Lecture 4 - Methods of Circuit Analysis

Electric Circuits (Fall 2015) Pingqiang Zhou

Lecture 4

- Methods of Circuit Analysis

10/8/2015

Reading: Chapter 3

Lecture 4 1


Outline

• With Ohm’s and Kirchhoff’s law established, they may

now be applied to circuit analysis.

• Two techniques will be presented in this lecture:

Nodal analysis, which is based on KCL

Mesh analysis, which is based on 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

(using MATLAB for example)

Computational circuit analysis using PSpice will also be

introduced here.

Lecture 4 2


Nodal Analysis

• Instead of focusing on the voltages of the circuit elements, one

may look at the voltages at the nodes of the circuit.

• Given a circuit with n nodes, the nodal analysis is accomplished

via three steps:

1. Select a node as the reference (i.e., ground) node. Define the

node voltages (except reference node and the ones set by the

voltage sources). Voltages are relative to the reference node.

2. Apply KCL at nodes with unknown voltage, expressing current in

terms of the node voltages (using the I-V relationships of branch

elements).

Special cases: floating voltage sources.

3. Solve the resulting simultaneous equations to obtain the unknown

node voltages.

Lecture 4 3


Apply Nodal Analysis

• Let’s apply nodal analysis to this circuit to see

how it works.

• This circuit has a node that is designed as

ground. We will use that as the reference node

(node 0). The remaining two nodes are

designed 1 and 2 and assigned voltages v1

and v2.

• Now apply KCL to each node:

At node 1

At node 2

2121 iiII

322 iiI

Lecture 4 4

Three variables, but only two equations.


Apply Nodal Analysis II

• We can now use Ohm’s law to express the

unknown currents i1, i2, and i3 in terms of node

voltages 𝑣1, 𝑣2.

• In doing so, keep in mind that current flows

from high potential to low. From this we get:

• The last step is to solve the system of

equations

233

3

23

2122

2

212

111

1

11

or0

or

or0

vGiR

vi

vvGiR

vvi

vGiR

vi

Substituting

back into the

node

equations3

2

2

212

2

21

1

121

R

v

R

vvI

R

vv

R

vII

232122

2121121

vGvvGI

vvGvGII

or

Lecture 4 5

2121 iiII

322 iiI

Why not keep using branch currents as variables?


Nodal Analysis: Exercise #1

6Lecture 4

R4V1 R2

+

- IS

R3R1


V2V

1R

2

R1

R4

R5

R3 I1

Va


Challenges: • Determine number of nodes needed

• Deal with different types of sources

Lecture 4 7


Nodal Analysis w/ “Floating Voltage Source”

8Lecture 4

A “floating” voltage source is one for which neither side is

connected to the reference node, e.g. VLL in the circuit below:

Problem: We cannot write KCL at nodes a or b because there is no

way to express the current through the voltage source in terms of Va-Vb.

Solution: Define a “supernode” – that chunk of the circuit containing

nodes a and b. Express KCL for this supernode. Incorporate voltage

source constraint into KCL equation.

R4R 2 I2

Va Vb

+-

VLL

I1



9Lecture 4

Eq’n 1: KCL at supernode

Substitute property of voltage source:

Supernode

R4R2 I2

Va Vb

+-

VLL

I1


Mesh Analysis

• Another general procedure for analyzing circuits is to use the

mesh currents as the 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 in one aspect: It can only apply to

circuits that is planar.▪ A planar circuit can be drawn such that there are no crossing

branches.Lecture 4 10

Mesh = Independent loop?


Example

• A loop is independent if it contains at least one branch

which is not a part of any other independent loop.

• A mesh is a loop that does not contain any other loop

within it

Lecture 4 11


Planar vs Nonpalanar

A nonplanar circuit: The branch with

the 13Ω resistor prevents the circuit

from being drawn without crossing

branches

A planar circuit: It can be redrawn

to avoid crossing branches

Lecture 4 12


Mesh Analysis Steps

• Mesh analysis follows these steps:

1. Assign mesh currents i1,i2,…in to the n meshes

2. Apply KVL to each of the n mesh currents.

3. Solve the resulting n simultaneous equations to get the

mesh currents

Lecture 4 13


Mesh Analysis Example

• The above circuit has two paths that are meshes (abefa and bcdeb)

The outer loop (abcdefa) is a loop, but not a mesh

• First, mesh currents i1 and i2 are assigned to the two meshes.

• Applying KVL to the meshes:

223213

123222

123131

213111 00

ViRRiR

iiRViR

ViRiRR

iiRiRV

Lecture 4 14


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, the current i2 is equal to -5A.

Lecture 4 15


Supermesh

• In this example, a 6A current source is shared between mesh 1

and 2.

• A supermesh is required because mesh analysis uses KVL, but

the voltage across a current source cannot be known in

advance.

The supermesh is formed by merging the two meshes.

The current source and the 2Ω resistor in series with it are removed.

Lecture 4 16


Supermesh

• Apply KVL to the supermesh

• We next apply KCL to the node in the branch where the two meshes

intersect.

• Note that the supermesh required using both KVL and KCL.

20146or0410620 21221 iiiii

612 ii

Lecture 4 17


Nodal Analysis by Inspection

• A faster way to construct a matrix for solving a circuit by

nodal analysis

It requires that all current sources within the circuit be

independent

• The equations for the example

𝐺1 + 𝐺2 −𝐺2−𝐺2 𝐺2 + 𝐺3

𝑣1𝑣2

=𝐼1 − 𝐼2𝐼2

Lecture 4 18


Exercise

• By inspection, obtain the node-voltage equations for the

circuit shown below

19Lecture 4


Mesh Analysis by Inspection

• There is a similarly fast way to construct a matrix for

solving a circuit by mesh analysis It requires that all voltage sources within the circuit be independent

𝑅1 + 𝑅3 −𝑅3−𝑅3 𝑅2 + 𝑅3

𝑖1𝑖2

=𝑉1−𝑉2

Lecture 4 20


Exercise

• By inspection, obtain the mesh-current equations for the

circuit shown below

21Lecture 4


Nodal vs. Mesh

• In principle both the nodal analysis and mesh analysis are

useful for any given circuit.

• What then determines if one is going to be more efficient

for solving a circuit problem?

• There are two factors that dictate the best choice:

The nature of the particular network

The information required

Lecture 4 22


Nodal Analysis if…

• If the network contains:

Many parallel connected elements

Current sources

Supernodes

Circuits with 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 computer

Easy to program

Lecture 4 23


Mesh Analysis when…

• If the network contains:

Many series connected elements

Voltage sources

Supermeshes

A circuit with 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.

Lecture 4 24


Example• Find 𝑣0 in the circuit,

which has

4 nodes

3 meshes

2 dependent sources

25

• Nodal

3 node-voltage equations

2 constraint equations

• Mesh

1 supermesh

4 constraint equations

Lecture 4


Circuit Analysis with PSpice

• PSpice is a common program used for circuit analysis.

• It is capable of determining all of the branch voltages and

currents if the numerical values for all circuit components

are known.

• Analysis using PSpice begins with drawing a schematic

view of the circuit.

Lecture 4 26


Application: DC Transistor Circuit

• Here we will use the approaches learned in

this lecture to analyze a transistor circuit.

• In general, there are two types of transistors

commonly used: Field Effect (FET) and

Bipolar Junction (BJT). This problem 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.

Lecture 4 27

https://en.wikipedia.org/wiki/Field-effect_transistor

https://en.wikipedia.org/wiki/Bipolar_junction_transistor


KCL and KVL for a BJT

• The currents from each terminal can

be related to each other as follows:

• The base and collector current can

be related to each other by the

parameter , which can range from

50-1000

• Applying KVL to the BJT gives:

CBE III

0 BCEBCE VVV

BC II

Lecture 4 28


DC model of a BJT

• A transistor has a few operating modes depending on the

applied voltages/currents. In this problem, we will be

interested in the operation in “active mode”.

This is the mode used for amplifying signals.

• The figure below shows 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.

Lecture 4 29


Setting up a BJT circuit

Below are three approaches to solving a transistor circuit. Note when

the equivalent model is used and when it is not.

Original circuit Mesh analysis

Nodal analysis PSpice analysis

Lecture 4 30

𝛽 = 150

𝑣𝐵𝐸 = 0.7𝑉


Summary

• Node Analysis Node voltage is the unknown

Solve by KCL

Special case: Floating voltage source using supernode

• Mesh Analysis: OPTIONAL Loop current is the unknown

Solve by KVL

Special case: Current source using supermesh

Lecture 4 31

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Lecture 4 - Methods of Circuit Analysis

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