UNIT I BASIC CIRCUIT CONCEPTS

Transcription

1 UNIT I BASIC CIRCUIT CONCEPTS Crcut elements Krchhoff s Law V-I Relatonshp of R,L and C Independent and Dependent sources Smple Resstve crcuts Networks reducton Voltage dvson current source transformaton. - Analyss of crcut usng mesh current and nodal voltage methods. Methods of Analyss

2 Resstance Methods of Analyss

3 Ohm s Law 3 Methods of Analyss

4 Resstors & Passve Sgn Conventon 4 Methods of Analyss

5 Example: Ohm s Law 5 Methods of Analyss

6 Short Crcut as Zero Resstance 6 Methods of Analyss

7 Short Crcut as Voltage Source (0V) 7 Methods of Analyss

8 Open Crcut 8 Methods of Analyss

9 Open Crcut as Current Source (0 A) 9 Methods of Analyss

10 Conductance 0 Methods of Analyss

11 Crcut Buldng Blocks Methods of Analyss

12 Branches Methods of Analyss

13 Nodes 3 Methods of Analyss

14 Loops 4 Methods of Analyss

15 Overvew of Krchhoff s Laws 5 Methods of Analyss

16 Krchhoff s Current Law 6 Methods of Analyss

17 Krchhoff s Current Law for Boundares 7 Methods of Analyss

18 KCL - Example 8 Methods of Analyss

19 Ideal Current Sources: Seres 9 Methods of Analyss

20 Krchhoff s Voltage Law - KVL 0 Methods of Analyss

21 KVL - Example Methods of Analyss

22 Example Applyng the Basc Laws Methods of Analyss

23 Example Applyng the Basc Laws 3 Methods of Analyss

24 Example Applyng the Basc Laws 4 Methods of Analyss

25 Resstors n Seres 5 Methods of Analyss

26 Resstors n Parallel 6 Methods of Analyss

27 Resstors n Parallel 7 Methods of Analyss

28 Voltage Dvder 8 Methods of Analyss

29 Current Dvder 9 Methods of Analyss

30 Resstor Network 30 Methods of Analyss

31 Resstor Network - Comments 3 Methods of Analyss

32 Delta Wye Transformatons 3 Methods of Analyss

33 Delta Wye Transformatons 33 Methods of Analyss

34 Example Delta Wye Transformatons 34 Methods of Analyss

35 35 Methods of Analyss

36 Methods of Analyss Introducton Nodal analyss Nodal analyss wth voltage source Mesh analyss Mesh analyss wth current source Nodal and mesh analyses by nspecton Nodal versus mesh analyss 36 Methods of Analyss

37 3. Nodal Analyss 37 Steps to Determne Node Voltages:. Select a node as the reference node. Assgn voltage v, v, v n- to the remanng n- nodes. The voltages are referenced wth respect to the reference node.. Apply KCL to each of the n- nonreference nodes. Use Ohm s law to express the branch currents n terms of node voltages. 3. Solve the resultng smultaneous equatons to obtan the unknown node voltages. Methods of Analyss

38 Fgure 3. Common symbols for ndcatng a reference node, (a) common ground, (b) ground, (c) chasss. 38 Methods of Analyss

39 Fgure Typcal crcut for nodal analyss Methods of Analyss

40 Methods of Analyss 40 3 I I I R v v lower hgher or 0 ) ( or or 0 G v R v v v G R v v G v R v

41 Methods of Analyss 4 3 R v R v v I R v v R v I I 3 ) ( ) ( G v v v G I v v G G v I I 3 I I I v v G G G G G G

42 Example 3. Calculus the node voltage n the crcut shown n Fg. 3.3(a) 4 Methods of Analyss

43 Example 3. At node v v v 0 43 Methods of Analyss

44 Example 3. At node v 4 v 4 5 v Methods of Analyss

45 Example 3. In matrx form: Methods of Analyss v v

46 Practce Problem 3. Fg Methods of Analyss

47 Example 3. Determne the voltage at the nodes n Fg. 3.5(a) 47 Methods of Analyss

48 Example 3. At node, 3 3 v v 4 3 x v v 48 Methods of Analyss

49 Example 3. At node v v x v 3 v 8 3 v Methods of Analyss

50 Example 3. v At node 3 v 4 3 v v 8 3 x ( v v ) 50 Methods of Analyss

51 Example 3. In matrx form: Methods of Analyss v v v

52 3.3 Nodal Analyss wth Voltage Sources 5 Case : The voltage source s connected between a nonreference node and the reference node: The nonreference node voltage s equal to the magntude of voltage source and the number of unknown nonreference nodes s reduced by one. Case : The voltage source s connected between two nonreferenced nodes: a generalzed node (supernode) s formed. Methods of Analyss

53 3.3 Nodal Analyss wth Voltage Sources v v v v v v v v Methods of Analyss 53 Fg. 3.7 A crcut wth a supernode.

54 54 A supernode s formed by enclosng a (dependent or ndependent) voltage source connected between two nonreference nodes and any elements connected n parallel wth t. The requred two equatons for regulatng the two nonreference node voltages are obtaned by the KCL of the supernode and the relatonshp of node voltages due to the voltage source. Methods of Analyss

55 Example 3.3 For the crcut shown n Fg. 3.9, fnd the node voltages. 7 0 v v 7 4 v v 0 55 Methods of Analyss

56 Example 3.4 Fnd the node voltages n the crcut of Fg Methods of Analyss

57 Example 3.4 v 3 At suopernode -, v 6 0 v v v v v 57 Methods of Analyss

58 Example 3.4 v v 3 v At supernode 3-4, 3 4 v3 v 6 v 3( v 4 v v 4 4 ) v Methods of Analyss

59 3.4 Mesh Analyss Mesh analyss: another procedure for analyzng crcuts, applcable to planar crcut. A Mesh s a loop whch does not contan any other loops wthn t 59 Methods of Analyss

60 Fg (a) A Planar crcut wth crossng branches, (b) The same crcut redrawn wth no crossng branches. Methods of Analyss

61 Fg. 3.6 A nonplanar crcut. 6 Methods of Analyss

62 Steps to Determne Mesh Currents:. Assgn mesh currents,,.., n to the n meshes.. Apply KVL to each of the n meshes. Use Ohm s law to express the voltages n terms of the mesh currents. 3. Solve the resultng n smultaneous equatons to get the mesh currents. 6 Methods of Analyss

63 Fg A crcut wth two meshes. Methods of Analyss

64 Apply KVL to each mesh. For mesh, For mesh, Methods of Analyss ) ( 0 ) ( V R R R R R V ) ( 0 ) ( V R R R R V R

65 Solve for the mesh currents. Use for a mesh current and I for a branch current. It s evdent from Fg. 3.7 that Methods of Analyss V V R R R R R R 3,, I I I

66 Example 3.5 Fnd the branch current I, I, and I 3 usng mesh analyss. 66 Methods of Analyss

67 Example 3.5 For mesh, For mesh, We can fnd and by substtuton method or Cramer s rule. Then, Methods of Analyss ) 0( ) 0( 4 6 3,, I I I

68 Example 3.6 Use mesh analyss to fnd the current I 0 n the crcut of Fg Methods of Analyss

69 Example 3.6 Apply KVL to each mesh. For mesh, For mesh, Methods of Analyss ) ( ) 0( ) 0( ) 4( 4 3 3

70 Example 3.6 For mesh 3, In matrx from Eqs. (3.6.) to (3.6.3) become we can calculus, and 3 by Cramer s rule, and fnd I 0. Methods of Analyss ) ( 4 ) ( ) ( 4, A, node At 0 ) ( 4 ) ( I I I

71 3.5 Mesh Analyss wth Current Sources Fg. 3. A crcut wth a current source. 7 Methods of Analyss

72 Case A Current source exst only n one mesh One mesh varable s reduced Case Current source exsts between two meshes, a super-mesh s obtaned. 7 Methods of Analyss

73 Fg. 3.3 a supermesh results when two meshes have a (dependent, ndependent) current source n common. 73 Methods of Analyss

74 Propertes of a Supermesh. The current s not completely gnored provdes the constrant equaton necessary to solve for the mesh current.. A supermesh has no current of ts own. 3. Several current sources n adjacency form a bgger supermesh. 74 Methods of Analyss

75 Example 3.7 For the crcut n Fg. 3.4, fnd to 4 usng mesh analyss. 75 Methods of Analyss

76 If a supermesh conssts of two meshes, two equatons are needed; one s obtaned usng KVL and Ohm s law to the supermesh and the other s obtaned by relaton regulated due to the current source Methods of Analyss

77 Smlarly, a supermesh formed from three meshes needs three equatons: one s from the supermesh and the other two equatons are obtaned from the two current sources. 77 Methods of Analyss

78 0 0 ) 8 ( ) 8 ( Methods of Analyss 78

79 3.6 Nodal and Mesh Analyss by Inspecton The analyss equatons can be obtaned by drect nspecton (a) For crcuts wth only resstors and ndependent current sources (b) For planar crcuts wth only resstors and ndependent voltage sources 79 Methods of Analyss

80 In the Fg. 3.6 (a), the crcut has two nonreference nodes and the node equatons Methods of Analyss (3.8) ) ( (3.7) ) ( I I I v v G G G G G G MATRIX G v v v G I v v G G v I I

81 In general, the node voltage equatons n terms of the conductance s Methods of Analyss 8 N N NN N N N N v v v G G G G G G G G G or smply Gv = where G : the conductance matrx, v : the output vector, : the nput vector

82 The crcut has two nonreference nodes and the node equatons were derved as R R R3 3 R3 R R 3 v v 8 Methods of Analyss

83 In general, f the crcut has N meshes, the mesh-current equatons as the resstances term s Methods of Analyss 83 N N NN N N N N v v v R R R R R R R R R or smply Rv = where R : the resstance matrx, : the output vector, v : the nput vector

84 Example 3.8 Wrte the node voltage matrx equatons n Fg Methods of Analyss

85 Example The crcut has 4 nonreference nodes, so G G , The off-dagonal terms are G G G G , 0, 0, G G 3 0., 4 G 3. 5, G, G G 3 G G , G , G Methods of Analyss

86 Example 3.8 The nput current vector n amperes The node-voltage equatons are Methods of Analyss , 0 3,, v v v v

87 Example 3.9 Wrte the mesh current equatons n Fg Methods of Analyss

88 Example 3.9 The nput voltage vector v n volts The mesh-current equatons are Methods of Analyss 88 6, 0, 6 6, 6 4 0, v v v v v

89 3.7 Nodal Versus Mesh Analyss 89 Both nodal and mesh analyses provde a systematc way of analyzng a complex network. The choce of the better method dctated by two factors. Frst factor : nature of the partcular network. The key s to select the method that results n the smaller number of equatons. Second factor : nformaton requred. Methods of Analyss

90 BJT Crcut Models 90 (a) dc equvalent model. (b) An npn transstor, Methods of Analyss

91 Example 3.3 For the BJT crcut n Fg.3.43, =50 and V BE = 0.7 V. Fnd v 0. 9 Methods of Analyss

92 Example 3.3 Use mesh analyss or nodal analyss 9 Methods of Analyss

93 Example Methods of Analyss

94 3.0 Summary. Nodal analyss: the applcaton of KCL at the nonreference nodes A crcut has fewer node equatons. A supernode: two nonreference nodes 3. Mesh analyss: the applcaton of KVL A crcut has fewer mesh equatons 4. A supermesh: two meshes 94 Methods of Analyss

95 UNIT II SINUSOIDAL STEADY STATE ANALYSIS 9 -Phasor Snusodal steady state response concepts of mpedance and admttance Analyss of smple crcuts Power and power factors Soluton of three phase balanced crcuts and three phase unbalanced crcuts -Power measurement n three phase crcuts. 95 Methods of Analyss

96 Snusodal Steady State Response

97 Snusodal Steady State Response. Identfy the frequency, angular frequency, peak value, RMS value, and phase of a snusodal sgnal.. Solve steady-state ac crcuts usng phasors and complex mpedances. 3. Compute power for steady-state ac crcuts. 4. Fnd Thévenn and Norton equvalent crcuts. 5. Determne load mpedances for maxmum power transfer.

98 The snusodal functon v(t) = V M sn t s plotted (a) versus t and (b) versus t.

99 The sne wave V M sn ( t + ) leads V M sn t by radan

100 Frequency f T Angular frequency T f sn z cos( z 90 o ) snt cos( t 90 o ) sn( t 90 o ) cost

101 Representaton of the two vectors v and v. In ths dagram, v leads v by 00 o + 30 o = 30 o, although t could also be argued that v leads v by 30 o. Generally we express the phase dfference by an angle less than or equal to 80 o n magntude.

102 Euler s dentty t f Acos t Acos f t In Euler expresson, A cos t = Real (A e j t ) A sn t = Im( A e j t ) Any snusodal functon can be expressed as n Euler form. 4-6

103 Applyng Euler s Identty The complex forcng functon V m e j ( t + ) produces the complex response I m e j (t + ).

104 The snusodal forcng functon V m cos ( t + θ) produces the steady-state response I m cos ( t + φ). The magnary snusodal nput j V m sn ( t + θ) produces the magnary snusodal output response j I m sn ( t + φ).

105 Re(V m e j ( t + ) ) Re(I m e j (t + ) ) Im(V m e j ( t + ) ) Im(I m e j (t + ) )

106 Phasor Defnton Tme functon : v t V ωt θ cos Phasor: V V V V Re( e Re( e j( t j( ) ) ) ) by droppngt

107 A phasor dagram showng the sum of V = 6 + j8 V and V = 3 j4 V, V + V = 9 + j4 V = Vs Vs = Ae j θ A = [9 + 4 ] / θ = tan - (4/9) Vs = o V.

108 Phasors Addton Step : Determne the phase for each term. Step : Add the phase's usng complex arthmetc. Step 3: Convert the Rectangular form to polar form. Step 4: Wrte the result as a tme functon.

109 Converson of rectangular to polar form v v t 0cos( t ( t) 0cos( t ) ) V V

110 cos 9 t t v s tan 9.96, 9.4) ( 3.06 V A Ae s j s j j j V V V

111 Phase relaton shp

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113 COMPLEX IMPEDANCE V L jl I L Z L jl L90 V L Z I L L

114 (a) In the phasor doman, (b) (a) a resstor R s represented by an mpedance of the same value; (b) a capactor C s represented by an mpedance /jc; (c) an nductor L s represented by an mpedance jl. (c)

115 V Ve jt I I C d V dt j C V C d Ve dt V I jt j C jcve Zc jt Zc s defned as the mpedance of a capactor The mpedance of a capactor s /jc. As I j C V, f v V cos t, then CV cos( t 90 )

116 As I M I j C V, CV M f v V M cos t, then CV M cos( t 90 )

117 I Ie jt d I V L L dt V j LI d V I Ie dt jt j L jlie Z L jt Z L s the mpedance of an nductor. The mpedance of a nductor s jl As V j C I, f I cos t, then v LI or I cos( t 90 ), and v LI cos t. cos( t 90 )

118 M M M M M M LI V t LI andv t I or t LI thenv t I f C j As, cos 90), cos( 90 ) cos(, cos I, V

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120 90 90 L L j Z Z C C j C j Z Z L L L L C C C C I V I V R RI R V Complex Impedance n Phasor Notaton

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122 Krchhoff s Laws n Phasor Form We can apply KVL drectly to phasors. The sum of the phasor voltages equals zero for any closed path. The sum of the phasor currents enterng a node must equal the sum of the phasor currents leavng.

123 05 ) 35.35cos(500 ) ( 05, I 50 ) cos(500 ) ( 75, I 50 ) cos(500 ) ( 5, 70.7 I I ) (50 00 t t v j V t t v j V t t v V Z V j j Z L C L L R R total total

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125 ) /( /00 / / j j Z j j j j j j j As j Zc R Z RC RC

126 Vc Vs v c Z L Z RC Z RC ( voltage dvson) j00 50 j50 50 j ( t) 0 cos (000t 35 ) 0cos 000t V

127 ) 35 ( cos ) ( t t j j j Z Z Vs I RC L

128 I I R R R C VC R ( t) 0. cos (000t 35 ) VC Zc 0 35 j00 ( t) 0. cos (000t ) A 0. 45

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130 Solve by nodal analyss V 0 V j0 V V j5 V V j j.5 eq() eq()

131 V ) cos( ) (0. () () () 0. 0.) (0., () () () t v j j V j V j eq eq by SolvngV V j V j Fromeq j V j V j j j j j j j j j As j V j V j V Fromeq eq j V V j V eq j j V V V

132 Vs= - j0, Z L =jl=j( )=j50 Use mesh analyss,

133 ) cos( ) ( ) 35 (0.08 V ) cos( ) ( ) 35 (0.08 )A cos( ) ( 50 0 ) ( 0 t t v R I V t t v Z I V t I j j I j I I j V V Vs R R L L L Z R

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141 AC Power Calculatons P V rms I rms cos PF cos v Q V rms I rms sn

142 apparent power V rms I rms P Q V I rms rms P I rms R P V Rrms R Q I rms X Q V Xrms X

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151 THÉVENIN EQUIVALENT CIRCUITS

152 The Thévenn voltage s equal to the open-crcut phasor voltage of the orgnal crcut. V t V oc We can fnd the Thévenn mpedance by zerong the ndependent sources and determnng the mpedance lookng nto the crcut termnals.

153 The Thévenn mpedance equals the open-crcut voltage dvded by the short-crcut current. Z t V I oc sc V I t sc I I n sc

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158 Maxmum Power Transfer If the load can take on any complex value, maxmum power transfer s attaned for a load mpedance equal to the complex conjugate of the Thévenn mpedance. If the load s requred to be a pure resstance, maxmum power transfer s attaned for a load resstance equal to the magntude of the Thévenn mpedance.

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161 6 UNIT III NETWORK THEOREMS (BOTH AC AND DC CIRCUITS) 9 Superposton theorem The venn s theorem - Norton s theorem -Recprocty theorem - Maxmum power transfer theorem.

162 9. Introducton Ths chapter ntroduces mportant fundamental theorems of network analyss. They are the Superposton theorem Thévenn s theorem Norton s theorem Maxmum power transfer theorem Substtuton Theorem Mllman s theorem Recprocty theorem

163 9. Superposton Theorem Used to fnd the soluton to networks wth two or more sources that are not n seres or parallel. The current through, or voltage across, an element n a network s equal to the algebrac sum of the currents or voltages produced ndependently by each source. Snce the effect of each source wll be determned ndependently, the number of networks to be analyzed wll equal the number of sources.

164 Superposton Theorem The total power delvered to a resstve element must be determned usng the total current through or the total voltage across the element and cannot be determned by a smple sum of the power levels establshed by each source.

165 9.3 Thévenn s Theorem Any two-termnal dc network can be replaced by an equvalent crcut consstng of a voltage source and a seres resstor.

166 Thévenn s Theorem Thévenn s theorem can be used to: Analyze networks wth sources that are not n seres or parallel. Reduce the number of components requred to establsh the same characterstcs at the output termnals. Investgate the effect of changng a partcular component on the behavor of a network wthout havng to analyze the entre network after each change.

167 Thévenn s Theorem Procedure to determne the proper values of R Th and E Th Prelmnary. Remove that porton of the network across whch the Thévenn equaton crcut s to be found. In the fgure below, ths requres that the load resstor R L be temporarly removed from the network.

168 Thévenn s Theorem. Mark the termnals of the remanng two-termnal network. (The mportance of ths step wll become obvous as we progress through some complex networks.) R Th : 3. Calculate R Th by frst settng all sources to zero (voltage sources are replaced by short crcuts, and current sources by open crcuts) and then fndng the resultant resstance between the two marked termnals. (If the nternal resstance of the voltage and/or current sources s ncluded n the orgnal network, t must reman when the sources are set to zero.)

169 Thévenn s Theorem E Th : 4. Calculate E Th by frst returnng all sources to ther orgnal poston and fndng the open-crcut voltage between the marked termnals. (Ths step s nvarably the one that wll lead to the most confuson and errors. In all cases, keep n mnd that t s the open-crcut potental between the two termnals marked n step.)

170 Thévenn s Theorem Concluson: 5. Draw the Thévenn equvalent crcut wth the porton of the crcut prevously removed replaced between the termnals of the equvalent crcut. Ths step s ndcated by the placement of the resstor R L between the termnals of the Thévenn equvalent crcut. Insert Fgure 9.6(b)

171 Two Expermental Procedures popular expermental procedures for determnng the parameters of the Thévenn equvalent network: Drect Measurement of E Th and R Th For any physcal network, the value of E Th can be determned expermentally by measurng the open-crcut voltage across the load termnals. The value of R Th can then be determned by completng the network wth a varable resstance R L.

172 Thévenn s Theorem Measurng V OC and I SC The Thévenn voltage s agan determned by measurng the open-crcut voltage across the termnals of nterest; that s, E Th = V OC. To determne R Th, a shortcrcut condton s establshed across the termnals of nterest and the current through the short crcut (I sc ) s measured wth an ammeter. Usng Ohm s law: R Th = V oc / I sc

173 7 3 Vth

174 7 4 Rth

175 Norton s Theorem Norton s theorem states the followng: Any two-termnal lnear blateral dc network can be replaced by an equvalent crcut consstng of a current and a parallel resstor. The steps leadng to the proper values of I N and R N. Prelmnary steps:. Remove that porton of the network across whch the Norton equvalent crcut s found.. Mark the termnals of the remanng two-termnal network.

176 Fndng R N : Norton s Theorem 3. Calculate R N by frst settng all sources to zero (voltage sources are replaced wth short crcuts, and current sources wth open crcuts) and then fndng the resultant resstance between the two marked termnals. (If the nternal resstance of the voltage and/or current sources s ncluded n the orgnal network, t must reman when the sources are set to zero.) Snce R N = R Th the procedure and value obtaned usng the approach descrbed for Thévenn s theorem wll determne the proper value of R N.

177 Fndng I N : Norton s Theorem 4. Calculate I N by frst returnng all the sources to ther orgnal poston and then fndng the short-crcut current between the marked termnals. It s the same current that would be measured by an ammeter placed between the marked termnals. Concluson: 5. Draw the Norton equvalent crcut wth the porton of the crcut prevously removed replaced between the termnals of the equvalent crcut.

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179 9.5 M Maxmum Power Transfer Theorem The maxmum power transfer theorem states the followng: A load wll receve maxmum power from a network when ts total resstve value s exactly equal to the Thévenn resstance of the network appled to the load. That s, R L = R Th

180 Maxmum Power Transfer Theorem For loads connected drectly to a dc voltage supply, maxmum power wll be delvered to the load when the load resstance s equal to the nternal resstance of the source; that s, when:r L = R nt

181 Recprocty Theorem The recprocty theorem s applcable only to sngle-source networks and states the followng: The current I n any branch of a network, due to a sngle voltage source E anywhere n the network, wll equal the current through the branch n whch the source was orgnally located f the source s placed n the branch n whch the current I was orgnally measured. The locaton of the voltage source and the resultng current may be nterchanged wthout a change n current

182 UNIT IV - TRANSIENT RESPONSE FOR DC CIRCUITS Transent response of RL, RC and RLC Laplace transform for DC nput wth snusodal nput.

183 Soluton to Frst Order Dfferental Equaton Consder the general Equaton dx( t) dt x( t) K s f ( t) Let the ntal condton be x(t = 0) = x( 0 ), then we solve the dfferental equaton: dx( t) dt x( t) K s f ( t) The complete soluton conssts of two parts: the homogeneous soluton (natural soluton) the partcular soluton (forced soluton)

184 The Natural Response / ) (, ) ( ) ( ) ( ) (, ) ( ) ( 0 ) ( ) ( t N N N N N N N N N e t x dt t x t dx dt t x t dx t x dt t dx or t x dt t dx Consder the general Equaton Settng the exctaton f (t) equal to zero, ) ( ) ( ) ( t f K t x dt t dx s It s called the natural response.

185 The Forced Response 0 ) ( ) ( ) ( t for F K t x F K t x dt t dx S F S F F Consder the general Equaton Settng the exctaton f (t) equal to F, a constant for t 0 ) ( ) ( ) ( t f K t x dt t dx s It s called the forced response.

186 The Complete Response ) ( ) ( ) ( / / x e F K e t x t x x t S t F N Consder the general Equaton The complete response s: the natural response + the forced response ) ( ) ( ) ( t f K t x dt t dx s Solve for, ) ( (0) ) ( (0) 0) ( 0 x x x x t x t for The Complete soluton: ) ( )] ( (0) [ ) ( / x e x x t x t / )] ( (0) [ t e x x called transent response () x called steady state response

187 TRANSIENT RESPONSE Fgure 5.

188 A general model of the transent Crcut wth swtched DC exctaton analyss problem Fgu re 5., 5.3

189 For a crcut contanng energy storage element Fgure 5.5, 5.6

190 (a) Crcut at t = 0 (b) Same crcut a long tme after the swtch s closed Fgure 5.9, 5.0 The capactor acts as open crcut for the steady state condton (a long tme after the swtch s closed).

191 (a) Crcut for t = 0 (b) Same crcut a long tme before the swtch s opened The nductor acts as short crcut for the steady state condton (a long tme after the swtch s closed).

192 Reason for transent response The voltage across a capactor cannot be changed nstantaneously. V C (0 ) V C (0 The current across an nductor cannot be changed nstantaneously. ) I L (0 ) I L (0 )

193 Example Fgure 5.,

194 Transents Analyss. Solve frst-order RC or RL crcuts.. Understand the concepts of transent response and steady-state response. 3. Relate the transent response of frstorder crcuts to the tme constant.

195 Transents The soluton of the dfferental equaton represents are response of the crcut. It s called natural response. The response must eventually de out, and therefore referred to as transent response. (source free response)

196 Dscharge of a Capactance through a Resstance 0, C R 0 c R C dvc dt t v C R t 0 Solvng the above equaton wth the ntal condton V c (0) = V

197 Dscharge of a Capactance through a Resstance C RC dv v C dt dv C t C dt t RCKse t st v C Ke R t v st Ke C st 0 t 0 0 v v C C v C t (0 t ) s RC Ke V K V t Ke e RC 0 / RC t RC

198 v C t V e t RC Exponental decay waveform RC s called the tme constant. At tme constant, the voltage s 36.8% of the ntal voltage. v C t RC t V ( e ) Exponental rsng waveform RC s called the tme constant. At tme constant, the voltage s 63.% of the ntal voltage.

199 RC CIRCUIT for t = 0 -, (t) = 0 u(t) s voltage-step functon Vu(t)

200 RC CIRCUIT R R C vu( t) v R dv dt Vu(t) RC C vc V, vu( t) V for t 0 C, C C dv dt C Solvng the dfferental equaton

201 Complete Response Complete response = natural response + forced response Natural response (source free response) s due to the ntal condton Forced response s the due to the external exctaton.

202 Fgure 5.7, 5.8 a). Complete, transent and steady state response b). Complete, natural, and forced responses of the crcut 5-8

203 Crcut Analyss for RC Crcut Apply KCL R R dv dt v C C s vr R v RC, R C C RC dv dt v s C v s s the source appled.

204 Soluton to Frst Order Dfferental Equaton Consder the general Equaton dx( t) dt Let the ntal condton be x(t = 0) = x( 0 ), then we solve the dfferental equaton: dx( t) dt x( t) K s x( t) f ( t) K s f ( t) The complete soluton consst of two parts: the homogeneous soluton (natural soluton) the partcular soluton (forced soluton)

205 The Natural Response / ) ( ) ( ) ( 0 ) ( ) ( t N N N N N e t x t x dt t dx or t x dt t dx Consder the general Equaton Settng the exctaton f (t) equal to zero, ) ( ) ( ) ( t f K t x dt t dx s It s called the natural response.

206 The Forced Response 0 ) ( ) ( ) ( t for F K t x F K t x dt t dx S F S F F Consder the general Equaton Settng the exctaton f (t) equal to F, a constant for t 0 ) ( ) ( ) ( t f K t x dt t dx s It s called the forced response.

207 The Complete Response ) ( ) ( ) ( / / x e F K e t x t x x t S t F N Consder the general Equaton The complete response s: the natural response + the forced response ) ( ) ( ) ( t f K t x dt t dx s Solve for, ) ( (0) ) ( (0) 0) ( 0 x x x x t x t for The Complete soluton: ) ( )] ( (0) [ ) ( / x e x x t x t / )] ( (0) [ t e x x called transent response () x called steady state response

208 Example Intal condton Vc(0) = 0V s C C C C C s R C R v v dt dv RC dt dv C R v v, C C C C v dt dv v dt dv

209 Example Intal condton Vc(0) = 0V and , (0) 00 t c c t c e v A A v As Ae v ) ( ) ( ) ( t f K t x dt t dx s ) ( ) ( ) ( / / x e F K e t x t x x t S t F N C v C dt dv

210 Energy stored n capactor dv p v Cv dt t t dv t pdt t Cv dt o o dt C v ( t ) v ( t ) o C t t o vdv If the zero-energy reference s selected at t o, mplyng that the capactor voltage s also zero at that nstant, then w c ( t) Cv

211 RC CIRCUIT Power dsspaton n the resstor s: p R = V /R = (V o /R) e W R V 0 o p R R( CV o dt V o ) e RC t - t /RC Total energy turned nto heat n the resstor t / RC 0 e / R RC 0 dt

212 RL CIRCUITS v Intal condton (t = 0) = I o R v L L d R dt Solvng 0 the R 0 L d dt dfferental equaton

213 RL CIRCUITS L R - V R + + V L - (t) Intal condton (t = 0) = I o L Rt o o t o I t o t I e I t t L R I t L R dt L R d dt L R d L R dt d o o / ) ( ) ( ln ln ln, 0

214 RL CIRCUIT (t) Power dsspaton n the resstor s: p R = R = I o e -Rt/L R - V R + R L + V L - Total energy turned nto heat n the resstor W R LI o It s expected as the energy stored n the nductor s I 0 o p R dt R ( L R I o ) e R 0 Rt / e L Rt 0 / L dt LI o

215 RL CIRCUIT Vu(t) R L + V L - (t) + _ Vu(t) k t R V R L sdes both Integratng dt R V Ld V dt d L R ) ln(, 0, ] ln ) [ln( ln 0, ) (0 / / t for e R V R V or e V R V t V R V R L V R L k thus L Rt L Rt where L/R s the tme constant

216 DC STEADY STATE The steps n determnng the forced response for RL or RC crcuts wth dc sources are:. Replace capactances wth open crcuts.. Replace nductances wth short crcuts. 3. Solve the remanng crcut.

217 UNIT V RESONANCE AND COUPLED CIRCUITS 9 Seres and parallel resonance ther frequency response Qualty factor and Bandwdth Self and mutual nductance Coeffcent of couplng Tuned crcuts Sngle tuned crcuts.

218 Any passve electrc crcut wll resonate f t has an nductor and capactor. Resonance s characterzed by the nput voltage and current beng n phase. The drvng pont mpedance (or admttance) s completely real when ths condton exsts. 8 In ths presentaton we wll consder (a) seres resonance, and (b) parallel resonance.

219 Consder the seres RLC crcut shown below. V = V M 0 V + _ R I L C The nput mpedance s gven by: Z R j( wl ) wc The magntude of the crcut current s; 9 I I R V m ( wl ) wc

220 Resonance occurs when, wl wc At resonance we desgnate w as w o and wrte; w o LC Ths s an mportant equaton to remember. It apples to both seres And parallel resonant crcuts. 0

221 Seres Resonance The magntude of the current response for the seres resonance crcut s as shown below. I V m R V m R Half power pont w w o w w Bandwdth: BW = w BW = w w

222 Seres Resonance The peak power delvered to the crcut s;. P V m The so-called half-power s gven when R I V m R We fnd the frequences, w and w, at whch ths half-power occurs by usng; R R ( wl ) wc

223 After some nsghtful algebra one wll fnd two frequences at whch the prevous equaton s satsfed, they are: an d w w R R L L LC R R L L LC The two half-power frequences are related to the resonant frequency by 3 wo w w

224 The bandwdth of the seres resonant crcut s gven by; R BW wb w w L We defne the Q (qualty factor) of the crcut as; Q wo L L R w RC R C o Usng Q, we can wrte the bandwdth as; BW w o Q 4 These are all mportant relatonshps.

225 An Observaton: If Q > 0, one can safely use the approxmaton; BW BW w wo and w wo These are useful approxmatons. 5

226 By usng Q = w o L/R n the equatons for w and w we have; and Q Q w w o Q Q w w o 6

227 An Example Illustratng Resonance: The 3 transfer functons consdered are: Case : s ks s 400 Case : s ks 5s 400 Case 3: s ks 0s 400 7

228 An Example Illustratng Resonance: The poles for the three cases are gven below. Case : Case : s s s j s j 400 ( 9.97)( 9.97) s s s j s j ( )( ) Case 3: 8 s s s j s j ( )( )

229 Comments: Observe the denomnator of the CE equaton. s R L s LC Compare to actual characterstc equaton for Case : o s s 400 w 400 w 0 rad/sec BW w o R rad/sec Q 0 L BW 9

230 Poles and Zeros In the s-plane: ( 3) () () x x x 0 jw axs s-plane Note the locaton of the poles for the three cases. Also note there s a zero at the orgn axs 3 0 x x ( 3) () () x -0

231 Comments: The frequency response starts at the orgn n the s-plane. At the orgn the transfer functon s zero because there s a zero at the orgn. As you get closer and closer to the complex pole, whch has a j parts n the neghborhood of 0, the response starts to ncrease. The response contnues to ncrease untl we reach w = 0. From there on the response decreases. 3 We should be able to reason through why the response has the above characterstcs, usng a graphcal approach.

232 Q = 0, 4, Ampltude w (ra d /s e c )

233 Next Case: Normalze all responses to at w o 0.9 Q = 0, 4, Ampltude w ( r a d /s e c )

234 Three db Calculatons: Now we use the analytcal expressons to calculate w and w. We wll then compare these values to what we fnd from the Matlab smulaton. Usng the followng equatons wth Q =, w, w w w o o Q Q 3 4 we fnd, w = 5.6 rad/sec w =.6 rad/sec

235 3 5 Parallel Resonance Consder the crcuts shown below: I R L C V I R L C V jwl jwc R V I jwc jwl R I V

236 Dualty I V R jwc jwl V I R jwl jwc We notce the above equatons are the same provded: 3 6 I R L R C V If we make the nner-change, then one equaton becomes the same as the other. For such case, we we say say the the one one crcut s s the the dual dual of of the the other. other.

237 What What ths ths means means s s s that that for for for all all the all the the equatons equatons we we have have we have derved derved for for the the parallel parallel resonant resonant crcut, crcut, crcut, we we can can we use can use use for for the the seres seres resonant resonant crcut crcut provded provded we we make make we make the the substtutons: R replaced be R L replaced by C 3 7 C replaced by L

238 3 8 Parallel Resonance Seres Resonance R w L Q O LC w O LC w O w RC Q o L R w w w BW BW ) ( RC w BW BW w w, w LC L R L R w w, LC RC RC w w, w, w, Q Q w w w o, Q Q w w w o

239 Example : Determne the resonant frequency for the crcut below. Z I N jwl( R ) jwc R jwl jwc ( w LRC ( w LC) jwl) jwrc 3 9 At resonance, the phase angle of Z must be equal to zero.

240 Analyss ( w LRC ( w LC) jwl) jwrc For zero phase; Ths gves; w LC wl wrc ( w LCR) ( w LC w R C 4 0 or w o ( LC R C )

241 Parallel Resonance Example : A parallel RLC resonant crcut has a resonant frequency admttance of x0 A seres - S(mohs). RLC resonant The Q crcut of the crcut has a resonant s 50, and frequency the resonant admttance frequency of s 0,000 x0 - S(mohs). rad/sec. Calculate The Q of the the crcut values s of 50, R, L, and and the C. resonant Fnd the frequency half-power s frequences 0,000 rad/sec. and Calculate the bandwdth. the values of R, L, and C. Fnd the half-power frequences and the bandwdth. 4 Frst, R = /G = /(0.0) = 50 ohms. w L O Second, from, we solve for L, knowng Q, R, and w o to Q fnd L = 0.5 H. Thrd, we can use R C Q w R O ,000x50 F

242 Example : (contnued) Parallel Resonance Fourth: We can use w BW wo Q x rad / sec and Ffth: Use the approxmatons; w = w o - 0.5w BW = 0, = 9,900 rad/sec w = w o - 0.5w BW = 0, = 0,00 rad/sec 4

243 Extenson of Seres Resonance Peak Voltages and Resonance: V L V S + _ V R L + _ + _ + R I C _ V C We know the followng: When w = w o =, V S and I are n phase, the drvng pont mpedance LC s purely real and equal to R. 4 3 A plot of I shows that t s maxmum at w = w o. We know the standard equatons for seres resonance apples: Q, w BW, etc.

244 Reflecton: A queston that arses s what s the nature of V R, V L, and V C? A lttle reflecton shows that V R s a peak value at w o. But we are not sure about the other two voltages. We know that at resonance they are equal and they have a magntude of QxV S. Irwn shows that the frequency at whch the voltage across the capactor s a maxmum s gven by; w w o Q max 4 4 The above beng true, we mght ask, what s the frequency at whch the voltage across the nductor s a maxmum? We answer ths queston by smulaton

245 Seres RLC Transfer Functons: The followng transfer functons apply to the seres RLC crcut. VC ( s) LC VS ( s) R s s L VL ( s) s VS ( s) R s s L LC LC 4 5 R s VR ( s) L VS ( s) R s s L LC

246 Parameter Selecton: We select values of R, L. and C for ths frst case so that Q = and w o = 000 rad/sec. Approprate values are; R = 50 ohms, L =.05 H, C = 5F. The transfer functons become as follows: 6 VC 4x0 V s 000s 4x0 S 6 VL s V s 000s 4x0 S VR 000s V s 000s 4x0 S 6

247 Smulaton Results Q =. 5 R s e s p o n s e f o r R L C s e r e s c r c u t, Q = V C V L. 5 Ampltude V R w ( r a d / s e c )

248 Analyss of the problem: Gven the prevous crcut. Fnd Q, w 0, w max, V c at w o, and V c at w max V S + _ V L + V + R L=5 mh R=50 C=5 F I + _ V C Soluton: w O LC 000rad 6 50x0 x5x0 / sec 4 8 Q w L O R 3 x0 x5x0 50

249 Problem Soluton: w w w MAX O Q V at w Q V x volts( peak) R O S o V C at w MAX Qx V S 4Q volts peak) 4 9 Now check the computer prntout.

250 Exnson of Seres Resonance Problem Soluton (Smulaton):.0e+003 * Maxmum

251 Smulaton Results: Q=0 R s e s p o n s e fo r R L C s e r e s c r c u t, Q = 0 0 Ampltude 8 6 V C V L 4 V R w ( r a d /s e c )

252 Observatons From The Study: 5 The voltage across the capactor and nductor for a seres RLC crcut s not at peak values at resonance for small Q (Q <3). Even for Q<3, the voltages across the capactor and nductor are equal at resonance and ther values wll be QxV S. For Q>0, the voltages across the capactors are for all practcal purposes at ther peak values and wll be QxV S. Regardless of the value of Q, the voltage across the resstor reaches ts peak value at w = w o. For hgh Q, the equatons dscussed for seres RLC resonance can be appled to any voltage n the RLC crcut. For Q<3, ths s not true.

253 Extenson of Resonant Crcuts Gven the followng crcut: I + _ C R L + V _ We want to fnd the frequency, w r, at whch the transfer functon for V/I wll resonate. The transfer functon wll exhbt resonance when the phase angle between V and I are zero. 5 3

254 The desred transfer functons s; Ths equaton can be smplfed to; V (/ sc)( R sl) I R sl / sc V R sl I LCs RCs Wth s jw 5 4 V R jwl I ( w LC) jwr

255 Resonant Condton: For the prevous transfer functon to be at a resonant pont, the phase angle of the numerator must be equal to the phase angle of the denomnator. num dem or, num wl tan R wrc, den. tan ( w LC) Therefore; wl R wrc ( w LC) 5 5

256 Resonant Condton Analyss: Cancelng the w s n the numerator and cross multplyng gves, L( w LC) R C or w L C L R C Ths gves, w r LC R L 5 6 Notce that f the rato of R/L s small compared to /LC, we have w r w o LC

257 Extenson of Resonant Crcuts Resonant Condton Analyss: What s the sgnfcance of w r and w o n the prevous two equatons? Clearly w r s a lower frequency of the two. To answer ths queston, consder the followng example. Gven the followng crcut wth the ndcated parameters. Wrte a Matlab program that wll determne the frequency response of the transfer functon of the voltage to the current as ndcated. I + _ C R L + V _ 5 7

258 8 R s e s p o n s e f o r R e s s ta n c e n s e r e s w t h L th e n P a r a lle l w th C rad/sec 4 R = o h m Ampltude R = 3 o h m s w ( r a d /s e c )