Page 1
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
[email protected]
Engineering 43
Chp 8 [3-4]Chp 8 [3-4]Magnetic Magnetic CouplingCoupling
Page 2
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Outline – Magnetic CouplingOutline – Magnetic Coupling
Mutual Inductance• Behavior of inductors sharing a common
magnetic field
Energy Analysis• Used to establish relationship between
mutual reluctance and self-inductance
Page 3
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Outline – Magnetic Coupling cont.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
Page 4
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Ideal TransformerThe Ideal Transformer Consider Now two Coils Wrapped Around a
Closed Magnetic (usually iron) Core.
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
A Area
Page 5
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ideal X-Former Physics/MathIdeal X-Former Physics/Math The Coils, N1 & N2, are
Flux-Linked: = N Then the Ratio of v1:v2
Next Apply Ampere’s Law (One of Maxwell’s Eqns)
2211 iNiNidlH encl
dt
Nd
dt
dv
dt
Nd
dt
dv
22
11
By Faraday’s Induction Law for Both Coils
2
1
2
1
2
1
N
N
dtd
dtd
N
N
v
v
Where• H Magnetic Field Strength
(Amp/m)
Page 6
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ideal X-Former Physics cont.1Ideal X-Former Physics cont.1 Ampere’s Law H=0 in Ampere’s Law
Now Manipulate Ampere’s Law Eqn The Path for the Closed
Line Integral is a path Within the Iron Core
If the Magnetic Core is IDEAL, then
2211 iNiNdlH
0H
1
2
2
1
2211 0
N
N
i
i
iNiN
or
0
0
211
211
22111
1
ivN
Niv
soiNiNN
v
Page 7
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ideal X-Former Physics cont.2Ideal X-Former Physics cont.2 The Ideal Xformer 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
But By Flux Linkage
0or
0
2211
22
12
1
211
iviv
iN
Nv
N
Niv
2121 NNvv So in the ideal Case
Ampere’s Law
Page 8
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ideal X-Former Circuit SymbolIdeal X-Former Circuit Symbol From The Device Physics;
The Ideal Xformer Eqns The Circuit Symbol
Since an Xformer is Two Coupled Inductors, Need the DOT Convention to Track Polarities
2
1
2
1
2211 0
N
N
v
v
iNiN
The Main practical Application for This Device:• TRANSFORM one AC
Voltage-Level to Another
Iron Core
Page 9
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Transformer ApplicationTransformer 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
Then the Circuit Symbol Usage as Applied to a Real Circuit
Pictorial Representation
Load
Src
Page 10
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Transformer Practical AppTransformer Practical App The Actual Ckt
Symbol As used On an Engineering Dwg
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
The Parallel (||) lines Between the Coils Signify the Magnetic Core
Page 11
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
208Vac, 80 kVA Input
208:115 Vac StepDown Xformer
208:24 Vac StepDown Xformer
Page 12
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Sign (Dot) ConventionsSign (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)
Page 13
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Phasor AnalysisPhasor Analysis In Practice, The Vast
Majority of Xformers are Used in AC Circuits
Recall The Symmetrical Ideal-Xformer Eqns
Illustration: InPut IMPEDANCE = V1/I1
These are LINEAR in i & v, so PHASOR Analysis Applies
Notice That This is an I/O Model; So the Eqns
1
2
2
1
2
1
2
1
2211
;
0
N
N
i
i
N
N
v
v
iNiN
1
2
2
1
1
2
2
1
2
1
2
1 ;
N
N
N
N
N
N
I
I
I
I
V
V
Page 14
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Phasor Analysis: Input ZPhasor Analysis: Input Z An I/O Xformer in
Phasor Domain Solve for V1
Now Apply Ohm’s Law to the Load, ZL
Now The Input Impedance Z1
2IZV2 L Now Sub for V2 and I2
From I/O Xformer Eqns
2
11
1
2122 N
N
N
NLL IZVIZV
1
2
2
11 IZV LN
N
LN
NZZ
I
V2
2
11
1
1
• For 10X stepDOWN (N1:N2 = 10:1) Z1 is 100X ZL
Page 15
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Phasor Analysis: Input Z cont.1Phasor Analysis: Input Z cont.1 An I/O Xformer Phasor
Domain Input Impedance
Thus ZL is Said to be REFLECTED to the Input Side (by [N1/N2]2)
For Future Reference
For a LOSSLESS Primary/Secondary Xformer the INput impedance is a Fcn ONLY of the LOAD Impedance, ZL
Ideal Xformer Phasor Eqns
LN
NZZ
I
V2
2
11
1
1
2*22
*
1
22
2
12
*111 S
N
N
N
NS
IVIVIV
ratio turns1
2 N
Nn
2121
212
1
SSn
nn
L
ZZ
IIV
V
Page 16
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical ExampleNumerical Example Given the Ckt Below
Find all I’s and V’s Using
Note: n = N2/N1 = 1/4
Game Plan• reflect impedance into
the primary side and make the transformer “transparent to Source”
Find I1 by Ohm
21 nLZZ
LZ
16321 jZ
5.1333.25.1342.51
0120
4181632
0120
211 jj
S
ZZ
VI
stepDOWNXformer
Page 17
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example cont.1Numerical Example cont.1 The Intermediate Ckt
About the Same Hassle-Factor; use ZI
Now Find V1 by Ohm or V-Divider
Next Determine SIGNS
LZ
16321 jZ
SVZZ
ZIZV
21
1111
Which is Easier?
07.1336.83
5.1333.257.2678.351V
1205.1342.51
16320120
5.1333.2)1632(
21
1
11
j
j
ZZ
Z
IZ
Z2
stepDOWNXformer
Page 18
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example cont.2Numerical Example cont.2 The Original Ckt
Compare Current Case to I/O Model• Voltage-2 is OPPOSITE
(NEGATIVE at Dot)
• Current-2 is OPPOSITE (INTO Dot)
Then In This Case
Recall Now the I/O Model Eqns
nnN
N
nnN
N
121
2
2
1
122
1
2
1 1
III
I
VVV
V
n
n
12
12 ;
II
VV
IN OUT
stepDOWNXformer
Page 19
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example cont.3Numerical Example cont.3 The Original Ckt The Output Voltage
Using the Signs as Determined by Dots and Polarities
93.166875.20
Quadrant 3rd720.433.20
88.1832.8125.0
13.0783.490.2525.0
2
112
V
VVV
j
j
n
5.16632.9
Quadrant 2nd176.2065.9
5439.0266.24
5.1333.244
2
11
2
I
II
I
j
jn
Then Output Current
stepDOWNXformer
Note: On Calculator aTan(–4.72/–20.33) = 13.07°• Recall RANGE of aTan = –90° to + 90°
Page 20
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
IllustrationIllustration Given the Ckt Below Find I1 & Vo
Note: n = N2/N1 = 2/1
Game Plan• reflect impedance into
the primary side and make the transformer “transparent to Src”
Again using
2n
21 nLZZ
5.01
12
22221 j
jZ
1Z
5.23 jZ i
stepUPXformer
Page 21
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Illustration cont.1Illustration cont.1 Given the Ckt Below Find I1 & Vo
Thus I1 by Ohm
2n
n
II 1
2
5.0122
012
121 jj
VS
ZZ
VI
81.3907.3
81.3991.3
0V121 AI
Next Find Vo by I2 and Ohm’s Law• Define I2 Direction per I/O
Model (V2 is ok)
81.3907.32
2
so and 2
10
201
2
V
L
IV
IZVI
I
Z2
stepUPXformer
Page 22
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt22
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Xformer Thevenin EquivalentXformer 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
00
121
2
III
I
n 11 SVV
112
11SOC
S nn
VVVV
VV
Page 23
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt23
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Xformer Thevenin Equiv. cont.1Xformer Thevenin Equiv. cont.1 Now find ZTH at
Terminals 2-2’
“Back Reflect” Impedance into SECONDARY
Thus the Thevenin Equivalent at 2-2’
THZ
12ZZ nTH
The Xformer has been “made Transparent” to the Secondary Side• Next: Find
Thevenin Equiv at Terminals 1-1’
Page 24
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt24
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Thevenin Equiv. from PrimaryThevenin Equiv. from Primary Given I/O XFormer Ckt
Find the The Thevenin Equivalent at 1-1’
Then The Open Ckt Voltage Depends on VS2
As in Open Ckt
Thevenin impedance will be the Secondary impedance reflected into the PRIMARY Ckt
nSOC 2VV
0 and 0 21 II
22
n
ZZTH
Page 25
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt25
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Primary v. Secondary TheveninPrimary v. Secondary Thevenin
Thevenin From Primary
Equivalent circuit reflecting into primary
Thevenin From Secondary
Equivalent circuit reflecting into secondary
The Base Ckt
Page 26
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Exmpl: Draw Thevenin Equiv’sExmpl: Draw Thevenin Equiv’s2n
Equivalent circuit reflecting into SECONDARY
Equivalent circuit reflecting into PRIMARY
Page 27
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt27
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ExampleExample Given the Ckt Below
Find I1
Note The Dot Locations
Note: n = N2/N1 = 2/1
Game Plan Find Thevenin Looking
from PRIMARY Side• Draw the Ckt
22 j
036
5.0)(22 j
036
1I
25.2
060361 j
I
66.38119.13
86.362016.3
0421I
V
nS 6
2
0122
V
1
1’
Page 28
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt28
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Safety ConsiderationsExample: 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
Page 29
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt29
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Safety Consid cont.1Example: 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!
Page 30
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt30
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Exmple Exmple Power Transmission 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 Voltage-Level to 400-765 kVac
Why? → Line SIZE (and others)
Page 31
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt31
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Exmpl – Power Xmission cont.1Exmpl – 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
Page 32
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt32
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Exmpl – Power Xmission cont.2Exmpl – Power Xmission cont.2 By Solid-State Physics
(c.f. ENGR-45)
• Where for the Wire– ρ Resistivity (Ω-m)
– l Length (m)
– A X-Section Area (m2)
In This Case
Then the Power Loss
By Power Rln for Resistive Ckts
Solve for d
Al Rwire
4
1609351002dA
mmil
MWMW Pwire 25.11225%5
4
25.11
22
2
d
lAlPVR
MWPRV P
wirewire
wirewireloss
Vd
V
lPd
P
V
d
l
11920 :Values Subbing
144 2
2
Page 33
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt33
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Exmpl – Power Xmission cont.3Exmpl – Power Xmission cont.3 Finally Solve for
Transmission Cable Diameter
md
md
00384.0500000
11920
128.015000
11920
500
15
d15 = 5.03”
• Pretty BIG & HEAVY
d500 = 0.15”
• MUCH Better
Page 34
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt34
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Summary: Ideal TransformerSummary: Ideal Transformer Consider Now two Coils Wrapped Around a Closed
Magnetic (usually iron) Core.
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
A Area
Page 35
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt35
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ideal X-Former Circuit SymbolIdeal X-Former Circuit Symbol The Ideal Xformer Eqns
• Faraday’s Law
The Circuit Symbol
As Two Coupled Inductors, Xformers Use the DOT Convention to Track Polarities
2
1
2
1
N
N
v
v
The Main practical Application for This Device:• TRANSFORM one AC
Voltage-Level to Another
Iron Core
02211 iNiN
• Ampere’s Law
Page 36
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt36
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Transformer ApplicationTransformer 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
Then the Circuit Symbol Usage as Applied to a Real Circuit
Pictorial Representation
Load
Page 37
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt37
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Sign (Dot) ConventionsSign (Dot) Conventions Have TWO Choices for
Polarity Definitions• Symmetrical
The Form of the Governing Equations
• INput/OUTput (I/O)
2
1
2
1
2211 0
N
N
v
v
iNiN
2
1
2
1
2211
N
N
v
v
iNiN
Page 38
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt38
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Phasor AnalysisPhasor 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
Illustration: InPut IMPEDANCE
Notice That This is an I/O Model; So the Eqns Yield
LN
NZZ
I
V2
2
11
1
1
(symm) 0
(I/O)
1
2211
2211
2
1
2
1
II
II
V
V
NN
NN
nN
N
Page 39
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt39
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard WorkWhiteBoard Work
Let’s Work This Nice Problem
Page 40
[email protected] • ENGR-43_Lec-10-2_Transformers.ppt40
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
208Vac, 80 kVA Input
208:115 Vac StepDown Xformer
208:24 Vac StepDown Xformer