2012/9/17

1

Methods of Analysis•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 resultingsimultaneous equations.– Ax = B or x = A-1B.

2012/9/17

2

Nodal Analysis•Circuit variables = 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.–UseOhm’s lawto express currents of resistors.–Apply KCL to each of the n-1 nodes.–Solve the resulting equations.

Earth ground Chassis ground

Case Study

2

21

2

1

322

221

232122

2121121

2333

23

21222

212

1111

11

322

2121

(4))2(

(3))1(

or0

or

or0

giveslawsOhm'Applying*

(2)givesKCLapplying2,nodeAt*

(1)givesKCLapplying1,nodeAt*

III

vv

GGGGGG

vGvvGI

vvGvGII

vGiR

vi

vvGiR

vvi

vGiR

vi

iiI

iiII

Assign vn

2012/9/17

3

Nodal Analysis with Voltage Sources

•If a voltage source is connectedbetween a nonreference node andthe reference node (or ground).–The node voltage is defined by the

voltage source.–Number of variables is reduced.–Simplified analysis.

Cont’d•If a voltage source is connected between two

nonreference nodes.–IS is difficult to define.–It’s difficult to solve the problem by using KCL.

V

I

VS

I-V curve of a voltage source

IS

- S

S

I

VV

2012/9/17

4

Cont’d•Analysis Strategy–The two nodes form a supernode (a closed boundary).–Eq. 1: Apply KCL to the supernode.–Eq. 2: Apply KVL to derive the relationship between the two

nodes.

Supernode

Case Study with Supernode

equations.3bysolvedvariables3

(3)5supernode,thetoKVLApplying

(2)6

08

042

supernode,thetoKCLApplying(1)V10

32

32

3121

3241

1

vv

vv

vvvv

iiii

v

2012/9/17

5

Example 1Supernode

212 vv72 21 ii

Example 2

Supernode Supernode

2021 vv xvvv 343

213 iii 5431 iiii

2012/9/17

6

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

other loop within it.

Mesh Analysis•Circuit variables = mesh currents–KCL must be satisfied. ( How ??? )

•Steps to analyze an n-mesh network–Assign mesh currents i1, i2,…, in.–UseOhm’s lawto express voltages of resistors.–Apply KVL to each of the n meshes.–Solve the resulting equations.

2012/9/17

7

Mesh Analysis•Applicable only for planar circuits.•An example for nonplanar circuits is shown

below.

Case Study

223213

123222

123131

213111

0givesKVLapplying2,meshFor

0givesKVLapplying1,meshFor

ViRRiR

iiRViR

ViRiRR

iiRiRV

2

1

2

1

323

331

VV

ii

RRRRRR

2012/9/17

8

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.

Cont’d•If a current source exists between two meshes.–VS is difficult to define.–It’s difficult to solve the problem by using KVL for each

mesh.

V

IIS

I-V curve of a current source

- S

S

V

II+VS_

2012/9/17

9

Cont’d•Analysis Strategy–A supermesh is resulted.–Eq. 1: Apply KVL to the supermesh.–Eq. 2: Apply KCL to derive the relationship between the

two mesh currents.

ExcludedSIii 12:KCL

Supermesh

Example 1

A2

064101,meshforKVLApplying

A5

1

21

2

iiii

i

21 ii

2012/9/17

10

Example 2

201460410620

supermesh,thetoKVLApplying

21

221

iiiii

A8.2A,2.36

0,nodetoKCLApplying

21

12

iiii

Supermesh

Example 3

•Applying KVL to the supermesh•Applying KCL to node P•Applying KCL to node Q•Applying KVL to mesh 4

4 variables solvedby 4 equations

Supermesh

=i2-5

=i2-3Io

2012/9/17

11

How to choose?•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.

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

2012/9/17

12

Bipolar Junction Transistors (BJTs)

1

1)(01

100)~(

V0.7

(KVL)0

(KCL)

EC

BE

BC

BE

BCEBCE

CBE

IIII

II

V

VVV

III

•Current-controlled devices

DC Equivalent Model of BJT

2012/9/17

13

Example of Amplifier Circuit

BC II