Systematic Circuit Analysis (T&R Chap 3)

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1 Systematic Circuit Analysis (T&R Chap 3) Nodevoltage analysis Using the voltages of the each node relative to a ground node, write down a set of consistent linear equations for these voltages Solve this set of equations using, say, Cramer s Rule Mesh current analysis Using the loop currents in the circuit, write down a set of consistent linear equations for these variables. Solve. This introduces us to procedures for systematically describing circuit variables and solving for them 45

2 Node voltages Nodal Analysis Pick one node as the ground node Label all other nodes and assign voltages,,, v N and currents with each branch i 1,, i M Recognize that the voltage across a branch B is the difference between the end node voltages A Thus v 3 = v C with the direction as indicated Write down the KCL relations at each node Write down the branch iv relations to express branch currents in terms of node voltages Accommodate current sources Obtain a set of linear equations for the node voltages v 5 v 4 C v C v 3 v 1 v 2 46

3 Nodal Analysis Ex 31 (T&R, 5th ed, p.72) Apply KCL R 1 i 0 i 1 i i S2 5 i 2 R 2 i S1 i 4 i 3 v C Write the element/branch eqns R 3 Reference node R 4 Substitute to get node voltage equations Solve for,, v C then i 0, i 1, i 2, i 3, i 4, i 5 47

4 Nodal Analysis Ex 31 (T&R, 5th ed, p.72) Apply KCL Node A: i 0 i 1 i 2 =0 Node B:i 1 i 3 i 5 =0 Node C: i 2 i 4 i 5 =0 Write the element/branch eqns i 0 =i S1 i 1 =G 1 ( ) i 2 =G 2 ( v C ) i 3 =G 3 i 4 =G 4 v C i 5 =i S2 Substitute to get node voltage equations Node A: (G 1 G 2 ) G 1 G 2 v C =i S1 Node B: G 1 (G 1 G 3 ) =i s2 Node C: G 2 (G 2 G 4 )v C =i S2 Solve for,, v C then i 0, i 1, i 2, i 3, i 4, i 5 & % G1 G2 G1 G2 i 0 R 1 R 3 i S2 R 2 i S1 i 4 i 3 i 5 Reference node G1 G1 G3 0 G2 0 G2 G4 #& "% i 1 v C i 2 # " = R 4 & % v C i S 1 i S 2 i S 2 # " 48

5 & % G1 G2 G1 G2 G1 G1 G3 0 Systematic Nodal Analysis G2 0 G2 G4 #& "% v C # " = & % i S 1 i S 2 i S 2 # " R 1 i 0 i 1 i i S2 5 i 2 R 2 i S1 i 4 i 3 v C Writing node equations by inspection Note that the matrix equation looks just like Gv=i for matrix G and vector v and i G is symmetric (and nonnegative definite) R 3 Reference node R 4 Diagonal (i,i) elements: sum of all conductances connected to node i Offdiagonal (i,j) elements: conductance between nodes i and j Righthand side: current sources entering node i There is no equation for the ground node the column sums give the conductance to ground 49

6 Nodal Analysis Ex. 32 (T&R, 5th ed, p.74) R/2 Node A: Conductances Source currents entering Node B: Conductances Node C: Conductances i S 2R 2R R/2 R Source currents entering Source currents entering v C 50

7 Nodal Analysis Ex. 32 (T&R, 5th ed, p.74) R/2 Node A: Conductances G/2 B 2G C =2.5G Source currents entering= i S Node B: i S 2R 2R R/2 R v C Conductances Node C: G/2 A G/2 C 2G ground =3G Source currents entering= 0 Conductances 2G A G/2 B G ground =3.5G Source currents entering= 0 & 2.5G 0.5G 2G #& va # & is 0.5G 3G 0.5G vb = 0 % 2G 0.5G 3.5G "% vc " % 0 # " 51

8 Nodal Analysis some points to watch 1. The formulation given is based on KCL with the sum of currents leaving the node 0=i total =G AtoB ( ) G AtoC ( v C ) G AtoGround i leavinga This yields 0=(G AtoB G AtoGround ) G AtoB G AtoC v C i enteringa (G AtoB G AtoGround ) G AtoB G AtoC v C =i enteringa 2. If in doubt about the sign of the current source, go back to this basic KCL formulation 3. This formulation works for independent current sources For dependent current sources (introduced later) use your wits 52

9 Nodal Analysis Ex. 32 (T&R, 5th ed, p. 72) 2KW i S1 1KW i S2 500W v O _ 53

10 Nodal Analysis Ex. 32 (T&R, 5th ed, p. 72) 2KW i S1 1KW i S2 500W v O _ Solve this using standard linear equation solvers Cramer s rule Gaussian elimination Matlab & % #& v "% v A B # " = & i % S1 i S 2 # " 54

11 Nodal Analysis with Voltage Sources Current through voltage source is not computable from voltage across it. We need some tricks They actually help us simplify things Method 1 source transformation R S Rest of the _ v S circuit v S R S R S Rest of the circuit Then use standard nodal analysis one less node 55

12 Nodal Analysis with Voltage Sources 2 Method 2 grounding one node _ vs Rest of the circuit This removes the variable (plus we know = v S ) simpler analysis But can be done once per circuit 56

13 Nodal Analysis with Voltage Sources 3 Method 3 Create a supernode Act as if A and B were one node KCL still works for this node Sum of currents entering supernode box is 0 Write KCL at all N3 other nodes (N2 nodes less Ground node) using and as usual Write one supernode KCL _ vs supernode Rest of the circuit Add the constraint =v S These three methods allow us to deal with all cases 57

14 Nodal Analysis Ex. 34 (T&R, 5th ed, p. 76) _ v S1 R 1 v 0 _ R 3 R 2 v S2 _ v S 1 R1 v S 2 R 1 R R 2 R 3 v 0 2 _ This is method 1 transform the voltage sources Applicable since voltage sources appear in series with Resist Now use nodal analysis with one node, A 58

15 Nodal Analysis Ex. 34 (T&R, 5th ed, p. 76) _ v S1 R 1 v 0 _ R 3 R 2 v S2 _ v S 1 R1 v S 2 R 1 R R 2 R 3 v 0 2 _ This is method 1 transform the voltage sources Applicable since voltage sources appear in series with Resist Now use nodal analysis with one node, A ( G G G ) v v A A = G v 1 1 S1 G 2 2 v S 2 G1v S1 G2v = G G G S

16 Nodal Analysis Ex. 35 (T&R, 5th ed, p. 77) v S i in _ R in 2R R/2 2R R/2 R v C What is the circuit input resistance viewed through v S? 60

17 Nodal Analysis Ex. 35 (T&R, 5th ed, p. 77) v S i in _ R in 2R R/2 2R R/2 R v C What is the circuit input resistance viewed through v S? = v S 0.5G 3G 0.5Gv C = 0 2G 0.5G 3.5v C = 0 Rewrite in terms of v S,, v C This is method 2 Solve 2.75v S 6.25v v S B =, v C = v S v S v C 11.75v i S in = = 2R R / R 10.25R R in = = 0.872R G 0.5Gv C = 0.5Gv S 0.5G 3.5Gv C = 2Gv S 61

18 Nodal Analysis Ex. 36 (T&R, 5th ed, p. 78) i 2 v S1 _ supernode i 3 v C R 2 R 3 i 4 R 1 _ i 1 v S2 R 4 v O reference Method 3 supernodes 62

19 Nodal Analysis Ex. 36 (T&R, 5th ed, p. 78) i 2 v S1 _ supernode i 3 v C i 1 R 2 R 3 R 1 _ v O v S2 R 4 reference Method 3 supernodes KCL for supernode: i1 i2 i3 i4 = 0 Or, using element equations G1 G2 ( ) G3 ( v C ) G 4 v C = 0 Now use = v S 2 i 4 (G " G # )v & (G ' G ( )v ) = (G # G ' )v # Other constituent relation v C = v S1 Two equations in two unknowns 63

20 Mesh Current Analysis Dual of Nodal Voltage Analysis with KCL Mesh Current Analysis with KVL Mesh = loop enclosing no elements Restricted to Planar Ccts no crossovers (unless you are really clever) Key Idea: If element K is contained in both mesh i and mesh j then its current is i k =i i i j where we have taken the reference directions as appropriate Same old tricks you already know R 1 R 2 Mesh A: v 0 v 1 v 3 =0 Mesh B: v 3 v 2 v 4 =0 v 0 i A i B v 1 v 2 v R S1 v 3 3 v 4 v S2 v 1 =R 1 i A v 2 =R 2 i B v 3 =R 3 (i A i B ) v 0 =v S1 v 4 =v S2 (R 1 R 3 )i A R 3 i B =v S1 R 3 i A (R 2 R 3 )i B =v S2 & % R1 R3 R3 R3 R2 R3 #& "% i A i B # " = & % v S 1 v S 2 # " 64

21 Mesh Analysis by inspection Matrix of Resistances R Diagonal ii elements: sum of resistances around loop Offdiagonal ij elements: resistance shared by loops i and j Vector of currents i As defined by you on your mesh diagram Ri = v S Voltage source vector v S Sum of voltage sources assisting the current in your mesh If this is hard to fathom, go back to the basic KVL to sort these directions out v S1 i C R 2 R 3 R v S2 1 i A i B R4 v 0 65

22 Mesh Analysis by inspection Matrix of Resistances R Diagonal ii elements: sum of resistances around loop Offdiagonal ij elements: resistance shared by loops i and j Vector of currents i As defined by you on your mesh diagram Ri = v S Voltage source vector v S Sum of voltage sources assisting the current in your mesh If this is hard to fathom, go back to the basic KVL to sort these directions out v S1 R 2 R 3 R v S2 1 i A i C i B R4 v 0 & R1 R 0 % R2 2 R 3 0 R R 3 4 R 2 R R 2 3 R 3 #& i i "% i A B C # " = & % v v S 2 v S 2 S1 # " 66

23 Mesh Equations with Current Sources Duals of tricks for nodal analysis with voltage sources 1. Source transformation to equivalent T&R, 5th ed, Example 38 p. 91 3KW 5V 5KW 3KW 5KW 4KW 2KW 4KW 2mA 5V i O i A i O 2KW 8V i B 1KW 1KW & 6000 % # & i "% i A B # " & 5 # = % 8" i A = ma i B = ma i O =i A i B = ma 67

24 Mesh Analysis with ICSs method 2 Current source belongs to a single mesh 3KW 5KW 5V i O 2KW 4KW i A i B i C 2mA Same example 1KW 2000i A 6000i A 11000i 2000i B B 4000i C i C = 5 = 0 = 2mA Same equations Same solution 68

25 Mesh Analysis with ICSs Method 3 Supermeshes easier than supernodes Current source in more than one mesh and/or not in parallel with a resistance 1. Create a supermesh by eliminating the whole branch involved 2. Resolve the individual currents last Supermesh R 1 i B R 2 Excluded branch i S1 i S2 R1 ( i B i A ) R2i B R4i C R3 ( i C i A) = 0 i A = i S1 i A R 3 i C R 4 v O i B i C = i S 2 69

26 Exercise from old midterm Set up mesh analysis equations Do not use any source transformation Supermesh i 2 i 3 = 3 10i 3 20i 3 20(i 2 i 1 ) = 0 Remaining mesh 20i 1 20(i 1 i 2 ) = 20 70

27 Summary of Mesh Analysis 1. Check if cct is planar or transformable to planar 2. Identify meshes, mesh currents & supermeshes 3. Simplify the cct where possible by combining elements in series or parallel 4. Write KVL for each mesh 5. Include expressions for ICSs 6. Solve for the mesh currents 71

28 Linearity & Superposition Linear cct modeled by linear elements and independent sources Linear functions Homogeneity: Additivity: f(kx)=kf(x) f(xy)=f(x)f(y) Superposition follows from linearity/additivity Linear cct response to multiple sources is the sum of the responses to each source 1. Turn off all independent sources except one and compute cct variables 2. Repeat for each independent source in turn 3. Total value of all cct variables is the sum of the values from all the individual sources 72

29 Turning off sources Voltage source Superposition Turned off when v=0 for all i a short circuit v S i _ i v i i v v v Current source Turned off when i=0 for all v an open circuit i S i v i v We have already used this in Thévenin and Norton equiv i v i v 73

30 Superposition v 0 Find using superposition v 0 = R 2 R 1 R 2 v S R 1 R 2 R 1 R 2 i S v 01 = R 2 R 1 R 2 v S v 02 = R 1R 2 R 1 R 2 i S 74

31 Where are we now? Finished resistive ccts with ICS and IVS Two analysis techniques nodal voltage and mesh current Preference depends on simplicity of the case at hand The aim has been to develop general techniques for access to tools like matlab Where to now? Active ccts with resistive elements transistors, opamps Life starts to get interesting getting closer to design Capacitance and inductance dynamic ccts Frequency response sdomain analysis Filters 75

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