Download Ch2 Basic Analysis Methods to Circuits

Engineering Circuit Analysis
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
2.2 Basic Nodal and Mesh Analysis
2.3 Useful Circuit Analysis Techniques
References: Hayt-Ch3, 4; Gao-Ch2;
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Key Words:
Equivalent Circuits Network
Equivalent Resistance,
Equivalent Independent Sources
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Circuits Network
о
о
Two-terminal Circuits Network
cо
о
d
о
о
a о
5
6
15
о
5
bо
о
I
I
a
о
+
a
о
N1
+
V
V
_
о
о
b
b
N2
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Resistance
How do we find I1 and I2?
I
I1
I2
R1
R2
1
1 1

R1 R2
V
R2
I
R1
R1  R2
V
R1
I2 
I
R2
R1  R2
I1 
V
-
V I
I1 + I2 = I
+
I
I
1 1
V V
  V   
R1 R2  R1 R2 
R1R2
R1  R2
Req 
I
R1 R2
R1  R2
V
V

R1 R2
Req
R1  R2
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Resistance
i(t)
i(t)
+
v(t)
-
+
v(t)
Req
-
Req is equivalent to the resistor network on the left in the sense that they
have the same i-v characteristics.
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Resistance
Condition : without knowing V&I . We only know Rs
Series and parallel Resistance
Method 1
n
1
1
＝
R eqp k 1 R k
n
Reqs   Rk
K 1
(source-free)
n
G   Gk
k 1
Condition : without knowing Rs . We only know V&I
a
Method 2
I
source
V
-free
b
Method 3
V
Ro  oc
I sc

Ro  Rab  V
I

source
Voc


source
Isc

Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Resistance
In practice , Vs-source voltage ≠ VL- Local
VL  Vs  I L R
Let Vs  10V;R  0.1
The IL-VL curve :
VL  0, RL  0 , Short Circuit (SC) . I SC  100A
IL(A)
100
0
10
VL  0, RL   , Open Circuit (OC) . VOC  10V
VOC
R

 0.1
O
VL(V)
I SC
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Resistance
P2.1

R2
R1
How do we find Rab?
a
R3
+
-
VS
Method 1
b
Ro  Rab  R1  R2  // R3 
I
Method 2
R2
R1
R1  R2 R3
R1  R2  R3
I
a

R3
V
V
R R R

 1 2 3V
R3 R1  R2 R1  R2 R3
V

Ro  Rab 
b
V R1  R2 R3

I R1  R2  R3
Method 3
VoC  
VS
R1  R2
R3  VS 
VS I  V / R
sc
s
3
R1  R2  R3
R1  R2  R3
Ro 
Voc
I sc
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Source Transformation
• Ideally:
– 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
• Practice:
– A real voltage source cannot supply arbitrarily large amounts of
current
– A real current source cannot have an arbitrarily large terminal
voltage
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Source Transformation
Rs
Vs
+
Is
-
Vs  Rs I s
Rs
Is 
Vs
Rs
Note: Consistency between the current source ref. direction and the
voltage Source ref. terminals.
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Source
How do we find I1 and I2?
Is1
Is2
R2
V
  I s1  I s 2 
R1
R1  R2
R1
V
  I s1  I s 2 
R2
R1  R2
I1
I2
+
I1 
R1
R2
V
I2 
-
I1  I 2  I s1  I s 2
1 1
V V
I s1  I s 2    V   
R1 R2
 R1 R2 
V  I s1  I s 2 
Ieq
R1 R2
R1  R2
Ch2 Basic Analysis Methods to Circuits
2.1 Equivalent Circuits
Equivalent Source
Series Voltage Source
n
+ VS1
-
+ VS2
-
+ VSn

-
+ VS

VS  VSk
k 1
-
n


R S   R Sk
k 1
parallel Current Source

IS1
IS2
ISn
I

RS=RS1// RS2//…// RSn
IS
n


I S   I Sk
k 1
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Key Words:
Branch Analysis, Nodal Analysis,
Mesh (Loop) Analysis
• Why?
– The analysis techniques previously (voltage divider, equivalent
resistance, etc.) provide an intuitive approach to analyzing circuits
– They are not systematic and cannot be easily automated by a
computer
• Comments:
– Analysis of circuits using node or loop analysis requires solutions
of systems of linear equations.
– These equations can usually be written by inspection of the circuit.
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Branch Analysis
P2.2
How do we find I1 and I2, I3?
R3=80
R1=0.5
I1
I2
R2=0.4
+
VS=14V
_
KCL
I3
KVL
Mesh 1: 14  0.5I1  0.4I 2  12  0
2  0.4 I 2
I1 
 4  0.8I 2
0.5
Mesh 2: 12  0.4I 2  80I3  0
I3 
E2=12V
12  0.4 I 2
 0.15  0.005I 2
80
4  0.8I 2  I 2  0.15  0.005I 2
I1  I 2  I 3
I 2  2.13A
I1  2.29A
Vs  E2  I 2 R2 2  0.4 I 2
E  I R 12  0.4 I 2
I3  2 2 2 

R3
80
R1
0.5
2  0.4 I 2
12  0.4 I 2
 I2 
I 2  2.14A I1  2.29A I 3  0.14A
0.5
80
I1 
I3  0.16A
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Branch Analysis

Suppose m branches, n nodals

write KCL equation for each independent node.
——(n-1) KCL equations

write KVL equation for each independent mesh/loop
——m-(n-1) KVL equations
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Branch Analysis
Here’s a quick example of a circuit that we will see later when we
model the operation of transistors. For now, let’s assume ideal
independent and dependent sources.

Ⅰ
We can write the following
equations:
Ⅰ:
Ⅱ:
Ⅱ
Ⅲ
Ⅲ:
i1  iC  iCC  0
iB  i2  i1  0
iE  iB  iC  0
iC   iB
Vo  iE RE  i2 R2  0
VCC  i1R1  i2 R2  0
iB
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
1) Choose a reference node
500
500
+
I1
500
1k
V
500
-
The reference node is called the ground node.
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
1) Choose a reference node
500
.
Ⅰ I4
I5
I1
500
.I
500
Ⅱ
+
I8
6
1k
V
-
0
KCL
Ⅰ:
Ⅱ:
Ⅲ:
.
I7 Ⅲ
500
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
2) Assign node voltages to the other nodes
500
V1 I
4
1
I1
I5
V2
2
500
I6
I8
1k
500
I 7 V3
3
500
0
V1, V2, and V3 are unknowns for which we solve using KCL.
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
3) Apply KCL to each node other than the reference-express
currents in terms of node voltages.
V1 I4 500
V2 500 I7 V3
1
I5
I1
2 I6
I8
1k
500
3
500
0
KCL •
•
•
Node ① : I 4  I5  I1  0
Node ② : I 6  I 4  I 7  0
Node ③ : I8  I 7  I 2  0
V1
V1  V2
I

5
500
500 ,
V V
V
I6  2 , I7  3 2
1K
500
I4 
,
,
I8 
V3
500
V1  V2
V
 1 0
500 500
V  V1 V2 V2  V3
Node ② : 2


0
500 1k 500
V3  V2
V
 3  I2  0
Node ③ :
500 500
Node ① :  I1 
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
4) Solve the resulting system of linear equations.
• Node 1:
1 
V
 1
V1 

  2  I1
 500 500  500
• Node 2:
• Node 3:
•
•

V1
1
1 
V
 1
 V2 


 3 0
500
 500 1k 500  500

V2
1 
 1
 V3 

  I2
500
 500 500 
V1 500
I1
1
V2
500 V3
2
500
3
1k
500
0
The left hand side of the equation:
– The node voltage is multiplied by the sum of conductances of all
resistors connected to the node.
– The neighbourly node voltages are multiplied by the conductance of the
resistor(s) connecting to the two nodes and to be subtracted.
The right hand side of the equation:
– The right side of the equation is the sum of currents from sources
entering the node.
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
4) Solve the resulting system of linear equations.
• Node 1:
1 
V
 1
V1 

  2  I1
 500 500  500
• Node 2:
• Node 3:

V1
1
1 
V
 1
 V2 


 3 0
500
 500 1k 500  500

V1 500
I1
1
V2
500 V3
2
500
V2
1 
 1
 V3 

  I2
500
 500 500 
3
1k
Matrix Notation(Symmetric)
1
 1

 500 500

1
 
500


0

1
500
1
1
1


500 1k 500
1

500


 V
 1   I1 

1
 V2    0 

500     
1
1  V3   I 2 

500 500 
0
500
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
4) Solve the resulting system of linear equations.
• Node 1:
• Node 2:
• Node 3:
1 
V
 1
V1 

  2  I1
 500 500  500

V1
1
1 
V
 1
 V2 


 3 0
500
 500 1k 500  500

V1 500
I1
1
V2
1 
 1
 V3 

  I2
500
 500 500 
G11V1+G12V2 +G13V3 =I11
G21V2+G22V2 +G23V3 =I22
G31V1+G32V2+G33V3=I33
V2
500 V3
2
500
3
1k
500
I2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
What if there are dependent sources?
Example:
V1
Ib
1
5mA
1k
V2
50
100Ib
2
1k
+
Vo
V
V1  V2
 5mA
1k 50
Node ①： 1 
1
 1

 1k 50
 1
100


 50 50
V  V1
V
 100 I b  2  0
50
1k
Node ② ：2
1

 V1  5mA 
50

1
100
1  V2   0 



50 50 1k 

Ib 
V1  V2
50
V2  V1
V V
V
 100 1 2  2  0
50
50 1k
Matrix is not symmetric due to the dependent source.
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
What if there are voltage sources?
0.7V
R1
I b V2
1k
2
1
V1
Node 2:
Node 3:
+
-
+
3k
-
V3
3
V4
R3
50
+
4
100Ib
R2
R4
Vo
1k
-
V2  V1 V2

 I b  0 Difficulty: We do not know I – the current
b
1k
3k
through the voltage source?
V3  V4
 Ib  0
V3  V4 V4
50


 100 I b  0
Node 4:
50 1k
Independent Voltage Source: V3  V2  0.7V
Equations: KCL at node 2, node 3, node 4, and V3  V2  0.7V
Unknowns: Ib, V1, V2 (V3),V4
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
What if there are voltage sources?
CURRENT
CONTROLLED
VOLTAGE
SOURCE
Io=?
V2  V1  2kIx
 V1  2kI x  V2  2V1
KCL AT SUPERNODE
Ix 
V1
V2
- 4mA + + 2mA + = 0
2k
2k
V1
2k
V1  V2  4(V )  3V2  8(V )
 IO 
V2 4
 mA
2k 3
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Nodal Analysis
Advantages of Nodal Analysis
•
•
•
•
•
Solves directly for node voltages.
Current sources are easy.
Voltage sources are either very easy or somewhat difficult.
Works best for circuits with few nodes.
Works for any circuit.
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh(Loop) Analysis
1) Identifying the Meshes
1k
V1
1k
+
-
+
Mesh 1
Mesh 2
-
V2
1k
Mesh: A special kind of loop that doesn’t contain any loops within it.
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh(Loop) Analysis
2) Assigning Mesh Currents
1k
V1
1k
+
-
1k
I1
+
I2
-
V2
3) Apply KVL around each loop to get an equation in terms of the
loop currents.
I1 ( 1k + 1k) - I2 1k = V1
For Mesh 1: -V1 + I1 1k + (I1 - I2) 1k = 0
For Mesh 2: (I2 - I1) 1k + I2 1k + V2 = 0
- I1 1k + I2 ( 1k + 1k) = -V2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh(Loop) Analysis
3) Apply KVL around each loop to get an equation in terms of the
loop currents.
1k
I1 ( 1k + 1k) - I2 1k = V1
- I1 1k + I2 ( 1k + 1k) = -V2
V1
+
-
1k
1k
I1
4) Solve the resulting system of linear equations.
 1k   I1   V1 
1k  1k




  1k

I

V
1
k


1
k


 2   2 
I2
+
-
V2
Ch2 Basic Analysis Methods to Circuits

 

+
5
5
2.2 Basic Nodal and Mesh Analysis
I1
US
_
I1
5
MeshI (Loop)
I I Analysis
I I
1
1
1
1
1
5
I1
R11I m1  R12 I m 2  R13 I m 3  VS 11
R21I m1  R22
 I m 2  R23 I m 3  VS 22
I
E
I1
S
R31I m1  R32 I m 2  R33 I m 3  VS 33
I3
Mesh 1: I m1R1  VS1  I m1  I m3 R6  VS 2  I m1  I m 2 R2  0 R3

Mesh 2: I m 2 R3  VS 3  I m 2  I m3 R5  VS 2  I m1  I m 2 R2  0
Mesh 3: I m 2  I m3 R5  I m3  I m1 R6  I m3 R4  VS 4  0

I2
_ VS1 +
R1
Im1
Mesh 1
R2
+
VS2
Mesh 2
Im2
_
+

_
VS3
I6
I5
R5
R6
Mesh 3 R
4

I4
Im3

_
+
VS4
( R1  R2  R6 ) I m1  R2 I m 2  R6 I m 3  VS 1  VS 2
R2 I m1  ( R2  R3  R5 ) I m 2  R5 I m 3  VS 2  VS 3
 R6 I m1  R5 I m 2  ( R4  R5  R6 ) I m 3  VS 4
R2
 R6
 R1  R2  R6
  I m1  VS1  VS 2 

  I   V  V 
R
R

R

R
R
2
2
3
5
5

  m2   S 2 S 3 
  R6
R5
R4  R5  R6   I m3   VS 4 
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh (Loop) Analysis
R11I m1  R12 I m 2  R13 I m3  VS11
R21I m1  R22 I m 2  R23 I m3  VS 22
R31I m1  R32 I m 2  R33 I m3  VS 33
P2.4(P2.2)
R3=80
R1=0.5
+
VS=14V
_
Mesh 1:
 R1  R2  I m1  R2 I m2  VS  12
R2=0.4
Im2
Im1
E2=12V
VS  I m1R1   I m1  I m 2  R2  12  0
Mesh 2:
12   I m 2  I m1  R2  I m 2 R3  0
 I m1 R2   R2  R3  I m 2  12
 R1  R2  R2   I m1  VS  12

 R




-12
R

R
I
12
2
3   m2  
2 
 2
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh (Loop) Analysis
What if there are current sources?
I1
US
I1
P2.5
I1
I1
I1
+
40V
_
• The current sources in this circuit will have whatever voltage is
necessary to make the current correct.
• We can’t
use KVL around the loop because we don’t know the

R 1 R1


P2.6
_ voltage.
Mesh 1
1
2
I=?
I
I20
1
I1
R1
50
R1
I1
I
Im1
30
m2
2A 
Im2
+
7V
_
+
V
Im1 -


3
 2A
Mesh 2
Im3
40  I m1  20   I m1  I m 2   30  0
I m 2  2A
I  I m1  I m 2
I m1 =
Super Mesh:
7  I m 2  2  I m3 1  0
Mesh 2:
1
I m1  I m3  2
2

Super Mesh
Mesh 1:
 Im2  Im1  1 Im2  2   Im2  Im3   3  0
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh (Loop) Analysis
What if there are current sources?
2k
The
Supermesh
surrounds
this source!
2mA
I3
+
12V
-
2k
I2
I1
I0
1k
The
Supermesh
does not
include this
source!
4mA
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh (Loop) Analysis
What if there are current sources?
2k
• The 4mA current source sets I2:
2mA
I3
I2 = -4mA
1k
• The 2mA current source sets a
constraint on I1 and I3:
2k
+
12V
4mA
I2
I1
I1 - I3 = 2mA
I0
• We have two equations and
three unknowns. Where is the
third equation?
- 12  2k  I 3  1k  I 3  I 2   2k  I1  I 2   0
I1  2k  I 2 1k  2k   I 3 1k  2k   12V
I 2  4mA ; I 3  0.8mA ; I1  1.2mA
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh (Loop) Analysis
What if there are current sources?

P2.6
+
7V
_
Mesh 1
1
+
V
Im1 -
2
Im2


Mesh 2: -7V+I m 2  2  I m3 1  0
3
Node 3: I m1  I m3  2A
 2A
Im3
2
Node 3

Mesh 2
Mesh 1:
-7V+  I m1  I m2  1   I m3  I m2   3  I m3 1  0
1
Ch2 Basic Analysis Methods to Circuits
2.2 Useful Circuit Analysis Techniques
Mesh(Loop) Analysis
Dependent current source.
Current sources not shared by meshes.
We treat the dependent source as a
conventional source.
Equations for meshes with current sources
We are asked for Vo. We only need to solve for I3 .
Replace and rearrange
Then KVL on the remaining loop(s)
Vx  2kI1

  I1  2 I 2  4mA
Vx  4k ( I1  I 2 ) 
8kI3 = 3 + 2kI1 ⇒I 3 =
And express the controlling variable,
Vx, in terms of loop currents
VO  6kI 3 
33
[V ]
4
11
mA
8
Ch2 Basic Analysis Methods to Circuits
2.2 Basic Nodal and Mesh Analysis
Mesh (Loop) Analysis
Advantages of Loop Analysis
•
•
•
•
Solves directly for some currents.
Voltage sources are easy.
Current sources are either very easy or somewhat difficult.
Works best for circuits with few loops.
Disadvantages of Loop Analysis
• Some currents must be computed from loop currents.
• Choosing the supermesh may be difficult.
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Key Words:
Linearity
Superposition
Thevenin’s and Norton’s theorems
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Linearity
• Linearity is a mathematical property of circuits that makes very
powerful analysis techniques possible.
• 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)
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Linearity
• Linearity leads to simple solutions:
– Nodal analysis for linear circuits results in systems of linear
equations that can be solved by matrices
V1
500
1
I1
500
V3
2
500
1
 1

 500 500

1
 
500


0

V2
3
1k
1
500
1
1
1


500 1k 500
1

500

500

 V   I 
 1   1 
1

 V2    0 
500 
1
1  V3   I 2 

500 500 
0
I2
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Linearity
• The relationship between current and voltage for a linear element
satisfies two properties:
– Homogeneity
– Additivity
*Real circuit elements are not linear, but can be approximated as linear
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Linearity
• Homogeneity:
– Let v(t) be the voltage across an element with current i(t) flowing
through it.
– In an element satisfying homogeneity, if the current is increased by
a factor of K, the voltage increases by a factor of K.
• Additivity
– Let v1(t) be the voltage across an element with current i1(t) flowing
through it, and let v2(t) be the voltage across an element with
current i2 (t) flowing through it
– In an element satisfying additivity, if the current is the sum of i1 (t)
and i2 (t), then the voltage is the sum of v1 (t) and v2 (t).
Example: Resistor: V = R I
– If current is KI, then voltage is R KI = KV
– If current is I1 + I2, then voltage is R(I1 + I2) = RI1 + RI2 = V1 + V2
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
I1
Superposition
US
_
 

R1 R1
R3=80
I1
I1
I1
I1
I1
I1
I2
R1
R1  R3 
R3
I2 
VS 
E2
R R  R1 R3  R2 R3
R R  R2 R3  R1 R3
1 2 


1 2 



I 
R1
I1
I 2
I 2
S
R1=0.5
+
VS=14V
_
I
R2=0.4
E2=12V
• Superposition is a direct consequence of linearity
• It states that “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.”
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
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 techniques.
– Nodal analysis
– Loop analysis
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
For the above case:
Zero out Vs, we have :
Zero out E2, we have :
I
I2’’
R1
R3
I2’
R1
R2
E2
RR
R1 / / R3  1 3
R1  R3
R R  R2 R3  R1R3
R2   R1 / / R3   1 2
R1  R3
E2  R1  R3 
I 2  
R1 R2  R2 R3  R1R3
+
Vs_
R3
R2
R2 / / R3 
E2
R1   R1 / / R3  
R2 R3
R2  R3
R1 R2  R2 R3  R1 R3
R2  R3
Vs  R2  R3 
I
R1 R2  R2 R3  R1 R3
I 2  I 
R3
Vs R3

R2  R3 R1 R2  R2 R3  R1 R3
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
P2.7
2k
4mA
12V
2mA
1k
I0
+
2k
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
P2.7
I 0    I 2  I1 
I1  2mA
2k
KVL for mesh 2:
 I 2  I1  1k  I 2  2k  0
2mA
I1
1k
Io
’
Mesh 2
I2
2k
1
2
I 2  I1   mA
3
3
 2

I 0    I 2  I1       2 
 3

4
  mA
3
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
P2.7
I 0   I 2
2k
4mA
I1
I 2 1k   I 2  I1   0  I 2  2k  0
1k
I’’0
KVL for mesh 2:
Mesh 2
I2
2k
I2  0
I o  0
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
P2.7
I o   I 2
12V
2k
KVL for mesh 2:
-
1k
I’’’0
+
I2
I 2 1k  12V  I 2  2k  0
2k
I2 
12
 4mA
1k  2k
Mesh 2
I o  4mA
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Superposition
P2.7
2k
4mA
12V
-
2mA
1k
+
2k
I0
I0 = I’0 +I’’0+ I’’’0 = -16/3 mA
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Thevenin’s theorem
• Any circuit with sources (dependent and/or independent) and
resistors can be replaced by an equivalent circuit containing a single
voltage source and a single resistor
• Thevenin’s theorem implies that we can replace arbitrarily
complicated networks with simple networks for purposes of analysis
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Thevenin’s theorem
Independent Sources
RTh
Voc
Circuit with independent sources
+
-
Thevenin equivalent circuit
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Thevenin’s theorem
No Independent Sources
RTh
Circuit without independent sources
Thevenin equivalent circuit
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Norton’s theorem
• Very similar to Thevenin’s theorem
• It simply states that any circuit with sources (dependent and/or
independent) and resistors can be replaced by an equivalent circuit
containing a single current source and a single resistor
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Norton’s theorem
Norton Equivalent: Independent Sources
Isc
Circuit with one or more
independent sources
RTh
Norton equivalent circuit
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Norton’s theorem
Norton Equivalent: No Independent Sources
RTh
Circuit without independent sources
Norton equivalent circuit
Ch2 Basic Analysis Methods to Circuits
• Motivation of applying the Thevenin’s theorem and Norton’s theorem:
– Sometimes, in a complex circuit, we are only interested in working out the voltage
/current or power being consumed by a single load (resistor);
– We can then treat the rest of the circuit (excluding the interested load) as a voltage
(current) source concatenated with a source resistor;
– Simplify our analysis.
• The Thevenin’s theorem:
– Given a linear circuit, rearrange it in the form of two networks of A and B
connected by two wires. Define Voc as the open-circuit voltage which appears
across the terminals of A when B is disconnected. Then all currents and voltage in B
will remain unchanged, if we replace all the independent current or voltage source
in A by an independent voltage source which is in series with a resistor (RTh).
• The Norton’s Theorem:
– Given a linear circuit, rearrange it in the form of two networks of A and B
connected by two wires. Define isc as the short-circuit current which appears across
the terminals of A when B is disconnected. Then all currents and voltage in B will
remain unchanged, if we replace all the independent current or voltage source in A
by an independent current source isc which is in parallel with a resistor (RN).
Ch2 Basic Analysis Methods to Circuits
• Equivalent transform between a Thevenin equivalent circuit and a Norton
equivalent circuit
RTH
vL


VL
vS
If

RL
iS
vL
vL
RL
RN 
Thevenin equivalent
Norton equivalent
RL
VvLL  vS 
RL  RTH

RN 
VL  is
  RL
 RL  RTHN 
RTH  RN
and
v
VLL
=
vs  is RN  is RTH
VvLL
Ch2 Basic Analysis Methods to Circuits
• Therefore, for a source transform:
Thevenin
Norton
Norton:
RN  RTH , is  vs / RTH
Thevenin :
RTH  RN , vs  is  RN
Ch2 Basic Analysis Methods to Circuits
•
Example 45 (P88) Find a Thevenin and Norton equivalent circuit for the following
circuit excluding RL
3
7
12V
6
A
T
RL
B
N
7
7
4A
3
6
RL
4A
2
RL
Ch2 Basic Analysis Methods to Circuits
9
Thevenin
RL
8V
N
T
2
8V
equivalent
7
RL
0.889A
9
RL
Norton
equivalent
Ch2 Basic Analysis Methods to Circuits
•
Application of Thevenin’s theorem when there are only independent sources.
step : 1) Determine v of two connection points between network A and network B.
oc
2) Determine R of two connection points by replacing the voltage source by a shortTH
circuit or the current source by a open-circuit.
Test the above case?
•
Similarly, for Norton’s Theorem
steps : 1) Determine iSC between the two connection points between network A and B.
2) Determine R by two connection points by replacing the voltage source by a shortN
circuit or the current source by an open circuit.
Ch2 Basic Analysis Methods to Circuits
• Practice 4.6 (P90)
4
5
9V
I 2  ?
2
4
6
5
4
9V
4
voc
VS  VOC 
9V
 4  2.571V
4  4  6
6
5
10
4
RTH
1
1 1
20
RTH  (

)  5    5  7.857
10 4
7
Ch2 Basic Analysis Methods to Circuits
Thevenin equivalent :
7.857
I 2
2.521V
2
2.571V
I 2 
 260.8mA
7.857  2
Ch2 Basic Analysis Methods to Circuits
•
When there are multiple independent source, we shall use “superposition” .
Example 4.7 (P91)
Find the Thevenin and Norton equivalent for the network excluding the
1k resistor.
2k
3k
4V
To determine
2mA
1k
voc,
when only 4v voltage source is functioning.
5K
4V
voc'
voc'  4V
Ch2 Basic Analysis Methods to Circuits
When only 2mA current source is functioning :
5k
3k
voc
2k
N
T
4V
2mA
Therefore,
vc  voc  v  v  4V  4V  8V
'
oc
''
oc
voc'' voc''  4V
Ch2 Basic Analysis Methods to Circuits
To determine RTH
5k
RTH  5k 
Thevenin equivalent:
5k   RTH
VS  8V
1k
RN  RTH  5k 
iS  vs / RTH  1.6mA
Ch2 Basic Analysis Methods to Circuits
Norton equivalent
vs  1.6mA
RN  5k 
1k
Try to look into this problem from the Norton approach.
(Figure out the Norton equivalent circuit first)
Ch2 Basic Analysis Methods to Circuits
When there are both independent source and dependent source.
—Dependent source cannot be “zero out” as far as its controlling variable is not zero.
—Similar as before, vs  voc
—But we cannot determine RTH ( RN ) directly, however, we can use RTH ( RN )  voc
Example 4.8 (P92) —
Determine the Thevenin equivalent of the following circuit
2k
4V
To determine
since
voc
voc  vx
3k
vx
4000

vx

,applying KVL to the supermesh:
VX
4V  (
)  2k   3k   0  vx  0
4000
vs  voc  vx  8V
/ isc
Ch2 Basic Analysis Methods to Circuits
To determine
isc
2k
3k
vx
 0 vx  0
4000
4V
iSC
4V

 0.8mA
5k
RTH  vOC / iSC 
8V
 10k
0.8mA
Therefore its Thevenin equivalent is:
RTH  10k 
vs  8V
Ch2 Basic Analysis Methods to Circuits
• When there are only dependent sources:
– VOC = 0
– RTH can be determined by implying a test (imaginary) voltage across the two
terminals.
3Ω
i
Example 4.9 (p93)
Open circuit : i  0 , 1.5i 
2
 0 , VOC  0
2  3
1.5i
To determine RTh, imagine an independent
current source xA. as :
i = -xA,
v
RTh =
x
Apply KCL:
3Ω
1.5( - x) - v v
+ + ( - x) = 0
3
2
i
1.5i
2Ω
v
xA
v = 0.6x V
RTh = 0.6Ω
Giving :
0.6Ω
2Ω
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Thevenin’s theorem
• Circuits with independent sources
– Compute the open circuit voltage, this is Voc
– Compute the Thevenin resistance (set the sources to zero – short
circuit the voltage sources, open circuit the current sources), and
find the equivalent resistance, this is RTh
• Circuits with independent and dependent sources:
– Compute the open circuit voltage
– Compute the short circuit current
– The ratio of the two is RTh
• Circuits with dependent sources only*
– Voc is simply 0
– RTh is found by applying an independent voltage source (V volts)
to the terminals and finding voltage/current ratio
* Not required by this course.
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Norton’s theorem
• Circuits with independent sources, w/o dependent sources
– Compute the short circuit current, this is Isc
– Compute the Thevenin resistance (set the sources to zero – short
circuit the voltage sources, open circuit the current sources), and
find the equivalent resistance, this is RN
• Circuits with both independent and dependent sources
– Find Voc and Isc
– Compute RN= Voc/ Isc
• Circuits w/o independent sources*
– Apply a test voltage (current) source
– Find resulting current (voltage)
– Compute RN
Ch2 Basic Analysis Methods to Circuits
2.3 Useful Circuit Analysis Techniques
Maximum power transfer
VL2
RL
PL 
; VL 
VTH
RL
RTH  RL
RTH
+
-
VTH

VL

RL
(LOAD)
SOURCE
RTH  RL  2RL  0 
PL 
RL
V2
2 TH
RTH  RL 
For every choice of RL we have a different power.
How do we find the maximum value?
Consider PL as a function of RL and find the
maximum of such function
2



dPL
R

R
 2 RL RTH  RL  
2
TH
L

 VTH 
4

dRL
RTH  RL 


RL*  RTH
The maximum power transfer theorem
The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent
resistance of the circuit.
The value of the maximum power that can be transferred is
2
VTH
PL (max) 
4 RTH
Ch2 Basic Analysis Methods to Circuits
Analysis methods Review

KVL, KCL, I — V
Combination rules
Node method
Mesh method
Superposition
Thévenin
Norton

Any circuits
linear circuits