Transcript 4991_Chapter_3_Fall_..

ECE 4991 Electrical and Electronic
Circuits
Chapter 3
Where are we?
• Chapter 2 - The basic concepts and practice at
analyzing simple electric circuits with sources
and resistors
• Chapter 3 – More harder networks to analyze and
the notion of equivalent circuits
• Chapter 4 – Capacitors and inductors added to
the mix
• Chapter 5 – Analyzing transient situations in
complex passive networks
• Chapter 8 – New subject – the wonders of
operational amplifiers as system elements
• Chapter 9 – Introduction to semiconductors – the
basics and diodes – more network analysis
• Chapter 10 – Bipolar junction transistors and how
they work – now you can build your own op amp
2
What’s Important in
Chapter 3
1.
2.
3.
4.
5.
Definitions
Nodal Analysis
Mesh Analysis
The Principle of Superposition
Thevenin and Norton Equivalent
Circuits
6. Condition for Maximum Power
Transfer
3
1. Definitions
•
•
•
•
•
•
•
•
Node voltages
Branch currents
“Ground”
KCL
Nodal Analysis
Mesh currents
KVL
Mesh Analysis
• Principle of
Superposition
• Equivalent circuit
• Thevenin theorem
• Norton theorem
• One-port networks
• Source loading
4
2. Nodal Analysis
• Used to “analyze” circuits
• Solve for currents, voltages, power,
etc., throughout circuits
• Applies KCL to nodes
– Often used in concert with Ohm’s Law
5
Node Method
• Find nodes – Identify ground node
• Label branch currents & node voltages
• Node voltages, if not defined by a voltage
source, are independent variables
• Write KCL for nodes
• Solve for unknowns
6
Working with Nodal
Analysis
1.
2.
3.
Reference Node Selection
(usually Ground)
Define remaining n-1 node
voltages.
Apply KCL to obtain Nodal
Expressions for node a and
b respectively:
R2
Node a
iS
iS
i1
R1
Node b
i2
i3
R3
is – i1 – i2 = 0
i2 – i3 = 0
Node c
7
Working with Nodal
Analysis (cont.)
4.
Apply Node Voltage Method
va  vc
i1 
R1
va  vb
i2 
R2
R2
Node a
iS
iS
i1
R1
vb  vc
i3 
R3
Node b
i2
i3
R3
Node c
8
Working with Nodal
Analysis (cont.)
• Substitution step leads to:
is 
va
v  vb
 a
 0
R1
R2
va  vb
v
 b  0
R2
R3
• Or equivalently to:
1
 1 
1 
  va    vb  is
 R1 R2 
 R2 
 1
 1 
1 
  va    vb  0
 R2 
 R2 R3 
9
Example 3.2
• Solving for all unknown currents and voltages in the circuit
of Figure 3.5
– R1=1kΩ, R2=2kΩ, R3=10kΩ, R4=2kΩ,
– I1=10mA, I2=50mA
R3
Node 2
R2
Node 1
I1
I2
R1
R4
10
Example 3.2 (cont.)
• Applying KCL at nodes 1 & 2:
v1  0 v1  v2 v1  v2
I1 


0
R1
R2
R3
v1  v2 v1  v2 v2  0


 I2  0
R2
R3
R4
• After rewriting:
1
 1
1
1 
1 
 


 v1   
 v2  I1
 R1 R2 R3 
 R2 R3 
 1
 1
1 
1
1 
 
 v1   
 v2   I 2
 R2 R3 
 R2 R3 R4 
11
Example 3.2 (cont.)
• Substituting actual values leads to
the system of equations:
1.6v1  0.6v2  10
 0.6v1  1.1v2  50
v1  13.57V
v2  52.86V
v1  v2
 3.93mA
10 k
v1

 13.57 mA
1k
iR3 
iR1
12
Working with Nodal
Analysis with Voltage
Sources
va
v
b
R1
Vs
R2
R3
13
Working with Nodal
Analysis
R4
R2
I
V
R1
R3
14
Working with Nodal
Analysis
R3
R1
V
I
R4
R2
R5
R6
15
Working with Nodal
Analysis
16
For Next Time
1. Sign onto Blackboard, if still have
not
2. Practice Nodal Analysis
3. Learn about rest of chapter 3,
particularly about mesh analysis
17
3. Mesh Analysis
• Also used to “analyze” circuits
• Solve for currents, voltages, power,
etc., throughout circuits
• Applies KVL to meshes
– Often used in concert with Ohm’s Law
18
Mesh Method
• Identify meshes and mesh currents
• For n meshes and m current sources, there
are n-m independent variables
• Write KVL for all meshes with unknown
mesh currents
• Solve for unknowns
I
19
Working with Mesh
Analysis
1.
Defining Meshes (use the rule consistently: e.g.,
clock wise)
v1
v3
_
_
R
+ R
+
1
+
+
Vs
DC
3
i1
v2R2
_
i2
R4
v4
_
20
Mesh Analysis
2. Apply KVL
v1
+
R1
i1
v2R2
_
•
•
v3
R3
_
+
+
Vs
DC
_ +
i2
R4
v4
_
Vs-v1-v2 = 0
v2-v3-v4= 0
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Mesh Analysis
• Apply Ohm law:
v1
+
R1
v3
R3
_
+
+
Vs
DC
_ +
i1
v2R2
i2
v4
R4
_
_
Vs  i1 R1  i1  i2 R2  0
i1  i2 R2  i2 R3  i2 R4  0
22
Mesh Analysis
• Rewriting the equations:
v1
+
R1
v3
R3
_
+
+
Vs
DC
_ +
i1
v2R2
_
i2
v4
R4
_
R1  R2 i1  R2i2  Vs
 R2i1  R2  R3  R4 i2  0
23
Working with Mesh
Analysis
R1
I
R2
R3
24
Working with Mesh
Analysis
R4
R2
I
V
R1
R3
25
Working with Mesh
Analysis
R6
R3
R1
I
R4
R2
R5
V
26
Working with Mesh
Analysis
27
For Next Time
1. Sign onto Blackboard, if still have
not
2. Keep practicing Nodal Analysis
3. Practice Mesh Analysis
4. Learn about rest of chapter 3,
particularly about equivalent
circuits
28
4. The Principle of
Superposition
•
•
When working with linear circuits, can
find the solution for each energy source
and combine the results (Principle of
Superposition)
Procedure:
–
Remove all but one energy source
•
•
–
–
–
V sources  short wires
I sources  opens
Solve the circuit
Repeat for a different energy source
Add up the solutions
29
SUPERPOSITION:
Section 3.5 Superposition Principle:
 The output of a circuit can be found by finding the contribution
from each source acting alone and then adding the individual
responses to obtain the total response.
iT = i1 + i2
V1
V2
+
+
-
V1
iT
R
+
i1
R
+
V2
+
-
i2
R
30
SETTING SOURCES EQUAL TO ZERO:
Voltage Source:
 In order to set a voltage source to zero, it is replaced by a
short circuit.
R1
VS
R2
iS
+
-
R1
Voltage source
set equal to zero
R3
R2
iS
R3
31
SETTING SOURCES EQUAL TO ZERO:
Current Source:
 In order to set a current source to zero, it is replaced by an
open circuit.
R1
VS
iS
+
-
R1
Current source
set equal to zero VS
+
-
R2
R3
R2
R3
32
SUPERPOSITION:
Example 10:
 Calculate VR using superposition:
400
200
+
5V
+
-
5mA
VR
250
-
33
SUPERPOSITION:
Example 10 cont.:
1. Turn off all independent sources except one and find response
due to that source acting alone.
Turning off voltage source:
400
200
+
Voltage source
set equal to zero
5mA
VR
250
-
34
SUPERPOSITION:
Example 10 cont.:
 VR due to current source only (VR1):
i1 = 5mA
400
Current divider
400 + 200 + 250
Vi = i1250 = 2.353mA250 = 0.588V
VR1 = -0.588V
VR1 = -Vi
400
200
-
+
VR
5mA
i1
-
250
Vi
+
35
SUPERPOSITION:
Example 10 cont.:
 VR due to voltage source only (VR2):
VR2 = 5V
250
400 + 200 + 250
= 1.471V
Voltage divider
VR2 = 1.471V
400
200
+
+
5V
+
-
VR
-
250
VR2
36
SUPERPOSITION:
Example 10 cont.:
VR = VR1 + VR2 = -0.588V + 1.471V
VR = 0.882V
400
200
+
5V
+
-
5mA
VR
250
37
THEVENIN and NORTON CIRCUITS:
Section 3.5
 Thevenin and Norton circuits deal with the concept of equivalent
circuits.
 Even the most complicated circuits can be transformed into
an equivalent circuit containing a single source and resistor.
 When viewed from the load, any network composed of ideal voltage
and current sources, and of linear resistors, may be represented by an
equivalent circuit consisting of an ideal voltage source VT in series
with an equivalent resistance RT.
A
A
RT
VT
+
-
Thevenin
Circuit
B
IN
RN
Norton
Circuit
B
38
THEVENIN and NORTON CIRCUITS:
Thevenin Equivalent Circuits:
 At this point you should be asking yourself two questions; how do
we calculate the Thevenin voltage and the Thevenin resistance?
Thevenin Voltage (VT):
The Thevenin voltage is equal to the open-circuit voltage at
the load terminals with the load removed.
R1
R3
+
VS
+
-
R2
VOC
RL
VT = VOC
39
THEVENIN Equivalent
CIRCUITS:
R1
R3
v
+
VS
+
-
i
R2
VOC
VT = VOC
-
Vs  R1i  R2i  0
Vs
i
R1  R2
R2
VT  v  R2i 
Vs
R1  R2
40
THEVENIN and NORTON CIRCUITS:
Thevenin Resistance (RT):
The Thevenin resistance is the equivalent resistance seen by
the load with all independent sources removed (set equal to
zero).
R1
VS
+
R3
R2
-
RL
RT  R1 || R2  R3 
R1
VS set equal
to zero
R3
R2

1
1
1

R1 R2
R1 R2
 R3
R1  R2
REQ = RT
41
 R3
THEVENIN and NORTON CIRCUITS:
Example 11:
Find the Thevenin equivalent circuit at terminals ‘A’ and ‘B’:
 Thevenin Voltage:
VT = VOC = VAB = V12 = io12
Using mesh analysis:
20
iO = 211.27mA
5
A
15V
+
-
10
12
io +
VOC
B
42
THEVENIN and NORTON CIRCUITS:
Example 11 cont.:
VT = VOC = 211.3mA12
VT = 2.54V
Thevenin Resistance:
Setting all sources equal to zero and looking back into the
circuit from terminals “A” and “B”:
RT = 5.92 
RT = [(20 10) + 5] 12
20
5
A
15V
+
-
10
12
io +
VOC
B
43
Check
THEVENIN and NORTON CIRCUITS:
Example 12:
Find the Thevenin equivalent circuit seen by the load RL:
VT: VT = VOC = V700 = 4mA700
4mA
VT = 2.8V
500
400
+
+
10mA
200
700
V700
-
VOC
RL
44
THEVENIN and NORTON CIRCUITS:
Example 12 cont.:
Thevenin Resistance: Set all sources equal to zero:
RT: RT = 500 + 700
RT = 1.2k
500
400
+
+
200
700
V700
-
VOC
RL
45
THEVENIN and NORTON CIRCUITS:
Example 12 cont.:
Thevenin Resistance:
iSC = 4mA
RT =
700
700 + 500
VT
iSC
= 2.33mA
RT = 1.2k
RT = 2.8V/2.33mA
400
10mA
200
Current Divider
500
4mA
700
+
iSC
V700
46
THEVENIN and NORTON CIRCUITS:
Norton Equivalent Circuits:
Norton Current (IN):
The Norton current is equal to the short-circuit current at the
load terminals with the load removed.
R1
VS
+
-
R3
iSC
R2
IN = iSC
47
Norton Equivalent Circuit
R1
VS
Mesh
Analysis:
R3
v
iSC
+
R2
i1
-
i2
R1  R2 i1  R2isc  Vs
 R2i1  R2  R3 isc  0
Node Analysis: Vs  v  v  v
R1
R2
R3
R2 R3
v
Vs
R1 R2  R1 R3  R2 R3
48
Norton Equivalent Circuit
v
R2
iN  isc 

Vs
R3 R1 R2  R1 R3  R2 R3
49
THEVENIN and NORTON CIRCUITS:
Norton Resistance (RN):
The Norton resistance is the equivalent resistance seen by the
load with all independent sources removed (set equal to zero).
R1
VS
R3
iSC
+
R2
-
R1
VS set equal
to zero
R3
R2
REQ = RN
50
THEVENIN and NORTON CIRCUITS:
Summary:
Thevenin Equivalent Circuit:
The Thevenin voltage is equal to the open-circuit voltage at the
load terminals with the load removed.
The Thevenin resistance is the equivalent resistance seen by the load
with all independent sources removed (set equal to zero) or VT/iSC.
VT = VOC = VAB (with load removed)
RT = VT/iSC = REQ as seen by RL
51
THEVENIN and NORTON CIRCUITS:
Summary:
Norton Equivalent Circuit:
The Norton current is equal to the short-circuit current at the load
terminals with the load removed.
The Norton resistance is the equivalent resistance seen by the load
with all independent sources removed (set equal to zero) or VT/IN.
IN = ISC (with load removed)
RN = RT VOC/IN = REQ as seen by RL
52
SOURCE TRANSFORMATION:
 Source transformation allows for the conversion of an ideal
voltage source in series with a resistor to an ideal current
source in parallel with a resistor and vice versa.
 As previously seen, any circuit can be transformed to its Thevenin or
Norton equivalent circuit at the load resistance RL. Therefore, a
voltage source in series with a resistor (Thevenin) can be
transformed to a current source in parallel with a resistor (Norton)
and the V-I characteristics at the terminals “A” “B” will be the same.
R1
VS
A
A
Rest
of
Circuit
+
B
IS
Source
Transformation
Rest
of
Circuit
R1
B
53
SOURCE TRANSFORMATION:
 Source transformation allows for the conversion of an ideal voltage
source in series with a resistor to an ideal current source in parallel
with a resistor and vice versa.
+
R1
A
-
VS = ISR1
Rest
of
Circuit
B
Source
Transformation
A
IS = VS/R1
R1
Rest
of
Circuit
B
54
SOURCE TRANSFORMATION:
 Source transformation allows for the conversion of an ideal voltage
source in series with a resistor to an ideal current source in parallel
with a resistor and vice versa.
R1
VS
+
R3
R4
R2
-
IS = VS/R1
Voltages across
and currents
through R2, R3,
and R4 are the
same for both
circuits!
R3
R1
R2
R4
55
SOURCE TRANSFORMATION:
Example 13:
Use source transformation and current divider rule to
calculate io:
1k
1.5k
iO
8V
+
-
2k
300
56
SOURCE TRANSFORMATION:
Example 13 cont.:
Converting the voltage source in series with the 1.5k resistor to a
current source in parallel with a resistor we have the following
circuit:
Same V-I characteristics
iO
1k
8V/1.5k
= 5.33mA
1.5k
2k
300
57
SOURCE TRANSFORMATION:
Example 13 cont.:
iO = 5.33mA
1/1.3k
1/1.5k + 1/2k + 1/1.3k
iO = 2.12mA
iO
1k
5.33mA
1.5k
2k
300
58