Propedeútico: Circuitos 2. Systematic Nodal Analysis. Dr. Arturo Sarmiento Reyes. INAOE Coordinación de Electrónica CA D Group. Mayo 16 Junio 10, 2016

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Propedeútico: Circuitos 2. Systematic Nodal Analysis Dr. Arturo Reyes INAOE Coordinación de Electrónica CA D Group Mayo 16 Junio 10, 2016 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 1/26

Contents Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 2/26

Nuestro suelo es uno de los principales productores De?... Pesimistas!! Quino, 10 Años con Mafalda, Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 3/26

Interconnection pattern It represents the TOPOLOGY of the circuit, and it is recast in the the A and C matrices KCL & KVL b i Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 4/26 n k

The set of components For each element C i v depi = F i (v indi, v indj, ) C i F F is the set of branch relationships!! Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 5/26

Description of the method It is based on the node-to-branch incidence. Hint: take a look back on the A matrix!! It makes use of the KCL expressed as Ai = 0 In plain words: the sum of currents in every node of the circuit is equal to zero i3 i4 i2 i1 i 1 + i 4 i 3 i 2 = 0 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 6/26

Description of the method i3 i4 i2 i1 i 1 + i 4 i 3 i 2 = 0 The directions of the branches must always comply with the signs of the currents in KCL. In the expression above, the branches entering the node are positive Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 7/26

Description of the method i6 i3 i7 1 i4 2 i5 v2 i2 v1 i1 1 i 1 + i 4 i 3 i 2 = 0 2 i 2 + i 5 i 6 i 7 = 0 Notice that i 2 appears in both equations with opposite signs Due to the branch-to-node incidence Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 8/26

The branch relationship is given as: i = f (u) The currents in KCL must be expressed in terms of branch voltage. It makes use of the KVL expressed as A t v = u Or: i = f (A t v) i.e. as a function of the node voltages. Node voltages are the final unknowns!! Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 9/26

NA compatible elements Current sources i.e. the value of the source. Linear conductors i = j i = gu = g(v + v ) where u is the branch voltage, and v + and v are the node voltages. Voltage controlled current source Linear resistors i 1 = g mu 2 = g m(v + 2 v 2 ) i = ( ) 1 r u = 1 r (v + v ) Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 10/26

Remember this, elements can be: voltage-controlled Ideal current source i = J value Linear conductor i = Gu Voltage-controlled current source i2 = g m u 1 current-controlled Ideal voltage source u = E value Linear resistor u = Ri Current-controlled current source i2 = βi 1 Current-controlled voltage source u2 = r m i 1 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 11/26

NA-compatible components NA compatible elements Current sources & resistors (considered as conductances) 1 1 Another NA-compatible component is the voltage-controlled current source i x = g mu y Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 12/26

Partitioning procedure P 1 Current sources P 2 Other NA-compatible i 1 = j i 2 = G 2u 2 G 2 = diag [g b ] u 1 = A t 1v u 2 = A t 2v Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 13/26

NA formulation [ G n ] [ v ] = [ j n ] where G n = A 2G 2A T 2 j n = A 1j G n results from a congruence transformation of G 2. Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 14/26

Product A 2A T 2 Since A 2 is a n b matrix The product A 2A T 2 is of order n n In fact, the matrix relationship G n = A 2G 2A T 2 transforms G 2 a b b space into a space G n of dimension n n Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 15/26

Product A 2A T 2 main diagonal The entries of row and transposed row occur in the same location The products are of the form o o or o o, i.e. of the same sign Therefore, the product yields an integer that is the number of branches touching node 1 Practice with the matrix of the example Demonstrate the structure off the main diagonal Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 16/26

NA formulation properties where Properties: [ G n ] [ v ] = [ j n ] G n = A 2G 2A T 2 j n = A 1j Elements on the main diagonal G ii = conductances connected to node i Elements off the main diagonal G ij = conductances connected between node i and j Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 17/26

n1 b1 n2 b2 b4 b3 b5 n3 n0 b6 b7 b8 n4 b 1 j 1 b 2 g 2 b 3 r 3 b 4 g 4 b 5 g 5 b 6 j 6 b 7 g 7 b 8 r 8 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 18/26

KCL n 1 n 2 n 3 n 4 n1 step-by-step: establishing KCL b1 b2 n2 b4 b3 b5 n3 n0 b6 b7 b8 n4 i 1 + i 2 + i 3 = 0 i 1 + i 4 = 0 i 2 + i 5 + i 6 + i 7 = 0 i 3 i 6 i 7 + i 8 = 0 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 19/26

KCL n 1 n 2 n 3 n 4 n1 b1 b2 n2 b4 b3 b5 n3 n0 step-by-step: branch relationships b6 b7 b8 n4 i 1 + i 2 + i 3 = 0 i 1 + i 4 = 0 i 2 + i 5 + i 6 + i 7 = 0 i 3 i 6 i 7 + i 8 = 0 b 1 j 1 i 1 = j 1 b 2 g 2 i 2 = g 2 (v 1 v 3 ) b 3 r 3 i 3 = 1 r 3 (v 1 v 4 ) b 4 g 4 i 4 = g 4 (v 2 ) b 5 g 5 i 5 = g 5 (v 3 ) b 6 j 6 i 6 = j 6 b 7 g 7 i 7 = g 7 (v 3 v 4 ) b 8 r 8 i 8 = 1 r 8 (v 4 ) Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 20/26

n1 KCL n 1 n 2 n 3 n 4 b1 b2 n2 b4 b3 b5 n3 n0 b6 step-by-step: substitute in KCL b7 b8 n4 b 1 j 1 i 1 = j 1 b 2 g 2 i 2 = g 2(v 1 v 3) b 3 r 3 i 3 = 1 r 3 (v 1 v 4) b 4 g 4 i 4 = g 4(v 2) b 5 g 5 i 5 = g 5(v 3) b 6 j 6 i 6 = j 6 b 7 g 7 i 7 = g 7(v 3 v 4) b 8 r 8 i 8 = 1 r 8 (v 4) j 1 + g 2 (v 1 v 3 ) + 1 r 3 (v 1 v 4 ) = 0 j 1 + g 4 (v 2 ) = 0 g 2 (v 1 v 3 ) + g 5 (v 3 ) + j 6 + g 7 (v 3 v 4 ) = 0 1 r 3 (v 1 v 4 ) j 6 g 7 (v 3 v 4 ) + 1 r 8 (v 4 ) = 0 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 21/26

n1 KCL n 1 n 2 n 3 n 4 b1 b2 n2 b4 b3 b5 n3 step-by-step: reorder by node voltages n0 b6 b7 b8 n4 b 1 j 1 i 1 = j 1 b 2 g 2 i 2 = g 2(v 1 v 3) b 3 r 3 i 3 = 1 r 3 (v 1 v 4) b 4 g 4 i 4 = g 4(v 2) b 5 g 5 i 5 = g 5(v 3) b 6 j 6 i 6 = j 6 b 7 g 7 i 7 = g 7(v 3 v 4) b 8 r 8 i 8 = 1 r 8 (v 4) (g 2 + 1 r 3 )v 1 (g 2 )v 3 ( 1 r 3 )v 4 + j 1 = 0 (g 4 )v 2 j 1 = 0 ( g 2 )v 1 + (g 2 + g 5 + g 7 )v 3 (g 7 )v 4 + j 6 = 0 ( 1 r 3 )v 1 (g 7 )v 3 + ( 1 r 3 + g 7 + 1 r 8 )v 4 j 6 = 0 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 22/26

n1 KCL n 1 n 2 n 3 n 4 b1 b2 n2 b4 b3 b5 n3 n0 step-by-step: separate the sources b6 b7 b8 n4 b 1 j 1 i 1 = j 1 b 2 g 2 i 2 = g 2(v 1 v 3) b 3 r 3 i 3 = 1 r 3 (v 1 v 4) b 4 g 4 i 4 = g 4(v 2) b 5 g 5 i 5 = g 5(v 3) b 6 j 6 i 6 = j 6 b 7 g 7 i 7 = g 7(v 3 v 4) b 8 r 8 i 8 = 1 r 8 (v 4) (g 2 + 1 r 3 )v 1 (g 2 )v 3 ( 1 r 3 )v 4 = j 1 (g 4 )v 2 = j 1 ( g 2 )v 1 + (g 2 + g 5 + g 7 )v 3 (g 7 )v 4 = j 6 ( 1 r 3 )v 1 (g 7 )v 3 + ( 1 r 3 + g 7 + 1 r 8 )v 4 = j 6 = 0 Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 23/26

KCL n 1 n 2 n 3 n 4 n1 b1 b2 n2 b4 b3 b5 n3 n0 b6 b7 b8 step-by-step: matrix form (g 2 + 1 r 3 ) 0 (g 2) ( 1 r 3 ) 0 (g 4) 0 0 (g 2) 0 (g 2 + g 5 + g 7) (g 7) ( 1 r 3 ) 0 (g 7) +( 1 r 3 + g 7 + 1 r 8 ) n4 b 1 j 1 i 1 = j 1 b 2 g 2 i 2 = g 2(v 1 v 3) b 3 r 3 i 3 = 1 r 3 (v 1 v 4) b 4 g 4 i 4 = g 4(v 2) b 5 g 5 i 5 = g 5(v 3) b 6 j 6 i 6 = j 6 b 7 g 7 i 7 = g 7(v 3 v 4) b 8 r 8 i 8 = 1 r 8 (v 4) v 1 v 2 v 3 v 4 = Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 24/26 j 1 j 1 j 6 j 6

Dealing with voltage sources i5 i3 i6 i Vx 1 2 i2 v2 Vx v1 i4 i1 1 i 1 + i 2 i 3 i Vx = 0 2 i 4 i 5 i 6 + i Vx = 0 Added eqn v 1 v 2 = V x The source V x contributes with an additional variable, namely its current The additional equation is V x branch relationship as function of node voltages Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 25/26

Exercises... I do love exercises!! Prof. Grammar, BBC World Service: English Lessons by Radio Dr. Arturo Reyes INAOE, Mayo, 2016 Propedeútico: Circuitos 2. Systematic Nodal Analysis 26/26