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
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TO MATCH SPEAKERS AND AMPLIFIER IT IS MUCH EASIER TO CONSIDER THIS EQUIVALENT CIRCUIT!
V TH + R TH
REPLACE AMPLIFIER BY SIMPLER “EQUIVALENT”
THEVENIN’S EQUIVALENCE THEOREM
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
i v O
_
a b
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B
R TH v TH
i v O
_
PART A
Thevenin Equivalent Circuit for PART A
v TH
Thevenin Equivalent Source
R TH
Thevenin Equivalent Resistance
a b
LINEAR CIRCUIT PART B
NORTON’S EQUIVALENCE THEOREM
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
i v O
_
a b
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B
i N R N
PART A
Norton Equivalent Circuit for PART A
i N
Thevenin Equivalent Source
R N
Thevenin Equivalent Resistance
i v O
_
a b
LINEAR CIRCUIT 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
i O
DEFINE
R TH
SPECIAL CASE :
v i O O
OPEN
i
v R O TH
CIRCUIT (
i
i SC
0 ) ;
All independent sources set to zero in A
v O i SC v O
v OC i
v O R TH
0
v OC R TH
i SC
v O
i SC
R TH
v OC
R TH i
v OC
i SC i SC
v OC R TH
HOW DO WE INTERPRET THIS RESULT?
i
i O
i SC
OUTLINE OF PROOF - version 2 LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
i v O
_
a 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 O
m
*
i
n
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
v OC
4. If Part B is a short circuit then Vo is zero. In this case 0
m
*
i SC
v OC
m
v OC i SC v O
R TH
R TH i
v OC
How do we interpret this?
THEVENIN APPROACH LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
i v O
_
a b
ANY PART B
v O
R TH i
v OC
For ANY circuit in Part B
v OC
+ _ R TH i
PART A MUST BEHAVE LIKE THIS CIRCUIT
+
v O
_ This is the Thevenin equivalent circuit for the circuit in Part A The voltage source is called the THEVENIN EQUIVALENT SOURCE The resistance is called the THEVENIN EQUIVALENT RESISTANCE
Norton Approach
v O
v OC
R TH i
i
v OC R TH
v O R TH
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
i v O
_
a b
ANY PART B
i SC R TH
Norton
i a
v O b
Norton Equivalent Representa tion for Part A
i SC
Norton Equivalent Source
v OC R TH
i SC
ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS
v OC
+ _ R TH i +
v O
_
i SC R TH
Norton
i a
v O b
Thevenin
i SC
v OC R TH
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 + -
R V V S a R I I S a
THE MODELS ARE EQUIVALENT S WHEN
R V V S
R I
RI S
R b b
Improved model for voltage source Improved model 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
I
0
Or one more source transformation R eq R 3 V eq
+ -
V
R TH V
R eq I eq
R 4
V
0
V eq
R eq I eq
3 current sources in parallel and three resistors in parallel
V
0
R
4
R
4
R
3
R eq V eq
RECAP OF SOURCE TRANSFORMATION
a a
+ -
R V V S R I I S b b
Improved model for voltage source Improved model for current source THE MODELS ARE EQUIVALENT S WHEN
R V V S
R I
RI S
R
Source Transformationcan be used to determine the Thevenin or Norton Equivalent...
WE NOW REVIEW SEVERAL EFFICIENT APPROACHES TO DETERMINE THEVENIN OR NORTON EQUIVALENT CIRCUITS
v
A General Procedure to Determine the Thevenin Equivalent
v TH
Open Circuit vo v oltage at a b ltage if Part B is removed
i SC
Short Circuit Current current through a b if Part B is replaced by a short circuit source
R TH
v TH
Thevenin Equivalent Resistance
i SC
1. Determine the Thevenin equivalent Remove part B and compute the OPEN CIRCUIT voltage
V ab
One circuit problem LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
i v OC
_ 0
a b
V ab
_
Second circuit problem 2. Determine the SHORT CIRCUIT current Remove part B and compute the SHORT CIRCUIT current
I ab
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A
v
i SC
0 _
a b I ab TH
v OC
,
R TH
v OC i SC
AN EXAMPLE OF DETERMINING THE THEVENIN EQUIVALENT
V S
I
+ -
SC R TH
R 1
I
I S Now for the short circuit current Lets try source superposition When the current source is open the current through the short circuit is When the voltage source is set to zero, the current through the short circuit is
2
I SC
I S S
To compute the Thevenin resistance we use
V TH I SC
V S R
1
V TH
R 2 a b
I SC I
To Part B
1
SC
V S R
1
Part B is irrelevant.
The voltage V_ab will be the value of the Thevenin equivalent source.
What is an efficient technique to compute the open circuit voltage?
(
V R TH
1 2
R
1
V
1
R
2
TH
)
V
V S TH
I V S R
1
S
I
0
S
NODE ANALYSIS
V TH V TH
R
1
R
2
R
2
R
1
R
1
R
2
R V S
2
R
1
R
1
R
2
R
2
V S R
1
I S I S
R TH
R
1
R
1
R
2
R
2
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 R 1 a a V S
+ -
I S R 2 To Part B R 1 R 2 R TH b b “Part B”
R TH
3
k
Since the evaluation of the Thevenin equivalent can be very simple, we can add it to our toolkit for the solution of circuits!!
R TH
4
k
“Part B”
LEARNING BY DOING
“PART B”
6
V V O
1
k
1
k
5
k
( 6
V
) 1 [
V
] 5
k
LEARNING EXAMPLE
The original circuit becomes...
1
V TH
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”)
V TH
3
R
6
TH
6 12 [
V
4 ]
k
8 [
V
]
For the open loop voltage the part outside the region is eliminated And one can apply Thevenin one more time!
For open loop voltage use KVL
R
1
TH
4
k
V
1
TH
4
k
* 2
mA
8
V
16
V
...and we have a simple voltage divider!!
V
0 8 8 8 16 [
V
] 8
V
Or we can use Thevenin only once to get a voltage divider
For the Thevenin resistance
R TH
8
k
“Part B” For the Thevenin voltage we have to analyze the following circuit METHOD??
Thevenin Equivalent of “Part A” Source superposition, for example Contribution of the voltage source
1
V OC
6 3
6 12
V
8
V
Contribution of the current source
2
V OC
( 2
k
2
k
) * ( 2
mA
) 8
V
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 “Part B” For the open circuit voltage we analyze the following circuit (“Part A”) ...
R TH
10 3
k
Loop Analysis I
2 6
V
2
mA
4
kI
1 2
k
(
I
1
I
2 ) 0
V OC I
1 4
k
*
I
1 2
k
*
I
2 6 2
I
2
mA
6 20 / 3 4
V
5 3
mA
32 / 3 [
V
]
The circuit becomes...
LEARNING EXTENSION: USE THEVENIN TO COMPUTE Vo “PART B”
I V OC R TH R TH
3
k
|| 6
k
2
k
9
kI V OC
18 [
V
] 3
kI
12
I
2
mA
6 [
V
]
RESULTING EQUIVALENT CIRCUIT
R TH
2
k
2
k
V TH
6
V
V O
4 4 4 ( 6
V
) 3 [
V
] 4
k
V O
LEARNING EXTENSION: COMPUTE Vo USING NORTON
I SC R N I SC
R TH
I N
3
k
12
V
3
k
2
mA
PART B
2
mA
COMPUTE Vo USING THEVENIN
V TH
PART B
4
k I R N
2
k I N V O
2
kI
2
k
R N R
N
6
k I N
V O
2 3 9 ( 2 ) 4 3 [
V
]
V TH
3
k
12 2
mA
0
R TH
3
k
4
k R TH
V TH
+ 2
k V O
V O
2 2 7 ( 6
V
) 4 3 [
V
]
SAMPLE PROBLEM Equivalent Resistance: Independent sources only
R TH
+ -
I
1
I
2
R TH V TH KVL This is what we need to get
R TH
3
R
|| 3
R
1 .
5
R
V TH
I
Equivalent Voltage: Node, loop, superposition…
1
I S
V S
5
R
(
I
1
I
2 )
RI
2
Do loops
0
V TH
RI
2 2
R
(
I
1
I
2 )
How about source superposition?
Opening the current source: Short circuiting the voltage source
V
1
TH
V S
2
I S
I
1
3R
I
2
R + 2R V 2 TH KVL
V
2
TH
_
I
1 5 6
I S
RI
1 2
RI
2
V TH
1 2
RI S
V
1
TH
V
2
TH I
2 1 6
I S
SAMPLE PROLEM All independent sources
R
All resistors are in parallel!!
TH
|| 2
k
, 4
k
, 8
k
|| 8 / 7
k
V TH
The circuit can be simplified ,,, An to compute Equivalent Source...
V TH
Voltage divider
V TH
8
k
8
k
( 8 / 6 )
k
( 6 24 / 6 )[
V
]
SOURCE TRANSFORMATION
V TH
THEVENIN EQUIVALENT FOR CIRCUITS WITH ONLY DEPENDENT SOURCES
aI a a
x
(
R
1
R
1
R
1
R
2
R
2
I x
)
R
2 0
I X
0
I x
0 0
V TH
0
A circuit with only dependent sources cannot self start.
(actually that statement has to be qualified a bit.
a
R
R
?
FOR ANY PROPERLY DESIGNED CICUIT WITH ONLY DEPENDENT SOURCES
V OC
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 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...
(
V P
)
WE MUST COMPUTE CURRENT SUPPLIED BY PROBE SOURCE
I P
I X
V P
R
1
aI X I X
V R P
2
I P
1
R
2 1
R
1
a R
1
R
2
V P
(
V P
)
R TH
V P I P R TH
1
R
2
1
R
1
V P
a R
1
R
2
V P
The value chosen for the probe voltage is irrelevant.
Oftentimes we simply set it to one
IF WE CHOOSE A CURRENT SOURCE PROBE
(
I P
)
We must compute the node voltage V_p
KCL V P
R
2 1
R
2
V P
aI X
1
R
1
R
1
I P a R
1
R
2 0
V P I X
V P R
2
I P
(
I P
)
R TH
V P I P
The value of the probe current is irrelevant. For simplicity it is often choosen as one.
LEARNING EXAMPLE
V
1 FIND THE THEVENIN EQUIVALENT
V P KCL
@
V
1 :
V
1 1
k
V
1 2
V X
2
k
Controlling variable:
V X V
1
V P
1
k V P
SOLVING THE EQUATIONS 0
V
1
Do we use current probe or voltage probe?
If we use voltage probe there is only one node not connected through source
I P V
1
V
2 4
V P
7 ,
P k V X
3 7
V P
V P
2
V X I P
1
k
15
V P
14
k
V X
1
k I P V P R TH
V P I P
14 15
k
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 TH
V P I P
@V_1
V P
I P
@V_2 Controlling variable “Conventional” circuit with dependent sources - use node analysis R TH A B Thevenin equivalent
(
I P
) 5
V
3 1 (
V
2 (
V
2 2
V
1 1 2
V
2 2
V
1
V
) 1 ) 5
V
2 0 6 3
V
2
V
1 6 [
V
2 (
V
1 ] 6 * * / / 2 5
V
2
V
2 ) 30 21 0 10 7 (
V P
V
2 ) (
I P
1
mA
)
R TH
V
2 1
mA
( 10 / 7 )
k
SAMPLE PROBLEM
R TH A B Thevenin equivalent
V P
MUST FIND
V AB I P
1
mA
I_1 = I_p/2 I_3=0 R_th = 2kOhms
R TH
V I P P
V P
1
mA
The resistance is numerically equal to V_p but with units of KOhm
V P
.
METHOD?
Loop analysis
V P
2
k I
* 1
I
V X
2000 1
k
* (
I
;
I
2
I
)
I P
2
k
* (
I
I
3 2 3
Controlling variable
V X
3 1
Voltage across current probe
2
k
*
I
1 ( )
I
3
1
k
* (
I
3
I
2
)
2
k
* (
I
3
I
2
I
1
)
4
k
*
I
2
0
(
I
3 )
I
1 ) 0
Thevenin Equivalent Circuits with both Dependent and Independent Sources
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A + -
R TH V TH
v O
_
i
We will compute open circuit voltage and short circuit current
a
For each determination of a Thevenin equivalent we will solve two circuits
b R TH
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
V OC I SC b V TH
V OC
The approach of setting to zero all sources and then combining resistances to determine the Thevenin resistance is in general not applicable!!
EXAMPLE Use Thevenin to determine Vo
“Part B” Guidelines to partition: “Part A” should be as simple as possible. 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
V
1
Short circuit current
V A
Options???
I
"
X
Constraint at super node KCL at super node
( 12
V
1
V OC
)
V OC
(
aI
1
X
12 ) 12 1
k
Equation for controlling variable Solve
V OC
4 ( 36
a
/ 1
k
)
I
1
X
V OC
2
k V
1 2
k
V OC
12
V V OC
2
k OC
0
V A
2
k
0
I SC
12 1
k
||
V
2
k
18
mA
Negative resistances for some “a’s”
R TH
V OC I SC
4 ( 2
a
/ 1
k
) [
k
]
Solution to the problem
R TH
(
a
2
k
)
Setting all sources to zero and combining resistances will yield an incorrect value!!!!
V OC
V
0 1
k
1 1
k k
R TH V TH
Find Vo using Thevenin
I X
Short Circuit Current
V
1 1
I
1
X
Open circuit voltage
V
1 Method???
Super node
I X
V TH
KVL KVL
V
1 2
k
1
mA
V TH
V
1 ( 3
V
6
k
1000
I X
)
V
1 0
0
V TH
( 3 / 8 )[
V
]
V
1 ( 3 / 4 )[
V
] Controlling variable
I X
V
1 2
k I SC
The equivalent circuit
R TH 1k +
V
1 1
1000
I
1
X
1
I X
V
1 1
V
1 1 0
I
1
X
0
V TH
+ -
2k
2
k I
KCL
SC
1
mA R TH
V OC I SC
( 3
V
) /( 6
k
) ( 3 / 4 )
k
0 .
5
mA V
0 2 1 2 ( 3 / 4 ) ( 3 / 8 )[
V
] The equivalent resistance cannot be obtained by short circuiting the sources and determining the resistance of the resulting interconnection of resistors
V O _
EXAMPLE: Use Thevenin to compute Vo
DON’T PANIC!!
Select your partition Now compute V_0 using the Thevenin equivalent “Part B”
R TH V
0 6
k
6
k
8
k
11 [
V
]
V TH
Open Circuit Voltage Use loops Loop equations
I
1
V
1
X
2000 ;
I
2 2
mA
Controlling variable
V X
1 4
k
(
I
1
I
2 )
V
1
X
KVL for V_oc
2
kI
1
V
OC
2
kI
1 2
k
* 4
k I
1 (
I
1 3 [
V I
] 2 )
2
k
*
I
1
4
4
mA mA
3
V
11
V I
2
I
1
I sc
Short circuit current
I
1 "
V x
3
V
;
I
2000 2
k
2 (
I
2
mA SC
I
1 ) 0
Controlling variable Loop equations
I
1
I SC
4
mA
3
V
Same as before
2
k
*
I
1 2
k
11
mA
2
Thevenin resistance
V X
" 4
k
* (
I
1
I
2 )
R TH
V OC I SC
11 [
V
] ( 11 / 2 )
mA
2
k
EXAMPLE v S
+ -
V TH
+ -
Linear Model for Transistor R 1 R 2 V x R TH g m V x a b
a R 3
V TH
b
I SC
The alternative for mixed sources
V TH
V OC
,
R TH
V OC I SC
Open circuit voltage
V TH
g m R
3
V x V x
R
1
R
2
R
2
v S
V TH
g m R R
3 1
R
2
R
2
v S
Short circuit current
I SC
g m V x
g m R
1
R
2
R
2
v S
Equivalent Resistance
R TH
V OC I SC
R
3
SAMPLE PROBLEM supernode
I
1
Short circuit current
I X
Mixed sources. Must compute Voc and Isc
V TH
Open circuit voltage KCL at super node
I
1
I X
2
The two 4k resistors are in parallel
I X I
1 0
I X
I X
0
V TH
12 [
V
]
KVL
I
KCL at supernode
SC
4
I X I SC
4
k
* (
I SC
/ 4 ) 12 [
V
] 6
k
*
I SC R TH
V TH I SC
12
V
( 12 / 7 )
mA
7
k
0
I SC
12
mA
7
R TH a FINAL ANSWER V TH
b
SAMPLE PROBLEM
V X
Single node
V TH
Short circuit current
V X I SC V b
Mixed sources! Must compute open loop voltage and short circuit current Open circuit voltage
V TH
V X
V b
KCL@Vx For Vx use voltage divider We need to compute V_x
V x
1 2
V S
2
R
V X
1
R
aV X
1
V X
( 1
R
2
R
4
aR
( 2
V S
) 2 3
V S R TH
V OC I SC
V TH I SC
4
R
( 1 2
aR
) 3
I
KCL again can give the short circuit current
SC
aV
1
X
R TH
V X
1 2
R V S
a
I SC
1 4
R
( 1 4
aR
2
aR
)
V S
FINAL ANSWER
V TH
b
LEARNING EXAMPLE FIND AND PLOT
R TH
,
V OC
, WHEN 0
R X
10
k
R
DATA TO BE PLOTTED
TH
4
k
||
R X
4 4
R X R X V OC
12 6 4
k R
X R X
Using EXCEL to generate and plot data
THEVENIN EQUIVALENT EXAMPLE Rx[kOhm] Voc[V] Rth[kOhm]
0 0.1
0.2
0.3
0.4
0.5
12 11.8537
11.7143
11.5814
11.4545
11.3333
0 0.097560976
0.19047619
0.279069767
0.363636364
0.444444444
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
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
0.52173913
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
14 12 10 8 6 4 2 0 0
USING EXCEL
Voc[V] Rth[kOhm] 2 4 6
Rx[kOhm]
8 10
LEARNING EXAMPLE FIND AND PLOT
R TH
,
V OC
, WHEN 0
R X
10
k
R
DATA TO BE PLOTTED
TH
4
k
||
R X
4 4
R X R X V OC
12 6 4
k R
X R X
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
i v O
_
a b i
LINEAR CIRCUIT with ALL independent sources set to zero PART A
V TH
a
v O b
_ THIS INTERPRETATION APPLIES EVEN WHEN THE PASSIVE ELEMENTS INCLUDE INDUCTORS AND CAPACITORS USUAL INTERPRETATION
R 2R - V X + aV X a 2R V TH
+ -
b
Thevenin
MAXIMUM POWER TRANSFER
Courtesy of M.J. Renardson
http://angelfire.com/ab3/mjramp/index.html
From PreAmp (voltage ) V TH + R TH To speakers R TH The simplest model for a speaker is a resistance...
V TH + SPEAKER MODEL BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER
MAXIMUM POWER TRANSFER R TH
P L
V L
2
R L
;
V L
R TH R L
R L V TH P L
R TH R L
R L
2 2
V TH
V TH
+
V L
SOURCE
R L (LOAD) For every choice of R_L we have a different power.
How do we find the maximum value?
Consider P_L as a function of R_L and find the maximum of such function
dP L dR L
2
V TH
R TH
R L
2
R TH
2
R L R L
4
R TH
3
R L
Technically we need to verify that it is indeed a maximum The value of the maximum power that can be transferred is
P L
(max)
2
V TH
4
R TH
Set the derivative to zero to find extreme points.
For this case we need to set to zero the numerator
R TH
R L
2
R L
0 *
R L
R TH
The maximum power transfer theorem The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent resistance of the circuit.
ONLY IN THIS CASE WE NEED TO COMPUTE THE THEVENIN VOLTAGE
LEARNING EXAMPLE a DETERMINE
R L
FOR MAXIMUM POWER TRANSFER b We need to find the Thevenin resistance at a - b.
The circuit contains only independent sources ....
R TH
4
k
|| 3
k
, 6
k
|| Resistance for maximum power transfer 6
k
If we MUST find the value of the power that can be transferred THEN we need the Thevenin voltage!!!
loop
1 :
I
1 2
mA
loop
2
I
1
kI
2
3
V
0
KVL
:
V OC
4
k I
2 *
I
3 [
V
9
k
] 1 3
I
1 1 6
k
*
I
2 1 3 [
mA
] 10 [
V
]
P MX
V
2
TH
4
R TH P MX
100 [
V
4 * 6
k
2
]
25 [
mW
6 ]
LEARNING EXAMPLE DETERMINE
R L
AND MAXIMUM POWER TRANSFERRE D c a 1. Find the Thevenin equivalent at a - b 2. Remember that for maximum power transfer
R L
R TH P MX
4
V TH R
2
TH
b d This is a mixed sources problem .... And it is simpler if we do Thevenin at c - d and account for the 4k at the end
loop
1 :
I
1 4
mA
1
I I
2 Now the short circuit current
I
2
loop
2 : 2
kI
'
X
2
kI
2 Controlling variable: 4
k
(
I
2
I
'
X
I
2
I
1 ) 0 1 4
mA
V OC
R TH
2
k I
"
X
0
I SC
4
mA
Remember now where the partition was made
R L
6
k P MX
4 8 2 * 6 [
mW
] 8 3 [
mW
]
LEARNING EXAMPLE EXAMINE POWER, OUTPUT VOLTAGE AND CURRENT AS FUNCTIONS OF RESISTANCE
P V IN
2
V IN
2
R
2
P V OUT
R
2
R
1
V IN
R
2
2
I
2
V IN
R
2
V OUT
2
R
2
R
2
V IN