Lecture 271 More Nodal Analysis

Lecture 272 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.

Lecture 273 Example Circuit 1k  V in + - 2k  +10V + - VoVo Common Collector (Emitter Follower) Amplifier

Lecture 274 Small Signal Equivalent 1k  5mA 100I b + - VoVo 50  IbIb 2k 

Lecture 275 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-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

Lecture 276 Reference Node 1k  5mA 100I b + - VoVo 50  IbIb 2k 

Lecture 277 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-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

Lecture 278 Assign Node Voltages 1k  5mA 100I b + - VoVo 50  IbIb 2k  1 V1V1 V2V2 2

Lecture 279 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-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

Lecture 2710 Node 1 1k  5mA 100I b + - VoVo 50  IbIb 2k  1 V1V1 V2V2 2

Lecture 2711 Node 2 1k  5mA 100I b + - VoVo 50  IbIb 2k  1 V1V1 V2V2 2

Lecture 2712 The Dependent Source We must express I b in terms of the node voltages: Equation from Node 2 becomes

Lecture 2713 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-express currents in terms of node voltages. 4.Solve the resulting system of linear equations.

Lecture 2714 System of Equations Node 1: Node 2:

Lecture 2715 Matrix Formulation Matrix is not symmetric due to the dependent source.

Lecture 2716 Solve Equation V 1 = 4.975V V 2 = 4.974V

Lecture 2717 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.

Lecture 2718 Another Small Signal Equivalent 1k  5V 100I b + - VoVo 50  IbIb 2k  + -

Lecture 2719 Do Nodal Analysis 1k  5V 100I b + - VoVo 50  IbIb 2k  V1V1 V2V2 V3V3

Lecture Node 1 1k  5V 100I b + - VoVo 50  IbIb 2k  V1V1 V2V2 V3V3

Lecture 2721 Matrix Equation

Lecture 2722 Solve Equation V 1 = 5V V 2 = 4.975V V 3 = 4.974V

Lecture 2723 A Linear Large Signal Equivalent 5V 100I b + - VoVo 50  IbIb 2k  1k  V

Lecture 2724 A Linear Large Signal Equivalent 5V 100I b + - VoVo 50  IbIb 2k  1k  V V1V1 V2V2 V3V3 V4V4

Lecture 2725 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

Lecture I b + - VoVo 50  IbIb 2k  1k  V 1 4 V1V1 V2V2 V3V3 V4V4

Lecture 2727 the Supernode

Lecture 2728 Matrix Equation

Lecture 2729 Solve Equation V 1 = 5V V 2 = 4.978V V 3 = 4.278V V 4 = 4.278V