Transcript Source Transformations

ECE 102
Engineering Computation
Chapter 20
Source Transformations
Dr. Herbert G. Mayer, PSU
Status 10/2/2015
For use at CCUT Fall 2015
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Syllabus


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



Goal
CVS With Rp Removed
CCS With Rs Removed
CVS to CCS Transformation
Detailed Sample
Conclusion
Exercises
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Goal
 The Node-Voltage and Mesh-Current Methods are
powerful tools to compute circuit parameters
 Cramer’s Rule is especially useful for a large
number of unknowns
 Sometimes a circuit can be transformed into another
one that is simpler, yet electrically equivalent
 Generally that will simplify computations
 We’ll learn a few source transformations here
 Method 1: remove parallel load from CVS
 Method 2: remove serial load from CCS
 Method 3: Transform CVS <-> CCS bilaterally
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CVS With Rp Removed
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CVS With Rp Removed
 Removing the load Rp parallel to the CVS has
no impact on externally connected loads RL
 Such loads RL—not drawn here— will be in
series with resistor R
 Removal of Rp decreases the amount of
current that the CVS has to produce, to
deliver equal voltage to both Rp and the
series of R and load RL
 This simplification is one of several source
transformations an engineer should look for,
before computing unknowns in a circuit
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CCS With Rs Removed
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CCS With Rs Removed
 Removing the load Rs in series with the CCS
has no impact on external loads RL
 Such a load RL—not drawn here— will be
parallel to resistor R
 Removal of Rs will certainly decrease the
amount of voltage the CCS has to produce,
to deliver equal current to both Rs in series
with R parallel to RL
 Such a removal is one of several source
transformations to simplify computing
unknown units in a circuit
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CVS to CCS Bilateral Transformation
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CVS to CCS Transformation
 A given CVS of Vs Volt with resistor R in series
produces a current iL in a load, connected externally
 That current also flows through connected load RL
iL = Vs / ( R + RL )
 A CCS of iS Ampere with parallel resistor R produces
a current iL in an externally connected load RL
 For the transformation to be correct, these currents
must be equal for all loads RL
iL = is * R / ( R + RL )
 Setting the two equations for iL equal, we get:
is = Vs / R
Vs = is * R
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Detailed Sample
 We’ll use these simplifications in the next example
to generate an equivalent circuit that is minimal
 I.e. eliminate all redundancies from right to left
 This example is taken from [1], page 110-111,
expanded for added detail
 First we analyze the sample, identifying all
# of Essential nodes ____
# of Essential branches ____
 Then we compute the power consumed or produced
in the 6V CVS
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Detailed Sample, Step a
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Detailed Sample
identify all:
# of Essential nodes __4__
# of Essential branches __6__
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Detailed
Sample, Step b
,
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Detailed
Sample, Step c
,
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Detailed
Sample, Step d
,
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Detailed
Sample, Step e
,
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Detailed
Sample, Step f
,
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Detailed
Sample, Step g
,
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Detailed
Sample, Step h
,
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Power in 6 V CVS
 The current through network h, in the direction of
the 6 V CVS source is:
i = ( 19.2 - 6 ) / ( 4 + 12 ) [ V / Ω ]
i = 0.825 [ A ]
 Power in the 6 V CVS, being current * voltage is:
P = P6V = i * V = 0.825 * 6
P6V = 4.95 W
 That power is absorbed in the 6 V source, it is not
being delivered by the 6 V source! It is delivered by
the higher voltage CVS of 19.2 V
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Conclusion
 Such source transformations are not always
possible
 Exploiting them requires that there be a
certain degree of redundancy
 Frequently that is the case, and then we can
simplify
 Engineers must check carefully, how much
simplification is feasible, and then simplify
 But no more 
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Exercise 1
 Taken from [1], page 112, Example 4.9,
part a)
 Given the circuit on the following page,
compute the voltage drop v0 across the
100 Ω resistor
 Solely using source transformations
 Do not even resort to KCL or KVL, just
simplify and then use Ohm’s Law
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Exercise 1
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Exercise 1
 We know that the circuit does not change,
when we remove a resistor parallel to a CVS

Only the power delivered by the CVS will change
 So we can remove the 125 Ω resistor
 We also know that the circuit does not
change, when we remove a resistor in series
with a CCS

Only the overall power delivered by the CCS will change
 So we can remove the 10 Ω resistor
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Exercise 1, Simplified Step 1
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Exercise 1, Cont’d
 Computation of v0 does not change with
these 2 simplifications
 If we substitute the 250 V CVS with an
equivalent CCS, we have 2 parallel CCS
 These 2 CCSs can be combined
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Exercise 1, Simplified Step 3
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Exercise 1, Cont’d
 Combine 2 parallel CCS of 10 A and -8 A
 And combine 3 parallel resistors: 25 || 100 ||
20 Ω = 10 Ω
 Yielding an equivalent circuit that is simpler,
and shows the desired voltage drop v0 along
the equivalent source, and equivalent
resistor
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Exercise 1, Simplified Step 2
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Exercise 1, Cont’d
 We can compute v0
v0 = 2 A * 10 Ω
v0 = 20 V
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Exercise 2, Compute Power of V 250
 Next compute the power ps delivered
(or if sign reversed: absorbed) by the
250 V CVS
 The current delivered by the CVS is
named is
 And it equals the sum of i125 and i25
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Exercise 2, Compute Power of V 250
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Exercise 2, Compute Power of V 250
is
is
is
is
=
=
=
=
i125 + i25
250/125 + (250 - v0)/25
250/125 + (250 - 20)/25
11.2 A
Power ps is i * v
ps = 250 * 11.2 = 2,800 W
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Exercise 3, Compute Power of 8 A CCS
 Next compute the power p8A delivered
by the 8 A CCS
 First we find the voltage drop across the
8 A CCS, from the top essential node
toward the 10 Ω resistor, named v8A
 The voltage drop across the 10 Ω
resistor is simply 10 Ω * the current, by
definition 8 A, named v10Ω
 That is v10Ω = 80 V
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Exercise 3, Compute Power of 8 A CCS
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Exercise 3, Compute Power of 8 A CCS
V0
V0
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v8A
v8A
=
=
=
=
=
v8A + v10Ω
20 V
8*10 + v8A
20 - 80
-60
Power p8A is i8A * v8A
p8A = i8A*v8A
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