Chapter 5 Handy Circuit Analysis Techniques 1 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. a linear circuit element has a linear voltagecurrent relationship: if i(t) produces v(t), then Ki(t) produces Kv(t) if i1(t) produces v1(t) and i2(t) produces v2(t), then i1(t) + i2(t) produces v1(t) + v2(t), resistors, sources are linear elements1 a linear circuit is one with only linear elements 1Dependent sources need linear control equations to be linear elements. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2 For the circuit shown, the solution can be expressed as: 0.7 0.2v1 ia 0.2 1.2 v 2 ib Question: How much of v1 is due to source A, and how much is because of source B? We use the superposition principle to answer. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3 If we define A as 0.7 0.2 A 0.2 1.2 then v1 1 ia 1 0 1 ia A A A Superposition: v 2 ib ib 0 the response is the sum of experiments A and B. Experiment B Experiment A Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 In a linear network, the voltage across or the current through any element may be calculated by adding algebraically all the individual voltages or currents caused by the separate independent sources acting “alone”, i.e. with all other independent voltage sources replaced by short circuits and all other independent current sources replaced by open circuits. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5 Leave one source ON and turn all other sources OFF: voltage sources: set v=0. These become short circuits. current sources: set i=0. These become open circuits. Find the response from this source. Add the resulting responses to find the total response. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6 Use superposition to solve for the current ix Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7 First, turn the current source off: 3 ix 0.2 69 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 8 Then, turn the voltage source off: 6 ix (2) 0.8 69 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 9 Finally, combine the results: ix ix ix 0.2 0.8 1.0 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 10 Determine the maximum positive current to which the source Ix can be set before any resistor exceeds its power rating. Solution on next slide Answer: Ix<42.49 mA Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 11 Maximum current magnitude in 100 resistor is 0.25 /100 50mA Maximum current magnitude in 64 resistor is 0.25 / 64 62.5mA Current from voltage source alone is 6/164 36.6mA flowing clockwise 64 Current in 100 from I x alone is I x flowing to the left. 164 64 64 Therefore 0.0366 I x 0.05 or 0.05 0.0366 I x 0.05 164 164 221.9mA I x 34.33mA Current in 64 from I x is 100 I x flowing downward. 164 100 100 I x 0.0625 or 0.0625 0.0366 I x 0.0625 164 164 0.1625 I x 0.04247 Therefore 0.0366 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 12 When applying superposition to circuits with dependent sources, these dependent sources are never “turned off.” Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 13 current source off voltage source off ix = ix’+ix’’=2 + (−0.6) = 1.4 A Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 14 Ideal voltage sources: a first approximation model for a battery. Why do real batteries have a current limit and experience voltage drop as current increases? Two car battery models: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 15 For the car battery example: VL = 12 – 0.01 IL This line represents all possible RL Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 16 The source has an internal resistance or output resistance, which is modeled as Rs short circuit current (when RL=0) open circuit voltage (when RL=∞) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 17 The source has an internal parallel resistance which is modeled as Rp short circuit current (when RL=0) open circuit voltage (when RL=∞) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 18 The sources are equivalent if Rs=Rp and vs=isRs Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 19 The circuits (a) and (b) are equivalent at the terminals. If given circuit (a), but circuit (b) is more convenient, switch them! This process is called source transformation. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 20 We can find the current I in the circuit below using source transformation, as shown. I = (45-3)/(5+4.7+3) = 3.307 mA Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 21 Thévenin’s theorem: a linear network can be replaced by its Thévenin equivalent circuit, as shown below: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 22 We can repeatedly apply source transformation on network A to find its Thévenin equivalent circuit. This method has limitations- not all circuits can be source transformed. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23 Disconnect the load. Find the open circuit voltage voc Find the equivalent resistance Req of the network with all independent sources turned off. Then: VTH=voc and RTH=Req Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 24 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 25 Norton’s theorem: a linear network can be replaced by its Norton equivalent circuit, as shown below: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 26 Replace the load with a short circuit. Find the short circuit current isc Find the equivalent resistance Req of the network with all independent sources turned off. Then: IN=isc and RN=Req Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 27 The Thévenin and Norton equivalents are source transformations of each other! RTH=RN =Req and vTH=iNReq Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 28 Find the Thévenin and Norton equivalents for the network faced by the 1-kΩ resistor. Answer: next slide the load resistor this is the circuit we will simplify Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 29 Thévenin Norton Source Transformation Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 30 One method to find the Thévenin equivalent of a circuit with a dependent source: find VTH and IN and solve for RTH=VTH / IN Example: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 31 Finding the ratio VTH / IN fails when both quantities are zero! Solution: apply a test source. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 32 v test v test (1.5i) 1 2 3 i 1 Solve: vtest =0.6 V, and so RTH = 0.6 Ω Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 33 What load resistor will allow the practical source to deliver the maximum power to the load? Answer: RL=Rs [Solve dpL/dRL=0.] [Or: pL=i(vs-iRs), set dpL/di=0 to find imax=vs/2Rs. Hence RL=Rs] Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 34 The following resistors form a Δ: The following resistors form a Y: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 35 this Δ is equivalent to the Y if this Y is equivalent to the Δ if Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 36 How do we find the equivalent resistance of the following network? Convert a Δ to a Y Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 37 use the Δ to Y equations use standard serial and parallel combinations Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 38