Transcript ESS2220 Electric and Electronic Circuits

ESS2220
Electric and Electronic Circuits
by
Prof. K.S. Chang-Liao
(by courtesy of Prof. T. K. Wang)
Rules to Abide by:
1. No food in the class,
while beverages and chewing gums are ok;
2. Use the back door, if you are late;
3. Don’t wear slippers,
while sandals are ok.
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Text Book: Electrical Engineering –
Principles and Applications
3rd -5th ed.
Author: Allan R. Hambley, Published by: Prentice Hall
Grading: Exams. I and II: 40%+ 40%,
Quizzes: 10%, HW: 10%
一分耕耘 一分收穫!
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Course Contents
I. Electric Circuits
Chap. 1: Introduction
Chap. 2: Resistive Circuits
Chap. 3: Inductance and Capacitance
Chap. 4: Transients
Chap. 5: Steady-State Sinusoidal Analysis
Chap. 6: Frequency Response
II. Electronics
Chap.10: Diodes
Chap.11: Amplifiers
Chap.12: Field Effect Transistors
Chap.14: Operational Amplifiers
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Chapter 1 Introduction
• Define current, voltage, and power, including their
units.
• Calculate power and energy, as well as determine
whether energy is supplied or absorbed by a circuit
element.
• State and apply basic circuit laws.
• Solve for currents, voltages, and powers in simple
circuits.
5
Chapter 1 Introduction
1.1 The History of Electricity
6
Chapter 1 Introduction
1.1 The History of Electricity
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Chapter 1 Introduction
1.1 The History of Electricity
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Chapter 1 Introduction
1.2 Circuits, Currents, and Voltages
1.2.1 Circuits
An electric circuit consists of various types of elements
connected by connectors.
(Note: “open” or “short” circuit)
(a) Open circuit. (b) Short circuit.
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Chapter 1 Introduction
1.2 Circuits, Currents, and Voltages
1.2.2 Currents
i(t) 
dq(t)
dt
t
q(t)   i(t)dt  q(t0 )
t0
Electrical current is the time rate of flow of electrical charge
through a conductor or circuit element. The units are amperes
(A), which are equivalent to coulombs per second (C/s).
We define the direction of positive charge flow as
positive current direction.
10
Chapter 1 Introduction
1.2.2 Currents
Actual direction and reference direction
of current flow.
e.g.,
i 1  2A, i 2  3A, i 3  5A
Direct current (dc) and alternating current (ac)
11
Chapter 1 Introduction
1.2.2 Currents
Notation for currents:
i ab  -iba
1.2.3 Voltages
Voltage is the energy transferred per unit of
charge that flows through the element:
V (volt) = J (joule)/C (coulomb)
Notice that voltage is measured across the
ends of a circuit element,
whereas current is a measure of charge
flow through the element.
12
Chapter 1 Introduction
1.2.4 Actual and Reference Polarities of Voltage
Voltages are assigned polarities that indicate the direction of energy
flow.
＋
－
v ab  -vba
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Chapter 1 Introduction
1.2.5 Voltmeters and Ammeters
(a) A direct-reading (analog) meter.
(b) A digital meter.
(a) An ideal ammeter measures the current flow through its
terminals and has zero voltage.
(b) An ideal voltmeter measures the voltage across its terminals
and has zero terminal current.
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Chapter 1 Introduction
1.2.5 Voltmeters and Ammeters
Ideal voltmeters (ammeters) act like open (short) circuits.
(a) An example circuit, (b) plus an open circuit and a short
circuit. (c) The open circuit is replaced by a voltmeter, and
the short circuit is replaced by an ammeter. All resistances
are in ohms.
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Chapter 1 Introduction
1.2.5 Voltmeters and Ammeters
(a) The correspondence between the color-coded probes of the
voltmeter and the reference direction of the measured
voltage. In (b) the + sign of vais on the left, while in (c) the +
sign of vb is on the right. The colored probe is shown here in
blue. In the laboratory this probe will be red. We will refer to
the colored probe as the “red probe.”
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Chapter 1 Introduction
1.2.5 Voltmeters and Ammeters
(a) The correspondence between the color-coded probes of
the ammeter and the reference direction of the measured
current. In (b) the current ia is directed to the right, while in
(c) the current ib is directed to the left. The colored probe is
shown here in blue. In the laboratory this probe will be red.
We will refer to the colored probe as the “red probe.”
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Chapter 1 Introduction
1.2.6 Switches
18
Chapter 1 Introduction
1.3 Power and Energy
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Chapter 1 Introduction
1.3 Power and Energy
p  vi  V  A  J/C  C/s  J/s  Watt
Passive reference configuration:
Current reference enters the positive
polarity of the voltage.
In this case, p=vi .
A positive p means energy is being
absorbed by the element; a negative
p means energy is being supplied by
the element.
p  vi
Note: a positive p for a battery means:
charging the battery.
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1.3 Power and Energy
Example 1.2
21
Chapter 1 Introduction
1.4 Kirchhoff’s Current Law
Node: a joint of two or more circuit elements
Kirchhoff’s Current Law (KCL):
The net current entering (leaving) a node is zero, or
N
i
n1
n
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Chapter 1 Introduction
1.4 Kirchhoff’s Current Law
Exercise 1.7
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Chapter 1 Introduction
1.5 Kirchhoff’s Voltage Law
Loop: a closed electrical path
Kirchhoff’s Voltage Law (KVL):
The algebraic sum of the voltage
equals zero for any loop in an
N
electrical circuit, or  v n  0
n 1
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Chapter 1 Introduction
1.5 Kirchhoff’s Voltage Law
Both Kirchhoff’s voltage and current laws are the result of
conservation of energy (i.e., power supplied = power absorbed).
KVL
KCL
 va A
i

vb B

i
i1

i
C vc
V

i3
i2

Vi 1  Vi 2  Vi 3 , or
- Vi 1  Vi 2  Vi 3  0
i1  i 2  i 3
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Chapter 1 Introduction
1.5 Kirchhoff’s Voltage Law
Two or more circuit elements are in parallel if both ends of one
element are connected directly to corresponding ends of others.
In the following figure, D,E,F are in parallel, but not B,D.
The voltage across parallel elements are equal in magnitude and
have the same polarity.
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Chapter 1 Introduction
1.5 Kirchhoff’s Voltage Law
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Chapter 1 Introduction
1.6 Basic Circuit Elements
1.6.1 Conductors
* Zero voltage drop in (ideal) conductors
* “short circuit”: two points are connected by conductor
All points connected by conductors can be considered as a single
point.
* “open circuit”: no connector between two pints in a circuit.
28
Chapter 1 Introduction
1.6.1 Conductors
A sudden increase of the current in a circuit may cause elements
(e.g., source) to burnout, sometimes even initiate a fire!
We can use fuses to prevent currents from becoming too large.
29
Chapter 1 Introduction
1.6.2 Voltage Sources
* Independent Voltage Sources maintain a specific voltage
across its terminals, independent of other elements in
the circuit.
f=1
30
Chapter 1 Introduction
1.6.2 Voltage Sources
* Dependent (or Controlled) Voltage Sources
voltage-controlled voltage source, current-controlled voltage source
31
Chapter 1 Introduction
1.6.3 Current Sources
* Independent Current Source forces a specific current to flow
through itself, independent of other elements in the circuit.
f=50
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Chapter 1 Introduction
1.6.3 Current Sources
* Dependent (or Controlled) Current Sources
voltage-controlled current source, current-controlled current source
33
Chapter 1 Introduction
1.6.4 Resistors and Ohm’s Law
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Chapter 1 Introduction
1.6.4 Resistors and Ohm’s Law
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Chapter 1 Introduction
1.6.4 Resistors and Ohm’s Law
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Chapter 1 Introduction
1.6.4 Resistors and Ohm’s Law
37
Chapter 1 Introduction
1.6.4 Resistors and Ohm’s Law
38
Chapter 1 Introduction
1.6.4 Resistors and Ohm’s Law
*Ohm’s Law: v  iR
Resistance R:
R ( Ω )  v (V) / i(A)
* Conductance: G ( Siemens S or Ω -1 )  1 /R, i  Gv
*
ρL
,
A
ρ( Ω  m) : resistivity
R
39
Chapter 1 Introduction
1.6.5 Conductor, semiconductor, insulator
ρL
,
A
ρ( Ω  m) : resistivity
R
1.6.6 Power
* Power: p  vi  i 2 R  v 2 /R
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Chapter 1 Introduction
1.7 Circuit Basics
* Calculate the current, voltage, and power for each element
p R  v R i R  10  2  20W
p S  -v S i S  -10  2  -20W
pS  pR  0
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Chapter 1 Introduction
Example 1.6

use
arbitrary reference
iy
ix
vx

KVL : - v S - v x  o  - 10 - v x  0
 v x  -10V
Ω' s Law : v x  -5ix  i x  2A
KCL : i x  i y  0  i y  -2A
p S  i y v S  -2A(10V) -20W
p R  -i x v x  -2A(-10V) 20W
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Chapter 1 Introduction
Example 1.7: Using KVL, KCL, and Ohm’s Law to Solve a Circuit
15 V
iy 
3A
5Ω
i x  0.5i x  i y
ix  2 A
v x  10i x  20 V
Vs  v x  15
Vs  35 V
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Chapter 1 Introduction
Additional Notes
KCL :
Currents entering the node,
i1 - i 2  i 3  i4 - i5  0
Currents leaving the node,
- i1  i 2 - i 3 - i4  i5  0
i1
i5
i4
i2
i3
Current entering  Current leaving,
i1  i 3  i4  i 2  i5
KCL :
Current entering the boundary:
- i1  i 2  i 3 - i4 - i5  0
Currents leaving the boundary:
i1 - i 2 - i 3  i4  i5  0
Current entering  Current leaving
i 2  i 3  i1  i4  i5
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Additional Example – Find current and voltage
45
Chapter 1 Introduction
Additional Example –
Using KVL, KCL, and Ohm’s Law to Solve a Circuit
15 V
iy 
3A
5Ω
v x  10i x  20 V
i x  0.5i x  i y
Vs  v x  15
ix  2 A
Vs  35 V
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Chapter 1 Introduction
Quiz: Determine the power absorbed (received) or supplied
by elements C and D.
(1) Element C:
KVL (loop w/ B, C, D):
6v4  0
v  2V
PC  7 A  v  14W
A power of 14W is supplied by element C.
(2) Element D:
KCL (node b):
7  ( 10 )  i  0  i  3 A
PD  ( 4V )  i  12W
A power of 12W is absorbed (received) by element D.
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Chapter 1 Introduction
Quiz: Exercises 1.14 and 1.15
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Chapter 1 Introduction
KCL : i 1  i 2
KVL : - v 1 - v 2  0
 v 1  25V  v 2  -25V
i 1  i 2  v 2 /R  -1A
KCL : i R  i S  2A
Ω' s Law : v R  i R R  80V
KVL : v S  v R  80V
p R  v 2 i 2  (-25V)(-1A)  25W
p S  -v S i S  -160W
p S  v 1 i 1  (25V)(-1A) -25W
p R  v R i R  160W
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Chapter 2 Resistive Circuits
1. Solve circuits (i.e., find currents and voltages of interest) by
combining resistances in series and parallel.
2. Apply the voltage-division and current-division
principles.
3. Solve circuits by the node-voltage technique.
4. Solve circuits by the mesh-current technique.
5. Find Thévenin and Norton equivalents and apply
source transformations.
6. Apply the superposition principle.
7. Draw the circuit diagram and state the principles
of operation for the Wheatstone bridge.
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Chapter 2 Resistive Circuits
2.1 Series and parallel Resistances
2.1.1 Series Resistances
KVL : v  v 1  v 2  v 3 (      v n )
 R1 i  R2 i  R3 i(      Rn i)
 Req i
 Req  R1  R2  R3 (     Rn )
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Chapter 2 Resistive Circuits
2.1.2 Parallel Resistances
KCL : i  i 1  i 2  i 3 (      i n )

v
v
v
v


(  
)
R1 R 2 R 3
Rn
v

Req
 Req  1/(
1
1
1
1


  
)
R1 R 2 R 3
Rn
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Chapter 2 Resistive Circuits
Example 2.1 – Find equivalent resistance
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Chapter 2 Resistive Circuits
Exercise 2.1 – Find equivalent resistance
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Chapter 2 Resistive Circuits
2.2 Simple Network Analysis – Example 2.2 - Find all i, v, p
i 2  2A, i 3  1A, p s  -(90) 3  -270W, p1  10  (3)2  90W
p2  120W, p3  60W; p s  p1  p2  p3  0
55
Chapter 2 Resistive Circuits
2.3 Voltage-Divider and Current-Divider Circuits
Voltage-division Principle
Current-division Principle
Req 
v i  Ri i
v total
 Ri
R1  R 2  R 3 (     R n )
 v total
Ri
Req
1
( 1 R1 
v  Req i total 
i1 
1
R2
)

R1 R 2
R1  R 2
R1 R 2
i total
R1  R 2
R2
v

i total
R 1 R1  R 2
R1
v
i2 

i total
R 2 R1  R 2
i 1 : i 2  R 2 : R1
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Chapter 2 Resistive Circuits
Example 2.4
Rx 
R2 R3
30  60

 20Ω
R2  R3 30  60
vx 
Rx
20
vs 
 100  25V
R1  R x
60  20
vs
100
is 

 1.25A
R1  R x 60  20
R2
30
i3 
is 
 1.25  0.417A
R2  R3
30  60
vx
25
or i 3 

 0.417A
R3 60
Find v x , i s, and i 3
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Chapter 2 Resistive Circuits
Example 2.5
Req 
i1 
R2 R3
30  60

 20Ω
R2  R3 30  60
Req
R eq  R1
is 
20
15  10A
20  10
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Chapter 2 Resistive Circuits – Additional Example
59
Chapter 2 Resistive Circuits
– Quiz: Find equivalent resistance
60
Chapter 2 Resistive Circuits
2.4 Node-Voltage Analysis
2.4.1 Basic Procedures
(1) Selecting the Reference Node
(one end of a voltage source is a good choice)
(2) Assigning Node Voltages
(Labeling the voltages at each of the other nodes)
v 1 (or v 10 ), v 2 (or v 20 ),  
61
Chapter 2 Resistive Circuits
2.4.1 Basic Procedures
(1) Selecting the Reference Node
(2) Assigning Node Voltages
v 1 (or v 10 ), v 2 (or v 20 ),   
(3) Finding Element Voltages in Terms
of the Node Voltage
Use KVL to determine element voltages
-v 2  v x  v 3  0
 v x  v 2 - v 3 (  v 23 )
-v 1 - v y  v 2  0
 v y  v 2 - v 1 (  v 21 )
-v 1 - v z  v 3  0
 v z  v 3 - v 1 (  v 31 )
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Chapter 2 Resistive Circuits
2.4.1 Basic Procedures
(1) Selecting the Reference Node
(2) Assigning Node Voltages
(3) Finding Element Voltages in Terms of the Node Voltage
Passive Reference Configuration
v x  v nk  v n - v k , v y  v kn  v k - v n
i nk
n
 vx -
vn
k
vk
- vy 
i kn
(note : v x  -v y )
i nk  v nk /R, i kn  v kn /R
(note : i nk  -ikn )
63
Chapter 2 Resistive Circuits
2.4.1 Basic Procedures
(1) Selecting the Reference Node
(2) Assigning Node Voltages
(3) Finding Element Voltages in Terms of the Node Voltage
node 2 :
(4) Writing KCL Equations
(v 2 - v 1 )
(v 2 - v 3 )
v2


R2
R4
R3  0
node 3 :
(v 3 - v 3 )
v3
(v 3 - v 2 )


R1
R5
R3  0
node 1 :
v1  v s
v x  v nk  v n - v k , v y  v kn  v k - v n
ink
n
 vx -
vn
k
vk
- vy 
ikn
(note : v x  -v y )
i nk  v nk /R, i kn  v kn /R
(note : i nk  -ikn )
64
Chapter 2 Resistive Circuits
Quiz – Exercises2.12b
Find i b
65
Chapter 2 Resistive Circuits
Exercise 2.12b
66
Chapter 2 Resistive Circuits
Example 2.8 (Node-voltage
Analysis)
Find i x
67
Chapter 2 Resistive Circuits
Quiz: Exercise 2.9 (Node-voltage Analysis)
68
Chapter 2 Resistive Circuits
2.4.2 Circuit with Voltage Sources
* Can’t write KCL for node containing voltage source.
* Super-node combines several node.
* KCL: the net current flowing through any closed boundary is zero.
v 1 v 1 - (-15) v 2 v 2 - (-15)



 0 ..........(1)
R2
R1
R4
R3
69
Chapter 2 Resistive Circuits
2.4.2 Circuit with Voltage Sources
v 1 v 1 - (-15) v 2 v 2 - (-15)



0
R2
R1
R4
R3
..........(1)
* To solve for v, we will need another equation; for the other
super-node:
v
v
(-15)- v 2 (-15)- v 1
- 1  (- 2 ) 

 0 ..........(  )  eq.(1) eq.( )
R2
R4
R3
R1
* We obtain two identical equations!
* We will obtain dependent equations, if we use all the nodes in
writing current equations.
* We can use KVL to obtain another independent equation:
- v 1 - 10  v 2  0
..........(2)
or just v 2  v 1  10
Solve eqs.(1)and (2) for v 1 and v 2 .
70
Chapter 2 Resistive Circuits
Exercise 2.11
71
Chapter 2 Resistive Circuits
2.4.3 Circuits with Controlled Sources – Example 2.9
KCL equations at each node:
(Controlling variable) In terms of node voltage,
Substitute back:
72
Chapter 2 Resistive Circuits
2.4.3 Circuits with Controlled Sources – Example 2.10
Super-node containing voltage source:
KCL:
For node 3:
For the reference node:
KVL (back to the super-node):
v 1 - v 2  0.5 v x
eq.(3)
To expressv x (controlling variable)in terms of node voltages :
v x  v 3 - v1
eq.(4)
 Solve for v 1 , v 2 , v 3 , and v x .
73
Chapter 2 Resistive Circuits
Summary of Sec. 2.4 – Node-Voltage Analysis
1. Select a reference node and assign variables for the
unknown node voltages. If the reference node is chosen
at one end of an independent voltage source, one node
voltage is known at the start, and fewer need to be
computed.
2. Write network equations. First, use KCL to write current
equations for nodes and supernodes. Write as many
current equations as you can without using all of the
nodes. Then if you do not have enough equations
because of voltage sources connected between nodes,
use KVL (super-node) to write additional equations.
3. If the circuit contains dependent sources, find
expressions for the controlling variables in terms of the
node voltages. Substitute into the network equations, and
obtain equations having only the node voltages as
unknowns.
4. Put the equations into standard form and solve for the
node voltages.
5. Use the values found for the node voltages to calculate
any other currents or voltages of interest.
74
Chapter 2 Resistive Circuits
Quiz – Exercises 2.13b
Find i y
75
Chapter 2 Resistive Circuits
Exercise 2.13b
76
Chapter 2 Resistive Circuits – Additional Example
v a  8.667V ,
v b  2V ,
v c  10V
Find the value of A
77
Chapter 2 Resistive Circuits – Additional Example
Find the unknown shown
78
Chapter 2 Resistive Circuits – Additional Example
Find the unknown shown
79
Chapter 2 Resistive Circuits – Additional Example
80
Chapter 2 Resistive Circuits – Additional Example
v 1  10V ,
v 2  14V ,
v 3  12V ,
Find i b and the value of r
81
Chapter 2 Resistive Circuits
2.5 Mesh-Current Analysis
2.5.1 Basic Procedures
* Branch –Current and Mesh-Current Analyses
* Mesh currents flow around closed paths, it automatically satisfy KCL.
Brench Currents i 1 , i 2 , and i 3
 R1 i 1  R 3 i 3  v A
KVL : 
 - R3 i 3  R2 i 2  -v B
KCL : i 1  i 2  i 3
Mesh Currents i 1 and i 2
 R1 i 1  R3 (i1 - i 2 )  v A
KVL : 
 - R3 (i1 - i 2 )  R2 i 2  -v B

 R1 i 1  R3 (i1 - i 2 )  v A

 - R3 (i1 - i 2 )  R2 i 2  -v B
82
Chapter 2 Resistive Circuits
2.5 Mesh-Current Analysis
* Procedures of Mesh-Current Analysis:
(1) Choosing mesh current;
(2) Writing KVL equations for each mesh;
(3) Solving for mesh currents.
This is a loop but not a mesh (current)
83
Chapter 2 Resistive Circuits
Example 2.12 – Mesh Current Analysis
84
Chapter 2 Resistive Circuits
2.5.2 Mesh Current in Circuits Containing Current Sources
* We usually avoid writing KVL equations containing current
sources, since we don’t know the voltage drop across a current
source.
mesh 2 : 10(i2 - i1 )  5i 2  10  0
•
Another equation is needed, we may use:
i1  2A
•
Write current source in terms of mesh currents
85
Chapter 2 Resistive Circuits
2.5.2 Mesh Current in Circuits Containing Current Sources
mesh 3 : 3i3  4(i3 - i 2 )  2(i3 - i1 )  0
We can’t write KVL equations for meshes 1 and 2.
We have for the current source:
i 2 - i 1  5A
Still, one more equation is needed.
We create a super-mesh containing meshes 1 and 2 and apply KVL:
i1  2(i1 - i 3 )  4(i2 - i 3 )  10  0
86
Chapter 2 Resistive Circuits
2.5.3 Circuits Containing Controlled Sources – Example 2.13
Super mesh containing meshes 1 and 2
eq. (1)
For the source current:
eq. (a)
We also know that:
eq. (b)
From eqs. (a) and (b), we have:
eq. (2)
Solving eqs. (1) and (2), we have:
87
Chapter 2 Resistive Circuits
2.5.3 Standard Mesh-Current Analysis Procedures
1. If necessary, redraw the network without crossing conductors
or elements. Then define the mesh currents flowing around
each of the open areas defined by the network. For
consistency, we usually select a clockwise direction for each
of the mesh currents, but this is not a requirement.
2. Write network equations, stopping after the number of
equations is equal to the number of mesh currents. First, use
KVL to write voltage equations for meshes that do not contain
current sources. Next, if any current sources are present,
write expressions for their currents in terms of the mesh
currents. Finally, if a current source is common to two
meshes, write a KVL equation for the supermesh.
3. If the circuit contains dependent sources, find expressions for
the controlling variables in terms of the mesh currents.
Substitute into the network equations, and obtain equations
having only the mesh currents as unknowns.
88
Chapter 2 Resistive Circuits – Additional Example
89
Chapter 2 Resistive Circuits – Additional Example
90
Chapter 2 Resistive Circuits – Additional Example
91
Chapter 2 Resistive Circuits – Additional Example
92
Chapter 2 Resistive Circuits
Quiz – Exercises 2.20 (b) and 2.21 (b)
Exercise 2.20 (b)
Exercise 2.21 (b)
93
Chapter 2 Resistive Circuits
Exercise 2.20 (b)
94
Chapter 2 Resistive Circuits
Exercise 2.21 (b)
95
Chapter 2 Resistive Circuits
2.6 Thevenin and Norton Equivalent Circuits
* To replace two-terminal circuits by simple equivalent circuits.
2.6.1 Thevenin Equivalent Circuits
* Thevenin theorem: Any two-terminal circuit consisting resistances
and sources can be expressed as an independent voltage source
in series with a resistance.
96
Chapter 2 Resistive Circuits
2.6.1 Thevenin Equivalent
Circuits
* The Thevenin voltage equals
the open circuit voltage of the
original circuit.
voc
Vt  v oc
* The short circuit is the same
for the original circuit and for
the Thevenin equivalent.
i sc 
i sc
Vt
, i.e.,
Rt
Vt
Rt 
i sc
97
Chapter 2 Resistive Circuits
2.6.1 Thevenin Equivalent Circuits – Example 2.14
(1) Vt  v oc
i1 
vs
R1  R 2
15

 0.10A
100  50
Vt  v oc  R2 i 1
 50  0.10  5V
(2) Rt 
v oc
i sc
vs
15
i sc 

 0.15A
R1 100
Rt 
v oc
5

 33.3Ω
i sc
0.15
98
Chapter 2 Resistive Circuits
2.6.2 Zeroing Sources to Find Thevenin Resistance
* In zeroing a voltage source, we reduce its voltage to zero,
that is, to short the source.
* In zeroing a current source, we reduce its current to zero,
that is, to open the source.
* We can find the Thevenin resistance by zeroing the sources in the
original network and then computing the resistance between the
terminals, if there is no dependent source.
99
Chapter 2 Resistive Circuits
Example 2.15 – Zero Sources to Find Thevenin Resistance
(1) Zero the sources: Rt  Req  R1 //R2  4 Ω
(2) Find the short-circuit current of the original circuit,
i 2  0, i 1  20V/5Ω  4A
i sc  4  2  6A
Vt  i sc Rt  6A 4 Ω  24V
100
Chapter 2 Resistive Circuits
Example 2.16 - Thevenin Equivalent of a Circuit
with a Dependent Source
?
(1) We can’t fine the Thevenin resistance by zeroing the sources.
(2) Find the open-circuit voltage first (by node-voltage analysis).
at node 1, i x  2ix  v oc 10 , also i x  (10 - v oc )/5  v oc  8.57V
(3) Consider the short-circuit condition
i x  10V/5Ω  2A, i sc  3ix  6A  Rt  Vt /i sc  1.43Ω
101
Chapter 2 Resistive Circuits
Quiz – Exercise 2.22 : Find the Thevenin Equivalent
Vt  v oc  5A  10Ω  50V
10
i sc  5A 
 1A
40  10
Vt
Rt 
 50Ω
i sc
(1) Zeroing the source
Rt  Req  10  40  50 Ω
(2) Find i sc of the original circuit
10
 1A
40  10
(3)V t  i sc Rt  1A  50Ω  50V
i sc  5A 
(1) Zeroing the source
Rt  Req  10  40  50 Ω
(2) Find v oc of the original circuit
V t  v oc  5A  10 Ω  50V
102
Chapter 2 Resistive Circuits – Additional Example
To cause ib  2mA, R  ?
103
Chapter 2 Resistive Circuits – Additional Example
104
Chapter 2 Resistive Circuits – Quiz
To cause i L  2 A, RL  ?
105
Chapter 2 Resistive Circuits – Additional Example
?
106
Chapter 2 Resistive Circuits
2.6.3 Norton Equivalent Circuits
* Norton’s Theorem: Any two terminal circuit consisting
resistances and sources can be expressed as an
independent current source in parallel with a resistance.
* The resistance in Norton equivalent is the same as Thevenin
resistance.
(1) If we zero the source, the two circuits are the same.
(2) The Norton current is equal to the short-circuit current.
107
Chapter 2 Resistive Circuits
2.6.3 Norton Equivalent Circuits
or Vt  Rt i sc
or v oc  Rt i sc
108
Chapter 2 Resistive Circuits
Example 2.17 - Norton Equivalent Circuits
(1) Find open-circuit voltage
at the top node,
v x v oc - 15
v oc


0
4
R1
R2  R3
since v x 
R3
v oc  0.25voc
R2  R3
 v oc  4.62V
(2) Find short-circuit current
Under short - circuit condition, v x  0
 the controlled source appears open
 i sc 
vs
15V

 0.75A(  I n )
R1 20Ω
 Rn  Rt 
v oc 4.62

 6.15Ω
i sc
0.75
109
Chapter 2 Resistive Circuits
Quiz – Exercise 2.25 - Find Norton Equivalent Circuits
110
Chapter 2 Resistive Circuits
Quiz – Exercise 2.25 - Find Norton Equivalent Circuits
111
Chapter 2 Resistive Circuits – Additional Example
Find i SC and Rt
112
Chapter 2 Resistive Circuits – Additional Example
Find i SC :
i 2  i SC , i a  i 1  i SC
 mesh 1 KVL : 3i1  2( i 1  i SC )  6 ( i 1  i SC )  10  0

 7i1  4 i SC  0

 mesh 2 KVL : 5i  6 ( i  i )  0
SC
1
SC


 - 6i1  11i SC  0
 i SC  1.13 A (  I n )
113
Chapter 2 Resistive Circuits – Additional Example
Find v OC and Rt :
i 2  0 , ia  i1
mesh 1 KVL : 3i1  2 i a  6 i a  10  0
 i 1  i a  1.43 A
v OC  6 i a  6  1.43  8.58V
 Rt  v OC / i SC  8.58 / 1.13  7.57 Ω
114
Chapter 2 Resistive Circuits – Additional Example
Find Rt by zeroing the source :
There is a dependentsource!
(1) Zero the indep.source,(2)Connect a current source iT .
Rt  v T i T
 i 2   iT
 i i i
1
2
 a
 mesh 1 KVL : 3i1  2 i a  6 i a  0
 mesh 2 KVL : 5i  v  6 i  0
2
T
a


 v T  7.57 iT
 Rt  v T iT  7.57 Ω
115
Chapter 2 Resistive Circuits
2.6.4 Source Transformations
In this situation, both circuits have the same open-circuit voltage
and short-circuit current:
v oc  Vt (in Tnevenin Equivalent)  I n Rt (in Norton Equivalent)
i sc  I n (in Norton Equivalent)  Vt /Rt (in Thevenin Equivalent)
“Source Transformation” can be used to simplify problems.
116
Chapter 2 Resistive Circuits
2.6.4 Example 2.18 – Using
Source Transformations
(1) Approach I - Transform current
source to voltage source
Vt  I n Rt  1A  10Ω  10V
R1 i 1  R2 i 1  10 - 20  0
i 1  0.667A
i 2  i 1  1  1.667A
(2) Approach II - Transform
voltage source to current
source
I n  Vt /Rt  4A
i total  1  4  5A
R1
i2 
i total  1.667A
R1  R 2
i 1  1A  i 2  i 1  0.667A
117
Chapter 2 Resistive Circuits – Additional Example
Use source transformation to find the voltage
118
Chapter 2 Resistive Circuits
2.6.5 Maximum Power Transfer
What is the maximum power which can be delivered to the load?
We replace the original circuit by its Thevenin Equivalent
Vt
iL 
,
Rt  Rl
2
V
t RL
p L  i L2 R L 
(Rt  R L ) 2
to find a maximum
dpL
 0  R L  Rt
dRL
 p L man
Vt 2

4Rt
When the load resistance equals the Thevenin resistance, it can
absorb the maximum power from a two-terminal circuit.
119
Chapter 2 Resistive Circuits
Example: Find the maximum
power can be transferred.
120
Chapter 2 Resistive Circuits
2.7 Superposition Principle
Suppose that we have a circuit composed of resistances,
linear dependent sources and n independent sources, the
superposition principle states that the total response is the
sum of the responses to each of the independent sources
acting individually: r  r  r  r      r
T
1
2
3
n
Linear dependent source
vT  v 1  v 2
121
Chapter 2 Resistive Circuits
2.7 Superposition Principle
at top node :
at top node :
at top node :
v T - v s1 v T

 ki x  i s2
R1
R2
(v1 - v s1 ) v 1

 ki x  0
R1
R2
v2
v
 2  ki x  i s2
R1 R 2
i x  v T /R2
i x  v 1 /R2
i x  v 2 /R2



(R2 v s1  R1 R2 i s2 )
vT 
(R1  R2  kR1 )
R2 v s1
v1 
(R1  R2  kR1 )
v2 
R1 R2 i s2
(R1  R2  kR1 )
vT  v 1  v 2
122
Chapter 2 Resistive Circuits
2.8 Wheatstone Bridge
* Wheatstone bridge is used to precisely measure unknown
resistance.
* When the bridge is in balance:
i g  0 and v ab  0
node a : i 1  i 3
node b : i 2  i 4
KVL : R1 i 1  R2 i 2 ,
R3 i 3  R x i 4 (i.e., R3 i 1  R x i 2 )

R3 R x
R

 R x  2 R3
R1 R 2
R1
123
Chapter 2 Resistive Circuits
Example 2.21 Wheatstone Bridge
124
Chapter 2 Resistive Circuits
2.8 Wheatstone Bridges
125
Chapter 2 Resistive Circuits
Additional Example: Wheatstone Bridge
126
Chapter 2 Resistive Circuits
Additional Example: Wheatstone Bridge
127
Chapter 2 Resistive Circuits - SUMMARY
128
Chapter 2 Resistive Circuits - SUMMARY
129
Chapter 2 Resistive Circuits - SUMMARY
130
Chapter 2 Resistive Circuits - SUMMARY
131