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ECE201 Lect-111 Nodal and Loop Analysis cont’d (8.8) Dr. Holbert March 1, 2006

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ECE201 Lect-112 Advantages of Nodal Analysis Solves directly for node voltages. Current sources are easy. Voltage sources are either very easy or somewhat difficult. Works best for circuits with few nodes. Works for any circuit.

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ECE201 Lect-113 Advantages of Loop Analysis Solves directly for some currents. Voltage sources are easy. Current sources are either very easy or somewhat difficult. Works best for circuits with few loops.

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ECE201 Lect-114 Disadvantages of Loop Analysis Some currents must be computed from loop currents. Does not work with non-planar circuits. Choosing the supermesh may be difficult. FYI: PSpice uses a nodal analysis approach

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ECE201 Lect-115 Where We Are Nodal analysis is a technique that allows us to analyze more complicated circuits than those in Chapter 2. We have developed nodal analysis for circuits with independent current sources. We now look at circuits with dependent sources and with voltage sources.

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ECE201 Lect-116 Example Transistor Circuit 1k +–+– V in 2k +10V + – VoVo Common Collector (Emitter Follower) Amplifier

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ECE201 Lect-117 Why an Emitter Follower Amplifier? The output voltage is almost the same as the input voltage (for small signals, at least). To a circuit connected to the input, the EF amplifier looks like a 180k resistor. To a circuit connected to the output, the EF amplifier looks like a voltage source connected to a 10 resistor.

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ECE201 Lect-118 A Linear Large Signal Equivalent 5V 100I b + – VoVo 50 IbIb 2k 1k +–+– + – 0.7V

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ECE201 Lect-119 Steps of Nodal Analysis 1.Choose a reference node. 2.Assign node voltages to the other nodes. 3.Apply KCL to each node other than the reference node; express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

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ECE201 Lect-1110 A Linear Large Signal Equivalent 5V 100I b + – VoVo 50 IbIb 2k 1k 0.7V 1 234 V1V1 V2V2 V3V3 V4V4 +–+– + –

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ECE201 Lect-1111 Steps of Nodal Analysis 1.Choose a reference node. 2.Assign node voltages to the other nodes. 3.Apply KCL to each node other than the reference node; express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

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ECE201 Lect-1112 KCL @ Node 4 100I b + – VoVo 50 IbIb 2k 1k +–+– 0.7V 1 234 V1V1 V2V2 V3V3 V4V4 5V + –

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ECE201 Lect-1113 The Dependent Source We must express I b in terms of the node voltages: Equation from Node 4 becomes

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ECE201 Lect-1114 How to Proceed? The 0.7V voltage supply makes it impossible to apply KCL to nodes 2 and 3, since we don’t know what current is passing through the supply. We do know that V 2 - V 3 = 0.7V

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ECE201 Lect-1115 100I b + – VoVo 50 IbIb 2k 1k 0.7V 1 4 V1V1 V2V2 V3V3 V4V4 +–+– + –

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ECE201 Lect-1116 KCL @ the Supernode

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ECE201 Lect-1117 Another Analysis Example We will analyze a possible implementation of an AM Radio IF amplifier. (Actually, this would be one of four stages in the IF amplifier.) We will solve for output voltages using nodal (and eventually) mesh analysis. This circuit is a bandpass filter with center frequency 455kHz and bandwidth 40kHz.

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ECE201 Lect-1118 IF Amplifier 4k 1V 0 + – V out 100pF 160 100pF 80k – + VxVx 100V x +–+– +–+–

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ECE201 Lect-1119 Nodal AC Analysis Use AC steady-state analysis. Start with a frequency of =2 455,000.

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ECE201 Lect-1120 Impedances 4k 1V 0 + – V out 160 80k – + VxVx 100V x -j3.5k +–+– +–+–

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ECE201 Lect-1121 Nodal Analysis 12 4k 1V 0 + – V out 160 80k – + VxVx 100V x -j3.5k +–+– +–+–

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ECE201 Lect-1122 KCL @ Node 1

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ECE201 Lect-1123 KCL @ Node 2

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ECE201 Lect-1124 Matrix Formulation

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ECE201 Lect-1125 Solve Equations V 1 = 0.0259V-j0.1228V = 0.1255V -78 V 2 = 0.0277V-j4.15 10 -4 V=0.0277V -0.86 V out = -100V 2 = 2.77V 179.1

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ECE201 Lect-1126 Class Examples Learning Extension E3.6 Learning Extension E8.13 Learning Extension E8.14(a)

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