Transcript Nodal and Loop Analysis cont'd (8.8)
Nodal and Loop Analysis cont’d (8.8)
Dr. Holbert March 1, 2006 ECE201 Lect-11 1
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.
ECE201 Lect-11 2
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.
ECE201 Lect-11 3
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 ECE201 Lect-11 4
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-11 5
Example Transistor Circuit +10V
V in
+ – 1k W 2k W +
V o
– Common Collector (Emitter Follower) Amplifier ECE201 Lect-11 6
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 W resistor.
• To a circuit connected to the output, the EF amplifier looks like a voltage source connected to a 10 W resistor.
ECE201 Lect-11 7
A Linear Large Signal Equivalent 5V + – 1k W
I b
0.7V
+ – 50 W 100
I b
2k W +
V o
– ECE201 Lect-11 8
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-11 9
A Linear Large Signal Equivalent 5V 1 + –
V
1
1k W 0.7V
I b
V
2
+ – 2
V
3
3 50 W 100
I b
V
4
4 2k W +
V o
– ECE201 Lect-11 10
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-11 11
KCL @ Node 4 5V 1 + –
V
1 1k W 0.7V
I b V
2 + – 2
V
3 3 50 W 100
I b V
4 4 2k W +
V o
–
V
3
V
4 50 W 100
I b
V
4 2 k W ECE201 Lect-11 12
The Dependent Source • We must express
I b
voltages:
I b
in terms of the node
V
1 1
V
2 k W • Equation from Node 4 becomes
V
3
V
4 50 W 100
V
1
V
2 1 k W
V
4 2 k W 0 ECE201 Lect-11 13
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-11 14
1 + –
V
1 1k W 0.7V
I b V
2 + –
V
3 50 W 100
I b V
4 4 2k W +
V o
– ECE201 Lect-11 15
KCL @ the Supernode
V
2
V
1 1 k W
V
3
V
4 50 W 0 ECE201 Lect-11 16
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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IF Amplifier 100pF 4k W 1V 0 + – 160 W 100pF –
V
x
+ 80k W 100
V
x
+ – +
V
out
– ECE201 Lect-11 18
Nodal AC Analysis • Use AC steady-state analysis.
• Start with a frequency of w =2 p 455,000.
ECE201 Lect-11 19
Impedances -j3.5k
W 1V 0 + – 4k W 160 W -j3.5k
W –
V
x
+ 80k W 100
V
x
+ – +
V
out
– ECE201 Lect-11 20
Nodal Analysis -j3.5k
W 1V 0 + – 4k W 1 160 W -j3.5k
W – 2 80k W
V
x
100
V
x
+ + – +
V
out
– ECE201 Lect-11 21
KCL @ Node 1
V
1 1 4k W V
V
1 1 6 0 W
V
1 100
V
x j
3.5k
W -
V
1
V
2
j
3.5k
W 0
V
x
V
1 1 4 k W 1 1 6 0 W
V
2
j
1 3.5k
W 1
j
3.5k
W
V
2 100
j
3.5k
W 1
j
3.5k
W 1 V 4k W ECE201 Lect-11 22
KCL @ Node 2 -
V
2
V
1
j
3.5k
W
V
2 100
V
x
8 0k W 0
V
x
V
2
V
1 1
j
3.5k
W
V
2 1
j
3.5k
W 101 8 0k W 0 ECE201 Lect-11 23
Matrix Formulation 1 4 k W 1 160 W
j
1 3.5k
W 2
j
3.5k
W
j
100 3.5k
1 W
j
3.5k
W 1
j
3.5k
W 101 80k W
V V
2 1 1 V 4k W 0 ECE201 Lect-11 24
Solve Equations
V
2
V
1
= 0.0259V-
j
0.1228V = 0.1255V
= 0.0277V-
j
4.15
10 -4 V=0.0277V -78 -0.86
V
out
= -100
V 2
= 2.77V 179.1
ECE201 Lect-11 25
Class Examples • Learning Extension E3.6
• Learning Extension E8.13
• Learning Extension E8.14(a) ECE201 Lect-11 26