ADDITIONAL ANALYSIS TECHNIQUES
LEARNING GOALS
REVIEW LINEARITYThe property has two equivalent definitions. We show and application of homogeneity
APPLY SUPERPOSITIONWe discuss some implications of the superposition property inlinear circuits
DEVELOP THEVENIN’S AND NORTON’S THEOREMSThese are two very powerful analysis tools that allow us tofocus on parts of a circuit and hide away unnecessary complexities
MAXIMUM POWER TRANSFERThis is a very useful application of Thevenin’s and Norton’s theorems
The techniques developed in chapter 2; i.e., combination series/parallel, voltage divider and current divider are special techniques that are more efficient than the general methods, but have a limited applicability. It is to our advantage to keep them in our repertoire and use them when they are more efficient.In this section we develop additional techniques that simplify
the analysis of some circuits. In fact these techniques expand on concepts that we have already introduced: linearity and circuit equivalence
THE METHODS OF NODE AND LOOP ANALYSIS PROVIDE POWERFUL TOOLS TODETERMINE THE BEHAVIOR OF EVERY COMPONENT IN A CIRCUIT
SOME EQUIVALENT CIRCUITSALREADY USED
LINEARITYTHE MODELS USED ARE ALL LINEAR.MATHEMATICALLY THIS IMPLIES THAT THEYSATISFY THE PRINCIPLE OF SUPERPOSITION
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IFF LINEAR ISMODEL THE
uu
TuTuuuT
Tuy
AN ALTERNATIVE, AND EQUIVALENT, DEFINITION OF LINEARITY SPLITS THESUPERPOSITION PRINCIPLE IN TWO.
yhomogeneit 2.
additivity
IFF LINEAR IS MODEL THE
uTuuT
uuTuTuuuT
Tuy
,,)(
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NOTICE THAT, TECHNICALLY, LINEARITY CANNEVER BE VERIFIED EMPIRICALLY ON A SYSTEM.BUT IT COULD BE DISPROVED BY A SINGLECOUNTER EXAMPLE.IT CAN BE VERIFIED MATHEMATICALLY FOR THEMODELS USED.
USING NODE ANALYSIS FOR RESISTIVECIRCUITS ONE OBTAINS MODELS OF THEFORM fAv
v IS A VECTOR CONTAINING ALL THE NODEVOLTAGES AND f IS A VECTOR DEPENDINGONLY ON THE INDEPENDENT SOURCES. IN FACT THE MODEL CAN BE MADE MOREDETAILED AS FOLLOWS
BsAv
HERE, A,B, ARE MATRICES AND s IS A VECTOROF ALL INDEPENDENT SOURCES
FOR CIRCUIT ANALYSIS WE CAN USE THELINEARITY ASSUMPTION TO DEVELOPSPECIAL ANALYSIS TECHNIQUES
FIRST WE REVIEW THE TECHNIQUESCURRENTLY AVAILABLE
A CASE STUDY TO REVIEW PAST TECHNIQUES
Redrawing the circuit may help usin recognizing special casesO VDETERMINE
SOLUTION TECHNIQUES AVAILABLE??
USING HOMOGENEITY
Assume that the answer is known. Can we Compute the input in a very easy way ?!!
1V
If Vo is given then V1 can be computed using an inverse voltage divider.
02
211 V
R
RRV
… And Vs using a second voltage divider
02
2141
4 VR
RR
R
RRV
R
RRV
EQ
EQ
EQ
EQS
Solve now for the variable Vo
The procedure can be made entirely algorithmic
1. Give to Vo any arbitrary value (e.g., V’o =1 )
2. Compute the resulting source value and call it V’_s
3. Use linearity. kkVkVVV SS ,'0''
0'
4. The given value of the source (V_s) corresponds to
'S
S
V
Vk
Hence the desired output value is
'0'
'00 VV
VkVV
S
S
This is a nice little toolfor special problems.Normally when there isonly one source and when in our judgementsolving the problembackwards is actuallyeasier
EQR
SOLVE USING HOMOGENEITY
][12 VVVout ASSUME
1I
OV
NOW USE HOMOGENEITY
][2][12
][1][6
VVVV
VVVV
outO
outO
mA1OI ASSUME
][31 VV
][5.0 mA
][5.1 mASV
][62][5.1 1 VVkmAVS
][5.0 mAmA2
____6
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YHOMOGENEIT USE
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LEARNING EXTENSION
Source Superposition
This technique is a direct application of linearity.
It is normally useful when the circuit has onlya few sources.
c i r c u i t
+ -
VS
IS
+
VL
_
IL
Due to Linearity
V aV a IL S S 1 2
V L1 Can be computed by setting the current
source to zero and solving the circuit
V L2 Can be computed by setting the voltage
source to zero and solving the circuit
FOR CLARITY WE SHOW A CIRCUIT WITH ONLY TWO SOURCES
SVBY ONCONTRIBUTI1LV
SIBY ONCONTRIBUTI2LV
Circuit with currentsource set to zero(OPEN)
1LI
1LV
Circuit with voltage sourceset to zero (SHORT CIRCUITED)
2LI
2LV
SOURCE SUPERPOSITION
= +
The approach will be useful if solving the two circuits is simpler, or more convenient, than solving a circuit with two sources
Due to the linearity of the models we must have2121LLLLLL VVVIII Principle of Source Superposition
We can have any combination of sources. And we can partition any way we find convenient
LEARNING EXAMPLE 1i
=
][6||33 kReq
eq
eq
R
vi
kR
2"2
][)3||3(6
+
Loop equations
Contribution of v1
Contribution of v2
WE WISH TO COMPUTE THE CURRENT
Once we know the “partial circuits”
we need to be able to solve them in
an efficient manner
Now we set to zero the current source
LEARNING EXAMPLE
Current division
Ohm’s law
Voltage Divider
+-
V 0"6 k
3k
3V ][6"0
'00 VVVV
][2 V
ionsuperposit source using Compute 0V
We set to zero the voltage source
LEARNING EXAMPLE
Set to zero voltage source
We must be able to solve each circuit in a very efficient manner!!!
Set to zero current source
1V
If V1 is known then V’o is obtained using a voltage divider
V1 can be obtained by series parallel reduction and divider
2I
The current I2 can be obtained using a current dividerand V”o using Ohm’s law
2k||4k
2k
6k
I2+
V "0
_
2m A
+-
2k
4k||8k+
V1_
)6(3/82
3/81 V
+
V 1_
6k
2k
+
V '0
_
][7
18
26
61
' VVkk
kVO
WHEN IN DOUBT… REDRAW!
mAkkkk
kkkI )2(
)4||2(62
)4||2(22
"'
2" 6
OOO
O
VVV
kIV
ionsuperposit source using Compute 0V
Sample Problem
1. Consider only the voltage source
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3. Consider only the 4mA source
2. Consider only the 3mA source
Current divider
mAI 5.102
003 I
mAIIII 30302010 Using source superposition
IONSUPERPOSIT SOURCE USING COMPUTE 0I
SUPERPOSITION APPLIED TO OP-AMP CIRCUITS
TWO SOURCES. WE ANALYZE ONE AT A TIME
2
1 1
1
O
RV V
R
2
2 2
1
1O
RV V
R
Principle of superposition
2 2
1 2 1 2
1 1
1O O O
R RV V V V V
R R
CONTRIBUTION OF V1.Basic inverter circuit
CONTRIBUTION OF V2Basic non-inverting amplifier
Notice redrawing for addedclarity
1I
1
1
2
2
3
3
211 I
IO
Linearity