Transcript Thevenin Equivalence - 義守大學 I

THEVENIN’S AND NORTON’S THEOREMS
These are some of the most
powerful analysis results to be
discussed.
They permit to hide information that
is not relevant and concentrate in
what is important to the analysis
Low distortion audio power amplifier
TO MATCH SPEAKERS AND
AMPLIFIER ONE SHOULD ANALYZE
THIS CIRCUIT
From PreAmp
(voltage )
To speakers
Courtesy of M.J. Renardson
RTH
http://angelfire.com/ab3/mjramp/index.html
TO MATCH SPEAKERS AND AMPLIFIER
IT IS MUCH EASIER TO CONSIDER THIS
EQUIVALENT CIRCUIT!
VTH
+
-
REPLACE AMPLIFIER
BY SIMPLER
“EQUIVALENT”
THEVENIN’S EQUIVALENCE THEOREM
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
_
PART A
Thevenin Equivalent Circuit
for PART A
vTH
Thevenin Equivalent Source
RTH
Thevenin Equivalent Resistance
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
b
PART B
NORTON’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

RN
i
Norton Equivalent Circuit
for PART A
Thevenin Equivalent Source
Thevenin Equivalent Resistance
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
a
LINEAR CIRCUIT
vO
PART A
RN
b
_
_
iN
a
vO

iN
i
b
PART B
Examples of Valid and Invalid Partitions
OUTLINE OF PROOF - version 1
If Circuit A is unchanged then the current should be the same FOR ANY Vo
USE SOURCE SUPERPOSITION
iO

All independent
sources set to
zero in A

i  iO  i SC
v
v
i SC
DEFINE RTH   O  i   O  i SC ; vO
R
iO
TH
SPECIAL CASE: OPEN CIRCUIT (i  0)
v
v
v
vO  vOC  0   OC  i SC  RTH  OC  i SC  OC
i SC
RTH
RTH
v
i   O  i SC  vO  vOC  RTH i HOW DO WE INTERPRET THIS RESULT?
RTH
OUTLINE OF PROOF - version 2
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

i
a
vO
_
b
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
1. Because of the linearity of the models, for any Part B the relationship
between Vo and the current, i, has to be of the form
v  m *i  n
O
2. Result must hold for “every valid Part B” that we can imagine
3. If part B is an open circuit then i=0 and... n  vOC
4. If Part B is a short circuit then Vo is zero. In this case
0  m * iSC  vOC
vOC
m
  RTH
iSC
vO  RTH i  vOC
How do we interpret this?
THEVENIN APPROACH
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

a
i
vO
b
_
vO  RTH i  vOC
RTH
vOC
+
_
ANY
PART B
For ANY circuit in Part B
This is the Thevenin equivalent
circuit for the circuit in Part A
+
i
vO
PART A MUST BEHAVE LIKE
THIS CIRCUIT
_
The voltage source is called the
THEVENIN EQUIVALENT SOURCE
The resistance is called the
THEVENIN EQUIVALENT RESISTANCE
Norton Approach
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
vO  vOC  RTH i  i 

i
i a

i SC
RTH
vO

Norton
b
vOC
 i SC
RTH
a
vO
_
vOC
v
 O
RTH RTH
b
ANY
PART B
Norton Equivalent
Representa tion for Part A
i SC Norton Equivalent Source
ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS
i a
RTH
vOC
+
_

+
i
vO
_
i SC
RTH

Norton
Thevenin
i SC
vO
b
vOC

RTH
This equivalence can be viewed as a source transformation problem
It shows how to convert a voltage source in series with a resistor
into an equivalent current source in parallel with the resistor
SOURCE TRANSFORMATION CAN BE A GOOD TOOL TO REDUCE THE
COMPLEXITY OF A CIRCUIT
Source transformation is a good tool to reduce complexity in a circuit ...
WHEN IT CAN BE APPLIED!!
“ideal sources” are not good models for real behavior of sources
A real battery does not produce infinite current when short-circuited
+
-
RV
VS
a
a
RI
b
IS
b
THE MODELS ARE EQUIVALENTS WHEN
RV  RI  R
VS  RI S
Improved model
Improved model
for voltage source for current source
Source Transformationcan be used to determine the Thevenin or Norton Equivalent...
BUT THERE MAY BE MORE EFFICIENT TECHNIQUES
EXAMPLE: SOLVE BY SOURCE TRANSFORMATION
In between the terminals we connect a current
source and a resistance in parallel
The equivalent current source will have the
value 12V/3k
The 3k and the 6k resistors now are in parallel
and can be combined
In between the terminals we connect a voltage
source in series with the resistor
The equivalent source has value 4mA*2k
The 2k and the 2k resistor become connected
in series and can be combined
After the transformation the sources can be combined
The equivalent current source has value 8V/4k
and the combined current source has value 4mA
Options at this point
1. Do another source transformation and get
a single loop circuit
2. Use current divider to compute I_0 and then
compute V_0 using Ohm’s law
PROBLEM Compute V_0 using source transformation
EQUIVALENT CIRCUITS
I0
Or one more source transformation
R eq
Veq
+
-
RTH
3 current sources in parallel and
three resistors in parallel
R3
VVeqTH Req I eq

R4
V0

Veq  Req I eq
V0 
R4
Veq
R4  R3  Req
RECAP OF SOURCE TRANSFORMATION
+
-
RV
VS
a
a
RI
b
IS
b
THE MODELS ARE EQUIVALENTS WHEN
RV  RI  R
VS  RI S
Improved model
Improved model
for voltage source for current source
Source Transformationcan be used to determine the Thevenin or Norton Equivalent...
WE NOW REVIEW SEVERAL EFFICIENT APPROACHES
TO DETERMINE THEVENIN OR NORTON EQUIVALENT
CIRCUITS
A General Procedure to Determine the Thevenin Equivalent
Open Circuit voltage
vTH
voltage at a - b if Part B is removed
i SC
Short Circuit Current
current through a - b if Part B is replaced
by a short circuit
RTH 
vTH
i SC
Thevenin Equivalent Resistance
1. Determine the
Thevenin equivalent
source
Remove part B and
compute the OPEN
CIRCUIT voltage Vab
One circuit problem
i 0 a
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A


vOC
Vab
_
b
_
Second circuit problem
2. Determine the
SHORT CIRCUIT
current
Remove part B and
compute the SHORT
CIRCUIT current I ab
vTH  vOC , RTH
vOC

i SC
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
i SC

a
v0
I ab
_
b
AN EXAMPLE OF DETERMINING THE THEVENIN EQUIVALENT
R1
VTH
a
VS
+
-
IS
R2
I SC
To Part B
What is an efficient technique to compute the
open circuit voltage?
b
VTH VTH  VS

 IS  0
R2
R1
Now for the short circuit current
Lets try source superposition
When the current source is open the
1
current through the short circuit is I SC
VS

R1
When the voltage source is set to zero,
2
the current through the short circuit is I SC
I SC
To compute the Thevenin resistance we
use
RTH
VS
1
1
(  )VTH 
 IS
R1 R2
R1
VTH 
 IS
VS
 IS 
R1
VTH

I SC
Part B is irrelevant.
The voltage V_ab will be the value of the
Thevenin equivalent source.
VTH
NODE
ANALYSIS
R2
RR
VS  1 2 I S
R1  R2
R1  R2

R1R2  VS
  I S 

R1  R2  R1

RTH 
R1 R2
R1  R2
For this case the Thevenin resistance can be computed as
the resistance from a - b when all independent sources have been
set to zero
Is this a
general
result?
Determining the Thevenin Equivalent in Circuits with Only INDEPENDENT SOURCES
The Thevenin Equivalent Source is computed as the open loop voltage
The Thevenin Equivalent Resistance CAN BE COMPUTED by setting to zero all the sources
and then determining the resistance seen from the terminals where the equivalent will be placed
R1
a
a
VS
+
-
IS
To Part B
R2
R1
R2
b
RTH
b
“Part B”
RTH  3k
Since the evaluation of the Thevenin
equivalent can be very simple, we
can add it to our toolkit for the
solution of circuits!!
RTH  4k
“Part B”
LEARNING BY DOING
5k
“PART B”
6V
VO 
1k
(6V )  1[V ]
1k  5k
LEARNING EXAMPLE
COMPUTE Vo USING THEVENIN
In the region shown, one could use source
transformation twice and reduce that part to
a single source with a resistor.
... Or we can apply Thevenin Equivalence
to that part (viewed as “Part A”)
RTH  4k
The original circuit becomes...
VTH 
And one can apply Thevenin one more time!
For open loop voltage use KVL

1
VTH

6
12[V ]  8[V ]
3 6
For the open loop voltage
the part outside the region
is eliminated
1
TH
R
 4k
1
VTH
 4k * 2mA 8V  16V
...and we have a simple voltage divider!!
V0 
8
16[V ]  8V
88
Or we can use Thevenin only once to get a voltage divider
“Part B”
For the Thevenin voltage we have to analyze the
following circuit
METHOD??
For the Thevenin resistance
RTH  8k
Source superposition, for example
Contribution of the voltage source
1
VOC

6
12V  8V
3 6
Contribution of the current source
Thevenin Equivalent of “Part A”
2
VOC
 (2k  2k ) * (2mA)  8V
Simple Voltage Divider
LEARNING EXAMPLE
USE THEVENIN TO COMPUTE Vo
You have the choice on the way to partition
the circuit.
Make “Part A” as simple as possible
Since there are only independent sources,
for the Thevenin resistance we set to zero all
sources and determine the equivalent resistance
RTH  2  (2 || 4)
“Part B”

10
k
3
For the open circuit voltage we analyze the
following circuit (“Part A”) ...
Loop Analysis
I 2  2mA
 6V  4kI1  2k ( I1  I 2 )  0
VOC
6  2I2
5
I1 
mA  mA
6
3
 4k * I1  2k * I 2  20 / 3  4V  32 / 3[V ]
The circuit becomes...
LEARNING EXTENSION: USE THEVENIN TO COMPUTE Vo

“PART B”
I
VOC

9kI  18[V ]  I  2mA
VOC  3kI  12  6[V ]
RESULTING EQUIVALENT CIRCUIT
RTH  2k 2k

RTH
RTH  3k || 6k  2k

VTH  6V
VO 
4
(6V )  3[V ]
44

4k
VO

LEARNING EXTENSION: COMPUTE Vo USING NORTON
4k
I
RN
IN
I SC
RN  RTH  3k
PART B
12V
I SC  I N 
 2mA  2mA
3k
COMPUTE Vo USING THEVENIN
2k
 RN

VO  2kI  2k 
I N 
 RN  6k

3
4
VO  2 (2)  [V ]
9
3
PART B
VTH
RTH

+
-
VTH
VTH  12
 2mA  0
3k
RTH  3k  4k
VO 
2k
VO

2
4
(6V )  [V ]
27
3
SAMPLE PROBLEM
Equivalent Resistance: Independent sources only
RTH
I2
RTH  3R || 3R  1.5 R

KVL
I1
Equivalent Voltage: Node, loop, superposition… Do loops
VTH
I 1 I S  VS  5 R( I1  I 2 )  RI 2  0

VTH   RI2  2R( I1  I 2 )
How about source superposition?
+
-
Opening the current source:
R TH
VTH
3R
R
I2
2R
_
VS

2
5
1
I1  I S I 2  I S
6
6
+
1
2
VTH
 RI1  2 RI2  RI S
2
KVL
1
2
VTH  VTH
 VTH
V2TH
This is what we need to get
V
Short circuiting the voltage source
IS I1
1
TH
SAMPLE PROLEM
All independent sources
All resistors are in parallel!!
RTH || 2k ,4k ,8k || 8 / 7k

VTH

,,, An to compute Equivalent Source...



 VTH 

Voltage divider
VTH  
The circuit can be simplified
8k
(6  24 / 6)[V ]
8k  (8 / 6) k
SOURCE
TRANSFORMATION

VTH

THEVENIN EQUIVALENT FOR CIRCUITS WITH ONLY DEPENDENT SOURCES
A circuit with only dependent sources cannot self start.
(actually that statement has to be qualified a bit.
What happens if a  R1  R2 ? )
 aI x  ( R1  R2 ) I X  0
 a  R1  R2 I x  0
 a  R1  R2  0  I x  0
FOR ANY PROPERLY DESIGNED CICUIT WITH ONLY DEPENDENT
SOURCES
VOC  0, I SC  0
This is a big simplification!!
But we need a special approach for the
computation of the Thevenin equivalent resistance
Since the circuit cannot self start we need to probe it
with an external source
VTH  0
The source can be either a voltage source or a
current source and its value can be chosen arbitrarily!
Which one to choose is often determined by the
simplicity of the resulting circuit
IF WE CHOOSE A VOLTAGE PROBE...
WE MUST COMPUTE CURRENT SUPPLIED BY
PROBE SOURCE
VP  aI X
IP  IX 
R1
VP
IX 
R2
 1
1
a 
VP
I   
(V P ) P  R
 2 R1 R1R2 
VP
IP
VP

 1
1
a 

VP
 
 R2 R1 R1 R2 
RTH 
(VP )
RTH
The value chosen for the probe voltage is irrelevant.
Oftentimes we simply set it to one
IF WE CHOOSE A CURRENT SOURCE PROBE
We must compute the node voltage V_p
(IP )
KCL
VP
IX 
VP VP  aI X

 IP  0
R2
R2
R1
 1
1
a 
  
VP  I P
 R2 R1 R1R2 
(IP )
RTH
VP

IP
The value of the probe current is irrelevant. For simplicity
it is often choosen as one.
LEARNING EXAMPLE
FIND THE THEVENIN EQUIVALENT
KCL@V1 :
V1
VP
V1 V1  2VX V1  VP


0
1k
2k
1k
Controlling variable:
VX  VP  V1
SOLVING THE EQUATIONS
4
3
V1  VP ,
VX  VP
7
7
VP VP  2VX VX
IP 


2k
1k
1k
Do we use current probe or voltage probe?
If we use voltage probe there is only one
node not connected through source
VP
IP
IP 
RTH 
15VP
14k
VP 14
 k
I P 15
Using voltage probe. Must compute
current supplied
LEARNING EXAMPLE Find the Thevenin Equivalent circuit at A - B
Only dependent sources. Hence V_th = 0
To compute the equivalent resistance we
must apply an external probe
We choose to apply a current probe R  VP
TH
IP
@V_1

VP

IP
“Conventional” circuit with dependent
sources - use node analysis
@V_2
(IP )
Controlling variable
3(V1  2V1 )  6V1  2(V1  V2 )  0
2(V2  V1 )  3V2  6[V ]
RTH
A
B
Thevenin equivalent
5V1  2V2  0 * / 2
 2V1  5V2  6 * / 5
30 10
V2  
21 7
(VP  V2 )  ( I P  1mA)  RTH 
V2
 (10 / 7)k
1mA
SAMPLE PROBLEM
RTH
A
B
Thevenin equivalent
 VP 
I P  1mA
I_1 = I_p/2
I_3=0
R_th = 2kOhms
MUST FIND VAB  VP . METHOD?
Loop analysis
I1 
RTH
VP VP


I P 1mA
The resistance is
numerically equal
to V_p but with
units of KOhm
VX
; I2  I P
2000
2k * I 3  1k * ( I 2  I 3 )  2k * ( I 3  I 2  I1 )  4k * ( I 3  I1 )  0
Controlling variable
VX  1k * ( I 3  I 2 )
Voltage across current probe
 VP  1k * ( I 3  I 2 )  2k * ( I 3  I 2  I1 )  0
Thevenin Equivalent
Circuits with both Dependent and Independent Sources
We will compute open circuit voltage and short circuit current
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
RTH
+
-
VTH
i

a
vO
For each determination of a Thevenin equivalent
we will solve two circuits
b
_
Any and all the techniques discussed should be readily
available; e.g.,
KCL, KVL,combination series/parallel, node, loop analysis,
source superposition, source transformation, homogeneity
a
RTH 
b
VTH  VOC
VOC
I SC
The approach of setting to zero all sources and then
combining resistances to determine the Thevenin
resistance is in general not applicable!!
Guidelines to partition:
EXAMPLE Use Thevenin to determine Vo
“Part A” should be as simple as possible.
“Part B”
After “Part A” is replaced by the Thevenin
equivalent we should have a very simple circuit
The dependent sources and their controlling
variables must remain together
Open circuit voltage
V1
Constraint at super node V1
Options???
(12  VOC )  (aI 1X ) 12  VOC VOC


0
1k
2k
2k
V
Equation for controlling variable I 1X  OC
2k
36
KCL at super node
Solve
Short circuit current
 VOC  12  V1  12  VOC
VA
I "X
VOC  
4  (a / 1k )
V
 A 0
2k
I SC  
RTH 
Solution to the problem
Negative resistances for some “a’s”
RTH (a  2k )
VOC
2

[k]
I SC 4  (a / 1k )
Setting all sources to zero and combining
resistances will yield an incorrect value!!!!
V0 
VOC
12V
 18mA
1k || 2k
1k
VTH
1k  1k  RTH
Find Vo using Thevenin
Open circuit voltage
Method???
Super node
V1

VTH
IX
IX
KVL
V11
 1000 I 1X
KCL
I SC
 VTH  1000 I X  V1  0
Controlling variable
IX 
I 1X
The equivalent circuit
R TH
VTH
V11
1
1

 V1  0  I X  0
2k
VOC
 (3 / 4)k
I SC
1k
+
I SC  1mA  (3V ) /( 6k )  0.5mA
RTH 
V1  (3 / 4)[V ]
VTH  (3 / 8)[V ]
I 1X
V11

V1
V  (3V )
 1mA  1
0
2k
6k
Short Circuit Current
KVL
+
-
V0 
2k
2
(3 / 8)[V ]
2  1  (3 / 4)
The equivalent resistance cannot be obtained by
short circuiting the sources and determining the
resistance of the resulting interconnection of resistors
VO
_
V1
2k
EXAMPLE: Use Thevenin to compute Vo
DON’T PANIC!!
Select your partition
Now compute V_0 using the Thevenin equivalent
“Part B”
V0 
RTH
6k
11[V ]
6k  8k
VTH
V X1
; I 2  2mA
Loop equations I1 
2000
Open Circuit Voltage Use loops
Controlling variable
VX1  4k ( I1  I 2 )
VX1  2kI1  2kI1  4k ( I1  I 2 )  I1  4mA
VOC  2k * I1  3[V ] 2k * 4mA  3V  11V
KVL for V_oc
Loop equations
Short circuit current
I1
I1 
I sc
I2
V x"
2000
; I 2  2mA
 3V  2k ( I SC  I1 )  0
Controlling variable
VX"  4k * ( I1  I 2 )
I1  4mA
I SC 
Same as before
3V  2k * I1 11
 mA
2k
2
Thevenin resistance
RTH 
VOC
11[V ]

 2k
I SC (11 / 2)mA
EXAMPLE
Linear Model for Transistor
R1
The alternative for mixed sources
a
VTH  VOC , RTH
R3VTH
Open circuit voltage
Vx

vS
R2
+
-
g mVx

VTH   gm R3Vx
b
I SC
R TH
a
VOC

I SC
R2
R3 R2
Vx 
v S  VTH   gm
vS
R1  R2
R1  R2
Short circuit current
VTH
+
-
I SC   gmV x   gm
b
R2
vS
R1  R2
Equivalent Resistance
RTH 
VOC
 R3
I SC
SAMPLE PROBLEM
Mixed sources. Must compute Voc and Isc
supernode
Open circuit voltage
KCL at super node

I1  I X  2 I X  0
The two 4k resistors are in parallel
VTH
I1

I1  I X
I X  0  VTH  12[V ]
KCL at supernode
Short circuit current
I SC  4 I X
4k * ( I SC / 4)  12[V ]  6k * I SC  0
KVL
IX
I SC
RTH 
VTH
12V

 7 k
I SC (12 / 7)mA
RTH
a
FINAL ANSWER
VTH


b
I SC 
12
mA
7
SAMPLE PROBLEM
VX
 VTH 
Mixed sources! Must compute open loop voltage and
short circuit current
Open circuit voltage
VTH  V X  Vb
Vb
For Vx use voltage divider
For Vb use KVL
VX 
R
2
(2VS )  VS
R  2R
3
Vb  2 R(aVX )  VS  (1  4aR / 3)VS
Short circuit current
Single node
I SC
VX
VTH  VX  (2 RaVX  VS )  (1  2 Ra)( 2VS / 3)  VS
1 4aR
V


VS
TH
We need to compute V_x
3
KCL@Vx V 1  2V
3VS
V X1
V X1  Vs
1
x
S
VX1 

 aVX 
0
2R
R
2R
4  2aR
KCL again can give the short circuit current
I SC 
RTH 
aV X1
V X1  VS

2R
RTH
VOC VTH 4 R(1  2aR)


I SC I SC
3
VTH
I SC  
1  4aR
VS
4 R(1  2aR)
a FINAL ANSWER


b
LEARNING EXAMPLE FIND AND PLOT RTH , VOC , WHEN 0  RX  10k
DATA TO BE PLOTTED
RTH  4k || RX 
4 RX
4  RX
Using EXCEL to generate and plot data
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
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
Rx[kOhm]
8
10
 RX 
VOC  12  6

 4k  RX 
LEARNING EXAMPLE FIND AND PLOT RTH , VOC , WHEN 0  RX  10k
DATA TO BE PLOTTED
RTH  4k || RX 
4 RX
4  RX
 RX 
VOC  12  6

 4k  RX 
Using MATLAB to generate and plot data
»
»
»
»
»
»
»
Rx=[0:0.1:10]'; %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,'bo', Rx,Rth,'md')
title('USING MATLAB'), %proper graphing tools
grid, xlabel('Rx(kOhm)'), ylabel('Volts/kOhms')
legend('Voc[V]','Rth[kOhm]')
A MORE GENERAL VIEW OF THEVENIN THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

a
i
USUAL INTERPRETATION
vO
b
_
R
i
LINEAR CIRCUIT with
ALL independent
sources set to zero
PART A

VTH 
2R
a

vO
_
b
THIS INTERPRETATION APPLIES
EVEN WHEN THE PASSIVE ELEMENTS
INCLUDE INDUCTORS AND CAPACITORS
- VX
a
+
aVX
2R
VTH
+
-
b
Thevenin
MAXIMUM POWER TRANSFER
Courtesy of M.J. Renardson
http://angelfire.com/ab3/mjramp/index.html
From PreAmp
(voltage )
To speakers
The simplest model for a
speaker is a resistance...
RTH
RTH
VTH
VTH
+
-
+
-
SPEAKER
MODEL
BASIC MODEL FOR THE
ANALYSIS OF POWER
TRANSFER
MAXIMUM POWER TRANSFER
RL
2
VL2
RL
P

V
PL 
; VL 
VTH L
2 TH
 RTH  RL 
RL
RTH  RL
R TH

+
-
VTH
VL

For every choice of R_L we have a different power.
How do we find the maximum value?
RL
(LOAD)
Consider P_L as a function of R_L and find the
maximum of such function
 RTH  RL 2  2 RL RTH  RL  
dPL
2

 VTH 
4

3
dRL


R

R
TH
L


SOURCE
Technically we need
to verify that it is
indeed a maximum
Set the derivative to zero to find extreme points.
For this case we need to set to zero the numerator
RTH  RL  2 RL  0 
*
RL
 RTH
The maximum
power transfer
theorem
The value of the maximum
The load that maximizes the power transfer for a circuit is
power that can be
equal to the Thevenin equivalent resistance of the circuit.
transferred is
VTH2
PL (max)
4RTH
ONLY IN THIS CASE WE NEED TO COMPUTE THE THEVENIN VOLTAGE
LEARNING EXAMPLE DETERMINE RL FOR MAXIMUM POWER TRANSFER
a
We need to find the Thevenin resistance
at a - b.
The circuit contains only independent sources ....
b
RTH  4k  || 3k ,6k || 6k
Resistance for maximum
power transfer
If we MUST find the value of the
power that can be transferred THEN
we need the Thevenin voltage!!!
loop1 : I1  2mA
loop2 : 3k ( I 2  I1 )  6kI 2  3V  0
I2  
3[V ] 1
1
 I1  [mA]
9k 3
3
KVL : VOC  4k * I1  6k * I 2  10[V ]
PMX
2
VTH

4 RTH
PMX
100[V 2 ] 25

 [mW ]
4 * 6k
6
LEARNING EXAMPLE DETERMINE RL AND MAXIMUM POWER TRANSFERRED
c
a
1. Find the Thevenin equivalent at a - b
2. Remember that for maximum power
transfer
V2
RL  RTH
d
This is a mixed sources problem
PMX 
TH
4 RTH
b
.... And it is simpler if we do Thevenin
at c - d and account for the 4k at the end
loop1 : I1  4mA
loop2 :  2kI X'  2kI 2  4k ( I 2  I1 )  0
1I
I2
'
Controlling variable: I X
 I2
I 2  I1  4mA  VOC  8[V ]
Now the short circuit current
I "X  0
I SC  4mA
RTH  2k
Remember now where the partition was made
RL  6 k
PMX
82
8

[mW ]  [mW ]
4*6
3
LEARNING EXAMPLE
PV
IN
PV
OUT
VIN2

2  R2
 VIN 
 R2 

R

R
 1
2 
VIN
I
2  R2
VOUT
EXAMINE POWER, OUTPUT VOLTAGE AND CURRENT
AS FUNCTIONS OF RESISTANCE
R2

VIN
2  R2
2