ECEN 301Discussion #8 – Network Analysis1 DateDayClass No. TitleChaptersHW Due date Lab Due date Exam 29 SeptMon8Network Analysis3.4 – 3.5 NO LAB 30 SepTue 1 OctWed9Equivalent Circuits3.6 2 OctThu NO LAB 3 OctFri Recitation HW 4 4 OctSat 5 OctSun 6 OctMon10Energy Storage3.7, 4.1 – 4.2 LAB 3 7 OctTue Schedule…

ECEN 301Discussion #8 – Network Analysis2 Personal Revelation Moroni 10:3-5 3 Behold, I would exhort you that when ye shall read these things, if it be wisdom in God that ye should read them, that ye would remember how merciful the Lord hath been unto the children of men, from the creation of Adam even down until the time that ye shall receive these things, and ponder it in your hearts. 4 And when ye shall receive these things, I would exhort you that ye would ask God, the Eternal Father, in the name of Christ, if these things are not true; and if ye shall ask with a sincere heart, with real intent, having faith in Christ, he will manifest the truth of it unto you, by the power of the Holy Ghost. 5 And by the power of the Holy Ghost ye may know the truth of all things.

ECEN 301Discussion #8 – Network Analysis3 Lecture 8 – Network Analysis Controlled Sources Superposition Source Transformations

ECEN 301Discussion #8 – Network Analysis4 Network Analysis uNetwork Analysis Methods: ÜNode voltage method ÜMesh current method ÜSuperposition ÙEquivalent circuits ÜSource transformation Thévenin equivalent Norton equivalent

ECEN 301Discussion #8 – Network Analysis5 Controlled (Dependent) Sources Node and Mesh Analysis

ECEN 301Discussion #8 – Network Analysis6 Dependent (Controlled) Sources uDiamond shaped source indicates dependent source uDependent sources are an important part of amplifiers + _ Source TypeRelationship Voltage controlled voltage source (VCVS) v s = av x Current controlled voltage source (CCVS) v s = ai x Voltage controlled current source (VCCS) i s = av x Current controlled current source (CCCS) i s = ai x vsvs isis

ECEN 301Discussion #8 – Network Analysis7 Controlled Sources uNetwork analysis with controlled sources: ÙInitially treat controlled sources as ideal sources ÙIn addition to equations obtained by node/mesh analysis there will be the constraint equation (the controlled source equation) ÙSubstitute constraint equation into node/mesh equations

ECEN 301Discussion #8 – Network Analysis8 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω R2R2 R 1 R 4 R 3 v in +–+– –+–+ R5R5 +v–+v– + v out – 2v

ECEN 301Discussion #8 – Network Analysis9 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω R2R2 + R 1 – +R 4 – +R 3 – v in +–+– –+–+ R5R5 +v–+v– iaia icic ibib + v out – 2v Choose mesh analysis – simpler than node analysis 1.Mesh current directions chosen 2.Voltage polarities chosen and labeled 3.Identify n – m (3) mesh currents  i a is independent  i c is independent 4.Apply KVL around meshes a, b, and c

ECEN 301Discussion #8 – Network Analysis10 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω v in +–+– R2R2 + R 1 – +R 4 – +R 3 – –+–+ R5R5 +v–+v– iaia icic ibib + v out – 2v 4.Apply KVL at nodes a, b, and c

ECEN 301Discussion #8 – Network Analysis11 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω R2R2 + R 1 – +R 4 – +R 3 – v in +–+– –+–+ R5R5 +v–+v– iaia icic ibib + v out – 2v 5.Solve the n – m equations

ECEN 301Discussion #8 – Network Analysis12 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω R2R2 + R 1 – +R 4 – +R 3 – v in +–+– –+–+ R5R5 +v–+v– iaia icic ibib + v out – 2v 5.Solve the n – m equations (Matrices)

ECEN 301Discussion #8 – Network Analysis13 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω R2R2 + R 1 – +R 4 – +R 3 – v in +–+– –+–+ R5R5 +v–+v– iaia icic ibib + v out – 2v 5.Solve the n – m equations (Matrices)

ECEN 301Discussion #8 – Network Analysis14 Controlled Sources uExample1: find the gain (A v = v out /v in ) ÙR 1 = 1Ω, R 2 = 0.5Ω, R 3 = 0.25Ω, R 4 = 0.25Ω, R 5 = 0.25Ω R2R2 + R 1 – +R 4 – +R 3 – v in +–+– –+–+ R5R5 +v–+v– iaia icic ibib + v out – 2v  Find the gain

ECEN 301Discussion #8 – Network Analysis15 Controlled Sources uExample2: Find v 1 Ùv s = 15V, R 1 = 8Ω, R 2 = 6Ω, R 3 = 6Ω, R 4 = 6Ω, i x = v x /3 R1R1 R3R3 R2R2 R4R4 vsvs +–+– ixix + v x –

ECEN 301Discussion #8 – Network Analysis16 Controlled Sources uExample2: Find v 1 Ùv s = 15V, R 1 = 8Ω, R 2 = 6Ω, R 3 = 6Ω, R 4 = 6Ω, i x = v x /3 +R1–+R1– +R3–+R3– R2R2 i2i2 i1i1 i3i3 +R4–+R4– i4i4 Node a Node b Node c vava vbvb vcvc vsvs +–+– ixix 1.Label currents and voltages (polarities “arbitrarily” chosen) 2.Choose Node c (v c ) as the reference node (v c = 0) 3.Define remaining n – 1 (2) voltages  v a is independent  v b is independent 4.Apply KCL at nodes a and b + v x –

ECEN 301Discussion #8 – Network Analysis17 Controlled Sources uExample2: Find v 1 Ùv s = 15V, R 1 = 8Ω, R 2 = 6Ω, R 3 = 6Ω, R 4 = 6Ω, i x = v x /3 +R1–+R1– +R3–+R3– R2R2 i2i2 i1i1 i3i3 +R4–+R4– i4i4 Node a Node b Node c vava vbvb vcvc vsvs +–+– ixix + v x – 4.Apply KCL at nodes a and b

ECEN 301Discussion #8 – Network Analysis18 Controlled Sources uExample2: Find v 1 Ùv s = 15V, R 1 = 8Ω, R 2 = 6Ω, R 3 = 6Ω, R 4 = 6Ω, i x = v x /3 +R1–+R1– +R3–+R3– R2R2 i2i2 i1i1 i3i3 +R4–+R4– i4i4 Node a Node b Node c vava vbvb vcvc vsvs +–+– ixix + v x – 5.Solve the n – 1 – m equations

ECEN 301Discussion #8 – Network Analysis19 Controlled Sources uExample2: Find v 1 Ùv s = 15V, R 1 = 8Ω, R 2 = 6Ω, R 3 = 6Ω, R 4 = 6Ω, i x = v x /3 +R1–+R1– +R3–+R3– R2R2 i2i2 i1i1 i3i3 +R4–+R4– i4i4 Node a Node b Node c vava vbvb vcvc vsvs +–+– ixix + v x – 5.Solve the n – 1 – m equations

ECEN 301Discussion #8 – Network Analysis20 The Principle of Superposition

ECEN 301Discussion #8 – Network Analysis21 Superposition Superposition: in a linear circuit containing N sources, each branch voltage and current is the sum of N voltages and currents ÙEach of which can be found by setting all but one source equal to zero and solving the circuit containing that single source When setting voltage sources to zero they become short circuits (v = 0) vsvs +–+– When setting current sources to zero they become open circuits (i = 0) isis

ECEN 301Discussion #8 – Network Analysis22 Superposition uExample3: use superposition to find v R Ùi s = 12A, v s = 12V, R 1 = 1Ω, R 2 = 0.3Ω, R 3 = 0.23Ω R1R1 R2R2 R3R3 vsvs +–+– isis +vR–+vR–

ECEN 301Discussion #8 – Network Analysis23 Superposition uExample3: use superposition to find v R Ùi s = 12A, v s = 12V, R 1 = 1Ω, R 2 = 0.3Ω, R 3 = 0.23Ω +R1–+R1– –R2+–R2+ i1i1 i2i2 R3R3 i3i3 isis + v R1 – 1.Remove all sources except i s Source v s is replaced with short circuit

ECEN 301Discussion #8 – Network Analysis24 Superposition uExample3: use superposition to find v R Ùi s = 12A, v s = 12V, R 1 = 1Ω, R 2 = 0.3Ω, R 3 = 0.23Ω +R1–+R1– –R2+–R2+ i1i1 i2i2 R3R3 i3i3 isis + v R1 – Node a

ECEN 301Discussion #8 – Network Analysis25 Superposition uExample3: use superposition to find v R Ùi s = 12A, v s = 12V, R 1 = 1Ω, R 2 = 0.3Ω, R 3 = 0.23Ω +R1–+R1– –R2+–R2+ i1i1 i2i2 R3R3 i3i3 + v R2 – 2.Remove all sources except v s Source i s is replaced with open circuit vsvs +–+–

ECEN 301Discussion #8 – Network Analysis26 Superposition uExample3: use superposition to find v R Ùi s = 12A, v s = 12V, R 1 = 1Ω, R 2 = 0.3Ω, R 3 = 0.23Ω +R1–+R1– –R2+–R2+ i1i1 i2i2 R3R3 i3i3 + v R2 – vsvs +–+– Node a

ECEN 301Discussion #8 – Network Analysis27 Superposition uExample3: use superposition to find v R Ùi s = 12A, v s = 12V, R 1 = 1Ω, R 2 = 0.3Ω, R 3 = 0.23Ω R1R1 R2R2 R3R3 vsvs +–+– isis +vR–+vR–

ECEN 301Discussion #8 – Network Analysis28 Source Transformation

ECEN 301Discussion #8 – Network Analysis29 Source Transformations Source transformation: a procedure for transforming one source into another while retaining the terminal characteristics of the original source vsvs +–+– RsRs a b isis RpRp a b Node analysis is easier with current sources – mesh analysis is easier with voltage sources.

ECEN 301Discussion #8 – Network Analysis30 Source Transformations uHow can these circuits be equivalent? vsvs +–+– RsRs a b Load +v–+v– i isis RpRp a b +v–+v– i

ECEN 301Discussion #8 – Network Analysis31 Source Transformations uHow can these circuits be equivalent? vsvs +–+– RsRs a b Load +v–+v– i isis RpRp a b +v–+v– i

ECEN 301Discussion #8 – Network Analysis32 Source Transformations uHow can these circuits be equivalent? vsvs +–+– RsRs a b Load +v–+v– i isis RpRp a b +v–+v– i

ECEN 301Discussion #8 – Network Analysis33 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω iaia R1R1 R3R3 R 2 ibib R5R5 R 4 i

ECEN 301Discussion #8 – Network Analysis34 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω iaia R1R1 R3R3 R 2 ibib R5R5 R 4 i

ECEN 301Discussion #8 – Network Analysis35 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω R3R3 R 2 ibib R5R5 R 4 i vsvs +–+– R s

ECEN 301Discussion #8 – Network Analysis36 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω R3R3 R EQ ibib R5R5 R 4 i vsvs +–+–

ECEN 301Discussion #8 – Network Analysis37 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω R3R3 ibib R5R5 R 4 i isis RpRp

ECEN 301Discussion #8 – Network Analysis38 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω R EQ ibib R5R5 R 4 i isis

ECEN 301Discussion #8 – Network Analysis39 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω ibib R5R5 R 4 i vsvs +–+– R S

ECEN 301Discussion #8 – Network Analysis40 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω ibib R5R5 R EQ i vsvs +–+–

ECEN 301Discussion #8 – Network Analysis41 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω R EQ i vsvs +–+– v s2 –+–+ R s

ECEN 301Discussion #8 – Network Analysis42 Source Transformations uExample4: find i using transformations Ùi a = 5A, i b = 2A, R 1 = 5Ω, R 2 = 5Ω, R 3 = 10Ω, R 4 = 10Ω, R 5 = 5Ω R EQ2 i vsvs +–+– v s2 –+–+