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Engineering 43
Chp 5.4
Maximum Power
Transfer
Bruce Mayer, PE
Regsitered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
ReCall Thévenin Equivalent
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
RTH


vTH

i
a
vO
b
_

i
a
LINEAR CIRCUIT
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
 vTH = Thevenin
Equivalent
VOLTAGE
Source
 RTH = Thevenin
Equivalent
SERIES
RESISTANCE
b
PART B
PART A
 Thevenin Equivalent Circuit for PART A
Engineering-43: Engineering Circuit Analysis
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Recall Norton Equivalent
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

vO
_

iN
RN
i
i
 iN = Norton
a LINEAR CIRCUIT
Equivalent
May contain
independent and
CURRENT
dependent sources
Source
controlling
b with their
variables
 RN = Norton
PART B
Equivalent
PARALLEL
RESISTANCE
a
LINEAR CIRCUIT
vO
_
b
PART B
PART A
 Norton Equivalent Circuit for PART A
Engineering-43: Engineering Circuit Analysis
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Example
Vb
VX
 VTH 
 Recognize Mixed sources
• Must Compute Open Circuit
Voltage, VOC, and Short
Circuit Current, ISC
 The Open Ckt Voltage
VTH  VX  Vb
 Use V-Divider to Find VX
VX 
R
2
(2VS )  VS
R  2R
3
 For Vb Use KVL
Vb  2R(aVX )  VS  (1  4aR / 3)VS
 Now VTH = Vx - Vb
Engineering-43: Engineering Circuit Analysis
4
VTH  VX  Vb
 VX  (2 RaV X  VS )  (1  2 Ra )( 2VS / 3)  VS
 Solve for VTH
VTH
1 4aR
 VX  Vb  
VS
3
 The Short Ckt Current
• Note that Shorting a-to-b
Results in a Single Large
Node
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Example cont
Single node
I SC
VX
 Need to Find Vx
 KCL at Single Node
Vx  2VS VX
V  Vs

 aVX  X
0
2R
R
2R
 Solving For Vx
3VS
VX 
4  2aR
 Then RTH
RTH 
VOC VTH 4 R(2  aR)


I SC I SC
3
 The Equivalent Circuit
RTH
 KCL at Node-b for ISC
VX  VS
I SC  aVX 
2R
1  4aR
I SC  
VS
4 R(2  aR)
Engineering-43: Engineering Circuit Analysis
5
4 R ( 2  aR )
3
a
VTH
1 4aR

VS
3


b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Numerical Analysis
 Using Excel Spreadsheet
FIND AND PLOT RTH , VOC
WHEN 0  RX  10k
 Short INDEPENDENT
Sources to Find RTH
RTH
4k  R X
 4k || RX 
4k  R X
VOC
 RX 
 12  6

4
k

R
X 

 And VOC by 12V Source
and V-Divider for V
across RX
Engineering-43: Engineering Circuit Analysis
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Numerical Analysis - Plot
RTH  4k || RX 
4k  R X
4k  R X
 RX 
VOC  12  6

4
k

R
X 

THEVENIN EQUIVALENT EXAMPLE
Rx[kOhm] Voc[V] Rth[kOhm]
12
11.8537
11.7143
11.5814
11.4545
11.3333
11.2174
11.1064
11
10.898
10.8
10.7059
10.6154
10.5283
10.4444
10.3636
10.2857
10.2105
10.1379
10.0678
0
0.097560976
0.19047619
0.279069767
0.363636364
0.444444444
0.52173913
0.595744681
0.666666667
0.734693878
0.8
0.862745098
0.923076923
0.981132075
1.037037037
1.090909091
1.142857143
1.192982456
1.24137931
1.288135593
Engineering-43: Engineering Circuit Analysis
7
USING EXCEL
14
12
10
Voc[V]
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
8
6
Voc[V]
4
Rth[kOhm]
2
0
0
2
4
6
8
10
Rx[kOhm]
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Numerical Analysis - Limits
Lim RTH 
R X 
Lim
R X 
4k  R X
 4 k
4k  R X
 RX
VOC  12  6
 4k  R X
Engineering-43: Engineering Circuit Analysis
8

  6V

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Plot 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 use
Voc = 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]')
Engineering-43: Engineering Circuit Analysis
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
USING MATLAB
12
Voc [V]
Rth [kOhm]
Voc (V), Rth (kOhm)
10
8
6
4
2
0
0
5
Engineering-43: Engineering Circuit Analysis
10
10
Rx (kOhm)
15
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Thevenin Theorem – General View
 Typical Interpretation
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

a
i
vO
b
_
 The General View
i
LINEAR CIRCUIT with
ALL independent
sources set to zero
PART A

2R
a

vO
_
VTH 
b
Engineering-43: Engineering Circuit Analysis
11
Looks Like
Series
a
Resistance
R
- VX
+
aVX
2R
VTH
+
-
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Thevenin 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
Engineering-43: Engineering Circuit Analysis
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Amplifier Driving Speaker
 Consider an Amplifier Circuit
connected to a Speaker
Speaker
a.k.a. the
“LOAD”
Driving
Circuit
a.k.a. the
“SOURCE”
Engineering-43: Engineering Circuit Analysis
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Circuit Simplification
 Thévenin’s Equivalent Circuit Theorem Allows
Tremendous Simplification of the Amp Ckt
RS
Thevenin
Engineering-43: Engineering Circuit Analysis
14
VS
+
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Maximum Power Transfer
 Consider The Amp-Speaker Matching Issue
RTH
VTH
From PreAmp
(voltage )
Engineering-43: Engineering Circuit Analysis
15
+
-
To speakers
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Maximum Power Xfer Cont
 The Simplest Model for a
Speaker is to Consider it
as a RESISTOR only
RTH
VTH
 Since the “Load” Does the
“Work” We Would like to
Transfer the Maximum
Amount of Power from the
“Driving Ckt” to the Load
+
-
SPEAKER
MODEL
BASIC MODEL FOR THE
ANALYSIS OF POWER
TRANSFER
• Anything Less Results in
Lost Energy in the Driving
Ckt in the form of Heat
Engineering-43: Engineering Circuit Analysis
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Maximum Power Transfer
 Consider Thevenin Equivalent
Ckt with Load RL
 Find Load Pwr by V-Divider
2
L
V
RL
PL 
; VL 
VTH
RL
RTH  RL
RL
2
PL 
V
2 TH
RTH  RL 
 For every choice of RL we
have a different power.
• How to find the MAXIMUM
Power value?
Engineering-43: Engineering Circuit Analysis
17
RTH
+
-
VTH

VL

RL
(LOAD)
SOURCE
 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Max Power Xfer cont
 Find Max Power Condition Using Differential Calculus
 RTH  RL   2 RL 
dPL
2

 VTH 
3

dRL


R

R
TH
L


 Set The Derivative
To Zero To Find
MAX or MIN Points
• For this Case Set To Zero
The NUMERATOR
RTH  RL  2RL  0
Engineering-43: Engineering Circuit Analysis
18
 Solving for “Best” (max)
Load
*
RL  RTH
 This is The Maximum
Power Transfer Theorem
• The load that maximizes the
power transfer for a circuit is
equal to the Thevenin
equivalent resistance
of the circuitBruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Max Power Quantified
 By Calculus we
Know RL for PL,max
R  RTH
*
L
 Recall the Power
Transfer Eqn
RL
2
PL 
V
2 TH
RTH  RL 
Engineering-43: Engineering Circuit Analysis
19
 Sub RTH for RL
RTH
2
PL,max 
V
2 TH
RTH  RTH 
PL,max
RTH
RTH 2
2

V 
VTH
2 TH
2
4RTH
2RTH 
 So Finally
PL ,max
2
TH
1V

4 RTH
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Max Pwr Xfer Example
a
b
 Determine RL for
Maximum Power
Transfer
 Need to Find RTH
• Notice This Ckt
Contains Only
INDEPENDENT
Sources
 Thus RTH By
Source Deactivation
RTH  4k  3k 6k  6k
 This is Then the RL For
Max Power Transfer
Engineering-43: Engineering Circuit Analysis
20
 To Find the AMOUNT of
Power Transferred Need
the Thevenin Voltage
 Then use RTH = 6kΩ
along with VTH
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Max Pwr Xfer Example cont
 To Find VTH Use Meshes
 The Eqns for Loops 1 & 2
I1  2mA
3k * I 2  I1   6k * I 2  3V  0
 Solving for I2
3[V ] 1
1
I2  
 I1  [mA]
9k 3
3
 Now Apply KVL for VOC
VOC  4k * I1  6k * I 2  10[V ]
Engineering-43: Engineering Circuit Analysis
21
 Recall PL 
RL
2
V
2 TH
RTH  RL 
 At Max: PL = PMX, RL = RTH
PMX
VTH2

4 RTH
PMX
100[V 2 ] 25

 [mW ]
4 * 6k
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Max Pwr Xfer
 Determine RL and Max
Power Transferred
 Find Thevenin Equiv.
At This Terminal-Set
 Recall for Max Pwr Xfer
VTH2
RL  RTH PMX 
4 RTH
 This is a MIXED Source
Circuit
• Analysis Proceeds More
Quickly if We start at c-d
and Adjust for the 4kΩ at
the end
Engineering-43: Engineering Circuit Analysis
22
c
d
 Use Loop Analysis
I1
I2
 Eqns for Loops 1 & 2
I1  4mA
4k ( I 2  I1 )  2kI X'  2kI2  0
6I 2  2I X'  4I1  16mA
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
a
b
Max Pwr Xfer cont
 The Controlling Variable
I X'  I 2
so
I 2  I1  4mA and
VOC  2kI2  8V
 Remember now the
partition points
c
 Now Short Ckt Current
• The Added Wire Shorts
the 2k Resistor
I  0  I SC  4mA
"
X
 Then RTH
RTH
VOC
8V


 2k
I SC 4mA
Engineering-43: Engineering Circuit Analysis
23
8V
d
a
b
 The RTH for ckt at a-b =
2kΩ+4kΩ; SoRL  6k
Pmax
82
8

[mW ]  [mW ]
4*6
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
Thevenin & Norton Summary
 Independent
Sources Only
 Mixed INdep
and Dep Srcs
• RTH = RN by
Source
Deactivation
• VTH
– = VOC or
– = RN·ISC
• IN
– = ISC or
– = VOC/RTH
Engineering-43: Engineering Circuit Analysis
24
• 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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
WhiteBoard Work
 Let’s Work Problem
5.109
 Find Pmax for Load RL
Engineering-43: Engineering Circuit Analysis
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt
What’s an “Algorithm”
 A postage stamp
issued by the USSR
in 1983 to
commemorate the
1200th anniversary
of Muhammad alKhowarizmi, after
whom algorithms
are named.
Engineering-43: Engineering Circuit Analysis
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-04-3b_Thevein-Norton_Part-b.ppt