A3- Alternating Current

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run el Technical CollegeDepartmentof Aeronautical Studies VOL 1SECT 1CHAP 5

CHAPTER 5ALTERNATING CURRENT

1, Generat ionof an e,m.f,

the li n e s of force , then an e.m.f. w i l l be induced or generated, in the conductor,DIFtECTION 0 MOTION"i'1IRECTION OFMkGNETIC FLUXFIELD,

/DIRECTION OF INDUCED e,m.f,' \ \ - ' ,, ,L%(& Fig. 5:1

.-- - 43xs-2, Fleming ' Right Hand Rule

*LC < , 'A'Ref err ing to conventio nal curre nt, this ru le i s used t o determine thedi re ct io n of the induced e.m.f. If the thumb, fo re fi ng er and middlefinger of the r ig ht hand are placed mutually a t r i gh t angles , then: -

i ) The f i r s t Finger indicates the direct i on of themagnetic Fieldi i ) The s e ~ o n zinger indicates induced Current or EMFi i i ) The ThuMb in d ica te s d i rec t ion of Motron-

SIMPLE ALTERNATOR


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VOL 1 . -7SECT 1CHAP 5

If t he above p r inc ip les a r e app l i ed to a s im ple a l t e rn a to r a sshown, then an a.c. waveform would be produced to pa ss throughthe load.When the loop cu t s the f i e ld a t r i gh t ang les , maximum e.m.f. i sproduced. Minimum e.m.f. i s produced when th e loo p i s t r a v e l l i n g -.pa ra l l e l t o the f i e l d , a f t e r 90' o f ro t a t ion . For intermediateangles something l e s s than maximum i s produced.

4. Sine Waveform\

If the voltage generated in t h e a l t e r n a t o r l o o p i s p l o t t e dgraphica l ly as the loop revolves , t he r e s u l t w i l l be as shown inFig. 5 3 . This i s known as a si ne curve, because th e volta ge a tany i n s t a n t v a r i e s as the s in e of the angle through which t he loopi s turned.

5. After th e loop has ro ta te d through the angle 8, from th e pos i t ion ofzero e m. f . h en t h e p e r i p h e r d z e l o c i y c an b e r e p r es e n te d by i;--' ,. . . t hV m/s, and by l ine AL, drawn a t r ig h t angles to t he p lane o f theloop. AL can be resolved into two components AM & AN.

Fig. 5 5L L CMLA = 9 0 " - PULL= MA0 = .8.ince

AM = AL Sin 8 = V Sin 8cA s V Sin B i s always a t r i gh t ang les to the f i e l d , i t gives aninduced e.m.f.

E = ELV Sin 8 Volts - - - (5:1)Since B, L and V are constant , then the induced e.m.f. i s propor t iona lto Sin 8.The e,m.f. generated i n one s id e of th e loop = BLV Si n 0 Volts.Total e,m.f. generated i n loop = 2 BLV Sin 0When 0 90, then E max = 2 BLV S i n 8 Volts - (5:2)


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'V0L 1SECT 1CHAP 5;Slip RingsIn order to take of f the output from the alt er na to r the ends ofthe loop are a t tached to two separate ins ula ted bra ss r ings , whichturn with the loop. Tho brushes under spring tension,bear onthe s l i p r ings , and these a r e at tached to the ex te rna l c i rc u i t .Terminal Voltage Under b a d Cond itionsIn general, as soon a s a load i s connected t o the term inals -ofthe al tern ator, a current w i l l f l a w . Since this current mustflow through the in te rn al resis tan ce of the loop, brushes, l b d se tc . there w i l l be an in te rna l vol tage drop across th is in te r na lresistance. ?he terminal voltage will thus be less than the rullvoltage developed.

!hey cycle (c) i s a complete cycle of alternating events.Frequency (f)?he number of cycles per second i s referred to as the frequencyof the a.c. The u n i t i s the Hertz (HZ).Angular Velocity W

Fig. 5:6I

The rad ian may be defin ed as 't he angle subtended a t the ce nt re ofthe c i rc le , by an arc equa l i n length t o the rad ius.One radian = ~ 7 . 3 ~2 T or approx 6.28 radians = 360'

As there are 2 radians in one rev., and th er e a r e f . revs/sec,the conductor moves through 2% f r a d i a n ~ / ~ e c .The ang le turned i n one second i s termed the angular velocity.


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VOL 1SECT 1-a!!! J11. Periodic Time. T.

This i s the time taken fo r one cycle, the u ni t i s the second s .

A t r n a i ~ a , ~ U ~ vf 50 HR;L T = ;O = 0.02 secs.

1 2 . Instantaneous value of an a.c. i s the value of curre nt or voltagea t a given in st an t of time dhring i t s cycle.

Angular velocity i s 2 71 f. radians/sec.If time taken by OA to rotate through an angle 0 adians i s t secs.,then

0 Angular veloci ty x kine - - - ( 5 . 4 )1 77'= 2 T f t = 2T' x 1x i2 = 30'

The instantaneous va lue of the e.m.f. a t w i l l be: -e = E.max. Sin 0 vol t s1

so e = Eonax. Sin w t vol t sSimilarly i = I.max. Sin w t Amps.


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VOL 1SECT 1CHAP5

13. Comparison of Phase

'Iko sin us oi da l vo lta ge s or cu rr en ts may have th e same frequencybut dif fer en t phase.

If two s in e waves such a s vl and v2 i n Fig. 5:8 have the samefrequency, but pass through zero a t di ff er en t insta nce s, theyare sa id to be OUT OF PHASE, and th e angle $ by which they areout of step i s c a l l e d the PHASE ANGLE. In Fig. 5 8 he sine wavevoltage v2 i s lea din g the s in e wave voltage vl by d radians, s incei t passes through zero first , $ radians before v . Of course i ti s only po ssi ble to specif'y the phase angle, betaeen two s in ewaves, i f they have the same frequency. If a s in e wave volt age (orcurrent) i s compared with another of the same frequency then i t i smore ac cura tely sp ec ifie d by the genera l expression:-

e = Sin (wt + 6) v o l t s - - - (5 :5)where: $ = Phase Angle (Radians)e = Instantaneous voltage

(volts )Em = Peak Voltagew = 2 n f (rads/se c)f = frequency (hertz)

A sine wave voltage of 2 9 volts peak value and frequency 50 Hzle ad s another si ne wave vo ltage of 300 v o l ts peak va lue and samefrequency by 45 degrees. Express thes e vo lta ge s mathematically,and represent them by phasors and waveform.Solution :Let the two volta ges by vl and v2 res pecti ve ly.

"1= 250 s i n (100-ii t + v o l t s

v2 = 300 sin (100.11 t )


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VOL 1 ,-SECT 1CHAP 5

P w a s .

Fig. 5 9

0 Average Value(a) Average Value o r &an VJue of a complete cyc le of al te rn at in gcurrent or voltage i s always zero because the curve i ssymmetrical about zero.(b ) The MEAN value taken over one h al f cycle, f o r a s in e wave, i s0.637 or 2 of the -mum or peak valueT

e. average = 0.637 E max Voltsi. average = 0.637 1 max Amps

If- the wave i s other than a sine wave, then the average can becalculated graphically by the mid ordinate rule .15. Eff ect ive value (r.m.s) of an a.c.

1

?This value of alternating current i s equal t o the va lue of doc. whichwould give the same power or heating effect in a r e s i s t o r a s t he @52&average power expended by th e a. c.Average he at ing ef fec t of a.c. shown in Fig. 5:10

3 2= % ( I max) R wat ts2Same heating effect of doc. = I R watts

2I ~ R = I -R watts2= + I watts

I = 0.707. I m a ~mps - - - ( !XI -A graph i l l u s rain g the above points i s shown - - - Fig. 5:10.The r .m.s . value may'also be defined as the square root of the meanvalue of the squares of the instantaneous values taken over onecomplete cycle.


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16. Form Factor . --POL 1SECT 1m_d_P~

The Form Factor i s an ind ic at io n of the shape of th e wave.If the value i s less than loll,he wave i s usual ly f l a t topped;i f g re ater t h e 1.11, the wave i s pea@.

r.m.s. valueForm Factor = Average value - - 0 (5:7)Ihe Form fa ct o r f o r a si ne wave over ha l f a cycle.

- 0.707. max value, - 0.637. maxvalue

For a rectangular wave= r.m.s.Average= 1-

because mx i rm u n r. m. s. nd average values are equal.17. Fourier Analysis

All periodically repeating waveforms can be shown t o be made up o fa number o f s in e waves which a re mathematically re la te d. Thelowest frequency i s called the f'undamental, harmonics aremultiples of the fundamental frequency. The a d , bth, 6th, etc.a re even harmonics, whereas the 3rd, 5th and 7th etc . ar e oddharmonics. Those ar e i l lu st ra te d i n Fig. 5 : l l and 5:12 respectively.

18. Pure Resistance i n a.c. ci rc u it s

Fig. 5 1 3


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Since the c i rcui t i s pure ly re s i s t ive , the cu r ren t a t any ins t an ti s proport ional t o the applied voltage, and there fore th e maximumand minimum values of voltage and current appear together, theyar e i n phase with each other.Ohm's I a w appl ies to the c i rcui t : -

The power dissipation i s :2 v2Power = V x I, o r I R, o r R Watts

An a.c . c ircuit w i t h a re s i s t iv e load can be t rea t ed a s a d.c.c i rc ui t for the purposes of calculations , the value of thefrequency does not matt=.19. Pure a.c. Inductive c i r c u i t

IN 4 d ~ e%DacrM& Fig. 5:14L ~ S: '.- - 13 -


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VOL 1 7SECT 1CHAP 5

When an a.c. flows i n an inductance, a back e.m.f. i s se t up i nthe coi l , and this back e.m.f. i s propor t iona l to the ra te ofchange of current. Since there i s no resistance in t h e c i r c u i tthe applied e.m.f. w i l l be used i n en ti re ly overcoming the backe.m.f. and i t i s there fore, equal and opposite to i t a t ev eryinstant . This must be so a s there ar e no lo4 sses due to theresistance in the ci rc ui t. Consider Fig. 5l-4

- - -A t O the current i s increasing a t i t s maximum ra te, so the back, e.m.f. must be a t a maxirmun and in opposition to the applied e.m.f.

A t A the current i s steady for an ins tan t and since there i s nor a t e of change the induced e.m.f. i s zero. The next half cyclecontinues as shown.THE C-NT IN TED3 CIRCUIT IS LAGGING ON THE APPLIED E .M.F. BY 90'. '7I N A PURE INDUCTANCE THE ANGLE OF LAG IS 'DIEREFORE 900. @

20. Wattless CurrentA s the curren t through the co il increases, energy i s put in to themagnetic field $LIZ joule but when i t decreases the energy i sreturned in fill to the c i rcui t . %ere i s there fore no energy l o s tand therefore no power dissipation. This i s termed a watt lesscurr ent.

21. Reactance XL Of XGReactance i s t h e a b i l i t y of a c i rcui t to res is t the passage of ana.c. without the di ss ip at io n of energy.A reactance can be inductive or capacitive,

22 . Inductive Reactance Y LThis i s the opposition offered to an a,c. by an inductance.Reactance i s therefore expressed as a ra t io V and i s measured i nohms. 7-It i s represented by the symbol XL.Reactance XL = WL or 2.11-f L ohms - - - (5:10)

and X L = V ohms therefore 1 = V a m P ST 2FpL


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POL 1SECT 1CHAP 523. Var ia tion of ~ e a cGG--&d Current wi th Frequency

x ohmsI amps

Fig. 5:15FREQUENCY Hz

'Ihe inductive reactance i s proportional to the frequency and thecurrent i s inversely proportional to the frequency.These relatio nshi ps ar e represented graphically by the s tr ai gh tl i n e and the hyperbola.

Lhductance and Resistance i n se r i e s

Fig. 5:16

!his i s a more practical circuit since some resistance must bepresent. ?he resistance shown may be the resistance of the coi lor an ex te rnal r e s i s to r .


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VOL 1SECT 1

- - c w w25. In t h i s t ype of c i r c u i t the same value of current flows through

R & L. To for ce t h is curr ent through the ci r cu& t the vol tageacross the inductor must lea d the c urre nt by 90 . There must b ea pad. across R and L, but they ar e out of phase, and the re su l ta ntpad. (appl ied vol tage) , w i l l be th e phasor or vec tor sum, and notthe ar i thmet ic sum of the two. The phasor diagram i s shown i nFig. 5:16.

.-. . Tne f a c to r J ~ v t a k e s he place o f re si st an ce i n ohms law, @;?:-- 4-0i s t he t o t a l opposi ti on t o cu r r en t in t he c i r cu i t andi s known as th e impedance.

- - - (5:12)- / -Applied Voltage V = 1.2. voltsPhase Angle Tan 0 - = WL-

26. Power i n the Ci rcu itAll th e power used by the c i rc u i t i s due to the res is tance and i sgiven by the vol ts drop across the res is t an ce t imes the cu rren tthrough i t .

True Power = I.VRbut VR a V Cos (8rue power = I.V. cos $ wat t s - - - ( 5 : a )

27. Apparent PowerThis i s the product of V and I, and i s the pobTer apparently beingsup plie d by the source. It i s measured i n K o o A e where

Apparent Power = V x I KVA - - - 5:15)


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VOL 1SECT 1a-u.P 5-- -28, Power Fac to r?he power factor i s th at fa cto r by which the cu rrent and appl iedvol tage V must be mult ipl ie d to o btain the t ru e power diss ipated.'h e Power Fa ct or l i e s between nought f o r a p ur el y REACTIVE c i r c u i tand u n i t y f o r a purely RESISTIVE circuit,

-rue -ower- - -Power Factor Cos O' = Apparent power (5:16)

,

29, Pure CapacitanceFig. 5:19 hped ance Trian gle

Fig. 5 ~ 2 0


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VOL 1SECT 1CHAP 5AL n 1 1 - - - 1 . .n b u ale ~ ~ r a r g c r ~ sncreasing a t i t s m a x b m race, and so thecurrent i s a t a maxianun. A t A the charge i s not changing, so thecurre nt i s no t changing; a t B the charge i s changing in theopposite direct ion a t i t s marrdmun rate, and again the current i s a tm;udmum.SINCE THE APPLCED VOLTAGE I S PROPORTIONAL TO 7HE CHARGE I T FOLLOWSTHAT THE CURRENT I S 90 OUT OF PHASE; IT IS LEADING THE VOLTAGE BY 90'

As the vol ta e i increasing, energy i s being s tored i n the8 5capacitor, (% CV joules) and when i t decreases again to zerothis energy i s returned in ta ct to the cir cui t . Thus the ne texpenditure of energy i s zero, and the current i s known a s awattless current. '7

31. Capacit ive ReactanceIf the frequency of the supply to a given capacitor i s increased,there i s le s s t ime t o charge the capacitor to the f u l l supplyvoltage. The r a t e of charging must there for e in cr ea se and thecurrent increases. The same e f f e c t i s noted i f the frequency i sheld steady and the capacitance increases. As the frequency orcapacitance increases, the curr ent increase s also.This i s equivalent to saying the opposition to the current hasdecreased.

- I o r Ixc - z c 2.n fC ohms. - - - 5:17)Also : x = ohms. IC I ".* = 'rn

32. Variation of Reactance and Current with Frequency

Fig. 5 2 1Capacitive reactance i s inversely proport ional t o the capacitanceand the frequency of supply voltage. The st ra ig ht li n e and thehyperbola show how the ca pa cit ive reactance and the cu rr en t due t o agiven applied voltage vary with frequency.

---.


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33. Capacitance and Resistance

Fig. 5:2Current flowing in this c i r c u i t w i l l cause a voltsodrop across theres i s t ance and a v o l t s d rop across C lagging by 90 .

2 2Applied Voltage V = VR + VC2= I ~ R ~~~x~~= 12(R2 + xc2)

'&is expression J R' + xCz i s the impedance Z of the c i r c u i t and i smeasured in ohms.Phase angle tan $ = Vc = I . x c = 1WGR - - -I.R. (5:19)T

W the power in t h e c i r c u i t i s due t o the resista nce, and i s givenby the vo lt s drop acros s the res ista nce , t imes the curr en t through i t .h e ower = 1.V bu t VR = V Cos #R

so True Power = I.V. Cos $ Watts


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From the above vector diagram Cos $ = VR = 131TI- E I

Thus the c ir c u it again has a power factor = Cos35. Power Factor of a Capacitor- - --- - -- -

Since there are cert ain losses i n a capacitor which can be rep-resented by a res is tance in ser ies , i t follows that in a prac t i ca lc ir c u it the angle between voltage and cur ren t can never be qu it e 90'.Zhe smaller the lo ss es the gr eat er the angle and the less the powerabsorbed.Zhe power dissipated by i t w i l l be 1.V. Cos d. Cos d = 5 and whenR i s small i t may be neglected i n the impedance, and zthen Cos d = fl = 1J c.R. - - -

7( 5 . 2 2 )

36. L, C and R in Series

Fig. 5.24I i s the reference phasorsince I i s common to a l lcomponents i n th e c ir cu it .Fig. 5.25

v c . I


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' VOL 1SECT 1a-Ifl 5

v = I z0 nwhere Z = J R ~ ( x ~ & % ) ~ - - - 5.23)

Phase Angle = tan -1 ReactanceResistance

Power Factor = Cos 0 = True PowerApparent Power

37. Di st rib ut io n of p.d.Current flowing through the circuit causes a volts drop across eachcoqonent. 'Ihe phase relationships to the current i s shown by thephasor diagram (Fig. 5. 25)VL and VC are measured as separate voltages, but the res ul tan tvoltage across the two components would be t h e i r d if fe re nc e. -

38. Ser ie s Resonance vk 4

Resonance i s defined a s the condition, i n a c ir cu i t containingreactances, when t he power fa ct o r i s unity. kder th is condi t ion:(a) The phase angle $ = 0(b) Ihe voltage applied i s in phase w i t h the current flowing(c) ]II. = XC, h a t i s the combined reactance i s zero.


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c i r c u i t behaves simply a s i f i t w a s a pureres i s tance(e) Ihe Ihpedance i s a ndnirmun Z = R( f ) Ihe power taken from th e supply i s s+ly tha t required toreplace the energy l o s t due t o the resista nce(g) m e c i r c u i t i s in a st a t e of voltage resonance.When th e inductance and capac itance a re known the resonant frequencycan be found.

2- where f r i s the resonant2- f r ) - Ec frequency

A ser i es resonant c i rc u i t i s termed an acceptor c ir c ui t because i taccepts current mst rea dil y a t the resonant frequency. An acceptorc i r c u i t i s also of te n cal led a voltage magnification ci rc u it becausea t resonance the vol ta e across the inductance (equal to the voltageacross the capacitance7 i s grea ter than the supply volt age V.39. L and R in P ar a l l e l I

A

Fig* 5.27


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VOL 1SECT 1CHAPIn general the currents i n each branch of t he ci rc u it w i l l not bein phase with each other. Current in R = V / ~ , and i t i s i n phasew i t h V. Current in L = v/tJz, lagging by 90'Total curren t i s found f r o m the phasor diagram.

, \ P - - - - (5.28)Phase Angle = tan-' @ = tan = tan

-1 v= tan h x ; )-1 RPhase Angle PI = tan b L )

40. C and R in Para l le l

' r- \.I- -4.. and i s in phasewi th Vurrent through R = g ,

*,', , ' -\..Current through C = V.wc, and leads V by 90'

Total Current I =I- A - - - 5.39)hpedance Z - - - (5.31)

\ Phase Angle Tan (8 = Ic = V x R = WcR - - -c-

(5.32)- ' .- i 7, - J-\ 5 vc 2 a,-', \-- n


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LL = -1- c - Fig. 5.29. , ./ - LC ' 4 ( I t r."?- - ...- k . - ~ o t a l urrent I&,,= A T ) - - (5.33)- \ , . I , L

/- " ? " t k . .Phase Angle = tan-'- < Dynamic resistance or Z =F-"\" (5 39)r.-\

This type of c i r c u i t when used in radio, i s re fe r red to a s are jec tor s ince i t s impedance i s a maximum and the resultant supplycu rren t a mininnun a t resonance. ?he rejector c ircui t i s in acondition of current resonance.


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VOL 1 ,-'SECT 1CHAP 5

Ihe dynamic impedance Z of the pa ra l le l c i r cu i t can a lso becalcul ated from

or Z = XceQ - - - (5.40)where Q = (see pa.ra.56)

45. Forced and Free OscillationsForced: These are oscillations in a system, which a r e maintainedby an ex te rn al supp ly of energy, and which have t he frequency ofthe external supply.The reso nant frequency of such a sys tem i s given by:i, For se r i e s c i r cu i t f r = 1 Hz

2 1 J T Eii. For para l le l c i rcui t fr =1 Hz - - -2 'c T2 (5.W)

46. Free Oscillations ar e osc illa tion s in a system having capacitance,inductance and resistance, of which the frequency i s solely dependenton the cons t in t s of the ci rc ui t. The frequency a t which fr e eosci l l a t ions occur i s termed the natural frequency, and i s given by:

\.=,./ L+ J. i ~ ( h ( r t pl d; c - /G cFrom this expression i t i s apparent that i f R i s small comparedwith L then: -

f (nat) = 1 = fr.*VT rFor a l l but the m c s t exact purposes f r = f (na t ) ; but theoret ical lyf ( n a t ) i s always hit++ than fr .

'J'47. Circulating and Feed currentsAlthough a t resonance the supply cu rren t in a para l le l tuned c i rcui ti s very l ow , there are large currents flowing i n t he c i rcu i t( I ~ I&, Since the se ar e 180 out of phase, the r e s u l t i s acurrent c i rcula t ing i n the c i rcui t , and th is i s known as thecircul at ing current .


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VOL 1SECT 1CHAP 5

In a pract ical c i rc u i t a supply current i s needed to overcome theresonance the circ ula ting current i s lar ge r than the feed current,

- . and so a parallel tuned circuit has the property of currentI

magnification.\- j 48.) ihe uses of k e d Circui ts

The chief use of the s er ie s ci rc u it i s in i t s high resonantcurrent and i t is used to tune aerials.

Source ofa1 ernat inge.m.f.Example o f SeriesResonant o rAcceptor Circuit.

-- Fig. 5.32 qP-!&a& T - ' 1 . *.\P"' \A " , < \ L'Another use fo r t he s er ie s c ir cu it would be t o bypass pa rt ic ul ar $;.frequencies not required by the ci rc ui t , th at i s i t a c t s l i k e afil ter in removing a particular frequency.49. The high impedance a t resonance of a pa r a ll e l tuned ci r c u it , i s a lsomade use of i n rad io, because fo r a given current i t w i l l havermcimm voltage developed across i t a t resonance. For this reason

i t i s used frequently as a load in r e f . amplifying stages. It canalso be used as a f i l t e r in se ri es with the supply so developingm a x h u voltage a t the resonant frequency and reje ct in g the undesiredfrequency.

50. Response Curves

fig. 5.33.


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' VOL 1SECT 1 - -\CHAP 5

- -If a graph of current against frequency i s plo t t ed fo r a se r i e sacceptor circuit, I i s rmxbum a t the resonant frequency, f a l li n goff above or belaw this fl-equency.&om th e above repr ese ntati ve curves the f a l l off in c i r c u i t inth e f i r s t curve i s steeper than in the second. This means thatthe f i r s t discriminates more sharply than the second.

clac, 1s- - e Cm -QpSddIIUL +m - bnv u o AIM^?A a u r u bn1 nm+q- - -r v u +L--l lc l+ lthe second.Response curves are d s o drawn of the c i r c u i t impedance pl o tt edagain st frequency.( 51.j P a p a e l C ircuit

As i n the se r i e s c i rcu i t , the sharper the peak, the more sel ec ti vethe c i rcui t . This follows, sinc e the ra t i o between Z a t resonanceand a t non-resonance i s greater.Effect of L and C on se lec t ivi ty

I

If the resistance of the cir cu it i s kept constant but the valuesof L and C are varied in the same proportion, (inc reas ing the L/cra ti o) , the resonant frequency w i l l not be a l te red s ince i f L i sdoubled, C i s halved, and both reactance s ar e doubled. A t a l l otherfrequencies except the resonant frequency, i f the LC r a t i o i s highthe impedance of the circuit increases more rapidly and selectivityi s improved.


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VOL 1SECT 15

Fig. 5.35

F i g . 6:59

These curves a re re pre se nta tiv e only.53. Effect of Altering R only

If the L/C r a t i o i s kept constant and the re si st an ce i s increased,the current a t resonance decreases in accordance with ohms law.The response curve becomes f l a t t e r and s e le c ti vi ty decreases.However, th is i s not q ui te the same as the reduction with the L/Cr a t io , since this does not cause a reduction of current a t resonance.54. Eff ec ts of Departure from Resonance

'Ihe resonant frequency of a parallel tuned circuit can be alteredby varying the L and C values. For a constant LC r a t i o t he re i s onefrequency a t which the c ir c u it i s resonant, and above o r below t h i sfrequency the impedance of the circuit decreases and the supplycurrent increases. A t resonance the inpedance i s greates t , bu t themore resistanc e i n the ci rcu it , the le ss the res ult ant impedancebecomes.

55. Q Bctor (Magnif icat ion ~actoy)Series Circuit!he voltage across L & C for series resonance can be many timesthe applied voltage. Thus the s e ri e s resonant ci r c u i t has theproperty of magnifying the voltage. The extent of this magnificationdepends on the value of L & C since VL = I(x~) nd VC = I(xc).Since the voltage across L & C i s the same a t resonance o nly thevoltage across the inductance w i l l be considered.

VIhe r a t i o L/V i s termed the Q of the c i rcu i t ; the greater theinductive voltage, the gre at er the Q.


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VOL 1 4-7SECT 1CHAP 5;--- ---

If &L i s increased to increase Q, then &must be increase d i norder t o main ta in the c i r cu i t resonant, -itnd this i nc r eases t heL/c r a t i o .From a c i r c u i t with a high Q, a h ig h s e l e c t i v i t y i s obtained.Because of th is , th e Q of a c i r c u i t may be defin ed a s a measureof i t s s e l e c t i v i t y .

, 56. Pa ra l le l Circui tIn a p a r a l l e l c i r c u i t a t resonance t h e c i r c u i t h a s t h e p ro p e rt yof current magnification.

Circulating Current :-c _ _ -= supply curren t : I (5.44)In t h e p a r a l l e l c i r c u i t t he Q f a c t o r i s again a measure of i t ss e l e c t i v i t y and t h i s i s give n by:

57. Coupled GhcuitsIt i s very f requent ly necessary to t ran sfe r energy from one c i r c u i tto another, and when two ci r c u i t s ar e arranged so th a t this takesplace, they ar e s ai d to be coupled.There are two types:( i ) ~ h d .oupling( i i ) Direct coupl ing(i) ~ t u a loupling

Fig. 5.36

The a l t e rna t ing f l ux i n L1 in ks wi th the turns i n L2, and se ts upan emf such tha t both ci rc u i t s osci l la te .Because circuit 1 s coupled to c i rc u i t 2, the impedance ofc i r c u i t 1appears to a l te r , and i t can be shown th at th e eff ec tiv eimpedance of circuit 1 s dependent on the degree of coupling,known a s th e coupling f ac to r.


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VOL 1SECT 1a-I.AP 5

Ih fact the degree of coupl ing i s usu al ly expressed i n terms ofthe Bupl ing Factor 'kt, such that:-- - - (5.46)

where M = k t u a l h d u c an ce (Henrys )L1 & L2 = Self Inductance of coi ls(Henrys)'+ 58. Dir ec t Coupling

This type can be subdivided in to two d i s t in c t groups :(v ! ( i ) Coupling by an impedance common to b o t h c i r c u i t s

Resis tor can bereplaced bycapaci tor or.inductor.

Fig* 5.37

Diagram 5.37 Component 1. - In this case th e vo l t s drop across Ra c t s a s an app l ied vol tage t o c i r c u i t 2. In order to increa se thecoupling, increase Re- w ; - , \fl.L.r /*!L,-Diagram 5.37 Component 2. - Again th e vo lt s drop across L i s thevol tage appl ied to c i rc u i t 2. To incre ase the coupling inc rea se L. w;c\, , - L \c.\J----- - - - --Magram 5.37 Component 3. - This form of co upl ing may be termedcommon cap ac ito r, o r bottom ca pa ci ty coupling. In order to increasethe coupling decrease the capacitance. w:L\ lw '- ----- c a c t s( i i ) Coupling by Independent hpe dan ce

> e

Fig. 5.38


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8

* VOL 1S E C T 1 --CHAP 5 -

Tm 3 Q f l a2ei7gjj- is L a + m d Lo cne second ciFig. J ,JU\ ,J r c nby means of a current feed; i t i s necessary to decrease- . R toincrease the capl ing.- -- - .In diagram 2, where a capacitor i s used, coupling i s increased byincreasing C ( i e d e c r e a s i n g c ) .The coupling shown i n diagram 2 i s used widely in r ad io c i r cu i t s ,and i s known as top capacity coupling. Another capacitor i sB L -l 1 1 7 . . I . -r e g n e n b l yu b r o u u c e a lnto tne lower ranch, 111 e l t h e ror diagram 2, R or C could be replaced by an ind uctor a s shown i n 3 .

\ N.B. In the above t he coupling components a re not a p a rt of ei th ertuned circuit.\59. Wf ec t o f Degree of CouplingAs i n the case with mutual cdupling, i t can be shown that thevar ia t ion in the amount of coupling of any of the above ci r c u i t sr e s u l t s in varying response curves. The greater the degree ofcoupling, the ti gh te r they ar e said t o be coupled.

' Response curves f o r Coupled Ci rcui ts

. k - 0 corcr~at C-HL

Fig. 5.39,60. If primary and secondary response curves are plo t ted for acoupled c ir c u it having a c o m n frequency band, and various valuesof coupling, the re s u l t s would be as shown.

For weak couplings both curves have a max, current a t the resonantfrequency. A s the coupling i s increased, th e peak primarycurrent decreases and the secondary peak increases, u n t i l a t aspecial value of k th e pIlimary curve develops a s l i g h t double humpeffect , w h i l s t the secondary curve has a sharp peak and maximumheight. 'h is value of k i s c a l l ed t he c r i t i c a l coupl ing.If this value i s exceeded the secondary curve also develops a doublehump, and th e gre at er the value of k the further from the resonantfrequency these humps become. Ihus, c i r c u i t s a re sa id t o be over-coupled i n this condition.


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VOL 1SECT 1CHAP 5- -61. !he usefulness of the se curves l i e s in t he f ac t , t h a t by va ry ingthe coupl ing the c i r cu i t s can be made very selec t iv e, o r byovercoupling they may be made to respond t o a band of fre qu en cie s.

62. Banhidth and Bandpass Effects

The bandwidth of a tuned ci r c u i t i s most f requent ly def ined as thefrequency range between the poi nts on eit he r sid e of the resonancecurv e where th e power i s reduced t o a half of the peakIn terms of curre nt or vol tage, this means a reduction v Lthe aak value; or '707 peak value, because th e power i s proport ionalt o 15 o r ~ 2 .Thus the frequency range between points X and Y i s the Bandwidth Be

63. Filter c i r c u i t sF i l t e r c i r c u i t s a r e four terminal networks designed to pa ss ace rt ai n requ ired band of frequencies from inp ut to outpu t terminals,or to f i l t e r of f , or a t tenu ate the remainder of s i gn al f requenciespresen t a t t he inpu t t erminals .. Such f i l t e r c i r c u i t s use a s t h e i rba sis the fa ct s tha t the reactances of inductors and capaci tors , andthe impedances of acc eptor and rej ec to r tuned ci r c u i t s a l t e r wi thfrequency. They are thus made up from the reactive elements ofinduc tanc e and cap aci tanc e.Fi l t e r c i rc ui t s t ake fou r main forms:i ) High p as s f i l t e r si i ) Low p a s s f i l t e r si i i ) Bandpass f i l t e r si v ) Bands top f i l t e r s


36/39

VOL 1 .-\SECT 1CHAP 5

--- .\

64. Highpass F i l t e r s

F!rom a l l the numerous in pu t frequencies these f i l t e r s pass on tothe output terminals, a l l those above a certain cut off frequency,and f i l t e r of f or att enu ate th e remaining lower frequencies.65. I h e ci rc ui t of Flg. 5.43 shows a simple high pass f i l t e r . C. passesthe higher frequencies easily, onto the output terminals. Itoffers high reactance to the lower frequencies, but L offers lowreactance to these so they are f i l t er ed off through i t . L offershigh reactance to the required high frequencies and thus does notattenuate them appreciably. A typical attentuation/frequencygraph for a simple high pass f i l t e r i s shown.

1 .Fig. 5.u66. In pract ice a number of these f i l t e r c i rc ui ts ar e used i n successiona s shown. This improves the att en ua tio n of the lower frequ enciesand so th e cu t o ff reg ion becomes more abrupt and c l e a r l y defined.


37/39

W L 1SECT 1CHAP 5

FMm all the various input frequencies the f i l t e r passes on to theoutput terminals all those below a c er ta in cu t off frequency, andf i l t e r o f f , o r a t t enua te the r e s t of the higher f requencies . Asimple l o w pass filter i s shown.68. L passes the lower frequencies e a si ly onto the output terminals.

It off ers high reactance to the higher frequencies, but C of fe r slow reactance to these, so they are f i l t er e d off th rough i t . Coff ers high *reactance to the required low frequen cies and t h isdoes not attenuate them appreciably.

A typical attenuation frequency curve for a simple l o w pass f i l t e ri s shown. h ractice a number of these fi l ter c i r cu i t s a r e usedi n s u c c e ss i o ~. This improves th e att en ua tio n of the high erfrequencies, and so the c ut o ff region becomes m r e sha rply defined.

69. Bandpass F i l t e r sm -

Fig. 5.46

From a l l the numerous inp ut frequencies thes e f i l t e r s pass on to theoutput terminals a certain narrow band, and f i l t e r o f f o r a tt en ua te ,the remaining frequencies above an@ below this. A simple bandpassf i l t e r i s shown. Rejector c ir c u i t and acceptor c i rc u i t L, C, a retuned to the same frequency, the centre frequency of theLre&iredband. NO mutual coupling exists between L and L2.1

1- dI . . , _ < .- 37 - " % . F H' . , 3 *: . ,.- ..-3qA2-+.- * -r&4;eT;. -C . I - , - d r f d ~


38/39

VOL 1 . .SECT 15

70. The acceptor circuit offers low inpedance to the resonantfrequencies, and pass es them on t o t he outpu t term inals. Itoffe rs high inpedance t o a l l the other input frequencies, but there jec to r c i r cu i t of f e rs low impedance to these and so they aref i l t er ed of f through i t .

Fig 6 . 79.i( a If - 1 L a

A typical attenuation frequency curve i s shown for a simplebandpass filter. A more p r ac t i c a l c i r cu i t i s shown, and th i s w i l lgive more cl ea rl y defined cut off regions.

71. Bad s to p f i l t e r s

T; * T 3Fig. 5.48

INPUT O m ~ q

lhese f i l t e r s pass on to the output teroLinals all frequenciesexcept a ce rt a in narrow band which i s at tenuated or f i l te re d off .Ihe c ir c ui t shows a simple bandstop f i l t e r .Acceptor circuit 5 C1 and re jector c i rcu i t L2 C2 are tuned tothe same frequencies ; th e midpoint frequency of t he unwanted band.No mutual coupling exists between % and L2.


39/39

VOL 1SECT 1CHAP 5

Ihe re jec to r c i t c u i t off er s low impedance t o all the requiredfreq uenc ies, and so pass es them on to the output terminals. Itof fe rs high impedance to the unwanted band of f requ encie s, b utacceptor c i rcui t 5 C of fe r s a low impedance to these, and sothey are f i l ter ed 'off through i t .Ihe acceptor circuit offers high impedance to the wanted frequencies,-ana so, aoes not atte nu ate tnem app rec lao ~y .

i... -. ...

Elrunel Technical College,&is t01GSB/PJR JUNE 1983

Uz.

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A3- Alternating Current