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Chapter 5. Additional analysis techniques

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2Contents

1. Introduction

2. Superposition

3. Thevenin’s and Norton’s theorems

4. Maximum power transfer

5. Application Examples

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31. IntroductionExamples of equivalent circuits

To simply solution procedures, the number of nodes or loops should be minimized by replacing the original circuits with equivalent ones.

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4Linearity

)(1 t)(1 ti Linear systemL

)()( 21 tBtA )()( 21 tBitAi Linear systemL

)(2 tiLinear systemL

A system satisfying the above statements is called as a linear system. Resistors, Ca-pacitors, Inductors are all linear systems. An independent source is not a linear sys-tem.

All the circuits in the circuit theory class are linear systems!

)(2 t

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5

R1

1k

Resistor

Example of linear system

R1

1k

-)(1 t

)(1 ti

-)(2 t

)(2 ti

C1

1n

111 )()( Rtit

t

diC

t0

22 )(1

)(

C1

1n

Capacitor

input

output

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For the circuit shown in the figure, determine the output voltage Vout using linearity.

Example 5.1

First, arbitrarily assume that the output voltage is Vout = 1 [V].

][12 VVVout ][5.02

22 mA

k

VI ][34 221 VVIkV

][13

11 mA

k

VI ][5.1210 mAIII ][62 1 VVIkV oo

][21:6:12: VVVVV actualout

actualoutouto

For the arbitrary assumption that Vout = 1 [V], the source voltage Vo should be 6 V. Then from linearity, the actual output voltage should satisfy the following relation.

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Circuit

SV

SI

LV

LI

Circuit

SV

1,LV

1,LI

Circuit

SI

2,LV

2,LI

+=

2,1, LLL III

2,1, LLL VVV

2. Superposition

Circuit with voltage sourceset to zero (Short circuited)

Circuit with current source set to zero(Open circuited)

Source superposition

• Superposition is utilized to simplify the original linear circuits.• If a voltage source is eliminated, it is replaced by a short circuit connected

to the original terminals.• If a current source is eliminated, it is replaced by an open circuit.

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8Example 5.2

To provide motivation for this subject, let us examine the simple circuit below, inwhich two sources contribute to the current in the network. The actual values of the sources are left unspecified so that we can examine the concept of superposition.

+=

06)(3

0)(33

2212

2111

ikiik

iikik06)(3

0)(33

212

2111

ikiik

iikik06)(3

0)(33

2212

211

ikiik

iikik

2

1

2

1

93

36

i

i

kk

kk

21

21

2

1

2

1

2

3

15

1

21

13

15

1

kki

i

093

36 1

2

1 i

i

kk

kk

1

1

2

1 3

15

1

ki

i

22

1 0

93

36

i

i

kk

kk

2

2

2

1

215

1

ki

i

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9Example 5.3Let us use superposition to find Vo in the circuit in Fig. 5.3a.

+=

][46

][3

2][2

61

31

61

VkIV

mAmA

kk

kI

oo

o

][2][363

6VV

kk

kVo

][6 VVVV ooo

063)(2)(1

][2

22121

1

IkIIkIIk

mAI

][6],[12 VVmAI o

To verify the solution, we can apply the loop analysis technique.

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10Example 5.4Consider now the network in Fig. 5.4a. Let us use superposition to find Vo’.

+=

][7

18

26

6

][7

24

23

83

8

6

1

1

Vkk

kVV

Vk

k

k

V

o

][7

30][26||

3

10VmAkkVo

][7

48VVVV ooo

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11Thevenin’s and Norton’s theorems

+-

a

b

a

b

Single port network

ThTh ViRBAi

if

)(

1. To find B, measure voltage with i=0. (open circuit voltage)

2. To find A, measure the variation of υ with i changing. (impedance)

NorNor IYDCi

igi

)(

1. To find D, measure current with υ = 0. (short circuit current)

2. To find C, measure the variation of i with υ changing. (admittance)

SVVR

SIIR

i

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12How to construct Thevenin’s equivalent circuit

Circuit

SV

SI

OCV

0I

=

V

Circuit Ω

(1) Measure open circuit voltage (VOC) with a volt meter.

(2) Measure resistance (RTH) with sources suppressed. →Voltage sources short circuited and current sources open circuited.

short

open

Input resistance of a voltmeter is infinite.

Ohm meter

OCV

THR

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13How to construct Norton’s equivalent circuit

Circuit

SV

SI =

A

Circuit Ω

(1) Measure short circuited current (ISH).

(2) Measure resistance (RTH) with sources suppressed. →Voltage sources short circuited and current sources open circuited.

short

open

Input resistance of an ammeter is zero.

Ohm meter

SHI THR

SHI

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14Example 5.6

Let us use Thévenin’s and Norton’s theorems to find Vo in the network below.

][6963

6V

kk

kVo

][6][3

61

31

1VmA

kk

Vo

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15Example 5.7

Let us use Thévenin’s theorem to find Vo in the network in Fig. 5.9a.

][8][1263

61

VVkk

kVOC

][4

63

632

1

kkk

kkkRTh

][16][2482

VmAkVOC ][41

kRTh

][81616

8V

k

kVO

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Determine the Thévenin equivalent of the network in Fig. 5.11a at the terminals A-B.

Example 5.9 : Circuits containing only dependent sources→ Apply an external voltage or current source between the terminals.

01

1

2

2

1111

k

V

k

VV

k

V x

11 VVx

02)1(25 11 VV

7

3,

7

41 xVV

][14

15

7

3

2

3

2

1

1

21

1mA

kkkk

V

k

VI xx

o

][15

14][1 k

I

VR

oTh

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17Example 5.10

Let us determine at the terminals A-B for the network in Fig. 5.12a.

0312

2000 2111

k

VV

k

V

k

IV x

0123

212

mk

V

k

VVk

VI x 1

1

0252263 212111 VVVVVV

06526322 21212 VVVVV

][7

10

1][

7

10 22 k

m

VRVV Th

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18Example 5.11 : Circuits containing both independent and dependent sources

In these types of circuits we must calculate both the open-circuit voltage and short-circuit current to calculate the Thévenin equivalent resistance.

KCL for the super-node around the 12-V source is

Let us use Thévenin’s theorem to find Vo in the network in Fig. 5.13a.

022

12

1

200012

k

V

k

V

k

IV OCOCxOC

k

VI OC

x 2

036612244 OCOCOCOC VVVV ][6 VVOC

][18][

32

12mA

kISC

0xI

][3

1 k

I

VR

SC

OCTh

][7

18)6(

7

3)6(

31

11

1V

kkk

kVo

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19Example 5.12Let us find Vo in the network in Fig. 5.14a using Thévenin’s theorem.

][22 mAI

20001xV

I

)(4 21 IIkVx

][882 VVVV xxx

][1132 1 VIkVOC

mV

kVV

Ik xx

xSC 2

20004,0

200023

][8 VVx

][2

115.14

2

3

2000mAmm

k

VI x

SC

][2][

211

11 k

mAI

VR

Sc

OCTh

][4

3311

26

6V

kk

kVo

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Determine Vo in the circuit in Fig. 5.16a using the repeated application of source transformation.

Example 5.14

kkk

kkRmA

kI ThSC 2

63

63],[4

3

12

kkkRVkmV ThOC 422],[824

kRmAk

I ThSC 4],[24

8

kRVkmV ThOC 4],[1644

][816844

8Vk

kkk

kVO

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214. Maximum Power Transfer

Equivalent circuit of a signal source

LSLL RR

iRiP

,2

SR

2

2

)( LS

LL RR

RP

We want to determine the value of RL that maximizes this quantity. Hence, we differentiate this expression with respect to RL and equate the derivative to zero.

0)(

)(

)(

)(2)(

32

4

22

LS

LS

LS

LSLLS

L

L

RR

RR

RR

RRRRR

R

P

SL RR Maximum power transfer condition :

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22Example 5.16Let us find the value of RL for maximum power transfer in the network in Fig. 5.20a and the maximum power that can be transferred to this load.

kkk

kkkRR ThL 6

36

364

][3

1

2,036)(3

2

1212

mAI

mIIkIIk

][102864 21 VIkIkVOC

][17.4612

102

mWkk

PL

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23Example 5.17

Let us find RL for maximum power transfer and the maximum power transferred to this load in the circuit in Fig. 5.21a.

][82

,02

44

2000

VVk

VI

k

Vm

k

IV

OC

OCx

OCxOC

][40 mAII SCxx

kI

VR

SC

OCTh 2

][67.2612

82

mWkk

PL