Transcript bYTEBoss ECE201Lect-12

Equivalence/Linearity (5.1);
Superposition (5.2, 8.8)
Dr. Holbert
March 6, 2006
ECE201 Lect-12
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Equivalent Sources
• An ideal current source has the voltage
necessary to provide its rated current.
• An ideal voltage source supplies the current
necessary to provide its rated voltage.
• A real voltage source cannot supply
arbitrarily large amounts of current.
• A real current source cannot have an
arbitrarily large terminal voltage.
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A More Realistic Source Model
i(t)
Rs
vs(t)
+
–
+
The
v(t)
Circuit
–
The Source
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I-V Relationship
The I-V relationship for this source model is
v(t) = vs(t) - Rs i(t)
v(t)
i(t)
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Open Circuit Voltage
• If the current flowing from a source is zero,
then the source is connected to an open
circuit.
• The voltage at the source terminals with i(t)
equal to zero is called the open circuit
voltage:
voc(t)
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Short Circuit Current
• If the voltage across the source terminals is
zero, then the source is connected to a short
circuit.
• The current that flows when v(t) equals zero
is called the short circuit current:
isc(t)
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voc(t) and isc(t)
v(t)
voc(t)
isc(t)
i(t)
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voc(t) and isc(t)
• Since the open circuit voltage and the short
circuit current determine where the I-V line
crosses both axes, they completely define
the line.
• Any circuit that has the same I-V
characteristics is an equivalent circuit.
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Equivalent Current Source
i(t)
+
The
is(t)
Rs
v(t)
Circuit
–
vs (t )
i s (t ) 
Rs
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Source Transformation
Rs
Vs
+
–
Is
Rs
Vs
Is 
Rs
V s  Rs I s
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Source Transformation
• Equivalent sources can be used to simplify
the analysis of some circuits.
• A voltage source in series with a resistor is
transformed into a current source in parallel
with a resistor.
• A current source in parallel with a resistor is
transformed into a voltage source in series
with a resistor.
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Averaging Circuit
1kW
1kW
+
V1
+
–
Vout
1kW
+
–
V2
–
How can source transformation make analysis
of this circuit easier?
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Source Transformations
1kW
1kW
+
V1
+
–
Vout
1kW
+
–
V2
–
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Source Transformations
1kW
V1 /1kW
+ 1kW
1kW
V2 /1kW
Vout
–
Which is a single node-pair circuit that
we can use current division on!
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Linearity
Linearity leads to many useful properties of
circuits:
– Superposition: the effect of each source
can be considered separately.
– Equivalent circuits: any linear network
can be represented by an equivalent
source and resistance (Thevenin’s and
Norton’s theorems).
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Linearity
• More important as a concept than as an analysis
methodology, but allows addition and scaling of
current/voltage values
• Use a resistor as for example (V = R I):
– If current is KI, then new voltage is
R (KI) = KV
– If current is I1 + I2, then new voltage is
R(I1 + I2) = RI1 + RI2 = V1 + V2
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Class Example
• Learning Extension E5.1
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Superposition
“In any linear circuit containing multiple
independent sources, the current or voltage at
any point in the circuit may be calculated as
the algebraic sum of the individual
contributions of each source acting alone.”
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The Summing Circuit
1kW
1kW
+
V1
+
–
Vout
1kW
+
–
V2
–
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Superposition
1kW
1kW
1kW
+
V1
+
–
V’out
1kW
+
1kW
+
–
V’’out
1kW
+
–
–
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V2
Use of Superposition
V’out = V1/3
V’’out = V2/3
Vout = V’out + V’’out = V1/3 + V2/3
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How to Apply Superposition
• To find the contribution due to an individual
independent source, zero out the other
independent sources in the circuit.
– Voltage source  short circuit.
– Current source  open circuit.
• Solve the resulting circuit using your
favorite technique(s).
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Problem
2kW
4mA
12V
– +
2mA
1kW
2kW
I0
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2mA Source Contribution
2kW
2mA
1kW
2kW
I’0
I’0 = -4/3 mA
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4mA Source Contribution
2kW
4mA
1kW
2kW
I’’0
I’’0 = 0
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12V Source Contribution
12V
2kW
– +
1kW
2kW
I’’’0
I’’’0 = -4 mA
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Final Result
I’0 = -4/3 mA
I’’0 = 0
I’’’0 = -4 mA
I0 = I’0+ I’’0+ I’’’0 = -16/3 mA
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Superposition Procedure
1. For each independent voltage and current source (repeat the
following):
a) Replace the other independent voltage sources with a
short circuit (i.e., V = 0).
b) Replace the other independent current sources with an
open circuit (i.e., I = 0).
Note: Dependent sources are not changed!
c) Calculate the contribution of this particular voltage or
current source to the desired output parameter.
2. Algebraically sum the individual contributions (current
and/or voltage) from each independent source.
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Class Example
• Learning Extension E5.2
• Learning Extension E8.15(a) & (b)
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