Download Online material

EE 1105: Introduction to EE
Freshman Seminar
Lecture 5: Thevenin Equivalent,
Norton Equivalent, Delta-Wye,
Wye-Delta Conversion
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Thévenin’s Theorem
• A linear circuit can be represented at its output terminals
as an equivalent circuit consisting of a voltage source
Vth in series with a resistor Rth.
• Vth is determined when no load is applied on the output.
• Rth is determined by deactivating all independent
sources in circuit.
A

A
Network
1
•
B
•
Network
2
Application: Coupled networks.
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
RTH
+
_
Network
2
VTH

B
Example – Note, this does not work
with dependent sources
I2
V3
A
_+
R1
_+
R2
V1
V2
_
+
A
R3
I1
R1
R4
R3
R2
R4
B
B
Place a voltmeter across terminals A-B and
read the voltage. We call this the open-circuit
voltage.
No matter how complicated Network 1 is, we read
one voltage.
We call this voltage VAB =VTHEVENIN = VTH
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Deactivate independent sources
Place an ohmmeter across A-B and read
the resistance.
We call this the Thevenin equivalent
Resistance RTH
Alternate Method: Shortcircuit
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Example: Voltage Divider
Rth
R1
Vth
VOFF =
VAMPL =
FREQ =
Vs
OFF =
AMPL =
REQ =
RL
R2
0
0
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Norton Equivalent
RTh
a
a
=
VTh
IN
RTh
b
VTh  RTh I N
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
b
VTh
IN 
RTh
Wye/Delta Conversion (1/2)
b
b
Rc
R2
a
R1
a
R3
Rb
c
c
Wye
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Ra
Delta
Wye/Delta Conversion (2/2)
R1 R2  R1 R3  R2 R3
Ra 
R1
R1 R2  R1 R3  R2 R3
Rb 
R2
R1 R2  R1 R3  R2 R3
Rc 
R3
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Delta/Wye Conversion
Rb Rc
R1 
Ra  Rb  Rc
Ra Rc
R2 
Ra  Rb  Rc
Ra Rb
R3 
Ra  Rb  Rc
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Effects of D-Y & Y-D Conversions
• D-Y Conversion eliminates
a loop but adds a node.
• Y-D Conversion eliminates
a node but adds a loop.
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Resistive Wye Circuit
Eqn 1
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Resistive Wye Circuit
Eqn 2
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Resistive Delta Circuit
Eqn 3
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Resistive Delta Circuit
Eqn 4
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Delta-Wye Derivation
Use Eqn 4 to solve for delta voltages:
Eqn 5
Simplifying:
Eqn 6
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Delta-Wye Derivation
Equating the row 1, column 2 term of eqn 2 to
the row 1, column 2 term of eqn 6:
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Delta-Wye Derivation
Equating the row 1, column 1 term of eqn 2 to
the row 1, column 1 term of eqn 6:
Removing R3:
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Delta-Wye Derivation
Similarly, it can be shown that:
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Wye-Delta Derivation
Use Eqn 2 to solve for wye currents:
Eqn 15
Simplifying:
Eqn 16
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Wye-Delta Derivation
Equating the row 1, column 2 term of eqn 4 to
the row 1, column 2 term of eqn 16:
Then
Equating the row 1, column 1 term of eqn 4 to
the row 1, column 1 term of eqn 16:
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Wye-Delta Derivation
Removing R3:
Similarly, it can be shown that:
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Y-D Conversion Example
a
10 
e
Find Rab
20 
400 
30 
c
b
12 
20 
f
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
Several
Methods:
d
We choose to
120  eliminate
node c,
then simplify.
Y-D Conversion Calculations
20 12  20  30  12  30
Red 
 100 
12
20 12  20  30  12  30
Rdf 
 60 
20
20 12  20  30  12  30
Ref 
 40
30
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
New Circuit
a
10 
e
100 
400 
40 
60 
d
120 
b
20 
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
f
Further Simplifications
10 
a
e
80 
40 
20 
b
Rab  10 
40 
f
40  80  40 
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
40  80  40
d
 20  60 
Acknowledgements: Dr. Bill Dillon
Homework 5 due next class
Next Time:
Exam – Midterm 1, closed book, no calculator,
multiple choice.
Dan O. Popa, Intro to EE, Freshman Seminar, Spring 2015
26