Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Chapter 5 Handy Circuit Analysis Techniques.

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Presentation transcript:

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Chapter 5 Handy Circuit Analysis Techniques

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.2  a linear circuit element has a linear voltage- current relationship:  if i(t) produces v(t), then Ki(t) produces Kv(t)  if i 1 (t) produces v 1 (t) and i 2 (t) produces v 2 (t), then i 1 (t) + i 2 (t) produces v 1 (t) + v 2 (t),  resistors, sources are linear elements 1  a linear circuit is one with only linear elements 1 Dependent sources need linear control equations to be linear elements.

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.3 For the circuit shown, the solution can be expressed as: Question: How much of v 1 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.4 If we define A as then Experiment B Experiment A Superposition: the response is the sum of experiments A and B.

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. Answer: i 3 (t) = sin t V 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 i x Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.7

First, turn the current source off: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.8

Then, turn the voltage source off: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.9

Finally, combine the results: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.10

Determine the maximum positive current to which the source I x can be set before any resistor exceeds its power rating. Answer: I x <42.49 mA Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.11

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

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.14 current source off voltage source off i x = i x ’ +i x ’’ =2 + (−0.6) = 1.4 A

 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: V L = 12 – 0.01 I L Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.16 This line represents all possible R L increase R L

The source has an internal resistance or output resistance, which is modeled as R s Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.17 short circuit current (when R L =0) open circuit voltage (when R L =∞)

The source has an internal parallel resistance which is modeled as R p Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.18 short circuit current (when R L =0) open circuit voltage (when R L =∞)

The sources are equivalent if R s =R p and v s =i s R s 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)/( ) = 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 v oc  Find the equivalent resistance R eq of the network with all independent sources turned off. Then: V TH =v oc and R TH =R eq 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 i sc  Find the equivalent resistance R eq of the network with all independent sources turned off. Then: I N =i sc and R N =R eq 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! R TH =R N =R eq and v TH =i N R eq 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 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.29 the load resistor this is the circuit we will simplify

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.30 NortonThévenin Source Transformation

One method to find the Thévenin equivalent of a circuit with a dependent source: find V TH and I N and solve for R TH =V TH / I N Example: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.31

Finding the ratio V TH / I N 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

Solve: v test =0.6 V, and so R TH = 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: R L =R s [Solve dp L /dR L =0.] [Or: p L =i(v s -iR s ), set dp L /di=0 to find i max =v s /2R s. Hence R L =R s ] Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.34

 The following resistors form a Δ: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.35  The following resistors form a Y:

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.36 this Δ is equivalent to the Y if this Y is equivalent to the Δ if

 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

Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.38 use the Δ to Y equations use standard serial and parallel combinations