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1 Chapter 5. Additional analysis techniques EMLAB Contents 2 1. Introduction 2. Superposition 3. Thevenin’s and Norton’s theorems 4. Maximum power transfer 5. Application Examples EMLAB 1. Introduction 3 Examples of equivalent circuits To simply solution procedures, the number of nodes or loops should be minimized by replacing the original circuits with equivalent ones. EMLAB Linearity 4 1 (t ) i1 (t ) Linear system L i2 (t ) Linear system L 2 (t ) Linear system L A1 (t ) B 2 (t ) Ai1 (t ) Bi2 (t ) A system satisfying the above statements is called as a linear system. Resistors, Capacitors, Inductors are all linear systems. An independent source is not a linear system. All the circuits in the circuit theory class are linear systems! EMLAB 5 Example of linear system R1 output Resistor 1 (t ) - 1k R1 i1 (t ) 1k 1 (t ) i1 (t ) R1 input 2 (t ) - Capacitor C1 1n C1 i2 (t ) 1n 1 2 (t ) C t i 2 ( ) d 0 EMLAB 6 Example 5.1 For the circuit shown in the figure, determine the output voltage Vout using linearity. First, arbitrarily assume that the output voltage is Vout = 1 [V]. Vout V2 1 [V ] I1 V1 1 [mA] 3k I2 V2 0.5 [mA] 2k I 0 I1 I 2 1.5 [mA] V1 4k I 2 V2 3 [V ] Vo 2k I o V1 6 [V ] For the arbitrary assumption that Vout = 1 [V], the source voltage Vo should be 6 V. Then from linearity, the actual output voltage should satisfy the following relation. actual actual Vo : Vout 12 : Vout 6 : 1 Vout 2 [V ] EMLAB 7 2. Superposition Source superposition VS VS VL IS I L,2 I L ,1 IL Circuit Circuit with voltage source set to zero (Short circuited) = Circuit VL ,1 Circuit with current source set to zero(Open circuited) + Circuit VL , 2 IS I L I L ,1 I L , 2 VL VL ,1 VL , 2 • • • Superposition is utilized to simplify the original linear circuits. If a voltage source is eliminated, it is replaced by a short circuit connected to the original terminals. If a current source is eliminated, it is replaced by an open circuit. EMLAB Example 5.2 8 To provide motivation for this subject, let us examine the simple circuit below, in which two sources contribute to the current in the network. The actual values of the sources are left unspecified so that we can examine the concept of superposition. = 1 3k i1 3k (i1 i2 ) 0 3k (i2 i1 ) 6k i2 2 0 6k 3k i 1 1 i2 15k + 1 3k i1 3k (i1 i2 ) 0 3k (i2 i1 ) 6k i2 0 6k 3k 3k i1 1 9k i2 2 3 1 1 1 1 2 15k 2 31 2 2 2 1 3k i1 1 9k i2 0 i 1 1 i2 15k 31 1 3k i1 3k (i1 i2 ) 0 3k (i2 i1 ) 6k i2 2 0 6k 3k 3k i1 0 9k i2 2 i 1 1 i2 15k 2 2 2 EMLAB Example 5.3 9 Let us use superposition to find Vo in the circuit in Fig. 5.3a. = + Io 1 6k 2 [mA] 1 1 3k 6k Vo I o 6k 4 [V ] 2 [mA] 3 Vo 6k 3 [V ] 2 [V ] 3k 6k Vo Vo Vo 6 [V ] To verify the solution, we can apply the loop analysis technique. I1 2 [mA] 1k ( I1 I 2 ) 2k ( I1 I 2 ) 3 6k I 2 0 I 2 1 [mA], Vo 6 [V ] EMLAB Example 5.4 10 Consider now the network in Fig. 5.4a. Let us use superposition to find Vo’. + = 8k 3 24 [V ] V1 6 8k 2k 7 3 6k 18 Vo V1 [V ] 6k 2 k 7 30 10 Vo k || 6k 2 [mA] [V ] 7 3 Vo Vo Vo 48 [V ] 7 EMLAB Thevenin’s and Norton’s theorems 11 i a RV + - VS b a b RI IS Single port network f (i ) Ai B RThi VTh 1. To find B, measure voltage with i=0. (open circuit voltage) 2. To find A, measure the variation of υ with i changing. (impedance) i g (i ) i C D YNor I Nor 1. To find D, measure current with υ = 0. (short circuit current) 2. To find C, measure the variation of i with υ changing. (admittance) EMLAB How to construct Thevenin’s equivalent circuit 12 (1) Measure open circuit voltage (VOC) with a volt meter. VS I 0 VOC Circuit Input resistance of a voltmeter is infinite. RTH V = IS VOC (2) Measure resistance (RTH) with sources suppressed. →Voltage sources short circuited and current sources open circuited. short Circuit Ω Ohm meter open EMLAB How to construct Norton’s equivalent circuit 13 (1) Measure short circuited current (ISH). VS Input resistance of an ammeter is zero. I SH Circuit A = IS I SH RTH (2) Measure resistance (RTH) with sources suppressed. →Voltage sources short circuited and current sources open circuited. short Circuit Ω Ohm meter open EMLAB 14 Example 5.6 Let us use Thévenin’s and Norton’s theorems to find Vo in the network below. Vo Vo 1 1 1 3k 6k 6k 9 6 [V ] 3k 6k 3 [mA] 6 [V ] EMLAB 15 Example 5.7 Let us use Thévenin’s theorem to find Vo in the network in Fig. 5.9a. VOC1 6k 12 [V ] 8 [V ] 3k 6k VOC2 8 4k 2 [mA] 16 [V ] VO RTh1 2k 3k 6k 4 [ k ] 3k 6k RTh1 4 [k] 8k 16 8 [V ] 16k EMLAB Example 5.9 : Circuits containing only dependent sources 16 → Apply an external voltage or current source between the terminals. Determine the Thévenin equivalent of the network in Fig. 5.11a at the terminals A-B. V1 V1 2Vx V1 1 0 1k 2k 1k V1 Vx 1 V1 Io 5V1 2(1 V1 ) 2 0 4 3 , Vx 7 7 Vx 1 2Vx 1 3 3 15 [mA] 1k 1k 2k 2k 7k 14 RTh 1 [V ] 14 [ k ] Io 15 EMLAB 17 Example 5.10 Let us determine at the terminals A-B for the network in Fig. 5.12a. V1 2000 I x V1 V1 V2 0 2k 1k 3k V2 V1 V2 1m 0 3k 2k Ix V1 1k 3V1 6V1 2V1 2V2 5V1 2V2 0 2V2 2V1 3V2 6 2V1 5V2 6 0 V2 10 V 10 [V ] RTh 2 [k] 7 1m 7 EMLAB Example 5.11 : Circuits containing both independent and dependent sources 18 In these types of circuits we must calculate both the open-circuit voltage and short-circuit current to calculate the Thévenin equivalent resistance. Let us use Thévenin’s theorem to find Vo in the network in Fig. 5.13a. KCL for the super-node around the 12-V source is VOC 12 2000 I x VOC 12 VOC 0 1k 2k 2k V I x OC 2k I x 0 4VOC 24 VOC 12 VOC 6VOC 36 0 I SC Vo 12 2 [ k ] 3 1k 18 [mA] 1 1k 1k k 3 ( 6 ) RTh VOC 6 [V ] VOC 1 [ k ] I SC 3 3 18 ( 6) [V ] 7 7 EMLAB 19 Example 5.12 Let us find Vo in the network in Fig. 5.14a using Thévenin’s theorem. I 2 2 [mA] I1 Vx 2000 Vx 4k ( I1 I 2 ) Vx 2Vx 8 Vx 8 [V ] VOC 2k I1 3 11 [V ] V V 3 2k I SC x 0, Vx 4k x 2m 2000 2000 Vx 8 [V ] I SC RTh Vo Vx 3 11 4m 1.5m [mA] 2000 2k 2 VOC 11 2 [ k ] I Sc 11 [mA] 2 6k 33 11 [V ] 6k 2 k 4 EMLAB 20 Example 5.14 Determine Vo in the circuit in Fig. 5.16a using the repeated application of source transformation. I SC 12 3k 6k 4 [mA], RTh 2k 3k 3k 6k VOC 4m 2k 8 [V ], RTh 2k 2k 4k I SC 8 2 [mA], RTh 4k 4k VOC 4m 4k 16 [V ], RTh 4k VO 8k 16k 8 [V ] 4k 4k 8k EMLAB 21 4. Maximum Power Transfer RS PL i 2 RL , i PL Equivalent circuit of a signal source RS RL 2 RL ( RS RL ) 2 We want to determine the value of RL that maximizes this quantity. Hence, we differentiate this expression with respect to RL and equate the derivative to zero. 2 PL 2 ( RS RL ) 2 RL ( RS RL ) RL ( RS RL ) 4 2 Maximum power transfer condition : ( RS RL ) 0 ( RS RL )3 RL RS EMLAB 22 Example 5.16 Let us find the value of RL for maximum power transfer in the network in Fig. 5.20a and the maximum power that can be transferred to this load. RL RTh 4k 6k 3k 6k 6k 3k 3k ( I 2 I1 ) 6k I 2 3 0, I1 2m I2 1 [mA] 3 VOC 4k I1 6k I 2 8 2 10 [V ] 2 10 PL 6k 4.17 [mW ] 12k EMLAB 23 Example 5.17 Let us find RL for maximum power transfer and the maximum power transferred to this load in the circuit in Fig. 5.21a. VOC 2000 I x V V 4m OC 0, I x OC 4k 2k 2k VOC 8 [V ] I x 0 x I SC 4 [mA] RTh VOC 2 k I SC 2 8 PL 6k 2.67 [mW ] 12k EMLAB