lecture 291 more loop and nodal analysis. lecture 292 advantages of nodal analysis solves directly...
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Lecture 29 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.
Lecture 29 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.
Lecture 29 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.
Lecture 29 5
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 both nodal and mesh analysis.
• This circuit is a bandpass filter with center frequency 455kHz and bandwidth 40kHz.
Lecture 29 7
Analysis
• Use AC steady state analysis.
• Start with a frequency of =2 455,000.
• Nodal or mesh analysis?
Lecture 29 10
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
Lecture 29 12
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
Lecture 29 13
Solve Equations
V1 = 0.0259V-j0.122V = 0.1247V-78
V2 = 0.0277V-j4.1510-4V=0.0277V -0.86
Vout = -100V2 = 2.7V 179.2
Lecture 29 18
KVL Around Loop 2
V1V100k80
k9.15160
32
3212
x
j
II
IIII
0k80100 32 IIVV xx
32101
k80IIV
x
Lecture 29 20
KVL Around Loop 3
015.9k-k0815.9k- 23233 IIIII jj
015.9k80k15.9k
15.9k80k-
3
2
I
I
jj
j