ECE2262 Electric Circuits. Chapter 5: Circuit Theorems

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1 ECE2262 Electric Circuits Chapter 5: Circuit Theorems 1

2 Equivalence Linearity Superposition Thevenin s and Norton s Theorems Maximum Power Transfer Analysis of Circuits Using Circuit Theorems 2

3 5. 1 Equivalence Two circuits are equivalent if they have the same i-v characteristics at a specified pair of terminals Our aim is to simplify analysis replacing complicated sub-circuits by simpler equivalent circuits 3

4 V 1 V 2? I 2 I 1? 4

5 Source Transformation A source transformation allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa. a a R L! R L b b The current in R L The current in R L v i L = s R i L = i s R + R L R + R L These circuits are equivalent if these resistor currents are the same v s R + R L = R R + R L i s! 5 i s = v s R v s = Ri s

6 The resistance in parallel with the voltage source The resistance in series with the current source These circuits are equivalent with respect to terminals a,b since they produce the same voltage and current in any resistor R L inserted between nodes a,b. 6

7 Example Find the power P 6V = 6V! i 6V We could use the node-voltage method with 3 node-voltage equations Let s use the source transformation strategy by reducing the circuit in a way that preserves the identity of the branch containing the 6 V source.! 7

8 Step 1 Step 2 20! 5! = 4! and 4! " 8A = 32V = 20! in series with 32 V source! 32V 20! = 1.6A 8

9 Step 3 Step 4 30! 20! = 12! and 12! "1.6A = 19.2V The current in the direction of the voltage drop across the 6 V source is i 6V = 19.2! 6 16 = 0.825A! P 6V = 6V! i 6V = 4.95 W (absorbing) 9

10 Example Find v 0 and P 250V, P 8 A Remove the resistors 125! and 10!, they do not influence the formula for v 0! equivalent circuit source transformation 10

11 = 10! v 0 = 2A! 10" = 20V 11

12 P 250V! need I 250V I 250V = ! v 0 25 = ! = 11.2 P 250V = 250!11.2 = 2800 W (delivering) 12

13 P 8 A! need V 8 A v 0!10 " 8! V 8 A = 0! V 8 A = 20! 80 =!60V P 8 A = V 8 A! 8A = "480 W (delivering) 13

14 5. 2 Linearity = additivity + homogeneity 2! i x v 4! v 2 1 v 3 i s v s 2i x 1! The node-voltage method v 1 : v 1 = v s v 2 : v 2! v i! + i = 0! 3 x s 4 v v! v =!i 2 3 s v 1!v 3 2 v 3 : v v! v 3 1 2! i s = 0!! 1 2 v v 3 = i s 14

15 v s i s " \$ \$ # \$ / 4 1 / 4!1!1 / / 2 % " ' \$ ' \$ &' \$ # v 1 v 2 v 3 % " ' \$ ' = \$ ' \$ & # v s!i s i s % ' ' ' & " \$ \$ \$ # v 1 v 2 v 3 % ' ' = ' & " \$ \$ # \$ 1 0 0!5 / 3 44/ 3 8 0/ 3 1/ / 3 % " ' \$ ' \$ &' \$ # v s!i s i s % ' ' ' & x! y! Ay = x! y = A!1 x 15

16 Additivity: If x 1! y 1 and x 2! y 2 then x 1 + x 2! y 1 + y 2 Homogeneity: If x! y then!x!!y for any number!! v s! v! i s ( v s,i s ) = ( 4V,2A) v = 3V ("8V,"4A) v = 16

17 Additivity: If x 1! y 1 and x 2! y 2 then x 1 + x 2! y 1 + y 2 Homogeneity: If x! y then!x!!y for any number!! Linearity: x 1! y 1 and x 2! y 2 then! 1 x 1 +! 2 x 2!! 1 y 1 +! 2 y 2 for any numbers! 1,! 2 17

18 Example Let I 0 = 1mA, find the corresponding I The true value of I = 6mA 18

19 I 3 I 2 I 0 = 1mA CD: I 0 = I 2 = 2 3 I 2! I 2 = 3 2 I 0 = 3 2 ma = 4k!! V I = I 2! 4 = 6V I 3 = I = I 2 + I 3 = = 2mA V I = 1 2 ma Hence for the assumed I 0 = 1mA we have I = 2mA, then by the linearity if I = 6mA = 3! 2! 3!1 = 3mA! I 0 19

20 Example: Circuits 1 and 2 below are identical except for the voltage sources. Assuming that I 1 = 5A then the value of I 2 is? + I 2 20

21 5. 3 Superposition Superposition Principle: Let f 1,...f N be set of independent source strengths (v S,i S ) in a linear resistive circuit. Then, any electrical response y in the circuit ( v,i) can be expressed as y = k 1 f 1 + k 2 f k N f N, where k 1,..,k N are constants coefficients, unique for each response 21

22 In any linear circuit containing multiple independent sources, the current or voltage at any point in the network may be calculated as the algebraic sum of the individual contributions of each source acting alone. The principle of superposition allows us to reduce a complicated multisource problem to several simple problems. Each problem contains only a single independent source. = 0 = 0 22

23 23

24 Example i x 2! 4! v v 2 1 v 3 i s v s 2i x 1! If v s = 0 (short circuit)! v 1 ' = 0, v 2 ' =! 4 3 i s, v 3 ' = 2 3 i s If i s = 0 (open circuit)! v 1 '' = v s, v 2 '' =! 5 3 v s, v 3 '' = 1 3 v s It is clear that v 1 = v 1 ' + v 1 '', v 2 = v 2 ' + v 2 '', v 3 = v 3 ' + v 3 '' 24

25 25

26 Example 5.3 Find V 0 (a) Inactivate the 3V source! Response to 2 ma CD: I 0 = ! 2m = 2 3 ma! V ' 0 = I 0! 6k = 4V 26

27 (b) Inactivate the 2mA source! Response to 3 V V 0 '' = ! 3V = 2 V By the superposition principle: V 0 = V 0 ' + V 0 '' = = 6 V. 27

28 Example 5.4 Find V 0 28

29 (a) Inactivate the 2mA source ( ) 4 = 8 / 3k! VD: V 1 = 8 / 3 8 / 3+ 2! 6V = 24 7 V VD: V 0 ' = ! V 1 = 18 7 V 29

30 (b) Inactivate the 6 V source 2 4 = 4 / 3k! V 0 ''! = \$ " # 3% & 6 ' 2m = 30 7 V By the superposition principle: V 0 = V 0 ' + V 0 '' = = 48 7! 6.86V 30

31 Example Find v 0 : circuit with dependent sources (a) Response to the 10 V source 31

32 v! ' = ( ' "0.4 # v ) '! #10! v! = 0! 0.4! v " ' = 0 v 0 ' = !10 = 8V 32

33 (b) Response to the 5 A source KCL at a : v '' v '' 0 20! 0.4 " v '' # = 0! 5v 0 ''! 8v " '' = 0 KCL at b: 0.4! v " '' + v # 2i '' b " 10 # 5 = 0! 4v! '' + v b " 2i! '' = 50! v! '' = v b " 2i! '' & i! '' = "v 0 '' / 5 33

34 Hence we have '' '' '' 5v 0! 8v " = 0 and 4v! + v b " 2i! '' = 50! v! '' = v b " 2i! '' & i! '' = "v 0 '' / 5! " # \$% 9v 0 '' 2v 0 ''! 40v b = 0 + 5v b = 50! v 0 '' = 16 V & v b = 18 / 5V By the superposition principle: v 0 ' + v 0 '' = = 24 V 34

35 Example Superposition Applied to Op-Amp Circuits 35

36 Contribution of V 1 This is a basic inverting circuit: V 01 =! R 2 R 1 V 1 36

37 Contribution of V 2! This is a basic non-inverting circuit: V 02 = 1+ R \$ 2 " # % & V 2 Principle of Superposition: V 0 = V 01 + V 02 =! R " 2 V R % 2 R 1 # \$ & ' V 2 R 1 R 1 37

38 38

39 39

40 Warning: Power does not obey superposition 40

41 5. 4 Thevenin s and Norton s Theorems 41

42 Linear Circuit A a + v! b i LOAD B Any Circuit!V Th + i " R Th + v = 0 v = V Th! i " R Th the load draws current i and results in voltage v 42

43 Thevenin Equivalent of Circuit A A a + v! b i LOAD B Any Circuit v = V Th! i " R Th The values of V Th (Thevenin voltage) and R Th may be either positive or negative R Th - the Thevenin resistance is a quantity in a mathematical model - it is not a physical resistor 43

44 A a + v! b i LOAD B Any Circuit v = V Th! i " R Th How to calculate the Thevenin Voltage V Th from v? Assume i = 0 (open circuit circuit - no external load)! v = V Th V Th = v oc 44

45 Norton Equivalent of Circuit A I N A I N = V Th R Th R Th a + v! b i LOAD B Any Circuit How to calculate the Norton Current I N from v? v v = V Th! i " R Th! = V Th! i R Th! R Th I N v = V Th! i " R Th Assume v = 0 (short circuit)! I N = i sc 45

46 R Th - The Thevenin Resistance 1. The most general way for obtaining R Th is to use R Th = V Th I N, where V Th = v oc - open circuit voltage I N = i sc - short circuit current 46

47 2. The Thevenin resistance R Th can be determined directly by a source suppression method without finding the Thevenin voltage and Norton current. This applies directly to circuits that contain only independent sources. This is a result of the linearity property of the circuit. R Th = V Th = V!! Th I N I N!!! - scaling applied to independent sources!! 0 - the Thevenin resistance remains unchanged even in the limit case when all independent sources are suppressed to zero. 47

48 (1) Replace all independent voltage sources in the circuit by short circuits and all independent current sources by open circuits A Independent Sources Deactivated a b R Th (2) If the remaining circuit contains no dependent sources, then R Th is the equivalent resistance, which can be determined by using series/parallel resistor combinations. 48

49 A. Thevenin/Norton Equivalent for Circuits with Independent Sources Example Find a Thevenin/Norton Equivalent 1. Open circuit voltage at a! b: v ab = v 1! the voltage across the 3 A source KCL : v 1! v 1 20! 3 = 0! v 1 = 32V! v oc = 32 V 49

50 2. Short circuit current at a! b: KCL: v 2! v 2 20! 3+ v 2 4 = 0! v 2 = 16 V i sc = 16 4 = 4A! R Th = v oc i sc = 32 4 = 8! 50

51 3. The Thevenin Equivalent 4. The Norton Equivalent 51

52 5. The Thevenin resistance! Direct Method R Th R Th = = 8! 52

53 6. The circuit with load 24! V 24! = ! " 32 = 24V I 24! = 24V 24! = 1A! P 24! = 12 " 24 = 24W 53

54 Example Open circuit voltage V oc = 3+ V 1, V 1 = ( 2!10 "3 )! 3k = 6V! V oc = 9V 54

55 2. Short circuit current I sc = 2m + I 1 = 2m + 3V 3k = 2m +1m = 3mA R Th = V oc I sc = 9V 3m = 3k 55

56 3. R Th R Th = 3k!! 56

57 4. The circuit with load V 0 = ! 9 = 6V! P 6k! = 62 6k = 6mW 57

58 Example 5.6 Find V 0 12V 58

59 1. Open circuit voltage and R Th V oc1 = V 6k = 6!12 = 8V R Th = = 4k! 59

60 2. Circuit with load 60

61 3. Second Iteration 61

62 Open circuit voltage and R Th V oc2 = 8 + 2m! 4k = 16V R Th = 4k! 62

63 4. Circuit with load 16V V 0 = !16 = 8V 63

64 Example 5.7 Find V 0 64

65 1. Open circuit voltage and R Th Mesh-Current Method:!6 + 4k " I 1 + 2k " ( I 1! I 2 ) = 0 # \$ % I 2 = 2m! I 1 = 5 / 3 ma KVL: V oc = 4k! I 1 + 2k! I 2 " V oc = = 32 3 V 65

66 R Th R Th = = 10 3 k! Note: I sc = V oc R Th = 32 / 3 10 / 3 = 3.2mA 66

67 2. Thevenin Equivalent with Load Thevenin V 0 = ! 32 3 = 48 7 V = V 67

68 B. Thevenin and Norton Equivalent for Circuits with Dependent Sources Valid and Invalid Partitions We cannot split the dependent source and its controlling variable when we break the circuit to find the equivalent Thevenin/Norton circuits 68

69 B1 Thevenin/Norton Equivalent for Circuits with Only Dependent Sources: Test Source Approach I 0 R Th = 1V I 0 V 0 R Th = V 0 1mA 69

70 Example 5.8 Find R Th I 0 R Th = 1V I 0 V oc =? Thevenin equivalent? 70

71 I 0 = I 1 + I 2 + I 3 KVL (big loop):!v 1! V x +1 = 0! V 1 = 1! V x KCL (at V 1 ): V 1 1k + V! 2V 1 x 2k + V 1!1 1k = 0! V x = 3 7 V I 0 = I 1 + I 2 + I 3! I 1 = V x 1k = 3 7 ma, I = 1! 2V x 2 1k! I 0 = ma! R Th = 1 V I 0 = k! 71 = 1 7 ma, I 3 = 1 2k = 1 2 ma

72 Example 5. 9 Find R Th V oc =? Thevenin equivalent? 72

73 V 1 V 2 R Th = V 2 1 ma KCL: V 1 : V 1! 2000I x 2k + V 1 1k + V! V 1 2 3k = 0! I x = V 1 1k V 2 : V 2! V 1 3k + V 2 2k!1m = 0! V 2 = 10 7 V! R Th = V 2 1 ma = 10 7 k! 73

74 B2 Thevenin/Norton Equivalent for Circuits with Both Independent and Dependent Sources Example 5.10 Find V 0 74

75 V oc +12 supernode 1. Open circuit voltage KCL at the supernode: ' ( V oc +12) I x 1k + V oc +12 2k + V oc! 2k I x ' = 0 Since I x ' = V oc 2k! V oc =!6V! V Th = "6V 75

76 2. Short circuit current and R Th! I sc =! k =! 12 2 / 3k =!18mA R Th = V oc I sc =!6V!18mA = 1 3 k! 76

77 3. Circuit with Load Thevenin V o = ! ("6) = " 18 7 V = V 77

78 Example Find Thevenin Equivalent 1. Open circuit voltage V Th = v ab = v 25! = v ( i x = 0)! V Th = (!20i) " 25 =!500i i = 5! 3v 2k! = 5! 3V Th 2k! V Th =!5V 78

79 2. Short circuit current and R Th! v = 0! i sc =!20i, i = 5 2k = 2.5mA! i sc R Th = V Th i sc =!5!50m = 100" =!20 " 2.5 =!50mA Thevenin 79

80 Find the Thevenin Equivalent R Th Using a Test Source The equivalent method is to first deactivate all independent sources and then apply either a test voltage source or a test current source The Thevenin resistance R Th is calculated as R Th = V test I 0 R Th = V 0 I test 80

81 Example Deactivate the independent sources and excite the circuit by a test source Note: Use, e.g., v T = 1V 81

82 R Th = v T i T i T = v T i i =!3v T 2k =! 3 2 v ma T! i T = v T ! # " 3 2 v ma & T \$ % ' ( = = v T 25! v T! i T v T = 1 25! = 1 100! R Th = v T i T = 100! 82

83 83

84 5. 5 Maximum Power Transfer Circuit analysis plays an important role in the analysis of systems designed to transfer power from a source to a load. The efficiency of the power transfer: power utility systems are a good example of this type because they are concerned with the generation, transmission, and distribution of large quantities of electric power. If a power utility system is inefficient, a large percentage of the power generated is lost in the transmission and distribution processes, and thus wasted. Power transferred: communication and instrumentation systems are good examples because in the transmission of information, or data, via electric signals, the power available at the transmitter or detector is limited. Thus, transmitting as much of this power as possible to the receiver, or load, is desirable. In such applications the amount of power being transferred is small, so the efficiency of transfer is not a primary concern. 84

85 We consider maximum power transfer in systems that can be modeled by a purely resistive circuit. ( V Th, R Th ) We wish to determine the value of of R L that permits maximum power delivery to R L for the given circuit represented by the Thevenin Equivalent 85

86 Represent the Resistive Network by its Thevenin Equivalent P load = i 2 R L =! " # V Th R Th + R L 2 \$ % & R L ' max RL 86

87 P load =! " # V Th R Th + R L 2 \$ % & R L P load R L 87

88 Maximum is reached at the point where dp load dr L = 0! P load = " # \$ V Th R Th + R L 2 % & ' R L dp load dr L 2 d = V Th dr L!# " \$# R L ( ) 2 R Th + R L %# & '# ( ) 2! 2R L ( R Th + R L ) ( ) 4 " 2 R = V Th + R L Th \$ # \$ R Th + R L % ' &' = 0 88

89 Condition for Maximum Power Transfer R L = R Th The Maximum Power Delivered to R L P load =! " # V Th R Th + R L 2 \$ % & R L! P =! V Th max " R L =R Th # 2R Th 2 \$ % & R Th P max = V Th 2 4R Th 89

90 Example Find R L that results in max. power and the corresponding max. power that can be delivered to R L. V Th = ! 360 = 300 V R Th = = 25! 90

91 i L! max v L! max P load =! " # V Th R Th + R L 2 \$ % & R L =! 300 " # 25 + R L 2 \$ % & R L P load! R L = 25! for the maximum power transfer with 91 R L 2! 300 \$ P max = " % 25 = 900W # &

92 Example Find R L that results in max. power and the corresponding max. power that can be delivered to R L. 92

93 1. R Th R Th = 4k + 3k 6k = 6k This is the resistance for maximum power transfer If we wish to find the value of the power that can be transferred then we need the Thevenin voltage! 93

94 2. V oc * loop 1: I 1 = 2mA * loop 2: 3k( I 2! I 1 ) + 6kI 2 + 3V = 0! I 2 = 1 3 ma * KVL: V oc = 6kI 2 + 4kI 1! V oc = 10V * P max = V 2 Th = 4R Th ! 6k = 25 6 mw 94

95 Example Find R L that results in max. power and the corresponding max. power that can be delivered to R L. 95

96 1. Open circuit voltage V oc V oc! 2000I x ' KCL at the supernode: V oc! 2000I x ' 3k +1k + V oc 2k! 4m = 0 and I ' x = V oc 2k! V oc = 8V 96

97 2. Short circuit current I sc and R Th I x '' = 0! dependent source is zero! I sc = 4mA R Th = 8V 4mA = 2k! < Obtain by the test source approach 97

98 3. Circuit with Load 6 P load ==! 8V " # 6k + R L 2 \$ % & R L is maximized by R L = 6k! The maximum power transfer: P max =! 8V " # 6k + 6k 2 \$ % & 6k = 8 3 mw 98

99 Example Plot V out, I, P in, P out and P out / P in as a function of R 2 i L! max v L! max P L! max 99

100 * V out = R 2 R 1 + R 2 V in ; * I = V in R 1 + R 2 * P in = I! V in = V in 2 R 1 + R 2 * P out = I! V out =! " # V in R 1 + R 2 \$ % & 2 R 2 * efficiency = P out P in = R 2 R 1 + R 2 = R 2 / R 1 1+ R 2 / R 1!! 1+! 100

101

102 * As R 2! then V out! V in = 5 V * As R 2! then I! * small R 2! V out is small, large R 2! I is small! * maximum power transfer ( R 2 / R 1 = 1) does not correspond to max. efficiency * At R 2 / R 1 = 1! efficiency = P out P in = 0.5 (50%) * The fact that the eff. is higher for R 2 > R 1 is due to the fact that a higher percentage of the source power is transferred to the load (more \$ for MH), but the value of the load power is lower since the total circuit resistance goes up maximum power transfer! 2R 1 < R 1 + R 2 102

103 103

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