Introduction to Circuit Theory
Methods of Analysis
2012-09-12
Jieh-Tsorng Wu
National Chiao-Tung University
Department of Electronics Engineering
Outline
1.
2.
3.
4.
5.
Nodal Analysis
Nodal Analysis with Voltage Sources
Mesh Analysis
Mesh Analysis with Current Sources
Transistors
3. Methods of Analysis
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Methods of Circuit Analysis
 Nodal Analysis
 Based on KCL.
 Mesh Analysis
 Based on KVL.
 Linear algebra is applied to solve the resulting simultaneous equations.
3. Methods of Analysis
Circuit Theory; Jieh-Tsorng Wu
3
Nodal Analysis
 Circuit variables are node voltages.
 KVL is automatically satisfied.
 Steps to analyze an n-node network
 Select a reference node (as ground), 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.
Earth ground
3. Methods of Analysis
Chassis ground
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Nodal Analysis Example
Node 1 KCL:
I1  I 2  i1  i2
(1)
Node 2 KCL:
I 2  i2  i3
(2)
Ohm's law:
v 0
i1  1
R1
i2 
v1  v2
R2
or i2  G2  v1  v2 
i3 
v2  0
R3
or i3  G3v2
(1)  I1  I 2  G1v1  G2  v1  v2 
(3)
(2)  I 2  G2  v1  v2   G3v2
(4)
G  G2
 1
 G2
3. Methods of Analysis
or i1  G1v1
G2   v1   I1  I 2 

G2  G3  v2   I 2 
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Circuit Theory; Jieh-Tsorng Wu
Matrix Equation
 a b   v1   I1 

Av  I
 c d  v    I 

 2  2
I
 v1 
1  1 


A
v  A 1I
v 
I 
 2
 2
a b 
1  d b 
1


A
A

ad  bc  c a 
c d 
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Nodal Analysis Example
v1  v3 v1  v2

 3v1  2v2  v3  12
4
2
v v v
v v
ix  i2  i3  1 2  2 3  2   4v1  7v2  v3  0
2
8
4
v v v v
v v
i1  i2  2ix  1 3  2 3  2  1 2  2v1  3v2  v3  0
4
8
2
3  i1  ix
 3
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Nodal Analysis with Voltage Sources
If a voltage source is connected between a non-reference
node and the reference node (or ground).
 The node voltage is defined by the voltage source.
 Number of variables is reduced.
If a voltage source is connected between two
non-reference nodes.
 The two nodes form a supernode.
 Use one voltage variable for both nodes.
The voltage difference between these two
nodes is known.
 Apply KCL to the supernode.
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Supernode
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Super Node Example 1
v1  10
v2  v3  5
i1  i4  i2  i3

3. Methods of Analysis
v1  v2 v1  v3 v2 v3

 
2
4
8 6
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Super Node Example 2
v2  v1  2
2  i1  i2  7  2 
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v1 v2
  7  3v1  22
2 4
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Super Node Example 3
v1  v2  20
v3  v4  3vx  v4  3(v1  v4 )  3v1  2v4
v3  v2 v1  v4 v1


6
3
2
v1  v4 v3  v2 v4 v3

 
3
6
1 4
10  i3  i1  i2
 10 
i1  i3  i4  i5

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Mesh
A mesh is a loop which does not contain any other loops within it.
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Mesh Analysis
 Circuit variables are mesh current.
 KCL is automatically satisfied.
 Steps to analyze an n-mesh network
 Assign mesh currents i1, i2, …, in to the n meshes.
 Use Ohm’s law to express voltages of resistors..
 Apply KVL to each of the n meshes.
 Solve the resulting equations.
 Mesh analysis is only applicable to a circuit that is planar. A planar circuit is one
that can be drawn in a plane with no branches crossing on another.
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Planar and Nonplanar Circuits
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Mesh Analysis Example
V1  R1 I1  R3 I 3  0   V1  R1i1  R3 (i1  i2 )  0  ( R1  R3 )i1  R3i2  V1
 R3 I 3  R2 I 2  V2  0   R3 (i1  i2 )  R2i2  V2  0   R3i1  ( R2  R3 )i2  V2
 R1  R3
 R
3

 R3   i1   V1 

R2  R3  i2   V2 
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Mesh Analysis Example
24  10(i1  i2 )  12(i1  i3 )  0
 11i1  5i2  6i3  12
10(i2  i1 )  24i2  4(i2  i3 )  0
  5i1  19i2  2i3  0
12(i3  i1 )  4(i3  i2 )  4 I o  0 I o  i1  i2
  i1  i2  2i3  0
 11 5 6   i1  12 
 5 19 2  i    0 

 2  
 1 1 2   i3   0 
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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.
If a current source exists between two meshes.
 The two nodes form a supermesh.
 Use one current variable for both meshes. The current difference between these
two meshes is known.
 Apply KVL to the supermesh.
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Mesh Analysis with Current Source Example
i1  i2
i2  5
10  4i1  6(i1  i2 )  0  i1  2 A
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Supermesh Example
i2  i1  5
 i2  i1  5
i2  i3  3I o
I o  i4
 i3  i1  3i4  5
2i1  4i3  8(i3  i4 )  6i2  0
8(i4  i3 )  2i4  10  0
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Supermesh Example
i1  i2  3
 i1  i2  3
6  2(i1  i3 )  4(i2  i3 )  8i2  0  7i2  3i3  0
2i3  4(i3  i2 )  2(i3  i1 )  0   3i2  4i3  3
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Nodal Analysis or Mesh Analysis
Select the method that results in the smaller number of equations.
 Nodal Analysis
 More parallel-connected elements, voltage sources, or supernodes.
 Nnode < Nmesh
 If node voltages are required.
 Mesh Analysis
 More series-connected elements, current sources, or supermeshes.
 Nmesh < Nnode
 If branch currents are required.
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Transistors
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An npn Bipolar Junction Transistor (BJT)
VCE  VCB  VBE
(KVL)
I E  I B  IC
(KCL)
Transistor in active mode
VBE  0.7 V VCB  0
IC   I B
I E  1    I B
IC   I E


1
  100 0    1
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Amplifier Example 1
4  200k  I B  0.7  0

IB 
4  0.7
 165  A
200k
IC    I B  50  I B  8.25 mA
6  100 I C  vo  0
3. Methods of Analysis

vo  6  100 I C  5.175 V
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Amplifier Example 2
V1
V 2
 1
 0 V1  0.7

I B  9.5  A
200k 100k
v o  16  1k  I C  16  1k    I B  16  1k 150  9.5  14.575 V
IB 
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