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AIHT 1
Nodal and Loop Analysis cont’d
AIHT 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.
AIHT 3
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.
AIHT 4
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
AIHT 5
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.
AIHT 6
Example Transistor Circuit
1k+–
Vin
2k
+10V
+
–Vo
Common Collector (Emitter Follower)
Amplifier
AIHT 7
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.
AIHT 8
A Linear Large Signal Equivalent
5V100Ib
+
–
Vo
50
Ib
2k1k+–
+ –
0.7V
AIHT 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.
AIHT 10
A Linear Large Signal Equivalent
5V 100Ib
+
–
Vo
50
Ib
2k
1k
0.7V
12 3 4
V1 V2 V3 V4
+–
+ –
AIHT 11
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.
AIHT 12
KCL @ Node 4
k2100
50443 V
IVV
b
100Ib
+
–
Vo
50
Ib
2k
1k+–
0.7V
12 3 4
V1V2 V3 V4
5V
+ –
AIHT 13
The Dependent Source
• We must express Ib in terms of the node voltages:
• Equation from Node 4 becomes
k1
21 VVIb
0k2k1
10050
42143
VVVVV
AIHT 14
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
AIHT 15
100Ib
+
–
Vo
50
Ib
2k
1k
0.7V
14
V1V2 V3 V4
+–
+ –
AIHT 16
KCL @ the Supernode
050k1
4312
VVVV
AIHT 17
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.
AIHT 18
IF Amplifier
4k
1V 0
+
–
Vout
100pF
160
100pF
80k
–
+
Vx 100Vx
+–
+–
AIHT 19
Nodal AC Analysis
• Use AC steady-state analysis.
• Start with a frequency of =2 455,000.
AIHT 20
Impedances
4k
1V 0
+
–
Vout160
80k
–
+
Vx 100Vx
-j3.5k
-j3.5k
+–
+–
AIHT 21
Nodal Analysis
1 24k
1V 0
+
–
Vout160
80k
–
+
Vx 100Vx
-j3.5k
-j3.5k
+–
+–
AIHT 22
KCL @ Node 1
03.5k-3.5k-
100
0614k
V1 21111
jj
x VVVVVV
4k
V1
3.5k
1
3.5k-
100
3.5k-
1
3.5k-
1
061
1
k4
1
2
1
jj
jj
V
V
2VV x
AIHT 23
KCL @ Node 2
00k8
100
3.5k-212
x
j
VVVV
00k8
101
3.5k-
1
3.5k
121
jj
VV
2VV x
AIHT 24
04k
V1
80k
101
3.5k
1
3.5k
13.5k
1
3.5k
100
3.5k
2
160
1
k4
1
2
1
V
V
jj
jjj
Matrix Formulation
AIHT 25
Solve Equations
V1 = 0.0259V-j0.1228V = 0.1255V-78
V2 = 0.0277V-j4.1510-4V=0.0277V -0.86
Vout = -100V2 = 2.77V 179.1
