Ch5 Additional Analysis Techniques

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Ch5 Additional Analysis Techniques
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ADDITIONAL ANALYSIS TECHNIQUES
LEARNING GOALS
REVIEW LINEARITY The property has two equivalent
definitions. We show and application of
homogeneity
APPLY SUPERPOSITION We discuss some implications
of the superposition property in linear circuits
DEVELOP THEVENINS AND NORTONS THEOREMS These
are two very powerful analysis tools that allow
us to focus on parts of a circuit and hide away
unnecessary complexities
MAXIMUM POWER TRANSFER This is a very useful
application of Thevenins and Nortons theorems
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THE METHODS OF NODE AND LOOP ANALYSIS PROVIDE
POWERFUL TOOLS TO DETERMINE THE BEHAVIOR OF EVERY
COMPONENT IN A CIRCUIT
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
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SOME EQUIVALENT CIRCUITS ALREADY USED
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LINEARITY
THE MODELS USED ARE ALL LINEAR. MATHEMATICALLY
THIS IMPLIES THAT THEY SATISFY THE PRINCIPLE OF
SUPERPOSITION
FOR CIRCUIT ANALYSIS WE CAN USE THE LINEARITY
ASSUMPTION TO DEVELOP SPECIAL ANALYSIS TECHNIQUES
FIRST WE REVIEW THE TECHNIQUES CURRENTLY AVAILABLE
NOTICE THAT, TECHNICALLY, LINEARITY CAN NEVER BE
VERIFIED EMPIRICALLY ON A SYSTEM. BUT IT COULD BE
DISPROVED BY A SINGLE COUNTER EXAMPLE. IT CAN BE
VERIFIED MATHEMATICALLY FOR THE MODELS USED.
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A CASE STUDY TO REVIEW PAST TECHNIQUES
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The procedure can be made entirely algorithmic
USING HOMOGENEITY
1. Give to Vo any arbitrary value (e.g., Vo 1 )
2. Compute the resulting source value and call it
V_s
Assume that the answer is known. Can we Compute
the input in a very easy way ?!!
If Vo is given then V1 can be computed using an
inverse voltage divider.
This is a nice little tool for special
problems. Normally when there is only one source
and when in our judgement solving the
problem backwards is actually easier
And Vs using a second voltage divider
Solve now for the variable Vo
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SOLVE USING HOMOGENEITY
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LEARNING EXTENSION
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Source Superposition
This technique is a direct application of
linearity.
It is normally useful when the circuit has only a
few sources.
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FOR CLARITY WE SHOW A CIRCUIT WITH ONLY TWO
SOURCES
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The approach will be useful if solving the two
circuits is simpler, or more convenient, than
solving a circuit with two sources
We can have any combination of sources. And we
can partition any way we find convenient
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LEARNING EXAMPLE
WE WISH TO COMPUTE THE CURRENT


Once we know the partial circuits we need to be
able to solve them in an efficient manner
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LEARNING EXAMPLE
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LEARNING EXAMPLE
We must be able to solve each circuit in a very
efficient manner!!!
If V1 is known then Vo is obtained using a
voltage divider
V1 can be obtained by series parallel reduction
and divider
Set to zero current source
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Sample Problem
1. Consider only the voltage source
2. Consider only the 3mA source
3. Consider only the 4mA source
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THEVENINS AND NORTONS 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

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Low distortion audio power amplifier
TO MATCH SPEAKERS AND AMPLIFIER ONE SHOULD
ANALYZE THIS CIRCUIT
Courtesy of M.J. Renardson
http//angelfire.com/ab3/mjramp/index.html
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THEVENINS EQUIVALENCE THEOREM
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NORTONS EQUIVALENCE THEOREM
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OUTLINE OF PROOF - version 1
If Circuit A is unchanged then the current should
be the same FOR ANY Vo
USE SOURCE SUPERPOSITION
HOW DO WE INTERPRET THIS RESULT?
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OUTLINE OF PROOF - version 2
2. Result must hold for every valid Part B that
we can imagine
4. If Part B is a short circuit then Vo is zero.
In this case
How do we interpret this?
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THEVENIN APPROACH
This is the Thevenin equivalent circuit for the
circuit in Part A
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Norton Approach
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ANOTHER VIEW OF THEVENINS AND NORTONS THEOREMS
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
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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
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EXAMPLE SOLVE BY SOURCE TRANSFORMATION
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PROBLEM Compute V_0 using source transformation
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RECAP OF SOURCE TRANSFORMATION
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A General Procedure to Determine the Thevenin
Equivalent
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Three different types of Circuits

• Circuit contains only independent sources
• Circuit Contains only dependent sources
• Circuit contains both independent and dependent
  sources

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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
Since the evaluation of the Thevenin equivalent
can be very simple, we can add it to our toolkit
for the solution of circuits!!
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LEARNING BY DOING
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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)
For the open loop voltage the part outside the
region is eliminated
The original circuit becomes...
And one can apply Thevenin one more time!
...and we have a simple voltage divider!!
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Or we can use Thevenin only once to get a voltage
divider
For the Thevenin voltage we have to analyze
the following circuit
METHOD??
Source superposition, for example
Thevenin Equivalent of Part A
Simple Voltage Divider
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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
For the open circuit voltage we analyze the
following circuit (Part A) ...
The circuit becomes...
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LEARNING EXTENSION USE THEVENIN TO COMPUTE Vo
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LEARNING EXTENSION COMPUTE Vo USING NORTON
COMPUTE Vo USING THEVENIN
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SAMPLE PROBLEM
Equivalent Resistance Independent sources only
Equivalent Voltage Node, loop, superposition
Do loops
How about source superposition?
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SAMPLE PROBLEM
All independent sources
All resistors are in parallel!!
The circuit can be simplified
Voltage divider
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THEVENIN EQUIVALENT FOR CIRCUITS WITH ONLY
DEPENDENT SOURCES
A circuit with only dependent sources cannot self
start.
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
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IF WE CHOOSE A VOLTAGE PROBE...
WE MUST COMPUTE CURRENT SUPPLIED BY PROBE SOURCE
The value chosen for the probe voltage is
irrelevant. Oftentimes we simply set it to one
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IF WE CHOOSE A CURRENT SOURCE PROBE
We must compute the node voltage V_p
The value of the probe current is irrelevant. For
simplicity it is often choosen as one.
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LEARNING EXAMPLE
FIND THE THEVENIN EQUIVALENT
Do we use current probe or voltage probe?
If we use voltage probe there is only one node
not connected through source
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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
_at_V_1
_at_V_2
Conventional circuit with dependent sources -
use node analysis
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SAMPLE PROBLEM
Loop analysis
Controlling variable
Voltage across current probe
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Thevenin EquivalentCircuits with both Dependent
and Independent Sources
We will compute open circuit voltage and short
circuit current
For each determination of a Thevenin equivalent
we will solve two circuits
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
The approach of setting to zero all sources and
then combining resistances to determine the
Thevenin resistance is in general not applicable!!
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Guidelines to partition
EXAMPLE Use Thevenin to determine Vo
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
Constraint at super node
Open circuit voltage
Options???
KCL at super node
Equation for controlling variable
Solve
Short circuit current
Negative resistances for some as
Solution to the problem
Setting all sources to zero and
combining resistances will yield an incorrect
value!!!!
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Find Vo using Thevenin
Open circuit voltage
Method???
Short Circuit Current
The equivalent circuit
KCL
The equivalent resistance cannot be obtained
by short circuiting the sources and determining
the resistance of the resulting interconnection
of resistors
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EXAMPLE Use Thevenin to compute Vo
DONT PANIC!!
Select your partition
Now compute V_0 using the Thevenin equivalent
Use loops
Open Circuit Voltage
Loop equations
Controlling variable
Loop equations
Short circuit current
Controlling variable
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EXAMPLE
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SAMPLE PROBLEM
Mixed sources. Must compute Voc and Isc
The two 4k resistors are in parallel
Short circuit current
FINAL ANSWER
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SAMPLE PROBLEM
Mixed sources! Must compute open loop voltage
and short circuit current
Open circuit voltage
For Vx use voltage divider
For Vb use KVL
Short circuit current
We need to compute V_x
KCL_at_Vx
KCL again can give the short circuit current
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A MORE GENERAL VIEW OF THEVENIN THEOREM
THIS INTERPRETATION APPLIES EVEN WHEN THE PASSIVE
ELEMENTS INCLUDE INDUCTORS AND CAPACITORS
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MAXIMUM POWER TRANSFER
Courtesy of M.J. Renardson
http//angelfire.com/ab3/mjramp/index.html
The simplest model for a speaker is a
resistance...
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MAXIMUM POWER TRANSFER
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
Set the derivative to zero to find extreme
points. For this case we need to set to zero the
numerator
Technically we need to verify that it is indeed a
maximum
The maximum power transfer theorem
The value of the maximum power that can
be transferred is
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
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LEARNING EXAMPLE
The circuit contains only independent sources ....
If we MUST find the value of the power that can
be transferred THEN we need the Thevenin
voltage!!!
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LEARNING EXAMPLE
This is a mixed sources problem
Now the short circuit current
Remember now where the partition was made
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FIND THE THEVENIN EQUIVALENT FOR THE UNKNOWN
ELEMENT USING A RESISTOR AND A VOLTMETER
LEARNING BY APPLICATION
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LEARNING BY DESIGN
CURRENT OVERLOAD SENSOR
THIS POINT MUST GO HIGH WHEN CURRENT EXCEEDS 9A
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LEARNING BY DESIGN