Transcript Engineering 43 Chp 5.3a Thevenin Norton

Engineering 43
Chp 5.3a
Thevenin
Norton
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
2
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
3
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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 Equivalent
• The Equivalent Ckt in
RED “Looks” The
Same to the Speakers
As Does the
Complicated Circuit
Engineering-43: Engineering Circuit Analysis
4
RTH
VTH
+
-
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
RESISTANCE
b
PART B
PART A
 Thevenin Equivalent Circuit for PART A
Engineering-43: Engineering Circuit Analysis
5
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
 iN = Norton
Equivalent
CURRENT
Source
 RN = Norton
Equivalent
Parallel
RESISTANCE
b
PART B
PART A
 Norton Equivalent Circuit for PART A
Engineering-43: Engineering Circuit Analysis
6
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
 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
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
RTH
vO

 iO
Engineering-43: Engineering Circuit Analysis
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 Then By Ohm’s Law
i  iO  iSC or
vO
i
 iSC
RTH
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
vOC

 The Open Ckt Case Suggests
v
v
 iSC  OC
and
 RTH  OC
RTH
iSC
 Also recall
vO
vO
vOC
i
 iSC  

 vO  vOC  RTH i
RTH
RTH RTH
 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
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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  n (Linear Response)
2. Result must hold for “every valid Part B”
3. If part B is an open circuit then i=0 and... n  vO  vOC
4. If Part B is a short circuit then vO is zero. In this case
vOC
0  m  iSC  n  m  
  RTH or vO   RTH i  vOC
iSC
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
Examine Thevenin Approach
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

i
a
ANY
PART B
vO
_

For ANY Part-B Circuit

The Thevenin Equiv
Ckt for PART-A →
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
11
vOC
+
_
i
+
vO
_
PART A MUST BEHAVE LIKE
THIS CIRCUIT
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
12
i a
RTH
vO

Norton
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
15
 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
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
PROBLEM Find VO using source transformation
Norton
Norton
EQUIVALENT CIRCUITS
I0
Or one more source transformation
R eq
Veq
+
-
RTH
VVeqTH Req I eq

R4
V0

Engineering-43: Engineering Circuit Analysis
17
3 current sources in parallel and
three resistors in parallel
R3
Veq  Req I eq
V0 
R4
Veq
R4  R3  Req
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
Source Xform Summary
 These Models are Equivalent
+
-
RV
VS
a
RV  RI  R
a
RI
b
IS
VS  RI S
I S  R VS
b
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
18
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
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RTH
vOC

i SC
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
Graphically...
1. Determine the
Thevenin equivalent
source
2. Determine the
SHORT CIRCUIT
current
Remove part B and
compute the SHORT
CIRCUIT current I ab
vTH  VAB  vOC
RTH
Engineering-43: Engineering Circuit Analysis
i 0 a
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
Remove part B and
compute the OPEN
CIRCUIT voltage Vab
 Then
20
One circuit problem

vOC
_
b

Vab
_
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
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
b
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
21
 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
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
Example – Find Thevenin cont
R1
VS
+
-
VTH
IS
R2
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
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 Find the Total Short Ckt
Current

VS 
I SC   I S  
R1 

 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
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
23
R1
R2
RTH
b
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
Thevenin w/ Indep. Sources cont
“Part B”
RTH
 Since the evaluation of
the Thevenin equivalent
 3k can be very simple, we
can add it to our toolkit for
the solution of circuits
RTH  2k  3k 6k  4k
“Part B”
Engineering-43: Engineering Circuit Analysis
24
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
6V
5k
Engineering-43: Engineering Circuit Analysis
25
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
26
 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
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
27
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
28
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
29
8
V0 
16[V ]  8V
88
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
31
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
32
10
 2  2 4  k
3
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
6  2I 2
5
I1 
mA  mA
6
3
 Finally The Equivalent
Circuit
 Then VOC
VOC  4k * I1  2k * I 2
VOC  20 / 3V   4V   32 / 3[V ]
Engineering-43: Engineering Circuit Analysis
33
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
34
RTH
RTH  3k || 6k  2k
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
RTH  2k 2k


VTH  6V

4k
Engineering-43: Engineering Circuit Analysis
35

4
VO 
(6V )  3[V ]
44
VO

Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
CALCULATE Vo USING NORTON
4k
I
RN
IN
I SC
 RN

VO  2kI  2k 
I N 
 RN  6k

3
4
VO  2 (2)  [V ]
9
3
RN  RTH  3k
PART B
12V
I SC  I N 
 2mA  2mA
3k
COMPUTE Vo USING THEVENIN
2k
PART B
VTH
RTH

+
-
2k

VTH
VTH  12
 2mA  0  VTH  12  6 RTH  3k  4k
3k
Engineering-43: Engineering Circuit Analysis
36
VO 
VO
2
4
(6V )  [V ]
27
3
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
37
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
38
 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
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt
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
39
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt