Circuit Theory I
Methods of Analysis
Assistant Professor Suna BOLAT
Eastern Mediterranean University
Department of Electric and Electronic Eng.
Ref2: Anant Agarwaland Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007.
1
MIT OpenCourseWare(http://ocw.mit.edu/), Massachusetts Institute of Technology
The road so far...
• Method 1: Basic KVL, KCL method of Circuit analysis
• Method 2: Apply element combination rules
• Method 3: Nodal & Mesh analysis
– Particular application of KVL, KCL
2
Nodal & Mesh analysis
• Select reference node (ground) from which voltages are
measured.
• Label voltages of remaining nodes with respect to ground.
• These are the primary unknowns.
• Write KCL for all but the ground node, substituting device laws
and KVL.
• Solve for node voltages.
• Back solve for branch voltages and currents (i.e., the secondary
unknowns)
3
Nodal analysis
• Steps to Determine Node Voltages:
1. Select a node as the reference node. Assign voltage v1,
v2, …vn-1 to the remaining n-1 nodes. The voltages are
referenced with respect to the reference node.
2. Apply KCL to each of the n-1 non-reference nodes. Use
Ohm’s law to express the branch currents in terms of node
voltages.
3. Solve the resulting simultaneous equations to obtain the
unknown node voltages.
4
Nodal analysis
• Steps to Determine Node Voltages:
reference node
(a) common ground, (b) ground, (c) chassis.
5
Nodal Analysis
• Typical circuit for nodal analysis
voltages v1 and v2 are assigned with respect to the reference node
(i.e ground).
Reference Node
6
 Apply KCL to each nonreference node.
I1  I 2  i1  i2
I 2  i2  i3
vhigher  vlower
i
R
v1  0
i1 
or i1  G1v1
R1
v1  v2
i2 
or i2  G2 (v1  v2 )
R2
v2  0
i3 
or i3  G3v2
R3
7
• Substituting element relations into KCL
v1 v1  v2
 I1  I 2  
R1
R2
v1  v2 v2
I2 

R2
R3
 I1  I 2  G1v1  G2 (v1  v2 )
I 2  G2 (v1  v2 )  G3v2
G1  G2


  G2
 G2   v1   I1  I 2 

G2  G3  v2   I 2 
8
Example 3.1
• Calculate the node voltages in the circuit shown below.
9
Example
• Calculate the node voltages in the circuit shown
below.
10
Example 3.2
Determine the voltages at nodes 1, 2 and 3.
11
Reminder: Cramer’s rule
12
Nodal Analysis with Voltage Sources
• Case 1: The voltage source is connected between a
nonreference node and the reference node: The
nonreference node voltage is equal to the magnitude of
voltage source and the number of unknown nonreference
nodes is reduced by one.
• Case 2: The voltage source is connected between two
nonreference nodes: a generalized node (supernode) is
formed.
13
• A circuit with a supernode.
14
Supernode
• A supernode is formed by enclosing a (dependent or
independent) voltage source connected between two
nonreference nodes and any elements connected in
parallel with it.
• The required two equations for regulating the two
nonreference node voltages are obtained by the KCL of the
supernode and the relationship of node voltages due to
the voltage source.
15
Example
• Find the node voltages of the circuit below.
2  7  i1  i2  0
v1 v2
v1 v2
2  7    0    5  2v1  v2  20
2 4
2 4
v1  v2  2
2v1  v2  20
v1  v2  2
i1
i2
22
=  7.33V
3
22
16
 v2  
+2=  =  5.33V
3
3
3v1  22  v1  
16
Example
• Find the node voltages of the circuit below.
17
Example (continued...)
• Apply KCL to the two supernodes.
At supernode 1-2:
18
Example (continued...)
• Apply KCL to the two supernodes.
At supernode 3-4:
19
Example (continued...)
• Apply KVL around the loops:
Loop1:
Loop2:
Loop3:
20
Example (continued...)
• You can use Matlab to solve large matrix equations:
5 1  1  2   v1  60
4 2  5  16 v   0 

 2    
1  1 0
0  v3  20

   
3
0

1

2

 v4   0 
v1  26.67 V
v2  6.67 V
v3  173.33 V
v4  46.67 V
We now use MATLAB to solve the matrix Equation. The
Equation on the left can be written as
AV =B →
V= B/A= A-1B
>> A=[5 1 -1 -2
4 2 -5 -16
1 -1 0 0
3 0 -1 -2];
>> B= [60 0 20 0]';
>> V=inv(A)*B
Matlab Code
V=
26.6667
6.6667
173.3333
-46.6667
21
Mesh Analysis
1. Mesh analysis: another procedure for analyzing circuits,
applicable to planar circuits.
2. A Mesh is a loop which does not contain any other loops
within it.
3. Nodal analysis applies KCL to find voltages in a given circuit,
while Mesh Analysis applies KVL to calculate unknown
currents.
22
Mesh Analysis
•
•
A circuit is planar if it can be drawn on a plane with no branches crossing
one another. Otherwise it is nonplanar.
The circuit in (a) is planar, because the same circuit that is redrawn(b) has
no crossing branches.
23
Mesh Analysis
•
A nonplanar circuit.
24
Mesh Analysis
• Steps to Determine Mesh Currents:
1. Assign mesh currents i1, i2, .., in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to
express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get
the mesh currents.
25
Mesh Analysis
• A circuit with two meshes.
26
Mesh Analysis
• A circuit with two meshes.
• Apply KVL to each mesh. For mesh 1,
 V1  R1i1  R3 (i1  i2 )  0
( R1  R3 )i1  R3i2  V1
• For mesh 2, R i  V  R (i  i )  0
2 2
2
3 2
1
 R3i1  ( R2  R3 )i2  V2
27
Mesh Analysis
• Solve for the mesh currents.
 R1  R3
  R3
 R3   i1   V1 

R2  R3  i2   V2 
• Use i for a mesh current and I for a branch current.
It’s evident from the circuit that:
I1  i1 , I 2  i2 ,
I 3  i1  i2
28
Example 3.5
• A circuit Find the branch currents I1, I2, and I3 using mesh
analysis.
29
Example 3.5
• For mesh 1,
 15  5i1  10(i1  i2 )  10  0
3i1  2i2  1
• For mesh 2,
6i2  4i2  10(i2  i1 )  10  0
i1  2i2  1
• We can find i1 and i2 by substitution method or Cramer’s
rule. Then,
I1  i1 ,
I 2  i2 ,
I 3  i1  i2
30
Example 3.6
• Use mesh analysis to find the current Io in the circuit below.
31
Example 3.6
• Apply KVL to each mesh. For mesh 1,
 24  10(i1  i2 )  12(i1  i3 )  0
11i1  5i2  6i3  12
• For mesh 2,
24i2  4(i2  i3 )  10(i2  i1 )  0
 5i1  19i2  2i3  0
32
Example 3.6
• For mesh 3,
4 I 0  12(i3  i1 )  4(i3  i2 )  0
At node A, I 0  I1  i2 ,
4(i1  i2 )  12(i3  i1 )  4(i3  i2 )  0
 i1  i2  2i3  0
• In matrix from:
 11  5  6  i1  12
 5 19  2 i2    0 
  1  1 2  i   0 

 3   
• we can calculate i1, i2 and i3 by Cramer’s rule, and find Io.
33
Mesh analysis with current source
• Consider the following circuit with a current source.
34
Mesh analysis with current source
• Possibe cases of having a current source.
 Case 1
• Current source exist only in one mesh
mesh 2:
mesh 1:
• One mesh variable is reduced
 Case 2
• Current source exists between two meshes, a super-mesh is
obtained.
35
Mesh analysis with current source
• Possibe cases of having a current source.
• A supermesh is considerred when two meshes have a
(dependent/independent) current source in common.
36
Mesh analysis with current source
• Applying KVL in the supermesh below:
•
Apply KCL at node 0 above:
•
KVL in the supermesh above:
37
Mesh analysis with current source
• Properties of a Supermesh
1. The current is not completely ignored
•
provides the constraint equation necessary to solve for the
mesh current.
2. A supermesh has no current of its own.
3. Several current sources in adjacency form a bigger
supermesh.
4. A supermesh requires the application of both KVL and
KCL.
38
Mesh analysis with current source
•
•
Apply KVL in the supermesh (mesh 1 + mesh 2 + mesh 3):
•
Apply KCL at node P:
•
Apply KCL at node Q:
•
Apply KVL in mesh 4:
4 equations for 4 variables. Using Cramer’s Rule
1
0

 1

0
 4  i1   0 
0  4 5  i2   5

1 0
0  i3   5 
   
1  1 3  i4   0 
3
6
39
Nodal & Mesh analysis
• The analysis equations can be obtained by direct
inspection
a) For circuits with only resistors and
independent current sources. (nodal
analysis)
b) For planar circuits with only resistors
and independent voltage sources.
(mesh analysis)
40
a) For circuits with only resistors and independent
current sources. (nodal analysis).
•
Consider the following example:
•
The circuit has two nonreference nodes and the node
equations:
•
In Matrix form:
41
•
In general, if a circuit with independent current sources has N
nonreference nodes, the node-voltage equations can be
written in terms of conductances as:
•
G is called the conductance matrix, v is the output vector, and i is
the input vector.
42
b) For planar circuits with only resistors and independent
voltage sources. (mesh analysis)
•
Consider the following example:
•
The circuit has two meshes and the mesh
equations :
•
In Matrix form:
43
•
In general, if a circuit has N meshes, mesh-current equations
can be expressed in terms of resistances as:
•
R is called the resistance matrix, i is the output vector, and v is the
input vector.
44
Example 3.8
Write the node-voltage matrix equations in the following
circuit.
45
Example 3.8
•
The circuit has 4 nonreference nodes, so the diagonal terms of G are:
1 1
1 1 1
G11    0.3, G22     1.325
5 10
5 8 1
1 1 1
1 1 1
G33     0.5, G44     1.625
8 8 4
8 2 1
•
The off-diagonal terms of G are:
1
G12    0.2, G13  G14  0
5
1
1
G21  0.2, G23    0.125, G24    1
8
1
G31  0, G32  0.125, G34  0.125
G41  0, G42  1, G43  0.125
46
Example 3.8
•
The input current vector i in amperes:
i1  3, i2  1  2  3, i3  0, i4  2  4  6
•
The node-voltage equations can be calculated by:
0
0
 0.3  0.2
 v1   3
 0.2 1.325  0.125  1
 v    3 

 2    
 0.125  v3   0
 0  0.125 0.5
 0 1
 0.125
1.625  v4   6

47
Example 3.9
Write the mesh-current equations in in the following circuit.
48
Example 3.9
•
The input voltage vector v in volts :
v1  4, v2  10  4  6,
v3  12  6  6, v4  0, v5  6
•
The mesh-current equations are:
 9
 2

 2
 0

 0
2 2
0
10  4  1
4
9
0
1
0
8
1
0 3
0  i1
 1  i2

0  i3
 3  i4

4  i5

 4
 6
    6
 0
   6
  
49
Nodal vs mesh analysis
• Both nodal and mesh analyses provide a systematic way of
analyzing a complex network.
• The choice of the better method is dictated by two factors.
 First factor: nature of the particular network.
 The key is to select the method that results in the
smaller number of equations.
 Second factor: information required.
50
Summary
• Both nodal and mesh analysis provide a systematic way of
analyzing a complex network.
• The choice of the better method is dictated by two factors.
1.
Nodal analysis: Apply KCL at the nonreference nodes.
(The circuit with fewer node equations)
2. A supernode: Voltage source between two nonreference
nodes.
3.
Mesh analysis: Apply KVL for each mesh.
(The circuit with fewer mesh equations)
4. A supermesh: Current source between two meshes.
51
Application: DC Transistor Circuits: BJT Circuit Models
(a)An npn transistor,
(b) dc equivalent model.
52
Example 3.13
For the BJT circuit in the figure =150 and VBE = 0.7 V. Find v0.
53
Example 3.13
• Use mesh analysis
or nodal analysis
54