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Transcript
Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D. Rowan Hall 238A [email protected] http://rowan.jkbeard.com September 5, 2006 Passive Sign Convention Always mark one terminal with a + sign Voltage is positive measured from the other terminal Current is positive going into the terminal marked with the + sign September 5, 2006 Networks I for M.E. iR Slide 2 Source Sign Convention Voltage sources marked with + and – signs inside a circle or diamond Current sources marked with an arrow inside a square, circle or diamond Positive Current out of a voltage source Voltage out of a current source September 5, 2006 Networks I for M.E. Slide 3 General Methodology Write the loop equations Pick small or simple loops Make all of them one direction – clockwise or counterclockwise Make sure that every trace is covered once Use node voltage equations Leverage supernodes – treat as single nodes Write the node equations -- Currents are positive into each node Write the Ohm’s Law equations Rearrange the equations to allow vector-matrix notation Check your work Use computer to solve the matrix equation TI-89 for smaller problems Matlab or Mathcad for matrices larger than 4 X 4 or so Note that free linear algebra packages are available for most HLLs September 5, 2006 Networks I for M.E. Slide 4 Approach Used Here for 4.4-7 Use voltage node notation Leverage supernodes Write Ohm’s Law equations first Knowns are Resistances Source voltages and currents Controlled source multipliers Unknowns are Node voltages Currents through voltage sources Voltages across current sources September 5, 2006 Networks I for M.E. Slide 5 Technique for Loop Equations Add voltage drops around the loop The sum of voltage drops around a closed loop must be zero Voltage drop is voltage at the present node minus voltage on the other terminal of the resistor or source Drop is positive through a resistor when the loop goes into the “+” terminal and out the “-” terminal and its current is positive Look at the equation with and without Ohm’s Law Supernodes will identify themselves Node voltages instead of currents often gives simpler equations September 5, 2006 Networks I for M.E. Slide 6 Voltage Drop Direction of loop VIN VOUT VDROP VIN VOUT September 5, 2006 Networks I for M.E. Slide 7 Technique for Node Equations Current into a node through each resistor is the voltage on the other side of each resistor divided by the resistance Subtract the voltage at that node times the sum of the reciprocals of all the resistors connected to the node Add currents through sources Current sources connected to the node Currents through voltage sources connected to the node Use Ohm’s Law to pose the currents through resistors in terms of the node voltages September 5, 2006 Networks I for M.E. Slide 8 Matrix Notation Is A way of organizing several linear equations. Nothing is changed from the original equations. TI-89, Matlab, and Mathcad can solve the problem for you from there. September 5, 2006 Networks I for M.E. Slide 9 Problem 4.4-7 iA 1 ivv RA iiv 2 2 RB K iv iB 3 Kvv VA VS 10 v iB 3 RD iD 1 V S iC K iv 8 K vv 3 R A 2 RB 8 RD 4 Waaaay too messy for a quiz! September 5, 2006 Networks I for M.E. Slide 10 Ohm’s Law V1 VS iA RA V3 VS iB RB V3 iD RD Use to make node voltages the unknowns September 5, 2006 Networks I for M.E. Slide 11 Loop Equations Loop 1: 0 V V S 2 V3 V3 0 0 V2 VS (Supernode with ground) Loop 2: K IV iB V3 VS VS V1 0 K IV V3 VS V3 V1 0 RB Loop 3: 0 V3 K IV iB KVV VA 0 K IV V3 V3 VS KVV V1 VS 0 RB September 5, 2006 Networks I for M.E. Slide 12 Node Equations V1 VS Node 1: ivv iiv 0 RA V1 VS V3 VS Node 2: iC 0 RA RB V3 V3 VS Node 3: iiv 0 RD RB September 5, 2006 Networks I for M.E. Slide 13 Rearranging into Matrix Format V1 Loop 2 Loop 3 Node 1 Node 2 Node 3 1 K vv 1 RA 1 RA 0 September 5, 2006 V3 iiv K iv 1 RB 0 K iv 1 RB 0 0 1 1 RB 0 1 1 RB RD 0 ivv iC K iv V S R B 0 0 K iv K vv Vs V1 0 0 RB V 3 VS ivv 1 0 RA iiv 1 1 iC V 0 -1 RA RB S VS 1 0 R B Networks I for M.E. Slide 14 Solution VA 5v VB 2.5v VD 12.5v iA 2.5a iB 0.3125a iD 3.125a V1 15v V2 10v V3 12.5v ivv 5.9375a iiv 3.4375a iC 2.8125a September 5, 2006 Networks I for M.E. Slide 15 Problem 4.4-7 Solution 13 2 16 a 10v 2 12 a 2 15v 15 5 16 a 15v September 5, 2006 5 16 8 2 12 v a 12 12 v 3 167 a 4 Networks I for M.E. 10v 3 18 a Slide 16 Intuitive Approach From Loop 2 plus Loop 3: KVV VA VA VS From Loop 2 Kiv VA 1 VB RB The rest is found from Ohm’s Law Good for many homework problems, but not for most real-world problems September 5, 2006 Networks I for M.E. Slide 17 Homework Problem Put problem 4.4-7 in Matlab and solve it with the matrix method. Reproduce the results given here. Save your code, print it out, and turn it in with the homework. Use your program to solve problem 4.4-7 with the parameters to the right Why can’t Kvv be 1.0? What is special about the circuit that allows simple intuitive solutions? September 5, 2006 Networks I for M.E. VS 15 v K iv 10 K vv 4 RA 10 RB 20 RD 5 Slide 18