Basic Analysis Method Circuit

8/22/2019 Basic Analysis Method Circuit

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Ch2Basic Analysis Methods to Circuits

2.1Equivalent Circuits2.2 Basic Nodal and Mesh Analysis2.3 Useful Circuit Analysis Techniques

References: Hayt-Ch3, 4; Gao-Ch2;

Circuits and Analog Electronics


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Ch2 Basic Analysis Methods to Circuits

2.1Equivalent Circuits

Key Words:

Equivalent Circuits NetworkEquivalent Resistance,

Equivalent Independent Sources


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Equivalent Circuits Network

Two-terminal Circuits Network

b

N1

I a

+

_

V

N2

I a

b

+

V

a

b

c d

6 5

15 5



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How do we findI1 andI2?

R1 R2 V

+

-

I1 I2I

I1 +I2 =I

21

2

1

1RR

RI

R

VI

21

1

2

2RR

RI

R

VI

2121

11

RRV

R

V

R

VI

21

21

21

11

1

RR

RRI

RR

IV

21

21

RR

RRReq

Equivalent Resistance

21

21

RR

RR

V

R

VI

eq



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i(t)

+

-

v(t)

i(t)

+

-

v(t)Req

Req is equivalent to the resistor network on the left in the sense that theyhave the same i-v characteristics.



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Method 2

IV

RR abo

a

b

V

I

source

-free

sc

oc

o

I

VR

source Voc

Method 3

Isc

source

Series and parallel Resistance

n

Kkeqs

RR1

n

k keqp 1 R

1

R

1

n

kkGG 1

Method 1 (source-free)

Condition : without knowing V&I. We only know Rs

Condition : without knowing Rs. We only know V&I



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In practice , Vs-source voltage VL- Local

L s LV V I R

Let 10V; 0.1sV R

TheIL-VL curve:

0 10

100

VL(V)

IL(A)

, Open Circuit (OC) .0,L L

V R

100ASCI 0, 0L LV R , Short Circuit (SC) .

10VOC

V

0.1OCOSC

VR

I



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P2.1

Method 1 321 //RRRRR abo

321

321

RRR

RRR

Method 2

V

R1

R2 R3

a

b

I

V

RRR

RRR

RR

V

R

VI

321

321

213

321

321

RRR

RRR

I

VRR

abo

VS

R1

R2 R3

a

b

+

-

sc

oc

o

I

VR

How do we findRab?

SS

S

oCV

RRR

RRVR

RRR

VV

321

213

321

Method 3

3/sc sI V R



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Source Transformation

Ideally:

An ideal current source has the voltage necessary to provide its

rated current

An ideal voltage source supplies the current necessary to provide

its rated voltage

Practice:

A real voltage source cannot supply arbitrarily large amounts of

current A real current source cannot have an arbitrarily large terminal

voltage



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Source Transformation

Vs

+

-

Rs

Is Rs

sss IRV

s

s

s

R

VI

Note: Consistency between the current source ref. direction and the

voltage Source ref. terminals.



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How do we findI1 andI2?

Equivalent Source

Is2 VR1 R2

+

-

I1 I2

Is1

21

2121

RR

RRIIV

ss

Ieq

21 1 21 1 2

s s

RVI I I

R R R

12 1 22 1 2

s s

RVI I I

R R R

1 2 1 2 s sI I I I

1 2

1 2 1 2

1 1s s

V VI I V

R R R R



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Equivalent Source

Series Voltage Source

+ + +- - -VS1 VS2 VSn

+ -VS

n

k

SkS VV1

n

k

SkS RR1

parallel Current Source

IS1 IS2 ISn

IS

I

n

k

SkS II1

RS=RS1//RS2////RSn



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2.2 Basic Nodal and Mesh Analysis

Key Words:

Branch Analysis,Nodal Analysis,

Mesh (Loop) Analysis

Why?

The analysis techniques previously (voltage divider, equivalentresistance, etc.) provide an intuitive approach to analyzing circuits

They are not systematic and cannot be easily automated by acomputer

Comments:

Analysis of circuits using node or loop analysis requires solutionsof systems of linear equations.

These equations can usually be written by inspection of the circuit.



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R3=80

R2=0.4

+

_VS=14V E2=12V

R1=0.5


Branch Analysis

I2

I3

I1

P2.2 How do we findI1 andI2, I3?KVL

Mesh 1:

Mesh 2:

1 214 0.5 0.4 12 0I I

2

1 2

2 0.44 0.8

0.5

II I

2 312 0.4 80 0I I

2

3 2

12 0.40.15 0.005

80

II I

2 2 24 0.8 0.15 0.005I I I

2 2.13AI 1 2.29AI 3 0.16AI KCL1 2 3I I I

2 2 2 21

1

2 0.4

0.5sV E I R II

R

2 2 2 23

3

12 0.4

80

E I R II

R

2 22

2 0.4 12 0.4

0.5 80

I I

I

2 2.14AI 1 2.29AI 3 0.14AI



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Branch Analysis

write KCL equation for each independent node.

(n-1) KCL equations

write KVL equation for each independent mesh/loopm-(n-1) KVL equations

Suppose m branches, n nodals



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Branch AnalysisHeres a quick example of a circuit that we will see later when we

model the operation of transistors. For now, lets assume ideal

independent and dependent sources.

We can write the followingequations:

: 1 0C CCi i i : 2 1 0Bi i i

: 0E B Ci i i C Bi i

2 2 0o E EV i R i R

1 1 2 2 0CCV i R i R

Bi



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Nodal Analysis

1) Choose a reference node

The reference node is called the ground node.

+

-

V 500

500

1k

500

500I1 I2



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:


Nodal Analysis

1) Choose a reference node

0

+

-

V 500

500

1k

500

500I1 I2

I4

I5 I6

I7

I8

KCL :

:


.. .


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Nodal Analysis

2) Assign node voltages to the other nodes

V1, V2, and V3 are unknowns for which we solve using KCL.

500

500

1k

500

500I1 I2

1 2 3

V1 V2 V3I4

I5 I6

I7

I8

0



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Nodal Analysis

3) Apply KCL to each node other than the reference-express

currents in terms of node voltages.

500

500

1k

500

500I1 I2

1 2 3

V1 V2 V3I4

I5 I6

I7

I8

0

500

214

VVI

5001

5

VI

26

1K

V

I 3 2

7500

V V

I

3

8500

V

I

, ,

, ,

0500500

1211

VVVI

0500k1500

32212

VVVVV

0500500 2323

I

VVV

Node:

Node:

Node

:

Node : Node:

Node:

4 5 1 0I I I 6 4 7

0I I I

8 7 2 0I I I

KCL



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Nodal Analysis

4) Solve the resulting system of linear equations.

Node 1:

Node 2:

Node 3:

12

1500500

1

500

1I

VV

0500500

1

k1

1

500

1

500

3

2

1

V

V

V

232

500

1

500

1

500IV

V

The left hand side of the equation:

The node voltage is multiplied by the sum of conductances of allresistors connected to the node.

The neighbourly node voltages are multiplied by the conductance of the

resistor(s) connecting to the two nodes and to be subtracted.

The right hand side of the equation:

The right side of the equation is the sum of currents from sources

entering the node.

500

500

1k

500

500

I1I2

1 2 3

V1 V2 V3

0



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Nodal Analysis


Node 1:

Node 2:

Node 3:

12

1500500

1

500

1I

VV

0500500

1

k1

1

500

1

500

3

2

1

V

V

V

232

500

1

500

1

500IV

V

500

500

1k

500

500

I1I2

1 2 3

V1 V2 V3

2

1

3

2

1

0

500

1

500

1

500

10

500

1

500

1

k1

1

500

1

500

1

0500

1

500

1

500

1

I

I

V

V

V

Matrix Notation(Symmetric)



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Nodal Analysis4) Solve the resulting system of linear equations.

Node 1:

Node 2:

Node 3:

12

1500500

1

500

1I

VV

0500500

1

k1

1

500

1

500

3

2

1

V

V

V

232

500

1

500

1

500IV

V

500

500

1k

500

500

I1I2

1 2 3

V1 V2 V3

G11V1+G12V2 +G13V3 =I11

G21V2+G22V2 +G23V3=I22

G31V1+G32V2+G33V3=I33



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Nodal AnalysisWhat if there are dependent sources?

1k5mA 100Ib

+

-Vo

50

Ib

1k

Example:21

V2V1

50

21 VVIb

0k150

10050

22112

VVVVV

0

mA5

k1

1

50

100

50

1

50

100

50

150

1

50

1

k1

1

2

1

V

V

Matrix is not symmetric due to the dependent source.

mA550k1

211

VVVNode 0

k1100

50212

VI

VVbNode



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Nodal AnalysisWhat if there are voltage sources?

100Ib

+

-

Vo

50

Ib

1k

1k

+-

+ -

0.7V

14

V1

V2 V3 V4

2 3

3k

R1

R2

R3

R4

Difficulty: We do not knowIbthe current

through the voltage source?

Equations: KCL at node 2, node 3, node 4, and

Unknowns:Ib, V1, V2(V3),V4

0k3k1

212

bI

VVVNode 2:

Node 3: 050

43

bI

VV

Independent Voltage Source:

Node 4: 0100k150

443

bI

VVV

3 20.7VV V

3 20.7VV V



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Nodal AnalysisWhat if there are voltage sources?

CURRENT

CONTROLLED

VOLTAGESOURCE

Io=?

xkIVV 212

KCL AT SUPERNODE0=

2+2+

2+4-

21

k

VmA

k

VmA

k

VIx

21

121 22 VVkIV x

)(421 VVV )(83 2 VV

mAkV

IO 3

4

22



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Nodal Analysis

Solves directly for node voltages.

Current sources are easy.

Voltage sources are either very easy or somewhat difficult.

Works best for circuits with few nodes.

Works for any circuit.

Advantages of Nodal Analysis



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Mesh(Loop) Analysis

1) Identifying the Meshes

Mesh 2

1k

1k

1k

V1+

-V2

+

-Mesh 1

Mesh: A special kind of loop that doesnt contain any loops within it.



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Mesh(Loop) Analysis

2) Assigning Mesh Currents

V1

1k

1k

1k

+

-V2

+

-I1 I2

3) Apply KVL around each loop to get an equation in terms of theloop currents.

-V1 +I1 1k + (I1 - I2) 1k = 0

(I2 - I1) 1k +I2 1k + V2 = 0

I1 ( 1k + 1k) - I2 1k = V1

-I1 1k +I2 ( 1k + 1k) = -V2

For Mesh 1:

For Mesh 2:



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Mesh(Loop) Analysis

3) Apply KVL around each loop to get an equation in terms of the

loop currents.


2

1

2

1

k1k1k1

k1k1k1

V

V

I

I

I1 ( 1k + 1k) - I2 1k = V1

-I1 1k +I2 ( 1k + 1k) = -V2

1k

1k

1k

+

-V2

+

-I1 I2

V1



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R1

R3

+_

VS1

R2 R6

R5 R4

VS2

VS3 VS4+

+

+_ _

_

I1

I2

I3 I5

I6

I4

33333232131

22323222121

11313212111

Smmm

Smmm

Smmm

VIRIRIR

VIRIRIR

VIRIRIR

1 2 6 2 6 1 1 2

2 2 3 5 5 2 2 3

6 5 4 5 6 3 4

m S S

m S S

m S

R R R R R I V V

R R R R R I V V

R R R R R I V

0

0

0

443613532

2212532332

2212631111

Smmmmm

mmSmmSm

mmSmmSm

VRIRIIRII

RIIVRIIVRI

RIIVRIIVRIMesh 1:

Mesh 2:

Mesh 3:

436542516

3235253212

2136221621

)(

)(

)(

Smmm

SSmmm

SSmmm

VIRRRIRIR

VVIRIRRRIR

VVIRIRIRRR

Im1 Mesh 1

Im2Mesh 2

Im3Mesh 3



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33333232131

22323222121

11313212111

Smmm

Smmm

Smmm

VIRIRIR

VIRIRIR

VIRIRIR

P2.4(P2.2)R3=80

R2=0.4

+

_VS=14V E2=12V

R1=0.5

Im1

Im2

Mesh 1:

1 1 1 2 2

1 2 1 2 2

12 0

12

S m m m

m m S

V I R I I R

R R I R I V

Mesh 2:

2 1 2 2 3

1 2 2 3 2

12 0

12

m m m

m m

I I R I R

I R R R I

1 2 2 1

2 2 3 2 2

12

12

m S

m

R R R I V

R R R I


-12


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2A

50

+

_40V

20

30

I



What if there are current sources?

The current sources in this circuit will have whatever voltage is

necessary to make the current correct.

We cant use KVL around the loop because we dont know thevoltage.

P2.5 I= ?

P2.6

2

+

_

7V3

1

2A

2

1

+

-

VIm1 Im2

1 1 2

2 1

1 2

40 20 30 0

2A =

m m m

m m

m m

I I I

I I

I I I

2 1 2 2 31 2 3 0m m m m mI I I I I

Mesh 1:

1 3 2m mI I Mesh 2:

2 37 2 1 0m mI I

Super Mesh:

Im2Mesh 1

Im3

Mesh 2Im1

Super Mesh



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1k

2k

2k

12V

+

- 4mA

2mA

I0I1 I2

I3

The

Supermesh

surrounds

this source!

The

Supermesh

does not

include thissource!



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Mesh (Loop) AnalysisWhat if there are current sources?

1k

2k

2k

12V+

-4mA

2mA

I0

I1 I2

I3

The 4mA current source setsI2:

I2 = -4mA

The 2mA current source sets a

constraint onI1 andI3:

I1 -I3 = 2mA

We have two equations and

three unknowns. Where is the

third equation?

3 3 2 1 2

1 2 3

2 3 1

12 2k 1k 2k 0

2k 1k 2k 1k 2k 12V

4mA ; 0.8mA ; 1.2mA

I I I I I

I I I

I I I


-


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2

+

_7V

3

1

2A

2

1




P2.6

+

-

V

Im2

Im3

2 3

1 3

-7V+ 2 1 0

2A

m m

m m

I I

I I

1 2 3 2 3-7V+ 1 3 1 0m m m m mI I I I I Mesh 1:

Mesh 2:

Node 3:

Mesh 1

Node 3


Im1

Mesh 2


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2.2 Useful Circuit Analysis Techniques

Mesh(Loop) Analysis

Dependent current source.

Current sources not shared by meshes.

We treat the dependent source as a

conventional source.

Then KVL on the remaining loop(s)

And express the controlling variable,

Vx, in terms of loop currents

Equations for meshes with current sources

mAIk Ik I8

11=2+3=8 313

][

4

336 3 VkIVO

We are asked for Vo. We only need to solve for I3.

Replace and rearrange

mAIIIIkV

kIV

x

x

42)(4

221

21

1



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Mesh (Loop) AnalysisAdvantages of Loop Analysis

Solves directly for some currents.

Voltage sources are easy.

Current sources are either very easy or somewhat difficult. Works best for circuits with few loops.

Disadvantages of Loop Analysis

Some currents must be computed from loop currents.

Choosing the supermesh may be difficult.



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2.3Useful Circuit Analysis Techniques

Key Words:

Linearity

Superposition

Thevenins and Nortons theorems



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Linearity


Linearity is a mathematical property of circuits that makes very

powerful analysis techniques possible.

Linearity leads to many useful properties of circuits:

Superposition: the effect of each source can be consideredseparately.

Equivalent circuits: Any linear network can be represented by an

equivalent source and resistance (Thevenins and Nortons

theorems)



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Linearity


Linearity leads to simple solutions:

Nodal analysis for linear circuits results in systems of linear

equations that can be solved by matrices

2

1

3

2

1

0

500

1

500

1

500

10

500

1

500

1

k1

1

500

1

500

1

0500

1

500

1

500

1

I

I

V

V

V

500

500

1k

500

500I1 I2

1 2 3V1 V2 V3



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Linearity


The relationship between current and voltage for a linear element

satisfies two properties:

Homogeneity

Additivity

*Real circuit elements are not linear, but can be approximated as linear



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Linearity


Homogeneity:

Let v(t) be the voltage across an element with current i(t) flowingthrough it.

In an element satisfying homogeneity, if the current is increased by

a factor ofK, the voltage increases by a factor ofK. Additivity

Let v1(t) be the voltage across an element with current i1(t) flowingthrough it, and let v2(t) be the voltage across an element withcurrent i2(t) flowing through it

In an element satisfying additivity, if the current is the sum ofi1

(t)and i2(t), then the voltage is the sum ofv1(t) and v2(t).

Example: Resistor: V=R I

If current isKI, then voltage is R KI=KV

If current isI1+I2, then voltage is R(I1+I2) =RI1+RI2 = V1+ V2



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Superposition is a direct consequence of linearity

It states that in any li near circui t containing mul tiple independent sources,the cur rent or voltage at any point in the circuit may be calculated as the

algebraic sum of the individual contr ibutions of each source acting alone.

Superposition


R3=80

R2=0.4

+

_VS=14V E2=12V

R1=0.5

I2

I 2

313221

31

323121

32

22

ERRRRRR

RRV

RRRRRR

RI

I

S

I



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Superposition


How to Apply Superposition?

To find the contribution due to an individual independent source, zero

out the other independent sources in the circuit.

Voltage source short circuit.

Current source open circuit.

Solve the resulting circuit using your favorite techniques.

Nodal analysis

Loop analysis



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SuperpositionFor the above case:


Zero out Vs, we have : Zero out E2, we have :

R1 R3

E2

R2

I2

R1 R3

E2

R2

I2

I

+

_Vs

1 31 3

1 3

1 2 2 3 1 32 1 3

1 3

2 1 3

2

1 2 2 3 1 3

/ /

/ /

R RR R

R RR R R R R R

R R RR R

E R RI

R R R R R R

2 32 3

2 3

1 2 2 3 1 31 1 3

2 3

2 3

1 2 2 3 1 3

3 32

2 3 1 2 2 3 1 3

/ /

/ /

s

s

R RR R

R R

R R R R R RR R R

R R

V R RI

R R R R R R

R V RI I

R R R R R R R R



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Superposition


P2.7

2k1k

2k

12V

+-

I0

2mA

4mA



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Superposition


P2.7

2k1k

2k

Io

2mA

0 2 1

1 2 A

I I I

I m

KVL for mesh 2:

2 1 2

2 1

1k 2k 0

1 2A

3 3

I I I

I I m

0 2 12 23

4A

3

I I I

m

I1 I2

Mesh 2



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Superposition


P2.7

2k1k

2k

I0

4mAI

1

I2

KVL for mesh 2: 2 2 1 21k 0 2k 0I I I I

2 0

0o

I

I

Mesh 2

0 2I I



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Superposition


P2.7

2k1k

2k

12V

+-

I0

I2

Mesh 2

2oI I

KVL for mesh 2:

2 21k 12V 2k 0I I

2

124 A

1k 2k

I m

4 AoI m


C 2 i A i Ci i


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Superposition


P2.7

I0 = I0 +I0+I0 = -16/3 mA

2k1k

2k

12V

+-

I0

2mA

4mA

2k1k

2k

12V

+-

I0

2mA

4mA


Ch2 B i A l i M h d Ci i


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Thevenins theorem


Any circuit with sources (dependent and/or independent) and

resistors can be replaced by an equivalent circuit containing a single

voltage source and a single resistor

Thevenins theorem implies that we can replace arbitrarily

complicated networks with simple networks for purposes of analysis


Ch2 B i A l i M th d t Ci it


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Thevenins theorem


Circuit with independent sources

RTh

Voc+

-

Thevenin equivalent circuit

Independent Sources




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Thevenins theorem


No Independent Sources

Circuit without independent sources

RTh

Thevenin equivalent circuit




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Nortons theorem Very similar to Thevenins theorem

It simply states that any circuit with sources (dependent and/or

independent) and resistors can be replaced by an equivalent circuit

containing a single current source and a single resistor




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Nortons theoremNorton Equivalent: Independent Sources

Circuit with one or more

independent sources

RTh

Norton equivalent circuit

Isc




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Nortons theoremNorton Equivalent: No Independent Sources

Circuit without independent sources

RTh

Norton equivalent circuit




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Motivation of applying the Thevenins theorem and Nortons theorem:

Sometimes, in a complex circuit, we are only interested in working out the voltage

/current or power being consumed by a single load (resistor);

We can then treat the rest of the circuit (excluding the interested load) as a voltage

(current) source concatenated with a source resistor;

Simplify our analysis.

The Thevenins theorem:

Given a linear circuit, rearrange it in the form of two networks of A and B

connected by two wires. Define Voc as the open-circuit voltage which appears

across the terminals of A when B is disconnected. Then all currents and voltage in B

will remain unchanged, if we replace all the independent current or voltage source

in A by an independent voltage source which is in series with a resistor(RTh).

The Nortons Theorem: Given a linear circuit, rearrange it in the form of two networks of A and B

connected by two wires. Define isc as the short-circuit current which appears across

the terminals of A when B is disconnected. Then all currents and voltage in B will

remain unchanged, if we replace all the independent current or voltage source in A

by an independent current source isc which is in parallel with a resistor(RN).




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Equivalent transform between a Thevenin equivalent circuit and a Norton

equivalent circuit

LR

THR

LV

Sv

Si

NR

LR

Lv

Lv

Lv

Thevenin equivalent Norton equivalent

L

L SL TH

Rv v

R R

TH NR R

Lv

Lv

If and

=

LTHL

N

sL RRR

R

iV

THsNss RiRiv


N

VL

VL VL



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Therefore, for a source transform:

Thevenin Norton:Norton Thevenin :

, /N TH s s THR R i v R ,

TH N s s N R R v i R




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Example 45 (P88) Find a Thevenin and Norton equivalent circuit for the following

circuit excluding LR

12V

3 7

6LR

A B

4A

3 6

7

LR

T N

4A 2

7

LR




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8V

2 7

LR

N T

8V

9

LR

0.889A 9 LR

Thevenin

equivalent

Norton

equivalent




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Application of Thevenins theorem when there are only independent sources.

step : 1) Determine of two connection points between network A and network B.

2) Determine of two connection points by replacing the voltage source by a short-

circuit or the current source by a open-circuit.

Similarly, for Nortons Theorem

steps : 1) Determine between the two connection points between network A and B.

2) Determine by two connection points by replacing the voltage source by a short-

circuit or the current source by an open circuit.

ocv

THR

NR


Test the above case?

SCi



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Practice 4.6 (P90)

9V

4

4

5

2

6

2 ?I

ocv

4

4

5

6

9V 9V 4 2.571V4 4 6

S OCV V

4

5

10THR

11 1 20( ) 5 5 7.85710 4 7

THR




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Thevenin equivalent :

2.521V

7.857

2

2I

2

2.571V260.8 A

7.857 2I m




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When there are multiple independent source, we shall use superposition .

Example 4.7 (P91) Find the Thevenin and Norton equivalent for the network excluding the

resistor.1k

2mA 1k

3k2k

4V

To determine ,

when only 4v voltage source is functioning.

ocv

'

ocv'

4ocv V

5K

4V




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When only 2mA current source is functioning :

2mA

2k

3k

ocvN T

''

ocv'' 4ocv V

4V

5k

Therefore,' '' 4V 4V 8V

c oc oc ocv v v v




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To determineTH

R

5THR k

5k

Thevenin equivalent:

8SV V

5TH

k R

1k

5N TH

R R k

/ 1.6S s TH

i v R mA




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Norton equivalent

1.6sv mA

5NR k 1k

Try to look into this problem from the Norton approach.

(Figure out the Norton equivalent circuit first)




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When there are both independent source and dependent source.

Dependent source cannot be zero out as far as its controlling variable is not zero.

Similar as before,But we cannot determine directly, however, we can use

s ocv v

( )TH NR R ( ) /TH N oc scR R v i

Example 4.8 (P92) Determine the Thevenin equivalent of the following circuit

xv

4000

xv

2k 3k

4V

To determine

since ,applying KVL to the supermesh:

ocv

oc xv v4V ( ) 2 3 0 0

4000X

x

Vk k v

8Vs oc x

v v v




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To determinesc

i

04000

xv 0xv 4V

2k 3k

Therefore its Thevenin equivalent is:

8sv V

10TH

R k

mA8.0k5

V4

SCi

k10mA8.0

V8/ SCOCTH ivR




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When there are only dependent sources:

VOC= 0

RTHcan be determined by implying a test (imaginary) voltage across the twoterminals.

To determineRTh, imagine an independent

current sourcexA. as :

3

2

i

1.5i

3

2

i

1.5i vxA

Open circuit :2

0 , 1.5 0 , 02 3

OCi i V

Example 4.9 (p93)

Apply KCL: Giving :

0.6


x

vRTh =i = -xA,

0=)(+2

+3

)(5.1x

vvx-

- --

v = 0.6x V

RTh = 0.6



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Thevenins theorem


Circuits with independent sources

Compute the open circuit voltage, this is Voc

Compute the Thevenin resistance (set the sources to zeroshort

circuit the voltage sources, open circuit the current sources), and

find the equivalent resistance, this isRTh

Circuits with independent and dependent sources:

Compute the open circuit voltage

Compute the short circuit current

The ratio of the two isRTh

Circuits with dependent sources only*

Voc is simply 0

RTh is found by applying an independent voltage source (V volts)

to the terminals and finding voltage/current ratio

* Not required by this course.




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Nortons theorem

Circuits with independent sources, w/o dependent sources

Compute the short circuit current, this isIsc

Compute the Thevenin resistance (set the sources to zeroshort

circuit the voltage sources, open circuit the current sources), and

find the equivalent resistance, this isRN

Circuits with both independent and dependent sources

Find VocandIsc

ComputeRN= Voc/ Isc

Circuits w/o independent sources* Apply a test voltage (current) source

Find resulting current (voltage)

ComputeRN




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Maximum power transfer

+

-

SOURCE

(LOAD)

RTH

VTH

RL

LV

TH

LTH

LL

L

LL V

RR

RV

R

VP

;

2

2

2 TH

LTH

LL V

RR

RP

For every choice of RL we have a different power.

How do we find the maximum value?

Consider PL as a function of RL and find themaximum of such function

4

2

2 2

LTH

LTHLLTHTH

L

L

RR

RRRRRV

dR

dP

02 LLTH RRR THL RR * The maximum power transfer theorem

The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent

resistance of the circuit.

The value of the maximum power that can be transferred is

TH

TH

L R

V

P 4(max)

2

C s c ys s e ods o C cu s



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Analysis methodsReview

Any circuits

linear circuits

KVL, KCL, IV

Combination rules

Node method

Mesh method

Superposition

Thvenin

Norton

y

Documents

Basic Analysis Method Circuit