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Methods of Analysis
Instructor: Chia-Ming TsaiElectronics Engineering
National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.
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Contents• Introduction
• Nodal Analysis
• Nodal Analysis with Voltage Sources
• Mesh Analysis
• Mesh Analysis with Current Sources
• Nodal Analysis vs. Mesh Analysis
• Applications
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Introduction
• Nodal Analysis– Based on KCL
• Mesh Analysis– Based on KVL
• Linear algebra is applied to solve the resulting simultaneous equations.– Ax=B, x=A-1B
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Nodal Analysis
• Circuit variables = node voltages
• Steps to determine node voltages– Select a reference node, assign voltages v1, v2,…, vn-1 for the remaining n-1 nodes
– Use Ohm’s law to express currents of resistors– Apply KCL to each of the n-1 nodes– Solve the resulting equations
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Symbols for Reference Node (Ground)
Used in this course
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Case Study
2
21
2
1
332
221
232122
322
2121121
2121
2333
23
21222
212
1111
11
gives KCL applying 2, nodeAt
gives KCL applying 1, nodeAt
or 0
or
or 0
gives law sOhm' Applying
I
II
v
v
GGG
GGG
vGvvGI
iiI
vvGvGII
iiII
vGiR
vi
vvGiR
vvi
vGiR
vi
Assign vn
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Nodal Analysis with Voltage Sources
• If a voltage source is connected between a nonreference node and the reference node (or ground)– The node voltage is defined by th
e voltage source– Number of variables is reduced– Simplified analysis
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Continued• If a voltage source is connect
ed between two nonreference nodes– The two nodes form a superno
de– Apply KCL to the supernode
(similar to a closed boundary)– Apply KVL to derive the relati
onship between the two nodes
Supernode
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Case Study with Supernode
equations 3by solved variables3
(3) 5
supernode, the toKVL Applying
(2) 2
0
2
0
22
supernode, the toKCL Applying
(1) V 10
32
32
3121
3241
1
vv
vv
vvvv
iiii
v
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Example 1
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Example 2
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What is a mesh?• A mesh is a loop that does not contain any
other loop within it.
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Mesh Analysis• Circuit variables = mesh currents
• Steps to determine mesh currents– Assign mesh currents i1, i2,…, in
– Use Ohm’s law to express voltages of resistors– Apply KVL to each of the n meshes– Solve the resulting equations
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Continued• Applicable only for planar circuits• An example for nonplanar circuits is shown be
low
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Case Study
223213
123222
123131
213111
0
gives KVL applying 2,mesh For
0
gives KVL applying 1,mesh For
ViRRiR
iiRViR
ViRiRR
iiRiRV
2
1
2
1
323
331
V
V
i
i
RRR
RRR
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Mesh Analysis with Current Sources
• If a current source exists only in one mesh– The mesh current is defined by the current
source– Number of variables is reduced– Simplified analysis
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Continued
Supermesh
SIii 12
Excluded
• If a current source exists between two meshes– A supermesh is resulte
d– Apply KVL to the super
mesh– Apply KCL to derive th
e relationship between the two mesh currents
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Example 1
A 2
06410
1,mesh for KVL Applying
A 5
1
21
2
i
iii
i
21 ii
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Example 2
20146
0410620
supermesh, the toKVL Applying
21
221
ii
iii
A 8.2 A, 2.3
6
0, node toKCL Applying
21
12
ii
ii
Supermesh
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Example 3
Supermesh
• Applying KVL to the supermesh• Applying KCL to node P• Applying KCL to node Q• Applying KVL to mesh 4
4 variables solved by 4 equations
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How to choose?• Nodal Analysis
– More parallel-connected elements, current sources, or supernodes
– Nnode < Nmesh
– If node voltages are required
• Mesh Analysis– More series-connected elements, voltage sourc
es, or supermeshes
– Nmesh < Nnode
– If branch currents are required
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Applications: Transistors
• Bipolar Junction Transistors (BJTs)• Field-Effect Transistors (FETs)
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Bipolar Junction Transistors (BJTs)
1
1) (0
1
100)~ (
V 0.7
(KVL) 0
(KCL)
EC
BE
BC
BE
BCEBCE
CBE
II
II
II
V
VVV
III
• Current-controlled devices
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DC Equivalent Model of BJT
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Example of Amplifier Circuit