Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-43_Lec-02c_Sp12_MaxPwr_SuperPosition.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

MaxPowerSuperPositi

on



OutLine: MaxPwr & SuperPose Work On WhtBd Student Suggest

HomeWork Problem Thevénin & Norton Review

• Example Problem (WhtBd) Maximum Power Transfer

Theorem Derivation• MaxPwr Application Examples

Thevénin & Norton Summary

sL

LLs

Ls

LS

Ls

S

Ls

LS

Ls

S

Ls

LS

LL

L

Ls

LS

Ls

LS

LLL

Ls

LS

LL

L

L

RR

RRR

RRRV

RRV

RRRV

RRV

RRRV

RRP

RRRV

RRR

VRR

VP

RRR

VVRVP

RP

2

2

2d

ddd

0

1

dd

0

3

2

2

2

3

2

2

2

2

2

2

222

2



OutLine: MaxPwr & SuperPose Linearity & Homogeneity

• Guess Solution, Work BackWards, Scale Guess

• Comparative Case Study SuperPosition → Activate & DeActivate

• Example Problem (WhtBd)



Thevénin’s Equivalence TheoremLINEAR C IRC U IT

M ay containindependent and

dependent sourceswith their contro ll ing

variablesPART A

LINEAR C IRC U ITM ay co ntain

independent anddependent sources

with their contro l l ingvariablesPART B

a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

THR

THv

PART A

Thevenin Equivalent Circuit for PART A

vTH = Thévenin Equivalent VOLTAGE Source

RTH = Thévenin Equivalent SERIES RESISTANCE



Norton’s Equivalence Theorem

Norton Equivalent Circuit for PART A

iN = Norton Equivalent CURRENT Source

RN = Norton Equivalent PARALLEL RESISTANCE

LINEAR C IRC U ITM ay contain

independent anddependent sources

with their co ntro l l ingvariablesPART A

LINEAR C IRC U ITM ay contain

independent anddependent so urces

with their co ntro l l ingvariablesPART B

a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

NRNi

PART A



Example: VOC, ISC, RTH = RN

Use Thevénin and Norton for find the OutPut Voltage in the Circuit Below

Recall: VTH = VOC & IN = ISC



OCV

A B

A BSCI



Now

Isc



Maximum Power Transfer Consider The Amp-Speaker Matching Issue

From PreAmp(voltage ) To speakers

+-

R T H

V T H



Maximum Power Xfer Cont The Simplest Model for a

Speaker is to Consider it as a RESISTOR only

Since the “Load” Does the “Work” We Would like to Transfer the Maximum Amount of Power from the “Driving Ckt” to the Load• Anything Less Results in

Lost Energy in the Driving Ckt in the form of Heat

+-

R T H

VT H SPEAKERM O DEL

BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER



Maximum Power Transfer Consider Thevenin Equivalent

Ckt with Load RL

Find Load Pwr by V-Divider

For every choice of RL we have a different power.• How to find the MAXIMUM

Power value?

+-

SOURCE

(LO AD )

R TH

V TH

R L

LV

THLTH

LL

L

LL V

RRRV

RVP

;

2

2

2 THLTH

LL V

RRRP

Consider PL as a

FUNCTION of RL and find the maximum of such a function have at left!• i.e., Take 1st Derivative

and Set to Zero



Max Power Xfer cont Find Max Power Condition Using Differential Calculus

Set The Derivative To Zero To Find MAX or MIN Points• For this Case Set To Zero

The NUMERATOR

32 2

LTH

LLTHTH

L

L

RRRRRV

dRdP

Solving for “Best” (Pmax) Load

THL RR *

This is The Maximum Power Transfer Theorem• The load that maximizes the

power transfer for a circuit is equal to the Thevenin equivalent resistanceof the circuit

02

023

2

*

LLTH

LTH

LLTHTH

RL

L

RRR

RRRRRV

dRdP

L



Max Power Quantified By Calculus we

Know RL for PL,max

THL RR *

Recall the Power Transfer Eqn

2

2 THLTH

LL V

RRRP

Sub RTH for RL

2

2max, THTHTH

THL V

RRRP

2

22

2max, 42 THTH

THTH

TH

THL V

RRV

RRP

So Finally

TH

THL R

VP2

max, 41



Max Pwr Xfer Example Determine RL for

Maximum Power Transfer

Need to Find RTH

• Notice This Ckt Contains Only INDEPENDENT Sources

Thus RTH BySource Deactivation

kkkkRTH 6634 This is Then the RL For

Max Power Transfer

To Find the AMOUNT of Power Transferred Need the Thevenin Voltage

Then use RTH = 6kΩ along with VTH

a b



Max Pwr Xfer Example cont To Find VTH Use Meshes The Eqns for Loops 1 & 2

Solving for I2

][10][2][8 31*62*4

*6*4 21

VVV

mAkmAk

IkIkVOC

mAI 21

03*6*3 212 VIkIIk

][31

31

][9][3

12 mAIkVI

Now Apply KVL for VOC

Recall

22 TH

LTH

LL V

RRRP

At Max: PL = PMX, RL = RTH

TH

THMX R

VP4

2

][625

6*4][100 2

mWkVPMX



Max Pwr Xfer Determine RL and Max

Power Transferred Find Thevenin Equiv.

At This Terminal-Set Recall for Max Pwr Xfer

a

b

THL RR TH

THMX R

VP4

2

This is a MIXED Source Circuit• Analysis Proceeds More

Quickly if We start at c-d and Adjust for the 4kΩ at the end

c

d

Use Loop Analysis

Eqns for Loops 1 & 2mAI 41

022)(4 2'

12 kIIkIIk X

mAIII X 16426 1'

2

2I1I



Max Pwr Xfer cont The Controlling Variable

Now Short Ckt Current• The Added Wire Shorts

the 2k Resistor

Remember now the partition points

VkIVmAII

IIIII

OC

X

82and4

so 426so

2

12

122

2'

mAIII

SC

XX

4020000 ""

Then RTH

kmAV

IVRSC

OCTH 2

48

V8

c

d The RTH for ckt at a-b =

2kΩ+4kΩ; So kRL 6*

][38][

6*482

max mWmWP

a

b



Thevenin & Norton Summary Independent

Sources Only• RTH = RN by

Source Deactivation

• VTH – = VOC or– = RN·ISC

• IN – = ISC or– = VOC/RTH

Mixed INdep and Dep Srcs• Must Keep

Indep & dep Srcs Together in Driving Ckt

• VTH = VOC

• IN = ISC

• RTH = RN = VOC/ ISC

DEPENDENT Sources Only• Must Apply V

or I PROBE– Pick One,

say IP = 1.00 mA, then Calculate the other, say VP

• VTH = IN = 0• RTH = RN

= VP/ IP



WhiteBoard Work Let’s Work this nice

Max Power Problem

Find Pmax for Load RL



Previous Equivalent Circuits [Independent Srcs]

• Vsrc’s in Series• Isrc’s in Parallel

Series & Parallel Resistors

The Complementary Configs are Inconsistent with Source Definitions



Linearity Models Used So Far Are All LINEAR

• Mathematically This Implies That They satisfy the principle of SUPERPOSITION

The Model T(u) is Linear IF AND ONLY IF

22112211 uTuTuuT

• For All Possible– Input Pairs: u1 & u2

– Scalars α1 & α2

AN Alternative, And Equivalent, Linearity & Superposition Definition• The Model T(u) is

Linear IF AND ONLY IF It Exhibits– ADDITIVITY– HOMOGENEITY



Linearity cont. Linearity

Characteristics• Additivity

NOTE • Technically, Linearity

Can Never Be Verified Empirically on a System

• But It Could Be Disproved by a SINGLE Counter Example.

• It Can Be Verified Mathematically For The Models Used

2121 uTuTuuT

uTuT )(

• Homogeneity– a.k.a. Scaling



Linearity cont. Using Node

Analysis For Resistive Circuits Yields Models Of The FormfAv

The Model Can Be Made More Detailed

BsAv • Where

– A and B are Matrices– s Is A Vector Of All

Independent Sources

For Ckt Analysis Use The Linearity Assumption To Develop Special Analysis Methods

• Where– v Is A Vector

Containing All The Node Voltages

– f Is a Vector Containing Only independent Sources



Past Techniques Case Study Find Vo

Redraw the Ckt to Reveal Special Cases• After Untangling

Solution Techniques Available?



Case Study cont. Loop Analysis for Vo Node Analysis

• Out → Positive

---

- -

- -3



Case Study cont. Series-Parallel

Resistor-Combinations

In other Words 213 || RRRREQ

By VOLTAGE Divider



Use Homogeneity Analysis Find Vo by

Scaling

1VEQR

If Vo is Given Then V1 Can Be Found By The Inverse Voltage Divider

OO VRRRVV

RRRV

2

2111

21

2

Now Use VS As a 2nd Inverse Divider

OEQ

EQ

EQ

EQS V

RRR

RRR

VRRR

V2

2141

4

Then Solvefor Vo

Assume That The Answer Is KNOWN • How to Find The Input

In A Very Easy Way ?



Homogeneity Analysis cont The Procedure Can Be Made

Entirely Algorithmic1. Give to Vo Any Arbitrary Value (e.g., V’o = 1V )2. Compute The Resulting Source Value

and Call It V’s3. Use linearity 0

'0

'''0 VkVVkVVV SSS

4. The given value of the source (Vs) corresponds to '

S

S

VVk

5. Then The Desired Output '0'

'00 VVVkVVS

S



Homogeneity Comment This is a Nice Tool For Special

Problems Normally Useful When

• There Is Only One Source• Best Judgment Indicates That Solving The

Problem BACKWARDS Is Actually Easier Than the Forward Solution-Path



Illustration Homogeneity Solve Using Homogeneity (Scaling)

Then By Ohm’s Law

1I

Assume • V’out = V2 = 1volt



Illustration Homogeneity cont Solve Using Homogeneity

Using Homogeneity• Scale from Initial

Assumption:

Again by Ohm’s Law

OV

Then ][2][12

:so ][1][6VVVVVVVV

outO

outO

Scaling Factor

VkVV

VV

VVk

outout

S

S

12

26

12

'

'



Source Superposition This Technique Is A

Direct Application Of Linearity

Normally Useful When The Circuit Has Only A Few Sources



Illustration Src Superposition Consider a Circuit With

Two Independent Sources: VS, IS

Calculated By Setting The CURRENT Source To ZERO (OPEN ckt) And Solving The Circuit

c i r c u i t

+ -VS

IS

+

VL

_

IL

V aV a IL S S 1 2

SVBY ONCONTRIBUTI1LV

SIBY ONCONTRIBUTI2LV

1LV

2LV

Calculated By Setting The VOLTAGE Source To ZERO (SHORT ckt) And Solving The Circuit

Now by Linearity



Illustration cont

1LI

1LV

2LI

2LV= +

Circuit With Current Source Set To Zero• OPEN Ckt

Circuit with Voltage Source set to Zero• SHORT Ckt By Linearity

2121LLLLLL VVVIII



Illustration cont.1LI

1LV

2LI

2LV= +

The Above Eqns Illustrate SUPERPOSITION

2121LLLLLL VVVIII This approach will be useful if

solving the two, 1-Src circuits is simpler, or more convenient, than solving a circuit with two sources

We can have any combination of sources. And we can partition any way we find convenient



Example Solve for i1

Loop equations

Contribution of v1

Contribution of v2

Once we know the “partial circuits” we need to be able to solve them in an efficient manner

=+

][6||33 kReq

215

][)3||3(6

22"2 k

vRvi

kR

eq

eq

Alternative for i1(t) By SuperPosition:

tititi ''1

'11

– Find i1’’ by I-Divider



Numerical Example

Find Vo By SuperPosition

Set to Zero The V-Src• i.e., SHORT it

Current division

Ohm’s law][460'

0 VkIV

mA 32mA

96

93mA 20 I

Contribution by Isrc →



Numerical Example cont. Find Vo By SuperPosition

Set to Zero The I-Src• i.e., OPEN it

+-

V 0"6k

3k

3V

][6240

"0

'00

VVVVV

YieldsVoltageDivider(UN-tangle)

Finally, Add by SuperPosition

By V-Divider

][2 VContribution by Vsrc



WhiteBoard Work Let’s Work this Nice

SuperPosition Problem

Find IO



Example SuperPosition Find Vo Using Source

SuperPosition Set to Zero The I-Src

• i.e., OPEN it

Set to Zero The V-Src• i.e., SHORT it



Example cont Define V1 on the V-Src ckt

If V1 is known then V’o is obtained using the 6&2 Voltage-Divider• V1 can be obtained by series

parallel reduction and divider

1V

+-

2k

4k||8k+

V1_

)6(3/82

3/81 V

+

V 1_

6k

2k

+

V '0

_

][7

1826

61

' VVkk

kVO



Numerical Example cont.2 Determine

• Current I2 By Current Divider• V”o Using Ohm’s Law

When in Doubt REDRAW

2I

2k ||4k

2k

6k

I2+

V "0

_

2m A

mAmAkkkk

kkkI752

)4||2(26)4||2(2

2

VVV

VVVVV

VkIV

O

OOO

O

857.6748

730

718

7306

"'

2"

The Current Division

Finally The SuperPosition Addition



Sample Problem Determine Io by Source

SuperPosition First Consider Only the

Voltage Source• Yields

Then

Second Consider Only the 3 mA I-Source

mAkVI 5.1

69

01

• Yields Current Divider



Sample Prob cont Determine Io by Source

SuperPosition By IO2 Current Divider

Third Consider 4mA Src The Current will Return on

the Path of LEAST Resistance; Thus

So by Source Superposition

mAkkmAI 5.1

63302

003 I

mAIIII 30302010



Illustration Use Source Superposition

to Determine Io

Open the Current Source

RVII S

O 15211

1I

By Equivalent Resistance

Next Short the V-Source

RRRVI S

3||361



Illustration cont Looks Odd & Confusing → REDRAW

Finally By Linearity

1

1

22

3

3

Now Use I-Divider

SO

SO

II

IRRRR

RI

52

6||322

2

2

2

22 2



All Done for Today

SuperPositionOf

Plane-PolarizedLight

Red & Green Light-Waves are Polarized in Perpendicular Planes

Cyan = Red + Green



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Appendix:Wheatsone

Bridge P2.103



The WheatStone Bridge

When the Wheatstone Bridge is Balanced:0and0 abg vi

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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Transcript of Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege