit? (GATE, 1990)

Solution. Current through galvanometer,

I -I _Zener voltagefsd - 2 - R2 +Ra

20 -6R2 + 500 = 200 x 10or

CorUl'1

R = 20 X 106

_ 500 = 99 5 k Q2 , 200 .

:tJGr dio

(a) (b)

wa

Fig. 3.25. Pertaining to Example 3.11.

As Zener diode current is not specified, let it be assumed zero.

.. II - 12 = Iz = 0 or II = 12 = 200 J,1A

Also 11= 25 - 20 = 200 X 10-6

Rl

R = 5 X 106 = 25 k.Q

1 200or

lUI Function of Zener diode is to provide a constant voltage to the galvanometer circuit,Whenever voltage across-this diode exceeds 20 V, it conducts and the excess current is shuntedaway from galvanometer G. So here diode D prevents overloading of the PMMC galvanometer,

3.1. Capacitor in the circuit of Fig. 3.2 (a) is initially charged with (a) Vo volts and (b) - Vo volts.For both these parts, determine the expressions for current -in the circuit and voltage across capacitor.Sketch the waveforms for current as well as capacitor voltage.

What is the final value of voltage across capacitor in each case?

[An§.

5 V

3oflar

(i

33 - Vo, 44.72 JlS, 62.61 A 510 V, - 28(J1• • (a) For the circuit h . .

t == 0, sketch the '. s 0:vn m FIg. 3.4 (a), the circuit is initially relaxed. If switch S is closedthese fu ti vanatlOns of z, vL, "c and v» as a function of time. Derive the expressions describinc Ions. '

(b) For part (a), Vs == 220 V, L = 4 mH, C = 51J.F.Find the diode conduction time and peak elilcurrent. Determine' also vc. vi: and VD after diode stops conducting.

[Ans. (b) 0.444ms, 7.778A, 440V, 0, -2203.4. For the circuit shown in Fig. 3.5 (a) ; Vo = 230 V, R = 25 Q and C = 10 IlF. If switch S is clo'

.at t = 0, determine expressions for the current in the circuit and voltage across capacitor C. Findtpeak value of diode current and energy lost in the circuit.

Derive the expressions used. [Ans. 9.2e-400Ot, _230e-4oOOt, 9.2A, Q.2645watt-t

3sion f

(i

value3

freewwave:

3.5. In the circuit shown in Fig. 3.26, switch S is open and a current of 20 A is flowing through! (ifreewheeling diode, R and L. If switch S is closed at t = 0, determine the expression for the cum volta!through the switch.

. 22 2 -10(:)[Ans ~(t) = - e

10mH

10n.

Fig. 3.26. Pertaining to Prob. 3.5.3.S. (a.) Describe how the energy trapped in an inductor can be recovered and returned to the SoUlI

(b) A' 23'0 V, L kW heater, 'fed through 'single-phase half-wave diode rectifier, has rated voltageits terminals. Find the ac input voltage, Find also PIV of diode and peak-diode current, ' ,

[.Ans. (b) 325.32 V, 460 V, 8.690f'lMMC

--0-3.7. (a) In the 'circuit shown in Fig. 3.27, a PMMCammeter is placed i~ series with dio~e and a P~MC

ltro,eter across the diode. Take PMMC mstruments Ideal.;~ d the readings on these instruments. Derive the e~res~.m ed for obtaining these readings. '

S1011S us "(b) If PMMG aIDllleter is replacedby MI ammeter, find

its ,reading.

4to~currs

[Ans. ~0.352 A, '0 V, 12.6812 A]

nc

f'lMMCFig. 3.27. Pertaining- to Pmb, 3,1,

~::~~:::~~~~:W~~"J1!Jf~~;;1;;~~';~~jf~Y,;.:?;>:7; .:'Ct'< ...~ [PNb.31

t - ~1.@a.Ci~ 3.8. (a) In the circuit of Fig. 3.28, ideal PMMCvol>tmeters are placed, one across capacitor and another

IQ.'· across diode as shown ..Find the voltmeter readings. Obtaine Co the expressions used for determining these readings.

(b) In case PMMC voltmeter 2 is replaced by MIvoltmeter, find its reading.

Os CiJ 3.9. A battery is charged by a single-phase half-waveo diode rectifier. The supply is 30 V, 50 Hz and the battery2 emf is constant at 6 V Find the resistance to be inserted in80

series with the battery to limit the charging current to 4 A.lSeq Take a voltage drop of 1 V across diode. Derive the expression used.~libi Draw waveform of voltage across diode and find its PIV.

[AnS 2~ [2Vm.cos61- (E + 1) (n - 2 61)], 2.5458 n, 48.42 V]3.10. (a) A single-phase half-wave uncontrolled rectifier is connected to RL load. Derive an expres-

sion for the load current in terms of Vm' Z, (J) etc.(b) For part (a), Vs = 230 Vat 50 Hz, R = 10n, L = 5 mH, extinction angle"; 210°. Find average

values of output voltage and output current. . [Ans. (b)193.172 V, 19.3172 AJ3.11. (aY A single-phase half-wave diode rectifier feeds power to (i) RL load and (ii) RL load with

freewheeling diode across it. Describe the working of this rectifier for both these parts with relevantwaveforms. and bring out the differences if any. Hence point out the effect of using a freewheeling diode':

(b) For part (a), Vs = 230 V at 50 Hz, R = 20 n, L = 1 H. Find the average values of the outputvoltage and output current with and without the use of a flywheeling diode..

[Ans. (b) With freewheeling diode: Vo = 103.52 V and Io = 5.176 A

,----{V2 }-----,

230V,50 Hz

lpP

D

Fig. 3.28. Pertaining to Prob. 3.8.

;hti1m:

Without freewheeling diode: Extinction angle ~ not known, so Yo, Io cannot be calculated J3.12 •. (a) For the circuit shown in Fig. 3.29, the output current _iois considered constant at Io because

oflarge L. Sketch the waveforms of vs, io, vo, in, ifd, and is·

(b) For the above circuit, find(i) average values of output voltage and output current,

(ii) average and rms values of freewheeling diode current,(iii) supply pf. [Ans. (b) (i) 103.52 V, 26.76 A (ii) 13.38 A, 18.925 A (iii) 0.63641agJ

is iD tot./- +D

2.0.

u 23QV.s ru 50 Hz FD VO

L

ltd '50V

'.

Fig. 3.29. Pertaining to PrOD.·3.12. Fig. 3.30. Pertaining to Prob. 3.13.

3.13. F0r the circuit shown in Fig. 3.30, Vs = 160 V, Vz = 40 V and zener diode current varies from~. tlll 4@rnA. Fiind the minimum and maximum values of Rl so as to allow voltage regulation for outputcurrent 10 = zero to its maximum value 10m. Also calculate 10m. [Ans. 3k 0., 30k 0., 36 rnA)

61

/.I.

I.it