No great work is ever done in a
hurry. To develop a great scientific
discovery, to print a great picture,
to write an immortal poem, to
Methods of become a minister, or a famous
Analysis general to do anything great
requires time, patience, and
perseverance. These things are
Chapter 3
done by degrees, “little by little.”
Overview
Two techniques will be presented:
✓ Nodal analysis based on Kirchhoff’s current law (KCL).
✓ Mesh analysis based on Kirchhoff’s voltage law (KVL).
Any linear circuit can be analyzed using these two techniques.
The analysis will result in a set of simultaneous equations
which may be solved by Cramer’s rule or computationally
© McGraw Hill 2
Nodal Analysis
• Given a circuit with n nodes, without voltage sources, the nodal
analysis is accomplished via three steps:
1. Select a node as the reference node (datum node). Assign voltages
v1,v2,…vn to the remaining n-1 nodes, voltages are relative to the
reference node.
2. Apply KCL to each non-reference node. Use Ohm’s law to express the
branch currents in terms of node voltages.
3. Solve the resulting n − 1 simultaneous equations to obtain the unknown
node voltages.
• The reference node is commonly referred to as the ground since its
voltage is by default zero.
© McGraw Hill 3
Applying Nodal Analysis
• This circuit has a ground node. We will
use it as the reference node (node 0).
• The remaining two nodes 1 and 2 are
assigned v1 and v2.
• Now apply KCL to each node.
At node 1,
I1 = I 2 + i1 + i2
At node 2,
I 2 + i2 = i3
© McGraw Hill 4
Apply Nodal Analysis
• Use Ohm’s law to express i1, i2, and i3 in terms
of node voltages.
• Keep in mind that current flows from high
potential to low.
• From this we get: v1 v1 − v2
I1 = I 2 + +
v1 − 0 R1 R2
i1 = or i1 = G1v1 Substituting
R1
back into the v1 − v2 v2
node I2 + =
v1 − v2
i2 = or i2 = G2 ( v1 − v2 ) R2 R3
R2 ⎯⎯⎯⎯⎯
→
equations
or
v2 − 0
i3 = or i3 = G3v2 I1 = I 2 + G1v1 + G2 ( v1 − v2 )
R3
I 2 + G2 ( v1 − v2 ) = G3v2
• The last step is to solve the equations.
© McGraw Hill 5
Apply Nodal Analysis
© McGraw Hill 6
Calculate the node voltages.
At node 1,
…(1)
At node 2,
…(2)
Use the elimination technique.
(1)+(2)
From (1)
© McGraw Hill 7
Calculate the node voltages.
Use Cramer’s rule
Negative i2 means that the current flows in the
opposite direction.
© McGraw Hill 8
Determine the voltages at the nodes.
At node 1,
…(1)
At node 2,
…(2)
At node 3,
…(3)
Use the elimination technique.
(1)+(3)
(2)+(3)
From (3)
© McGraw Hill 9
Determine the voltages at the nodes.
Use Cramer’s rule
© McGraw Hill 10
Determine the voltages at the nodes.
© McGraw Hill 11
Including voltage sources
1. Between the reference node and a non-
reference mode:
✓ Set the voltage at the non-reference
node to the voltage of the source.
✓ In the example circuit v1 = 10V.
2. Between two non-reference nodes.
✓ The two nodes form a supernode.
© McGraw Hill 12
Supernode
A supernode is formed by enclosing a
voltage source (dependant or independent)
connected between two non-reference
nodes and any elements connected in
parallel with it.
Why?
• Nodal analysis requires applying KCL.
Nodes 2 and 3 form
• The current through the voltage source
a supernode.
cannot be known in advance (Ohm’s law
does not apply). i1 + i4 = i2 + i3
• By lumping the nodes together, the v1 − v2 v1 − v3 v2 − 0 v3 − 0
+ = +
current balance can still be described. 2 4 8 6
© McGraw Hill 13
Analysis with a supernode
• To apply KVL to the supernode in the example, the circuit is redrawn
as shown.
• Going around this loop in the clockwise direction gives:
− v2 + 5 + v3 = 0 v2 − v3 = 5
• Note the following properties of a
supernode:
1. The voltage source inside the
supernode provides a constraint
equation for the node voltages.
2. A supernode has no voltage of its
own.
3. A supernode requires the application
of both KCL and KVL.
© McGraw Hill 14
Find the node voltages.
Applying KCL to the supernode
…(1)
Applying KVL …(2)
From (1) and (2),
© McGraw Hill 15
Find the node voltages.
At supernode 1-2,
…(1)
At supernode 3-4,
…(2)
For loop 1, …(3)
For loop 2,
…(4)
For loop 3,
…(5)
This is redundant
© McGraw Hill 16
Finding 4 voltages requires only four Eqs.
…(1’)
…(2’)
Using Cramer’s rule
© McGraw Hill 17
Mesh Analysis
Another general procedure is to use mesh currents as circuit variables.
Recall:
• A loop is a closed path with no node passed more than once.
• A mesh is a loop that does not contain any other loop within it.
Mesh analysis uses KVL to find unknown currents.
Mesh analysis is limited to planar circuits that can be drawn with no
crossing branches.
© McGraw Hill 18
Planar versus Nonpalanar
A nonplanar circuit:
The branch with 13 resistor prevents the
circuit from being drawn without crossing
branches
A planar circuit: It can be redrawn
to avoid crossing branches
© McGraw Hill 19
Mesh Analysis Steps
Mesh analysis follows these steps:
1. Assign mesh currents i1,i2,…in to n meshes.
2. Apply KVL to each mesh.
3. Solve the resulting n simultaneous equations to get the mesh currents.
© McGraw Hill 20
Mesh Analysis Example
• The circuit has two meshes.
• The outer loop (abcdefa) is a loop, but not a mesh.
• First, i1 and i2 are assigned to the two meshes.
• Applying KVL to the meshes:
−V1 + R1i1 + R3 ( i1 − i2 ) = 0 R2i2 + V2 + R3 ( i2 − i1 ) = 0
( R1 + R3 ) i1 − R3i2 = V1 − R3i1 + ( R2 + R3 ) i2 = −V2
© McGraw Hill 21
Mesh Analysis Example
© McGraw Hill 22
Find the branch currents I1, I2, and I3 using mesh analysis.
For mesh 1,
…(1)
For mesh 2,
…(2)
(2) into (1)
From (2)
© McGraw Hill 23
Use mesh analysis to find the current Io.
For mesh 1,
For mesh 2,
For mesh 3,
At node A
© McGraw Hill 24
© McGraw Hill 25
Mesh Analysis with Current Sources
• The presence of a current source makes the mesh analysis simpler in
that it reduces the number of equations.
• If the current source is located on only one mesh, the current for that
mesh is defined by the source.
• For example,
Here, i2 is equal to −5A. → i1 = -2A
© McGraw Hill 26
Supermesh
• Current sources (dependent or independent) that are shared by more
than one meshes need special treatment.
• The two meshes must be joined together, resulting in a supermesh.
• The supermesh is constructed by merging the two meshes and
excluding the shared source and any elements in series with it.
• A supermesh is required because mesh analysis uses KVL.
• But the voltage across a current source cannot be known in advance.
• Intersecting supermeshes in a circuit must be combined to form a
larger supermesh.
© McGraw Hill 27
Creating a Supermesh
• A 6A current course is shared between meshes 1 and 2.
• The supermesh is formed by merging the two meshes.
• The current source and the 2 resistor in series with it are removed.
© McGraw Hill 28
Supermesh Example
• Apply KVL to the supermesh.
− 20 + 6i1 + 10i2 + 4i2 = 0 or 6i1 + 14i2 = 20
• We next apply KCL to the node in the branch where the two meshes
intersect.
i2 = i1 + 6
• Solving these two equations we get
i1 = −3.2A i2 = 2.8A
© McGraw Hill 29
Find i1 to i4 using mesh analysis.
In the larger supermesh,
…(1)
At node P, …(2)
At node Q,
…(3)
In mesh 4,
…(4)
© McGraw Hill 30
Nodal Analysis by Inspection
• There is a faster way to construct a matrix for nodal
analysis.
• It requires that all current sources within the circuit be
independent.
• In general, for a circuit with N nonreference nodes, the
node-voltage equations may be written as:
G11 G12 G1N v1 i1
G G2 N v2 i2
21 G22 =
GN 1 GN 2 GNN vN iN
© McGraw Hill 31
Nodal Analysis by Inspection
• Each diagonal term Gjj on the conductance matrix is the sum of
conductances connected to the node indicated by the matrix index.
• The off diagonal terms Gjk are the negative of the sum of all
conductances connected between nodes j and k, j k.
• The unknown voltages are denoted as vk
• The sum of all independent current sources directly connected to node
k are denoted as ik. Current entering the node are treated as positive.
G11 G12 G1N v1 i1
G G2 N v2 i2
21 G22 =
GN 1 GN 2 GNN vN iN
© McGraw Hill 32
Mesh Analysis by Inspection
• There is a fast way to construct a matrix for mesh analysis.
• It requires that all voltage sources within the circuit be independent.
• In general, for a circuit with N meshes, the mesh-current equations
may be written as:
R11 R12 R1N i1 v1
R R22 R2 N i2 v2
21 =
RN 1 RN 2 RNN iN vN
© McGraw Hill 33
Mesh Analysis by Inspection
• Each diagonal term Rjj is the sum of resistances in the mesh indicated
by the matrix index.
• The off-diagonal terms Rjk are the negative of the sum of all
resistances in common with meshes j and k with j k.
• The unknown mesh currents in the clockwise direction are denoted as
ik.
• The sum taken clockwise of all voltage sources in mesh k are denoted
as vk. Voltage rises are treated as positive.
R11 R12 R1N i1 v1
R R22 R2 N i2 v2
21 =
RN 1 RN 2 RNN iN vN
© McGraw Hill 34
Write the node-voltage matrix equations
© McGraw Hill 35
By inspection, write the mesh-current equations.
© McGraw Hill 36
Selecting an Appropriate Approach
In principle both the nodal analysis and mesh analysis are
useful for any given circuit.
There are two factors that dictate the best choice:
• The nature of the particular network is the first factor.
• The second factor is the information required.
© McGraw Hill 37
Mesh analysis when…
• If the network contains:
✓ many series connected elements.
✓ voltage sources.
✓ supermeshes.
✓ fewer meshes than nodes.
• If branch or mesh currents are what is being solved for.
• Mesh analysis is the only suitable analysis for transistor
circuits.
• It is not appropriate for operational amplifiers because
there is no direct way to obtain the voltage across an op-
amp.
© McGraw Hill 38
Nodal analysis if…
• If the network contains:
✓ many parallel connected elements.
✓ current sources.
✓ supernodes.
✓ fewer nodes than meshes.
• If node voltages are what are being solved for.
• Non-planar circuits can only be solved using nodal
analysis.
• This format is easier to solve by computers.
© McGraw Hill 39
Transistors
© McGraw Hill 43
Application: DC transistor circuit
• In general, there are two types of transistors
commonly used: Field Effect (FET) and
Bipolar Junction (BJT).
• We will use a BJT.
• A BJT is a three terminal device, where the
input current into one terminal (the base)
affects the current flowing out of a second
terminal (the collector).
• The third terminal (the emitter) is the
common terminal for both currents.
© McGraw Hill 44
KCL and KVL for a BJT
• The currents from each terminal can be
related to each other as follows:
I E = I B + IC
• The base and collector current
can be related to each other
by the parameter , which can
range from 50-1000.
I C = I B
• Applying KVL to the BJT gives
VCE + VEB + VBC = 0
𝑉𝐶𝐸 = 𝑉𝐵𝐸 + 𝑉𝐶𝐵
© McGraw Hill 45
KCL and KVL for a BJT
© McGraw Hill 46
DC model of a BJT
• A transistor has a few operating modes depending on the applied
voltages/currents. In this problem, we are interested in the operation in
“active mode.”
• This is the mode used for amplifying signals.
• The equivalent DC model for a BJT in active mode.
Note that nodal analysis can only be applied to the BJT after using this model
© McGraw Hill 47
Find IB, IC, and vo.
For the input loop, KVL gives
Since VBE = 0.7 V in the active mode,
For the output loop, KVL gives
© McGraw Hill 48
Setting up a BJT circuit
Three approaches to solving a transistor circuit. Note when the equivalent
model is used and when it is not.
Mesh analysis
Original circuit
Nodal analysis PSpice analysis
© McGraw Hill 49
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