Methods of Analysis

Instructor: Chia-Ming TsaiElectronics Engineering

National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.

Contents• Introduction

• Nodal Analysis

• Nodal Analysis with Voltage Sources

• Mesh Analysis

• Mesh Analysis with Current Sources

• Nodal Analysis vs. Mesh Analysis

• Applications

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

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

Symbols for Reference Node (Ground)

Used in this course

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

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

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

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

Example 1

Example 2

What is a mesh?• A mesh is a loop that does not contain any

other loop within it.

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

Continued• Applicable only for planar circuits• An example for nonplanar circuits is shown be

low

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

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

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

Example 1

A 2

06410

1,mesh for KVL Applying

A 5

1

21

2

i

iii

i

21 ii

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

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

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

Applications: Transistors

• Bipolar Junction Transistors (BJTs)• Field-Effect Transistors (FETs)

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

DC Equivalent Model of BJT

Example of Amplifier Circuit