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Engineering 43
Chp 8 [3-4]
Magnetic Coupling
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Outline – Magnetic Coupling
 Mutual Inductance
• Behavior of inductors sharing a common
magnetic field
 Energy Analysis
• Used to establish relationship between
mutual reluctance and self-inductance
Engineering-43: Engineering Circuit Analysis
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Outline – Magnetic Coupling cont.
 Ideal Transformer
• Device modeling of components used to
change voltage and/or current levels
 Safety Considerations
• Important issues
for the safe
operation of circuits
with transformers
Engineering-43: Engineering Circuit Analysis
3
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
The Ideal Transformer
 Consider Now two Coils Wrapped Around a
Closed Magnetic (usually iron) Core.
A  Area
 The Iron Core Strongly confines the Magnetic Flux, ,
to the Interior of the Closed Ring
• All Turns, N1 & N2, of Both Coils are Linked by the Core Flux
– Again, this is a NONconductive (no wires) connection
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Ideal X-Former Physics/Math
 The Coils, N1 & N2, are
Flux-Linked:  = N
 By Faraday’s Induction
Law for Both Coils
d d  N1 
v1 

dt
dt
d d  N 2 
v2 

dt
dt
Engineering-43: Engineering Circuit Analysis
5
 Then the Ratio of v1:v2
v1 N1 d dt N1


v2 N 2 d dt N 2
 Next Apply Ampere’s
Law (One of
Maxwell’s Eqns)
 H  dl  i
encl
 N1i1  N 2i2
 Where
• H  Magnetic Field
Strength (Amp/m)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Ideal X-Former Physics cont.1
 Ampere’s Law
 H  dl  N i
11
 H=0 in Ampere’s Law
 N 2i2
N1i1  N 2i2  0
or
 The Path for the Closed
Line Integral is a path
Within the Iron Core
 If the Magnetic Core is
IDEAL, then
H 0
Engineering-43: Engineering Circuit Analysis
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i1
N2

i2
N1
 Now Manipulate
Ampere’s Law Eqn
v1
N1i1  N 2i2  0 so
N1
N2
v1i1 
v1i2  0
N1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Ideal X-Former Physics cont.2
 The Ideal Xformer
 But By Flux Linkage
v1  v2 N1 N2 
 So in the ideal Case
Ampere’s Law
N 2  N1 
 v2
i2  0
v1i1 
N1  N 2 
 But v•i = POWER, and
by the previous Eqn
the total power used by
the Xformer is ZERO
 Thus in Ideal Form,
a Transformer is
LOSSLESS
• Thus the INput Power =
OUTput Power
or v1i1  v2i2  0
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Ideal X-Former Circuit Symbol
 From The Device
Physics; The Ideal
Xformer Eqns
 The Circuit Symbol
N1i1  N 2i2  0
v1 N1

v2 N 2
 Since an Xformer is
Two Coupled Inductors,
Need the DOT
Convention to
Track Polarities
Engineering-43: Engineering Circuit Analysis
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Iron Core
 The Main practical
Application for This
Device:
• TRANSFORM one AC
Voltage-Level to Another
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Transformer Application
 When a Voltage is
Transformed, Give the
INput & OUTput sides
Special Names
• INput, v1, Side 
PRIMARY Circuit
• OUTput, v2, Side 
SECONDARY Circuit
 The Voltage Xformer
 Pictorial Representation
 Then the Circuit Symbol
Usage as Applied to a
Real Circuit
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Transformer Practical App
 The Actual Ckt
Symbol As used On
an Engineering Dwg
 The Parallel (||) lines
Between the Coils
Signify the
Magnetic Core
Engineering-43: Engineering Circuit Analysis
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 The Practical Symbol
does NOT Use DOTS
• NEMA has Established
Numbering Schemes
That are Functionally
Equivalent to the Dots
 The Multiple “Taps” on
the Primary Side Allow
The Transformation of
More Than One Voltage
Level
 See next Slide for a
REAL Xformer Design
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
208:115 Vac
StepDown Xformer
208:24 Vac
StepDown Xformer
208Vac, 80 kVA Input
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Sign (Dot) Conventions
 Have TWO Choices for
Polarity Definitions
• Symmetrical
 Thus The Form of the
Governing Equations
Will Depend on the
Assigned:
• DOT POSITION
• VOLTAGE POLARITY
• CURRENT DIRECTION
• INput/OUTput (I/O)
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Phasor Analysis
 In Practice, The Vast
Majority of Xformers are
Used in AC Circuits
 Recall The Symmetrical
Ideal-Xformer Eqns
 Illustration: InPut
IMPEDANCE = V1/I1
N1i1  N 2i2  0
v1 N1
i1
N2

;

v2 N 2
i2
N1
 These are LINEAR in
i & v, so PHASOR
Analysis Applies
Engineering-43: Engineering Circuit Analysis
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 Notice That This is an
I/O Model; So the Eqns
V1 N1
I1
N

;
 2
 I 2  N1
V2 N 2
I1
N2


I2
N1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Phasor Analysis: Input Z
 An I/O Xformer in
Phasor Domain
N2
N1
V2  Z L I 2  V1
 Z L I1
N1
N2
 Solve for V1
2
 N1 
 Z L I1
V1  
 N2 
 Now Apply Ohm’s Law
to the Load, ZL
V2  ZLI 2
 Now Sub for V2 and I2
From I/O Xformer Eqns
Engineering-43: Engineering Circuit Analysis
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 Now The Input
Impedance Z1
2
 N1 
V1
 Z L
 Z1  
I1
 N2 
• For 10X stepDOWN (N1:N2 =
10:1) Z1 is 100X ZL
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Phasor Analysis: Input Z cont.1
 An I/O Xformer Phasor
Domain Input
Impedance
2
 N1 
V1
 Z L
 Z1  
I1
 N2 
 For a LOSSLESS
Primary/Secondary
Xformer the INput
impedance is a Fcn
ONLY of the LOAD
Impedance, ZL
Engineering-43: Engineering Circuit Analysis
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 Thus ZL is Said to be
REFLECTED to the
Input Side (by [N1/N2]2)
 For Future Reference
*

N  N 
S1  V1I1*   V2 1  I 2 2   V2 I *2  S 2
 N 2  N1 
N2
n
 turns ratio
N1
 Ideal Xformer Phasor Eqns
V2
V1 
I 1  nI 2
n
ZL
Z1  2
S1  S 2
n
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Numerical Example
 Given the Ckt Below
Find all I’s and V’s
 Using
Z1  Z L n 2
stepDOWN
Xformer
Z1  32  j16
 Note: n = N2/N1 = 1/4
 Game Plan
• reflect impedance into
the primary side and
make the transformer
“transparent to Source”
Engineering-43: Engineering Circuit Analysis
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 Find I1 by Ohm
VS
1200
I1 

Z1  Z 2 32  j16  18  j 4
1200

 2.33  13.5
51.4213.5
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
ZL
Numerical Example cont.1
Z1I1  (32  j16)  2.33  13.5
 The Intermediate Ckt
ZL
Z2
Z1
32  j16
1200 
120
Z1  Z 2
51.4213.5
 About the Same
Hassle-Factor; use ZI
stepDOWN
Xformer
V1  35.7826.57  2.33  13.5
Z1  32  j16
 Now Find V1 by Ohm or
V-Divider
Z1
V1  Z1I1 
VS
Z1  Z 2
 83.3613.07
 Next Determine SIGNS
 Which is Easier?
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Numerical Example cont.2
V1 N1 1

  V2  nV1
V2 N 2 n
 The Original Ckt
I1
N
  2  n  I 2  I1 n
I2
N1
 Compare Current Case
to I/O Model
stepDOWN
Xformer
 Recall Now the I/O
Model Eqns
IN
OUT
• Voltage-2 is OPPOSITE
(NEGATIVE at Dot)
• Current-2 is OPPOSITE
(INTO Dot)
 Then In This Case
 V2  nV1 ;
Engineering-43: Engineering Circuit Analysis
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 I 2  I1 n
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Numerical Example cont.3
 The Original Ckt
stepDOWN
Xformer
 The Output Voltage
V2  nV1  0.25V1  0.2583.4913.07
 0.2581.32  j18.88
 20.33  j 4.720  3rd Quadrant
V2  20.875  166.93
 Then Output Current
 Using the Signs as
Determined by Dots
and Polarities
I1
 4I1   42.33  13.5
n
 42.266  j 0.5439
  9.065  j 2.176  2nd Quadrant
I 2  9.32166.5
I2  
 Note: On Calculator aTan(–4.72/–20.33) = 13.07°
• Recall RANGE of aTan = –90° to + 90°
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Illustration
 Given the Ckt Below Find I1 & Vo
n2
stepUP
Xformer
 Note: n = N2/N1 = 2/1
 Game Plan
2  j2  2
Z1 
 1  j 0.5
2
2 1
• reflect impedance into
the primary side and
make the transformer
“transparent to Src”
 Again using Z1  Z L n 2
Engineering-43: Engineering Circuit Analysis
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Z1
Zi  3  j 2.5
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Illustration cont.1
 Given the Ckt Below Find I1 & Vo
n2
Z2
stepUP
Xformer
 Thus I1 by Ohm
VS
12V0
I1 

Z 2  Z1 2  j 2  1  j 0.5
I1 
12V0
 3.07 A39.81
3.91  39.81
Engineering-43: Engineering Circuit Analysis
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I2 
I1
n
 Next Find Vo by I2 and
Ohm’s Law
• Define I2 Direction per
I/O Model (V2 is ok)
I
I 2  1 and V0 Z L I 2 so
2
I
V0  2  1  3.07V39.81
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Xformer Thevenin Equivalent
 Given I/O XFormer Ckt
 Find the The Thevenin
Equivalent at 2-2’
 First Find the OPEN Ckt
Voltage at 2-2’
• Note: The Dots &
Polarities Follow the
I/O Model
Engineering-43: Engineering Circuit Analysis
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I2  0 
  I1  0  V1  VS1
I1  nI 2 
V1  VS1 
  VOC  nVS1
V2  nV1 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Xformer Thevenin Equiv. cont.1
 Now find ZTH at
Terminals 2-2’
 “Back Reflect”
Impedance into
SECONDARY
ZTH  n Z1
2
Engineering-43: Engineering Circuit Analysis
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 Thus the Thevenin
Equivalent at 2-2’
ZTH
 The Xformer has been
“made Transparent” to
the Secondary Side
• Next: Find
Thevenin Equiv at
Terminals 1-1’
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Thevenin Equiv. from Primary
 Given I/O XFormer Ckt
 As in Open Ckt
I1  0 and I 2  0
ZTH 
 Thevenin impedance
will be the Secondary
impedance reflected
into the PRIMARY Ckt
Z2
n2
 Find the The Thevenin
Equivalent at 1-1’
 Then The Open Ckt
Voltage Depends on
VS2
V V n
OC
S2
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Primary v. Secondary Thevenin
 The Base Ckt
 Thevenin From Primary
 Thevenin From Secondary
 Equivalent circuit
reflecting into primary
 Equivalent circuit
reflecting into secondary
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Exmpl: Draw Thevenin Equiv’s
n2
Equivalent circuit reflecting
into SECONDARY
Equivalent circuit reflecting
into PRIMARY
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
2  j2
Example
360
 Given the Ckt Below
Find I1
1
 Note The Dot Locations
2  j 2 ()
1’
 Note: n = N2/N1 = 2/1
 Game Plan
 Find Thevenin Looking
from PRIMARY Side
• Draw the Ckt
Engineering-43: Engineering Circuit Analysis
27
I1 
0.5
I1
360
 120

 6V
n
2
VS2
I1 
360  60
2.5  j 2
420
 13.11938.66
3.2016  36.86
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Example: Safety Considerations
 Two Houses Powered
By DIFFERENT
XFormers
 Utility Circuit Breaker
X-Y OPENS: Powering
DOWN House-B
 The Well-Meaning
Neighbor Runs
Extension Cord
House-A → House-B
• This POWERS the
2nd-ary Side of the
House-B Pole
Xformer
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Example: Safety Consid cont.1
 Transformers are
BIdirectional Devices
• They can step-UP or
step-DOWN Voltages
 Thus the 120Vac/15A Extension Cord Produces
7200 Vac Across Terminals X-Z
 The Service Engineer (SE) Now Goes to the
Breaker to ReMake the Connection to House-B
 The SE expects ZERO Volts at X-Z; If She/He Does NOT
Check by DMM, then He/She Could Sustain a Potentially
FATAL 7.2 kV Electric-Shock!
Engineering-43: Engineering Circuit Analysis
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
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Exmple  Power Transmission
 At the Generating
Facility (e.g. Diablo
Canyon) Electricity is
Generated at
15-25 kVac
 But Xformers are used
to Set-UP the VoltageLevel to 400-765 kVac
 Why? → Line SIZE
(and others)
Engineering-43: Engineering Circuit Analysis
30
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Exmpl – Power Xmission cont.1

Case Study: Transmit
•
•
•
225 MW over
100 Miles of Wire
2 Conductors
•
•
95% Efficiency
Cu Wire w/ Resistivity
– ρ = 80 nΩ-m

Find the Wire
Diameter, d, for:
a) V = 15 kVac
b) V = 500 kVac
Engineering-43: Engineering Circuit Analysis
31
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Exmpl – Power Xmission cont.2
 By Solid-State Physics
(c.f. ENGR-45)
Rwire  l A
• Where for the Wire
– ρ  Resistivity (Ω-m)
– l  Length (m)
– A  X-Section Area (m2)
 In This Case
l  100mi  160935m
A  d 2 4
 Then the Power Loss
Engineering-43: Engineering Circuit Analysis
32
Pwire  5%  225MW  11.25MW
 By Power Rln for
Resistive Ckts
Ploss  V 2 Rwire  Pwire  11.25MW
 Rwire  V
2
Pwire  l A 
 Solve for d
4 l V 2

d 
2
d
P
d 4
4 lP 1
 V
1
Subbing Values : d  1920
V
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
l
2
Exmpl – Power Xmission cont.3
 Finally Solve for
Transmission Cable
Diameter
1
d15  1920
 0.128m
15000
1
d 500  1920
 0.00384m
500000
 d15 = 5.03”
• Pretty BIG & HEAVY
 d500 = 0.15”
• MUCH Better
Engineering-43: Engineering Circuit Analysis
33
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Summary: Ideal Transformer
 Consider Now two Coils Wrapped Around a Closed
Magnetic (usually iron) Core.
A  Area
 The Iron Core Strongly confines the Magnetic Flux, ,
to the Interior of the Closed Ring
• All Turns, N1 & N2, of Both Coils are Linked by the Core Flux
Engineering-43: Engineering Circuit Analysis
34
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Ideal X-Former Circuit Symbol
 The Ideal Xformer Eqns
 The Circuit Symbol
• Faraday’s Law
v1 N1

v2 N 2
• Ampere’s Law
N1i1  N2i2  0
 As Two Coupled
Inductors, Xformers
Use the DOT
Convention to Track
Polarities
Engineering-43: Engineering Circuit Analysis
35
Iron Core
 The Main practical
Application for This
Device:
• TRANSFORM one AC
Voltage-Level to Another
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Transformer Application
 When a Voltage is
Transfrormed, Give the
INput & OUTput sides
Special Names
• INput, v1, Side 
PRIMARY Circuit
• OUTput, v2, Side 
SECONDARY Circuit
 The Voltage Xformer
 Pictorial Representation
 Then the Circuit Symbol
Usage as Applied to a
Real Circuit
Engineering-43: Engineering Circuit Analysis
36
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Sign (Dot) Conventions
 Have TWO Choices for
Polarity Definitions
 The Form of the
Governing Equations
• Symmetrical
N1i1  N 2i2  0
v1 N1

v2 N 2
• INput/OUTput (I/O)
N1i1  N 2i2
v1 N1

v2 N 2
Engineering-43: Engineering Circuit Analysis
37
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
Phasor Analysis
 In Practice, The Vast
Majority of Xformers are
Used in AC Circuits
 Ideal-Xformer Eqns Are
LINEAR in i&v so
PHASOR Analysis
Applies
V1 N1 1


V2 N 2 n
N1I1  N 2I 2 (I/O)
N1I1  N 2I 2  0 (symm)
Engineering-43: Engineering Circuit Analysis
38
 Illustration: InPut
IMPEDANCE
 Notice That This is an I/O
Model; So the Eqns Yield
2
 N1 
V1
 Z L
 Z1  
I1
 N2 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
WhiteBoard Work
 Let’s Work This Nice
Problem
1:2
Is
I2=oA
1
-j1
j1
1


Ideal
Figure P8.60
Engineering-43: Engineering Circuit Analysis
39
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt
208:115 Vac
StepDown Xformer
208:24 Vac
StepDown Xformer
208Vac, 80 kVA Input
Engineering-43: Engineering Circuit Analysis
40
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt