Download 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
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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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Example Transistor Circuit
+10V
Vin
+
–
1kW
+
2kW
Vo
–
ECE201 Lect-11
Common Collector
(Emitter Follower)
Amplifier
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 180kW resistor.
• To a circuit connected to the output, the EF
amplifier looks like a voltage source
connected to a 10W resistor.
ECE201 Lect-11
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A Linear Large Signal Equivalent
0.7V
Ib
5V
+
–
1kW
+ –
50W
100Ib
2kW
+
Vo
–
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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
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A Linear Large Signal Equivalent
0.7V
1
5V
Ib V2
V1
+
–
1kW
2
+ –
V3
V4
3 50W
4
Vo
100Ib
2kW
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
0.7V
1
5V
Ib V2
V1
+
–
1kW
2
+ –
V3
V4
3 50W
100Ib
4
+
Vo
2kW
–
V3  V4
V4
 100 I b 
50W
2 kW
ECE201 Lect-11
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The Dependent Source
• We must express Ib in terms of the node
voltages:
V1  V2
Ib 
1 kW
• Equation from Node 4 becomes
V3  V4
V1  V2
V4
 100

0
50W
1 kW 2kW
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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
V2 - V3 = 0.7V
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0.7V
1
Ib V2
V1
+
–
1kW
+ –
V3
50W
100Ib
V4
4
Vo
2kW
ECE201 Lect-11
+
–
15
KCL @ the Supernode
V2  V1 V3  V4

0
1kW
50W
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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.
ECE201 Lect-11
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IF Amplifier
100pF
4kW
100pF
80kW
–
1V  0
+
–
160W
Vx
+
ECE201 Lect-11
+
100Vx
+
–
Vout
–
18
Nodal AC Analysis
• Use AC steady-state analysis.
• Start with a frequency of w=2p 455,000.
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Impedances
-j3.5kW
4kW
-j3.5kW
80kW
–
1V  0
+
–
160W
Vx
+
ECE201 Lect-11
+
100Vx
+
–
Vout
–
20
Nodal Analysis
-j3.5kW
4kW 1
1V  0
+
–
160W
-j3.5kW 2 80kW
–
Vx
+
ECE201 Lect-11
100Vx
+
+
–
Vout
–
21
KCL @ Node 1
V1  100Vx V1  V2
V1  1V
V1



0
4kW
160W - j3.5kW - j3.5kW
Vx  V2
 1

1
1
1

V1 



 4kW 160W - j 3.5kW - j3.5kW 
 100
1  1V
 
 V2 

 - j 3.5kW j 3.5kW  4kW
ECE201 Lect-11
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KCL @ Node 2
V2  V1 V2  100Vx

0
- j3.5kW
80kW
Vx  V2
 1 

1
101 
  V2 
  0
V1 

 j 3.5kW 
 - j 3.5kW 80kW 
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Matrix Formulation
1
2
 1
 4kW  160W  j 3.5kW

1


j 3.5kW
 100
1 

 1V 

V


j 3.5kW j 3.5kW
1
     4kW 
1
101  V2  

0



j 3.5kW 80kW 
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Solve Equations
V1 = 0.0259V-j0.1228V = 0.1255V-78
V2 = 0.0277V-j4.1510-4V=0.0277V  -0.86
Vout = -100V2 = 2.77V  179.1
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Class Examples
• Learning Extension E3.6
• Learning Extension E8.13
• Learning Extension E8.14(a)
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