Engineering 43
Super Node/Mesh
Thevénin/Norton
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
OutLine
 ReIterate NODE & LOOP/MESH
Analysis by way of Examples
 SuperNode Method
 SuperMesh Technique
 Loops vs Meshes; describe difference
 Loop & Node Compared
 Introduction to Thevenin & Norton
theorems
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
OutLine




Thevénin theorem: RTH=RN & VTH=VOC
Norton theorem: RN=RTH & IN=ISC
Source Transformation
Thevenin & Norton theorems for
• INdependent Source Circuits by
DeActivation
• Dependent Source Circuits by VOC/ISC
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Node Analysis (find V by kCl)
Example (On Board)
1. Define a GND Node
2. Label all Non-GND nodes not
connected to a Source; i.e.,
the UNknown Nodes
3. Write the CURRENT Law
Eqns at the Unknown Nodes
using Ohm’s Law;
I = ΔV/R
4. Clear Fractions by Multiplying
by the LCD
•
Be sure to Include UNITS
5. Work the Linear Algebra to
find Unknown Node Voltages
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop Analysis (find I by kVl)
1. Define a GND Node
2. Draw Current Loops or
Meshes to
•
•
NOT be Redandant
Pass Thru ALL Circuit Elements
Example (On Board)
3. Write the VOLTAGE Law
Eqns For each Loop/Mesh
usingNodes using Ohm’s
Law; ΔV=RI
4. Divide by Eqns by the LCM of
the I’s
•
Be sure to Include UNITS
5. Work the Linear Algebra to
find Unknown Loop/Mesh
Currents
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
ReCall Node (KCL) Analysis
 Need Only
ONE KCL Eqn
V2 V2  V3 V2  V1


0
6k
12k
12k
 The Remaining Eqns
From the Indep Srcs
 3 Nodes Plus the
Reference. In Principle
Need 3 Equations...
V1  12[V ]
V3  6[V ]
 Solving The Eqns
2V2  (V2  V3 )  (V2  V1 )  0
• But two nodes are connected
to GND through voltage
4V2  6[V ]  V2  1.5[V ]
sources. Hence those node
voltages are KNOWN
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
SuperNode Technique
SUPERNODE
 Consider This
Example
 Conventional Node
Analysis Requires All
Currents at a Node
@ V1
@ V2
IS
 But Have Ckt V-Src Reln
V1
 IS  0
6k
V
 I S  4mA  2  0
12k
 6mA 
V1  V2  6[V ]
 More Efficient solution:
 2 eqns, 3 unknowns...
Not Good (IS unknown)
• Recall: The Current thru the
Vsrc is NOT related to the
Potential Across it
Engineering-43: Engineering Circuit Analysis
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• Enclose The Source, and
All Elements In parallel,
Inside a Surface.
– Call That a SuperNode
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Supernode cont.
SUPERNODE
 Apply KCL to the
Surface
V1 V2
 6mA 

 4mA  0
6k 12k
• The Source Current
Is interior to the Surface
and is NOT Required
 Still Need 1 More
Equation – Look INSIDE
the Surface to Relate
V1 & V2
IS
 Now Have 2 Equations
in 2 Unknowns
 Then The Ckt Solution
Using LCD Technique
• See Next Slide
V1  V2  6[V ]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Now Apply Gaussian Elim
 The Equations
V1 V2

 6mA  4mA  0
6k 12k
(2) V1  V2  6[V ]
(1)
 Mult Eqn-1 by
LCD (12 kΩ)
 Use The V-Source
Rln Eqn to Find V2
V2  V1  6[V ]  4[V ]
SUPERNODE
IS
2V1  V2  24[V ]
V1  V2  6[V ]
 Add Eqns to Elim V2
3V1  30[V ]  V1  10[V ]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Is2
Find the node voltages
And the power supplied
By the voltage source
R3 I
V1
V2

R1
VS
V
R2
I s1
R1  R2  10k, R3  4k
VS  20[V ], I s1  10[mA], I s 2  6[mA]
V2 V1  20
V1
V
 2  10mA  0
10k 10k
  V1  V2  20[V ]
*10k  V1  V2  100[V ]
adding : 2V2  120[V ]
V1  100  V2  40[V ]
To compute the power supplied by the voltage
source We must know the current through it: @ node-1
IV 
Engineering-43: Engineering Circuit Analysis
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V1
V V
 6mA  1 2  5mA
10k
4k
P  20[V ]  5[mA]  100mW
BASED ON PASSIVE SIGN CONVENTION THE
Bruce Mayer, PE
POWER IS ABSORBED BY THE SOURCE!!
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Illustration using Conductances
 Write the Node Equations
• KCL At v1

 At The SuperNode
Have V-Constraint
•
v2 − v3 = vA 
 KCL Leaving Supernode

 Now Have 3 Eqns
in 3 Unknowns
• Solve Using Normal Techniques
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example
SUPERNODE
V3  12
 Find Io
 Known Node Voltages
V2  6V , V4  12V
 The SuperNode
V-Constraint
V1  V3  12V
or V1  V3  12V
 Now use KCL at SuperNode to Find V3
 Mult by 2 kΩ
LCD, collect
Terms to Find:
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example
SUPERNODE
 Find Io Using
Nodal Analysis
 Known Voltages for
Sources Connected
to GND
 Now Notice That V2 is
NOT Needed to Find Io
V1  6V , V4  4V
 The Constraint Eqn
• 2 Eqns in 2 Unknowns
3V2  2V3  2V
V3  V2  12V
 V2  V3  12V  3
 Now KCL at SuperNode
V2  6 V2 V3 V3  (4)





0
 2k
  2k
1
k
2
k
2
k


Engineering-43: Engineering Circuit Analysis
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and add eqns
-----------------5V3  38V  V3  7.6V
 By Ohm’s Law
IO 
V3
7.6V

 3.8mA
2k 2k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Dependent Sources
 Circuits With Dependent Sources
Present No Significant Additional
Complexity
 The Dependent Sources Are
Treated As Regular Sources
 As With Dependent CURRENT Sources
Must Add One
Equation For Each
Controlling Variable
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example – Dep Isrc
 Find Io by Nodal
Analysis
 Notice V-Source
Connected to the
Reference Node
V1  3V
 KCL At Node-2
V2  V1 V2

 2I x  0
3k
6k
 Controlling Variable
In Terms of
V2
I 
Node Potential x 6k
Engineering-43: Engineering Circuit Analysis
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 Sub Ix into KCL Eqn
V2  V1 V2
V2

2
0
3k
6k
6k
 Mult By 6 kΩ LCD
V2  2V1  0  V2  6V
 Then Io
V1  V2
IO 
 1mA
3k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Current Controlled V-Source
 Find Io
 Supernode Constraint
V2  V1  2kIx
 Controlling Variable in
Terms of Node Voltage
V1
Ix 
2k
 V1  2kIx  V2  2V1
 KCL at SuperNode
 4mA 
V1
V
 2mA  2  0
2k
2k
Engineering-43: Engineering Circuit Analysis
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 Multiply by LCD of 2 kΩ
V1  V2  4[V ]
 Recall  2V1  V2  0
 Then 3V2  8V  V2  8V 3
 So Finally
IO 
V2 4
 mA
2k 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
ReCall MESH Analysis
 Use Mesh Analysis
Mesh 1: I1  4mA
Mesh  2
 4kI1  12kI2  0
I2
I1
 Sub for I1 to Find I2
I2 
4 I1 16mA 4

 mA
12
12
3
 So Vo
VO  6k  I 2
VO  6k 1.333mA  8[V ]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
SuperMeshes
1. Create Mesh Currents
2. Write Constraint
Equation Due To Mesh
Currents SHARING
Current Sources
I 2  I 3  4mA
3. Write Equations For
Remaining Meshes
SUPERMESH
I1  2mA
4. Define A SuperMesh By
AVOIDNG The Shared
Current with a Carefully
Chosen Loop
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
SuperMeshes cont.
5. Write KVL For The SuperMesh as we do NOT know
the Voltage Across the 4 mA Current-Source
 6  1kI3  2kI2  2k ( I 2  I1 )  1k ( I 3  I1 )  0

We Now have 3 Eqns
in 3 Unknowns and the
Math Model is
Complete
•
SUPERMESH
Solve for I1, I2, I3 using
standard techniques
KVL
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
SuperNODE vs SuperMESH
 Use superNODE to
AVOID a V-source
current, IVs, in KCL
Eqns
 Use superMESH to
AVOID an I-source,
ΔVIs, in KVL Eqns
2mA
2K
1K
Vx
4mA
12V
2K
IOIO
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Shared Isrc – General Loop Approach
 Strategy
• Define Loop Currents
That Do NOT Share
Current Sources
– Even If It Means
ABANDONING Meshes
 For Convenience,
Begin by Using Mesh
Currents Until Reaching
Shared CURRENT
Source as V-across an
I-source is NOT Known
•
At That Point
Define a NEW Loop
Engineering-43: Engineering Circuit Analysis
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I3
I1
I2
 To Guarantee That The
New Loop Gives An
Independent Equation,
Must Ensure That It
Includes Components That
Are NOT Part Of Previously
Defined Loops
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
General Loop Approach cont.
 A Possible Approach
• Create a Loop by Avoiding
The Current Source
 The Eqns for Current
Source Loops
I1  2mA
I1
I2
I 2  4mA
 The Eqns for 3rd Loop (3 Eqns & 3 Unknowns)
I3
 6[V ]  1kI3  2k ( I 3  I 2 )  2k ( I 3  I 2  I1 )  1k ( I 3  I1 )  0
 The Loop Currents Obtained With This Method Are
Different From Those Obtained With A SuperMesh
• A SuperMesh used Previously Defined Mesh Currents
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example  Find V Across R’s
 For Loop Analysis Note
• Three Independent
Current Sources
• Four Meshes
• One Current Source
Shared By Two Meshes
I S1
• Three Loop Currents Can
Be Chosen Using Meshes
And Not Sharing Any
Source
28
R2
I2
IS2
R1
R3
 Careful Choice Of Loop
Currents Should Make
Only One Loop Equation
Necessary
Engineering-43: Engineering Circuit Analysis
I1
R4
I3
IS3 I
4
 Mesh Equations For
Loops With I-Sources
I1  I S 1
I2  IS2
I3  I S 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
+
VS
Example  Find V Across R’s cont.
 KVL for I4 Loop
VS  ( I 4  I 2 ) R3  ( I1  I 3  I 4 ) R1
 ( I 3  I 4 ) R4  0
 Solve For The
Current I4 Using The ISj
• Now Use Ohm’s Law To
Calc Required Voltages
V1  R1 ( I1  I 3  I 4 )
V2  R2 ( I 2  I1 )
V3  R3 ( I 2  I 4 )
V4  R4 ( I 3  I 4 )
Engineering-43: Engineering Circuit Analysis
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I1
I S1

V4


V
R2

R1
 V1 
R4
I3
I2
2
IS2
 V3 
R3
IS3 I
4
 Note that Loop-4 does
NOT pass thru ANY
CURRENT-Sources
• This AVOIDS the
UNknown potentials
across the I-sources
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
+
VS
Dependent Sources
 General Approach
• Treat The Dep.
Source As Though It
Were Independent
• Add One Equation
For The Controlling
Variable
 Example at Rt.: Mesh  Mesh-3 by KVL
Currents Defined by
 1kIx  2k ( I 3  I1 )  1k ( I 3  I 4 )  0
Sources I1  4mA
 Mesh-4 by KVL
VX
I2 
2k
Engineering-43: Engineering Circuit Analysis
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1k ( I 4  I 3 )  1k ( I 4  I 2 )  12V  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Dependent Sources cont.
 The Controlling
Variable Eqns
I x  I4  I2
Vx  2 k ( I 3  I1 )
 Combine Eqns,
Then Divide by 1kΩ
I1  4mA
I1  I 2  I 3  0
I 2  3I 3  2 I 4  8mA
 In Matrix Form
0   I1   4 
1 0 0
1 1  1 0   I   0 

 2   

0 1 3  2   I 3   8 

  

I
0

1

1
2

12

 4  

 I 2  I 3  2 I 4  12mA  Solve by Elimination
Engineering-43: Engineering Circuit Analysis
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or Linear-Algebra
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop & Node Compared
 Consider the Ckt
 Find Vo by NODE
Analysis
• ID Nodes
• Make a SuperNode
Engineering-43: Engineering Circuit Analysis
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 Vsrc to GND V4  4[V ]
 SuperNode
V1  V2  2Vx
Constraint
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop & Node Compared (2)
 Mult. By 1kΩ LCD
 KCL at SuperNode
V2  V2  V3  V1  V3  V1  4V  2V
or 2V1  2V2  2V3  6V
 The Node 3 KCL
V3  V2 V3  V1
 2mA 

0
1k
1k
 Mult. By 1kΩ LCD
 The SuperNode Eqn
 2mA 
V2 V2  V3 V1  V3 V1  4V



0
1k
1k
1k
1k
Engineering-43: Engineering Circuit Analysis
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V3  V2  V3  V1  2V
or  V1  V2  2V3  2V
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop & Node Compared (3)
 The Controlling Var.
 Thus 3 Eqns in
Unknowns V1, V2, V3
V1  V2  V3  3V
 V1  V2  2V3  2V
V1  3V2  0
Vx  V2
 In SuperNode Eqn
 Recall the GOAL
Vo  V1  V3
V1  V2  2Vx  V1  3V2  0
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop & Node Compared (4)
 Now by Loops
I1
I2
I4
I3
 The Mesh/Loop
Eqns
• Loop-1: I1  2mA
• Loop-3: I 3  2mA
• Loop-2:
 2Vx  1kI2  1k I 2  I 3   0
– Note I3 = –2 mA
 Start with 3 Meshes
 Add a General Loop
to avoid the Isrc
Engineering-43: Engineering Circuit Analysis
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• Loop-4:
1k I 3  I 4  I1   2Vx  1kI4  4V  0
– Note I1 = 2 mA
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop & Node Compared (5)
 The Controling Var
I1
I2
I4
I3
 As Before Vx = V2
 And V2 is related to
the net current From
Node-2 to GND
Engineering-43: Engineering Circuit Analysis
36
 By Net Current &
Ohm’s Law
Vx  1kI1  I 3  I 4 
 SubOut Vx in Loop-2
& Loop-4 Eqns:
 2Vx  1kI2  1k I 2  I 3   0
1k I 3  I 4  I1   2Vx  1kI4  4V  0
 After Subbing Find:
4kI4  8V
2kI2  4kI4  6V
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Loop & Node Compared (6)
 Summarize Loops
 Recall the GOAL
Vo  1kI2
I2
 Using The
Loop/Mesh Eqns
Find
I1  2mA I 2  1mA
I 3  2mA I 4  2mA
Engineering-43: Engineering Circuit Analysis
37
 General Comments
• Nodes (KCL) are
generally easier if we
have VOLTAGE
Sources
• Loops (kVL) are
generally easier if we
have CURRENT
Sources
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin’s & Norton’sTheorems
 These Are Some Of The Most Powerful
Circuit analysis Methods
 They Permit “Hiding” Information That Is
Not Relevant And Allow Concentration
On What Is Important To The Analysis
Engineering-43: Engineering Circuit Analysis
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Low Distortion Power Amp
 to Match
Speakers And
Amplifier One
Should Analyze
The Amp Ckt
From PreAmp
(voltage )
To speakers
Engineering-43: Engineering Circuit Analysis
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Low Dist Pwr Amp cont
 To Even STAND A CHANCE to Match the
Speakers & Amp We Need to Simplify the Ckt
• Consider a Reduced CIRCUIT EQUIVALENT
 Replace the OpAmp+BJT Amplifier Ckt
with a MUCH Simpler (Linear) Equivalent
• The Equivalent Ckt in
RED “Looks” The
Same to the Speakers
As Does the
Complicated Circuit
Engineering-43: Engineering Circuit Analysis
40
RTH
VTH
+
-
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin’s Equivalence Theorem
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
RTH


vTH

i
a
vO
b
_

i
a
LINEAR CIRCUIT
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
 vTH = Thevenin
Equivalent
VOLTAGE
Source
 RTH = Thevenin
Equivalent
Series (Source)
RESISTANCE
b
PART B
PART A
 Thevenin Equivalent Circuit for PART A
Engineering-43: Engineering Circuit Analysis
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Norton’s Equivalence Theorem
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

RN
a
vO
b
_

iN
i
i
a
LINEAR CIRCUIT
vO
_
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
b
 iN = Norton
Equivalent
CURRENT
Source
 RN = Norton
Equivalent
Parallel
(Source)
RESISTANCE
PART B
PART A
 Norton Equivalent Circuit for PART A
Engineering-43: Engineering Circuit Analysis
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Examine Thevenin Approach
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

i
a
vO
_

For ANY Part-B Circuit

The Thevenin Equiv
Ckt for PART-A →
b
ANY
PART B
RTH
vO  vOC  RTH i
•
•
V-Src is Called the THEVENIN
EQUIVALENT SOURCE
R is called the THEVENIN
EQUIVALENT RESISTANCE
Engineering-43: Engineering Circuit Analysis
43
vOC
+
_
i
+
vO
_
PART A MUST BEHAVE LIKE
THIS CIRCUIT
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Examine Norton Approach
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A


i
a
ANY
PART B
vO
_
b
In The Norton Case
vOC
vO
vO  vOC  RTH i  i 

RTH RTH
vOC
 iSC
RTH

The Norton Equiv Ckt
for PART-A →
•

i SC
The I-Src is Called The
NORTON EQUIVALENT SOURCE
Engineering-43: Engineering Circuit Analysis
44
i a
RTH
vO

Norton
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
b
Interpret Thevenin & Norton
i a
RTH
vOC
+
_
+
i
vO
_
i SC
RTH
RTH
vO

b
Norton
Thevenin
 In BOTH Cases

vOC
 RN 
iSC
 This equivalence can be viewed as a source transformation problem.
It shows how to convert a voltage source in series with a resistor into
an equivalent current source in parallel with the resistor
• SOURCE TRANSFORMATION CAN BE A GOOD TOOL
TO REDUCE THE COMPLEXITY OF A CIRCUIT
Engineering-43: Engineering Circuit Analysis
45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Source Transformations
 Source transformation is a good tool to reduce
complexity in a circuit ...WHEN IT CAN APPLIED
• “IDEAL sources” are NOT good models for the
REAL behavior of sources
– .e.g., A Battery does NOT Supply huge current When Its Terminals
are connected across a tiny Resistance as Would an “Ideal” Source
+
-
RV
VS
a
a
 These Models are
Equivalent When
b
RV  RI  R
RI
b
IS
Improved model
Improved model
for voltage source for current source
VS  RI S
 Source X-forms can be used to determine the
Thevenin or Norton Equivalent
• But There May be More Efficient Methods
Engineering-43: Engineering Circuit Analysis
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example  Solve by Src Xform
 In between the terminals we
connect a current source and a
resistance in parallel
 The equivalent current source
will have the value 12V/3kΩ
 The 3k and the 6k resistors
now are in parallel and can be
combined
Engineering-43: Engineering Circuit Analysis
47
 In between the terminals we
connect a voltage source in
series with the resistor
 The equivalent V-source has
value 4mA*2kΩ
 The new 2k and the 2k resistor
become connected in series
and can be combined
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Solve by Src Xform cont.
 After the transformation the
sources can be combined
 The equivalent current source
has value 8V/4kΩ = 2mA
 The Options at This Point
1. Do another source transformation and get a
single loop circuit
2. Use current divider to compute IO and then
Calc VO using Ohm’s law
Engineering-43: Engineering Circuit Analysis
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
PROBLEM Find VO using source transformation
Norton
Norton
EQUIVALENT CIRCUITS
I0
Or one more source transformation
R eq
Veq
+
-
RTH
3 current sources in parallel and
three resistors in parallel
R3
VVeqTH Req I eq
R4
Engineering-43: Engineering Circuit Analysis
49
Veq  Req I eq
V0 
R4
Veq
R4  R3  Req
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Source Xform Summary
 These Models are Equivalent
+
-
RV
VS
a
a
RI
b
IS
RV  RI  R
VS  RI S
b
I S  R VS
Improved model
Improved model
for voltage source for current source
 Source X-forms can be used to determine the
Thevenin or Norton Equivalent
 Next Review Several Additional Approaches To
Determine Thevenin Or Norton Equivalent Circuits
Engineering-43: Engineering Circuit Analysis
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Determine the Thevenin Equiv.
 vTH = OPEN CIRCUIT Voltage at A-B if Part-B
is Removed and Left UNconnected
 iSC = SHORT CIRCUIT Current at A-B if
Voltage at A-B is Removed and Replaced with
a Wire (a short)
 Then by R = V/I
Engineering-43: Engineering Circuit Analysis
51
RTH
vOC

i SC
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Graphically...
1. Determine the
Thevenin equivalent
source
2. Determine the
SHORT CIRCUIT
current
Remove part B and
compute the OPEN
CIRCUIT voltage Vab
Remove part B and
compute the SHORT
CIRCUIT current I ab
 Then
vTH  Vab  vOC
RTH
Engineering-43: Engineering Circuit Analysis
52
One circuit problem
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
i 0 a

Vab

vOC
_
_
b
Second circuit problem
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
i SC

v0
I ab
_
i N  I ab i SC
vOC

 RN
i SC
a
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
b
Thevenin w/ Indep. Sources
 The Thevenin Equivalent V-Source is computed
as the open loop voltage
 The Thevenin Equivalent Resistance CAN BE
COMPUTED by setting to zero all the INDEPENDENT
sources and then determining the resistance seen from
the terminals where the equivalent will be placed
R1
a
a
VS
+
-
IS
To Part B
R2
b
Engineering-43: Engineering Circuit Analysis
53
R1
R2
RTH
b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin w/ Indep. Sources cont
 Since the evaluation of
the Thevenin equivalent
 3k Resistance for
INdependent-Source-Only
circuits can be very
simple, we can add it to
our toolkit for the solution
of circuits
“Part B”
RTH
RTH  2k  3k 6k  4k
“Part B”
Engineering-43: Engineering Circuit Analysis
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example
 Find Vo Using Thevenin’s
Theorem
 Identify Part-B (the Load)
 Break The Circuit At the
Part-B Terminals
“PART B”
 DEactivate 12V Source to
Find Thevenin Resistance
• Produces a SHORT
6V
5k
Engineering-43: Engineering Circuit Analysis
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example cont.
 Note That RTH Could be
Found using ISC
I tot
I SC
 By Series-Parallel R’s
Req  6  6 2  7.5k
 Then Itot
I tot  12V 7.5k  1.6mA
Engineering-43: Engineering Circuit Analysis
56
 Then by I-Divider
6
I SC  1.6mA
 1.2mA
62
 Finally RTH
RTH  VOC I SC  6V 1.2mA  5k
• Same As Before
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example cont.2
 Finally the Thevenin
Equivalent Circuit
1[V ]
 And Vo By V-Divider
1k
VO 
(6V )  1[V ]
1k  5k
Engineering-43: Engineering Circuit Analysis
57
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Let’s do MQ-02e on Board
 Find: Vt & Rt
A
B
Engineering-43: Engineering Circuit Analysis
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example
 Use Thevenin To
Find Vo
 Have a CHOICE on
How to Partition the Ckt
• Make “Part-B” As Simple
as Possible
“Part B”
 Deactivate the 6V and
2mA Source for RTH
RTH
Engineering-43: Engineering Circuit Analysis
59
10
 2  2 4  k
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example cont
 For the open circuit
voltage we analyze the
circuit at Right (“Part A”)
 Use Loop/Mesh Analysis
I 2  2mA
 6V  4kI1  2k ( I1  I 2 )  0
 Finally The Equivalent
Circuit
6  2I 2
5
I1 
mA  mA
6
3
 Then VOC
VOC  4k * I1  2k * I 2
VOC  20 / 3V   4V   32 / 3[V ]
Engineering-43: Engineering Circuit Analysis
60
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
CALCULATE Vo USING NORTON
4k
I
RN
IN
I SC
 RN

VO  2kI  2k 
I N 
 RN  6k

RN  RTH  3k
PART B
12V
I SC  I N 
 2mA  2mA
3k
COMPUTE Vo USING THEVENIN
2k
3
 4
VO  2k  (2mA)  [V ]
9
 3
PART B
VTH
RTH

+
-
2k

VTH
VTH  12
 2mA  0  VTH  12  6 RTH  3k  4k
3k
Engineering-43: Engineering Circuit Analysis
61
VO 
VO
2
4
(6V )  [V ]
27
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
DEpendent & INdependent Srcs
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
RTH
+
-
VTH

i
a
vO
_
b
a
RTH 
VOC
I SC
b
VTH  VOC
Engineering-43: Engineering Circuit Analysis
62
 Find The Open Circuit Voltage And
Short Circuit Current
 Solve Two Circuits (Voc & Isc) For
Each Thevenin Equivalent
 Any and all the techniques may be
used; e.g., KCL, KVL, combination
series/parallel, node & loop analysis,
source superposition, source
transformation, homogeneity
 Setting To Zero All Sources And
Then Combining Resistances To
Determine The Thevenin Resistance
is, in General, NOT Applicable!!
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example
Vb
VX
 VTH 
 Recognize Mixed sources
• Must Compute Open Circuit
Voltage, VOC, and Short
Circuit Current, ISC
 The Open Ckt Voltage
VTH  VX  Vb
 Use V-Divider to Find VX
VX 
R
2
(2VS )  VS
R  2R
3
 For Vb Use KVL
Vb  2 R(aVX )  VS  (1  4aR / 3)VS
as : VX  2 3VS
 Now VTH = Vx − Vb
Engineering-43: Engineering Circuit Analysis
63
VTH  VX  Vb
 VX  (2 RaV X  VS )  (1  2 Ra )( 2VS / 3)  VS
 Solve for VTH
VTH
1 4aR
 VX  Vb  
VS
3
 The Short Ckt Current
• Note that Shorting a-to-b
Results in a Single Large
Node
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Va = Vb = V X
Single node
Example cont
I SC
VX
 Need to Find Vx
 KCL at Single Node
Vx  2VS VX
V  Vs

 aVX  X
0
2R
R
2R
 Then RTH
RTH 
 Solving For Vx
3VS
VX 
4  2aR
VOC VTH 4 R(2  aR)


I SC I SC
3
 The Equivalent Circuit
R TH
 KCL at Node-b for ISC
VX  VS
I SC  aVX 
2R
1  4aR
I SC  
VS
4 R(2  aR)
Engineering-43: Engineering Circuit Analysis
64
4 R ( 2  aR )
3
a
VTH

1 4aR
VS
3


b
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Illustration
“Part B”
 Use Thevenin to
Determine Vo
 Partition Guidelines
V1
• “Part-B” Should be as
Simple As Possible
• After “Part A” is replaced by
the Thevenin equivalent
should result in a very
 Constraint at SuperNode
simple circuit
V1  VOC  12  V1  12  VOC
• The DEpendent srcs and
their controlling variables
 KCL at SuperNode
MUST remain together
'
 Use SuperNode to Find
Open Ckt Voltage
Engineering-43: Engineering Circuit Analysis
65
(12  VOC )  (aI X ) 12  VOC VOC


0
1k
2k
2k
Where : a  2 k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Illustration cont
 The Controlling
Variable ' VOC
IX 
2k
VOC
 Solving 3 Eqns for 3
Unknowns Yields
 At Node-A find I "  VA  0
X
36
36
2k


4  (a / 1k )
4  (2k / 1k )  V =0 → The Dependent
A
 Now Tackle
Source is a SHORT
Short Circuit Current
• Yields Reduced Ckt
VA
Engineering-43: Engineering Circuit Analysis
66
1 || 2  2 3 k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Illustration cont
 Using the Reduced Ckt
12V
I SC  
 18mA
1k || 2k
 Now Find RTH
RTH 
VOC
6V
1

 k
I SC  18 mA 3
 Finally the Solution
RTH (a  2k )
 Note: Some ckts can
produce NEGATIVE RTH
 Setting All Sources To
Zero And Combining
Resistances Will Yield An
INCORRECT Value
Engineering-43: Engineering Circuit Analysis
67
VOC
1k
V0 
VTH
1k  1k  RTH
V0 
1
 6V    18 V
2.3333
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example
 Find Vo using Thevenin
 Define Part-A
 Find VOC using
SuperNode
IX
Super node
V1
 Apply KVL
 VTH 1000I X  V1  0

VTH
IX
KVL

V1
V1  (3V )
KCL :
 1mA 
0
2k
6k
V1  (3 / 4)[V ]
Engineering-43: Engineering Circuit Analysis
68
 With The Controlling
Variable Find
V1
IX 
2k
VTH  (3 / 8)[V ]
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example cont
 Next The Short
Circuit Current
V  1000 I
1
1
1
X
V11
MINUS
3V
1
V
I 1X  1  V11  2kI1X  1kI1X
2k
 Using VOC & ISC
 The ONLY Value That
Satisfies the Above eqns
RTH 
R TH
 KCL at Top Node
• Recall Dep Src is a SHORT
1k
+
750
VTH
I SC  1mA  (3V ) /(6k )  0.5mA
V0 
69
3 8 V  3 k
VOC

I SC 1 2  ma 4
 The Equiv. Ckt
I 1X  0  V11  0
Engineering-43: Engineering Circuit Analysis
I SC
I 1X
+
-
2k
375mV
2
(3 / 8)[V ]
2  1  (3 / 4) Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
VO
_
Note on Example
 The Equivalent Resistance CanNOT Be
Obtained By DeActivating The Sources
And Determining The Resistance Of The
Resulting InterConnection Of Resistors
• Suggest Trying it → Rth,wrong = 2.5 kΩ
– Rth,actual = 0.75 kΩ
Req
Engineering-43: Engineering Circuit Analysis
70
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
EXAMPLE: Find Vo By Thevenin
 Select Partition
 Use Meshes to Find VOC
“Part B”
KVL for V_oc
 By Dep. Src Constraint
VX'  2kI1  2kI1  4k ( I1  I 2 )  I1  4mA
 Now KVL on Entrance Loop
 In The Mesh Eqns
VX'
I1 
; I 2  2mA
2000
0  VOC  2k * I1  3[V ]
 The Controlling Variable
'


V
'
X
VX  4k I1  I 2   4k   I 2 
2k

Engineering-43: Engineering Circuit Analysis 
71
 Solve for VOC
VOC  2k  4mA  3V  11V
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Find Vo By Thevenin cont
 Now Find ISC
 The Mesh Equations
I1 
I1
I sc
"
x
V
; I 2  2mA
2000
 3V  2k ( I SC  I1 )  0
 The Controlling Variable
VX"  4k * ( I1  I 2 )
 Solving for I1 Find Again
I2
 Then Thevenin Resistance
VOC
11[V ]
RTH 

 2k
I SC (11 / 2)mA
 Use Thevenin To Find Vo
I1  4mA
 Find ISC by Mesh KVL
3V  2k * I1 11
I SC 
 mA
2k
2
Engineering-43: Engineering Circuit Analysis
72
RTH
VTH
V0 
6k
11[V ]  33 4 V 
2k  6k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
 The Merthod for Mixed
Sources
V
Illustration
VTH  VOC , RTH 
Linear Model for Transistor
R1
Vx

vS
R2
+
-
a

b
I SC
RTH
VTH
+
-
b
Engineering-43: Engineering Circuit Analysis
73
 
VTH  I R3 R3   g mVx R3
Vx 
RR
R2
vS  VTH   g m 3 2 vS
R1  R2
R1  R2
 For the Short Ckt Current
I SC
a
RTH
I SC
 The Open Ckt Voltage
R3VTH
g mVx
OC
VOC

I SC
R2
  g mVx   g m
vS
R1  R2
R2
 g m vS
R3
R1  R2

 R3
R2
 g m vS
R1  R2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
WhiteBoard Work
 Let’s Work This
Problem
2K
2K
+
6V
2K
Vo
2K
4mA
-
• Find Vo by Source
Transformation
Engineering-43: Engineering Circuit Analysis
74
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
All Done for Today
More on
Source
Xforms
Engineering-43: Engineering Circuit Analysis
75
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example
 Find Vo Using Thevenin’s Theorem
 Alternative: apply
Thevenin Equivalence
to that part (viewed as
“Part A”)
 Deactivating (Shorting) The
12V Source Yields
RTH  2  6 3  4k
 in the region shown,
could use source
transformation twice and  Opening the Loop at the
reduce that part to a
Points Shown Yields
single source with a
6
resistor.
VOC  VTH 
12[V ]  8[V ]
3 6
Engineering-43: Engineering Circuit Analysis
76
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example cont.
 Then the Original Circuit Becomes After “Theveninizing”

1
VTH

 For Open Circuit Voltage
Use KVL
1
VTH
 4k * 2mA  8V  16V
 Result is V-Divider for Vo
 Apply Thevenin Again
 Deactivating The 8V &
2mA Sources Gives
R1TH  4k
Engineering-43: Engineering Circuit Analysis
77
8
V0 
16[V ]  8V
88
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Example  Alternative
 Can Apply Thevenin only once to get a voltage divider
• For the Thevenin Resistance Deactivate Sources
“Part B”
RTH  8k
 For the Thevenin voltage Need to analyze this circuit
 Find VOC by
SuperPosition
Engineering-43: Engineering Circuit Analysis
78
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Thevenin Alternative cont.
 Open 2mA Source To find
Vsrc Contribution to VOC
1
OC
V
6

12V  8V
3 6
 Short 12V Source To find Isrc Contribution to VOC
2
VOC
 (2k  6k 3k ) * (2mA)  8V
 Thevenin Equivalent of “Part A”
 A Simple Voltage-Divider as Before
Engineering-43: Engineering Circuit Analysis
79
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example
 Find Vo
 To Start
I S1
V1
• Identify & Label All Nodes
• Write Node Equations
• Examine Ckt to Determine
Best Solution Strategy
 Notice
V0  V1  V2
 Need Only V1 and V2
to Find Vo
 Known Node Potential
@V3 : V3  VS1  12[V ]
Engineering-43: Engineering Circuit Analysis
80
V4
R1
IS2
R2
V2
 VO 
V3
R3
R4
IS3
+

VS 1
R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k
Is1 =2mA, Is2 = 4mA, Is3 = 4mA,
Vs1 = 12 V
 Now KCL at Node 1
V1  V2 V1
@ V1 :  I S1 

0
R1
R4
V1  V2 V1
 2[mA] 

0
1k
2k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example cont.
V4
 At Node 2
I S1
V V V V V V
V
@ V2 :  I S 3  2 1  2 3  2 4  0 1
R1
R3
R2
V  V V  12 V2  V4
 4[mA]  2 1  2

0
1k
1k
2k
IS2
R2
R1
V2
 VO 
V3
R3
R4
IS3
+

VS 1
 At Node 4
V4  V2
@ V4 : I S1  I S 2 
0
R2
R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k
Is1 =2mA, Is2 = 4mA, Is3 = 4mA,
Vs = 12 V
V4  V2
2[mA]  4[mA] 
0
2k
 To Solve the System of Equations Use
LCD-multiplication and Gaussian Elimination
Engineering-43: Engineering Circuit Analysis
81
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example cont.
 The LCDs
*2kΩ
V1  V2 V1
3V1  2V2  4[V ] (1)
 2[mA] 

0
1k
2k
*2kΩ
V2  V1 V2  12 V2  V4
 2V1  5V2  V4  32V ] (2)
 4[mA] 


0
1k
1k
2k
V4  V2
2[mA]  4[mA] 
0
2k
*2kΩ
 V2  V4  4[V ]
(3)
 Now Add Eqns (2) & (3) To Eliminate V4
 2V1  4V2  36[V ]  V1  2V2  18[V ]
(4)
 Now Add Eqns (4) & (1) To Eliminate V2
2V1  22[V ]  V1  11[V ]  BackSub into (4) To Find V2
11[V ]  2V2  18[V ]  V2  14.5[V ]  Find Vo by Difference Eqn
V0  V1  V2  11[V ]  14.5[V ]  3.5[V ]
Engineering-43: Engineering Circuit Analysis
82
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Complex SuperNode
supernode
 Write the Node Eqns
 Set UP
V2
R4
R2
V1
+ +
-
 Nodes Connected To
Reference Through A
Voltage Source
R5
+
-
R1
• Identify all nodes
• Select a reference
• Label All nodes
V3
V4
V5
R3
R6
 Eqn Bookkeeping:
•
•
•
•
 Voltage Sources In
Between Nodes And
Possible Supernodes
KCL@ V3
KCL@ SuperNode
2 Constraint Equations
One Known Node
• Choose to Connect V2 & V4
Engineering-43: Engineering Circuit Analysis
83
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
R7
Complex SuperNode cont.
 Now KCL at Node-3
supernode
V3  V2 V3  V4 V3


0
R4
R5
R7
 Now KCL at Supernode
V2
• Take Care Not to Omit
Any Currents
Vs2
R1
+
-
R2
V1
Vs1
R4
V3
Vs3
+ -
+
-
R5
V4
V5
R3
R6
V2  V1 V5  V1 V5 V4 V4  V3 V2  V3





0
R1
R2
R3 R6
R5
R4
 Constraints Due to Voltage Sources
V1  VS1
V2  V5  VS 2
V5  V4  VS 3
 5 Equations 5 Unknowns → Have to Sweat Details
Engineering-43: Engineering Circuit Analysis
84
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
R7
Dep V-Source Example
 Find Io by Nodal Analysis
 Notice V-Source
Connected to the
Reference Node
V3  6V
 SuperNode Constraint
V1  V2  2Vx
 Controlling Variable in
Terms of Node Voltage
Vx  V2  V1  3V2
Engineering-43: Engineering Circuit Analysis
85
 KCL at SuperNode
 Mult By 12 kΩ LCD
2(V1  6)  V1  2V2  V2  6  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Dep V-Source Example cont
 Simplify the LCD Eqn
3V1  3V2  18V
and 3V2  V1
 4V1  18V
V1  4.5V
 By Ohm’s Law
V1
9V
3
Io 

 mA
12k 24k 8
Engineering-43: Engineering Circuit Analysis
86
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example
Problem 3.46 (6th Ed)
Determine VO
 Select Soln Method
+
• Loop Analysis
VS
– 3 meshes
– One current source
+
-
I2
4k
I3
• Nodal Analysis
6k
2k
IS
– 3 non-reference nodes
– One super node
2k
_
I1
IS = 2mA, VS = 6V
• Both Approaches
Seem Comparable →  Write Loop Eqns for
Select LOOP Analysis
Meshes 1, 2, 3 by KVL
I1  I S
– Specifically Choose
 VS  4k ( I 2  I 3 )  2k ( I 2  I1 )  0
MESHES
 Select Mesh Currents
Engineering-43: Engineering Circuit Analysis
87
VO
2k ( I 3  I1 )  4k ( I 3  I 2 )  6kI3  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example cont
Problem 3.46 (6th Ed)
 We Seek Vo, Thus
Using Ohm’s Law
Need only Find I3
 Simplify: Divide Loop
Eqns by 1kΩ
• I Coeffs Become
NUMBERS
• Voltages Converted
to mA
 Note That I1 = IS and
Sub into Loop Eqns
Engineering-43: Engineering Circuit Analysis
88
Determine VO
+
VS
+
-
I2
4k
I3
6k
VO
2k
IS
2k
_
I1
IS = 2mA, VS = 6V
 Substitute into The
Remaining Loop Eqns
• See Next Slide
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example cont.2
Problem 3.46 (6th Ed)
Determine VO
 The Loop-2 and
Loop-3 Eqns
+
VS


6
I

4
I


2
I

(
6

4
)
mA
2
3
1
 2

1k


 4 I 2  12 I 3  2 I S  4mA 3 and Add
----------------------------32
28 I 3  10  2  4  3  I 3 
mA
28
VS
+
-
I2
4k
I3
2k
IS
2k
_
I1
IS = 2mA, VS = 6V
 Then by Ohm’s Law
V 8
48
VO  6kI3  6 k  mA  V
A 7
7
Engineering-43: Engineering Circuit Analysis
89
6k
VO
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Find Vo- Compare Mesh vs. Loop
 Using MESH Currents
 Using LOOP Currents
Treat The Dependent Source As One More Voltage Source
 Mesh-1 & Mesh-2
 Loop-1 & Loop-2
 2Vx  2kI1  4k ( I1  I 2 )  0
 3  6kI2  4k ( I 2  I1 )  0
Engineering-43: Engineering Circuit Analysis
90
 2Vx  2k ( I1  I 2 )  4kI1  0
 2Vx  2k ( I1  I 2 )  3  6kI2  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Compare Loop vs. Mesh cont.
 Using MESH Currents
 Using LOOP Currents
 Now Express The Controlling Variable In Terms Of
MESH or LOOP Currents
Vx  4kI1
Vx  4k ( I1  I 2 )
Solving
 2kI1  4kI2  0
 6kI1  6kI2  0
 4kI1  10kI2  3
 6kI1  8kI2  3
Engineering-43: Engineering Circuit Analysis
91
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Compare Loop vs. Node cont.
 Using MESH Currents
I1  3mA, I 2  1.5mA
 Using LOOP Currents
Solutions
Finally
I1  1.5mA, I 2  1.5mA
VO  6kI 2  9[V ]

Notice The Difference Between
MESH Current I1 and LOOP Current
I1 even Though They Are Associated
With The Same Path
Engineering-43: Engineering Circuit Analysis
92

The Selection Of LOOP Currents
Simplifies Expression for Vx and
Computation of Vo
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Dep Isrc Not Shared by Mesh
 Treat The Dependent
Source As A
Conventional Source
 Equations For Meshes
With Current Sources
 Then KVL on The
Remaining Loop (I3)
Engineering-43: Engineering Circuit Analysis
93
 Express The Controlling
Variable, Vx, In Terms Of
Loop Currents
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Dep Isrc Not Shared by Mesh
 Asked to Find Only Vo
• Need Only Determine I3
 The Dep Src Eqns
Vx

 Vx  2kI1 
2000
  I1  2 I 2  4mA

Vx  4 k ( I1  I 2 )
I1 
 From KVL Eqn for I3
 Thus
33
VO  6kI3  V
4
11
8kI3  3  2kI1  I 3  mA
8
Engineering-43: Engineering Circuit Analysis
94
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Find Vo Using Mesh Analysis
I1
 Draw the Mesh Currents
 Write KVL Mesh Eqns
For Mesh-1 & Mesh-2
 2kIx  2kI1  4k ( I1  I 2 )  0
I2
 Controlling Variable In
Terms Of Loop Currents
I x  I2
12  2kI2  4k ( I 2  I1 )  0
Engineering-43: Engineering Circuit Analysis
95
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Find Vo Using Mesh Analysis cont
I1
 Substitute & Collect
Terms
6kI1  6kI2  0
 4kI1  6kI2  12
2kI1
 12
Engineering-43: Engineering Circuit Analysis
96
I2
 Solve for I2
 24  6kI2  12  I 2  6mA
 Finally Vo
VO  2kI2  12[V ]
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
WhiteBoard Work
 Let’s Work This
Problem
1K
+
12V
1K
2IX
1K
1K
IO
IX
VO
-
 Find the OutPut
Voltage, VO
Engineering-43: Engineering Circuit Analysis
97
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Engineering-43: Engineering Circuit Analysis
98
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Outline of Theorem Proof
 Consider Linear Circuit → Replace vo with a SOURCE
 If Circuit-A is Unchanged Then The Current Should Be
The Same FOR ANY Vo (Source or Rat’sNest Generated)
 Use Source SuperPosition
• 1st: Inside Ckt-A OPEN all I-Src’s, SHORT All V-Srcs
– Results in io Due to vo
• 2nd: Short the External V-Src, vo
– Results in iSC Due to Sources Inside Ckt-A
Engineering-43: Engineering Circuit Analysis
99
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Theorem Proof Outline cont.
 Graphically the Superposition

iO
All independent
sources set to
zero in A

iSC
 Then The Total Current
i  iO  iSC
 Now DEFINE
using V/I
for Ckt-A RTH
vO

 iO
Engineering-43: Engineering Circuit Analysis
100
 Then By Ohm’s Law
i  iO  iSC or
vO
i
 iSC
RTH
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Theorem Proof Outline cont.2
 Consider Special Case Where Ckt-B is an OPEN (i =0)
For i  0 : vO  vOC
vOC
0
 iSC
RTH
 The Open Ckt Eqn Suggests
v
v
 iSC  OC
and
 RTH  OC
RTH
iSC
 Also recall
i 0
Think y = mx + b
 How Do To Interpret These Results?
• vOC is the EQUIVALENT of a single Voltage Source
• RTH is the EQUIVALENT of a Single Resistance which
generates a Voltage DROP due to the Load Current, i
101
vOC

vO
vO
vOC
i
 iSC  

 vO   RTH i  vOC
RTH
RTH RTH
Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Theorem Proof –Version 2
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

i
a
vO
_
b
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
1. Because of the LINEARITY of the models, for any Part
B the relationship between vO and the current, i, has to
be of the form vO  m  i  b (Linear Response)
2. Result must hold for “every valid Part B”
3. If part B is an open circuit then i=0 and... b  vO  vOC
4. If Part B is a short circuit then vO is zero. In this case
vOC
0  m  iSC  b  m  
  RTH or vO   RTH i  vOC
iSC
Engineering-43: Engineering Circuit Analysis
102
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example  Find Thevenin Equiv.
R1
VS
IS
+
-
VTH
R2
a
I SC
To Part B
b
 Find VTH by Nodal
Analysis: Iout = 0
VTH VTH  VS

 IS  0
R2
R1
(
V
1
1
 )VTH  S  I S
R1 R2
R1
VTH
R2
R1 R2

VS 
IS
R1  R2
R1  R2
VTH

R1 R2  VS
  I S 

R1  R2  R1

Engineering-43: Engineering Circuit Analysis
103
 Part B is irrelevant. The
voltage Vab will be the
value of the Thevenin
equivalent source.
 For Short Circuit Current
Use Superposition
 When IS is Open the
Current Thru the Short
I  VS R1
1
SC
 When VS is Shorted the
Current Thru the Short
I
2
SC
 IS
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example – Find Thevenin cont
R1
VS
+
-
VTH
IS
R2
 Find the Total Short Ckt
Current

VS 
I SC   I S  
R1 

a
I SC
To Part B
b
 Find Thevenin Resistance
RTH
VTH

I SC
 Then RTH
RTH
R1 R2

R1  R2
Engineering-43: Engineering Circuit Analysis
104
 To Find RTH Recall

R1R2  VS
  I S 
VTH 
R1  R2  R1

 In this case the Thevenin resistance
can be computed as the resistance
from a-b when all independent sources
have been set to zero
• Is this a GENERAL Result?
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example
 Find Vo Using
Thevenin’s
Theorem
“PART B”
 First, Identify
Part-B
 Deactivate
(i.e., Short Ckt)
6V & 12V Sources
to Find RTH
Engineering-43: Engineering Circuit Analysis
105
RTH
RTH  3k || 6k  2k
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Numerical Example cont.

 Use Loop Analysis to
Find the Open Circuit
Voltage
I
VOC
9kI  18[V ]  I  2mA
0  VOC  3kI  12  VOC  6[V ]
 The Resulting
Equivalent Circuit
 Finally the Output
4
VO 
(6V )  3[V ]
44
RTH  2k 2k


VTH  6V

4k
Engineering-43: Engineering Circuit Analysis
106

VO

Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example
I2

I1
KVL
+
-
Xform
RTH
VTH
VTH

 Deactivate Srcs for RTH
RTH
RTH  3R || 3R  1.5R
 Use Loops for VTH
I 1 I S
 VS  5R( I1  I 2 )  RI 2  0
VTH  RI 2  2R( I1  I 2 )
Engineering-43: Engineering Circuit Analysis
107
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
Example cont.
 OR, Use Superposition to Find
Thevenin Voltage
 First Open The Current Source
VS
 1R  2 R 
1
VTH  VS 

2
 1R  2 R  3R 
 Next Short-Circuit the
Voltage Source
• Using I-Divider
IS I1
+
R
V2TH
3R
I2
2R
_
KVL
5
I1  I S
6
1
I2  IS
6
Engineering-43: Engineering Circuit Analysis
108
 Find Isrc Contribution by
KVL 2
1
VTH  RI 1  2 RI 2 
 Add to Find Total VTH
1
VTH  VTH
 VTH2
VTH
VS RI S
 
2
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx
2
RI S
WhiteBoard Work
 Let’s Work This
Problem
2K
2K
+
6V
2K
Vo
2K
4mA
-
• Find Vo by Source
Transformation
Engineering-43: Engineering Circuit Analysis
109
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx