ECE201 Lect-151 Nodal Analysis (7.8) Dr. Stephen Goodnick October 3, 2003.

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Presentation transcript:

ECE201 Lect-151 Nodal Analysis (7.8) Dr. Stephen Goodnick October 3, 2003

ECE201 Lect-152 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.

ECE201 Lect-153 Example Transistor Circuit 1k  +–+– V in 2k  +10V + – VoVo Common Collector (Emitter Follower) Amplifier

ECE201 Lect-154 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.

ECE201 Lect-155 A Linear Large Signal Equivalent 5V 100I b + – VoVo 50  IbIb 2k  1k  +–+– + – 0.7V

ECE201 Lect-156 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.

ECE201 Lect-157 A Linear Large Signal Equivalent 5V 100I b + – VoVo 50  IbIb 2k  1k  0.7V V1V1 V2V2 V3V3 V4V4 +–+– + –

ECE201 Lect-158 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.

ECE201 Lect-159 Node 4 100I b + – VoVo 50  IbIb 2k  1k  +–+– 0.7V V1V1 V2V2 V3V3 V4V4 5V + –

ECE201 Lect-1510 The Dependent Source We must express I b in terms of the node voltages: Equation from Node 4 becomes

ECE201 Lect-1511 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

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

ECE201 Lect-1513 the Supernode

ECE201 Lect-1514 Class Example Learning Extension E3.6