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Nodal Analysis (3.1)

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Solves directly for node voltages. Current sources are easy. ... 4. Solve the resulting system of linear equations. Lecture 7. 8. A Linear Large Signal Equivalent ... – PowerPoint PPT presentation

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Title: Nodal Analysis (3.1)


1
Nodal Analysis (3.1)
  • Prof. Phillips
  • February 7, 2003

2
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.

3
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.

4
Example Transistor Circuit
10V
Common Collector (Emitter Follower) Amplifier
1kW

Vin

2kW
Vo

5
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 180kW resistor.
  • To a circuit connected to the output, the EF
    amplifier looks like a voltage source connected
    to a 10W resistor.

6
A Linear Large Signal Equivalent
0.7V
Ib


50W
2kW
1kW
5V

Vo
100Ib

7
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.

8
A Linear Large Signal Equivalent
0.7V
Ib
V2
V1
V3
V4

1

50W
1kW
2
3
4

5V
Vo
100Ib
2kW

9
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.

10
KCL _at_ Node 4
0.7V
Ib
V2
V1
V3
V4

1

50W
1kW
2
3
4

5V
Vo
100Ib
2kW

11
The Dependent Source
  • We must express Ib in terms of the node voltages
  • Equation from Node 4 becomes

12
How to Proceed?
  • The 0.7V voltage supply makes it impossible to
    apply KCL to nodes 2 and 3, since we dont know
    what current is passing through the supply.
  • We do know that
  • V2 - V3 0.7V

13
The Supernode!
0.7V
Ib
V2
V1
V3
V4

1

50W
1kW
4

Vo
100Ib
2kW

14
KCL _at_ the Supernode
15
Class Examples
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