Download - Bruce Mayer, PE Regsitered Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PERegsitered Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 5.4Chp 5.4Maximum Maximum

PowerPowerTransferTransfer



ReCall ThReCall Théévenin Equivalentvenin Equivalent

LINEAR C IRC U ITM ay contain

independent anddependent sources

with their contro ll ingvariablesPART A



with their contro ll 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 = Thevenin Equivalent VOLTAGE Source

RTH = Thevenin Equivalent SERIES RESISTANCE



Recall Norton EquivalentRecall Norton Equivalent

Norton Equivalent Circuit for PART A

iN = Norton Equivalent CURRENT Source

RN = Norton Equivalent PARALLEL RESISTANCE






with their contro ll ingvariablesPART B

a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

NRNi

PART A



ExampleExample Recognize Mixed sources

• Must Compute Open Circuit Voltage, VOC, and Short Circuit Current, ISC

The Open Ckt Voltage

Use V-Divider to Find VX

XV bV

SSX VVRR

RV

3

2)2(

2

bXTH VVV

SSXb VaRVaVRV )3/41()(2

THV

For Vb Use KVL

SSSXX

bXTH

VVRaVRaVV

VVV

)3/2)(21()2(

SbXTH VaR

VVV3

41

Solve for VTH

The Short Ckt Current• Note that Shorting a-to-b

Results in a Single Large Node Now VTH = Vx - Vb



Example contExample cont Need to Find Vx

KCL at Single Node

Then RTH

SCISingle node

XV

022

2

R

VVaV

R

V

R

VV sXX

XSx

aR

VV SX 24

3

Solving For Vx

KCL at Node-b for ISC

R

VVaVI SX

XSC 2

SSC VaRR

aRI

)2(4

41

3

)2(4 aRR

I

V

I

VR

SC

TH

SC

OCTH

The Equivalent Circuit

a

b

R TH

V TH

SVaR

3

41

3

)2(4 aRR



Numerical AnalysisNumerical Analysis Using Excel Spreadsheet

Short INDEPENDENT Sources to Find RTH

kR

VR

X

OCTH

100 WHEN

, PLOT AND FIND

X

XXTH Rk

RkRkR

4

4||4

And VOC by 12V Source and V-Divider for V across RX

X

XOC Rk

RV

4612



Numerical Analysis - PlotNumerical Analysis - Plot

X

XOC

X

XXTH

Rk

RV

Rk

RkRkR

4612

4

4||4

THEVENIN EQUIVALENT EXAMPLE

Rx[kOhm] Voc[V] Rth[kOhm]0 12 0

0.1 11.8537 0.0975609760.2 11.7143 0.190476190.3 11.5814 0.2790697670.4 11.4545 0.3636363640.5 11.3333 0.4444444440.6 11.2174 0.521739130.7 11.1064 0.5957446810.8 11 0.6666666670.9 10.898 0.734693878

1 10.8 0.81.1 10.7059 0.8627450981.2 10.6154 0.9230769231.3 10.5283 0.9811320751.4 10.4444 1.0370370371.5 10.3636 1.0909090911.6 10.2857 1.1428571431.7 10.2105 1.1929824561.8 10.1379 1.241379311.9 10.0678 1.288135593

USING EXCEL

0

2

4

6

8

10

12

14

0 2 4 6 8 10

Rx[kOhm]

Vo

c[V

]

Voc[V]

Rth[kOhm]



Numerical Analysis - LimitsNumerical Analysis - Limits

VRk

RV

kRk

RkR

X

XOC

R

X

XTH

R

Lim

Lim

X

X

64

612

44

4



Plot using MATLAB Script FilePlot using MATLAB Script File

ENGR43_Chp5_Rth_Voc_Analysis_MATLAB_0602.m

% ENGR43_Chp5_Rth_Voc_Analysis_MATLAB_0602.m% Bruce Mayer, PE% ENGR43 * 27Feb06%Rx = [0:0.1:20]'; %define the range of resistors to useVoc = 12-6*Rx./(Rx+4); %the formula for Voc. Notice "./"Rth = 4*Rx./(4+Rx); %formula for Thevenin resistance. plot(Rx,Voc,'bx', Rx,Rth,'mv')title('USING MATLAB'), grid, xlabel('Rx (kOhm)'), ylabel('Voc (V), Rth (kOhm)')legend('Voc [V]','Rth [kOhm]')



0 5 10 15 200

2

4

6

8

10

12USING MATLAB

Rx (kOhm)

Voc

(V

), R

th (

kOhm

)Voc [V]Rth [kOhm]



Thevenin Theorem – General ViewThevenin Theorem – General View Typical Interpretation

The General View




a

b_Ov

i

+-

R

2 R

- V X +

2 R

a V X

V TH

a

b

LINEAR C IRC U IT withALL independent

sources set to zeroPART A

a

b_Ov

i

THV

Looks Like Series

Resistance



Thevenin General - CommentsThevenin General - Comments

VTH Becomes the Sole Equivalent Power Source/Sink for the “Part-A” (a.k.a. Driving) Circuit• It’s Value is Set to Maintain The

Open Ckt Voltage at vo

This Interpretation Applies Even When The Passive Elements Include INDUCTORS and CAPACITORS



Amplifier Driving SpeakerAmplifier Driving Speaker Consider an Amplifier Circuit

connected to a Speaker

DrivingCircuita.k.a. the

“SOURCE”

Speakera.k.a. the

“LOAD”



Circuit SimplificationCircuit Simplification Thévenin’s Equivalent Circuit Theorem Allows

Tremendous Simplification of the Amp Ckt

Thevenin +

RS

VS



Maximum Power TransferMaximum Power Transfer Consider The Amp-Speaker Matching Issue

From PreAmp(voltage ) To speakers

+-

R T H

V T H



Maximum Power Xfer ContMaximum 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 TransferMaximum 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?

+-

SO U R CE

(L O A D )

R TH

V TH

R L

LV

THLTH

LL

L

LL V

RR

RV

R

VP

;

2

2

2 TH

LTH

LL V

RR

RP

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 contMax 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

RR

RRRV

dR

dP

02 LLTH RRR

Solving for “Best” (max) 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



Max Power QuantifiedMax Power Quantified

By Calculus we Know RL for PL,max

THL RR *

Recall the Power Transfer Eqn

2

2 TH

LTH

LL V

RR

RP

Sub RTH for RL

2

2max, TH

THTH

THL V

RR

RP

2

22

2max, 42TH

TH

THTH

TH

THL V

R

RV

R

RP

So Finally

TH

THL R

VP

2

max, 4

1



Max Pwr Xfer ExampleMax Pwr Xfer Example Determine RL for

Maximum Power Transfer

Need to Find RTH

• Notice This Ckt Contains Only INDEPENDENT Sources

Thus RTH BySource Deactivation

ab

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



Max Pwr Xfer Example contMax Pwr Xfer Example cont To Find VTH Use Meshes

The Eqns for Loops 1 & 2

Solving for I2

][10*6*4 21 VIkIkVOC

mAI 21

03*6*3 212 VIkIIk

][3

1

3

1

9

][312 mAI

k

VI

Now Apply KVL for VOC

Recall

22 TH

LTH

LL V

RR

RP

At Max: PL = PMX, RL = RTH

TH

THMX R

VP

4

2

][6

25

6*4

][100 2

mWk

VPMX



Max Pwr XferMax 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

VP

4

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 kIkIIIk X

mAIII X 16426 1'

2

2I1I



Max Pwr Xfer contMax Pwr Xfer cont The Controlling Variable

Now Short Ckt Current• The Added Wire Shorts

the 2k Resistor

Remember now the partition points

VkIV

mAII

II

OC

X

82

and4

so

2

12

2'

mAII SCX 40"

Then RTH

kmA

V

I

VR

SC

OCTH 2

4

8

V8

c

d

The RTH for ckt at a-b = 2kΩ+4kΩ; So kRL 6

][3

8][

6*4

82

max mWmWP

a

b



Thevenin & Norton SummaryThevenin & 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 WorkWhiteBoard Work

Let’s Work Problem 5.109

Find Pmax for Load RL



What’s an “Algorithm”What’s an “Algorithm” A postage stamp

issued by the USSR in 1983 to commemorate the 1200th anniversary of Muhammad al-Khowarizmi, after whom algorithms are named.

Download - Bruce Mayer, PE Regsitered Electrical & Mechanical Engineer BMayer@ChabotCollege

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