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Lecture 71 Nodal Analysis (3.1) Prof. Phillips February 7, 2003

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Lecture 72 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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Lecture 73 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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Lecture 74 Example Transistor Circuit 1k +–+– V in 2k +10V + – VoVo Common Collector (Emitter Follower) Amplifier

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Lecture 75 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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Lecture 76 A Linear Large Signal Equivalent 5V 100I b + – VoVo 50 IbIb 2k 1k +–+– + – 0.7V

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Lecture 77 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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Lecture 78 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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Lecture 79 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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Lecture 710 KCL @ Node 4 100I b + – VoVo 50 IbIb 2k 1k +–+– 0.7V 1 234 V1V1 V2V2 V3V3 V4V4 5V + –

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Lecture 711 The Dependent Source We must express I b in terms of the node voltages: Equation from Node 4 becomes

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Lecture 712 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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Lecture 713 100I b + – VoVo 50 IbIb 2k 1k 0.7V 1 4 V1V1 V2V2 V3V3 V4V4 +–+– + –

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Lecture 714 KCL @ the Supernode

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Lecture 715 Class Examples

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