MAGNETICALLY COUPLED NETWORKS LEARNING GOALS Mutual Inductance Behavior of inductors sharing a...
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Transcript of MAGNETICALLY COUPLED NETWORKS LEARNING GOALS Mutual Inductance Behavior of inductors sharing a...
MAGNETICALLY COUPLED NETWORKS
LEARNING GOALS
Mutual Inductance Behavior of inductors sharing a common magnetic field
The ideal transformer Device modeling components used to change voltage and/or current levels
BASIC CONCEPTS – A REVIEW
Magnetic field Total magnetic flux linked by N-turn coil
Ampere’s Law(linear model)
Faraday’sInduction Law
Assumes constant L and linear models!
Ideal Inductor
MUTUAL INDUCTANCE
Overview of Induction Laws
Magneticflux
webers)
linkageflux Total
( N
Li
If linkage is created by a current flowing through the coils…
(Ampere’s Law)
The voltage created at the terminals of the components is
dt
diLv (Faraday’s Induction Law)
Induced linkson secondcoil
2( )
One has the effect of mutual inductance
TWO-COIL SYSTEM (both currents contribute to flux)
Self-induced Mutual-induced
Linear model simplifyingnotation
THE ‘DOT’ CONVENTION
COUPLED COILS WITH DIFFERENT WINDING CONFIGURATION
Dots mark reference polarity for voltages induced by each flux
THE DOT CONVENTION REVIEW
Currents and voltages followpassive sign convention
Flux 2 inducedvoltage has + at dot
)()()(
)()()(
22
12
2111
tdt
diLt
dt
diMtv
tdt
diMt
dt
diLtv
LEARNING EXAMPLE
)(1 ti )(2 ti
))(( 2 tv
For other cases change polarities orcurrent directions to convert to thisbasic case
dt
diM
dt
diLtv 21
11 )(
dt
diL
dt
diMtv 2
21
2 )(
dt
diL
dt
diMv
dt
diM
dt
diLv
22
12
2111
LEARNING EXAMPLE
Mesh 1
LEARNING EXAMPLE - CONTINUED
Mesh 2 Voltage Terms
2212
2111
ILjMIjV
MIjILjV
Phasor model for mutually
coupled linear inductorsAssuming complex exponential sources
PHASORS AND MUTUAL INDUCTANCE
)()()(
)()()(
22
12
2111
tdt
diLt
dt
diMtv
tdt
diMt
dt
diLtv
LEARNING EXAMPLEThe coupled inductors can be connected in four different ways.Find the model for each case
CASE I
1V 2VILjMIjV
MIjILjV
22
11
Currents into dots
CASE 2
1V 2V
Currents into dots
I I
I I 21 VVV
21 VVV
1 1
2 2
V j L I j MIV j MI j L I
1 2( 2 ) eqV j L L M I j L I
ILMLjV )2( 21
eqL
M
Leq
of valuethe on
constraint physical a imposes 0
CASE 3Currents into dots
1I 2I
V V
21 III
221
211
ILjMIjV
MIjILjV
12 III
)(
)(
121
111
IILjMIjV
IIMjILjV
ILjIMLjV
MIjIMLjV
212
11
)(
)(
)/(
)/(
1
2
ML
ML
IMLLMLMjVMLL )()()2( 12221 I
MLL
MLLjV
221
221
CASE 4Currents into dots
1I 2I
V )( V 21 III
221
211
ILjMIjV
MIjILjV
2
1 2
1 2 2
L L MV j I
L L M
LEARNING EXAMPLE0V VOLTAGETHE FIND
1V
2V
2I
1123024 VI :KVL
022 222 IIjV- :KVL
)(62
)(24
212
211
IjIjV
IjIjV
CIRCUIT INDUCTANCEMUTUAL
20 2IV
SV
21
21
)622(20
2)42(
IjjIj
IjIjVS
1I
42/
2/
j
j
22)42(42 IjVj S
168
22 j
VjI S
j
j
j
VS816
2
j
VIV S
242 20
57.2647.4
3024 42.337.5
1. Coupled inductors. Define theirvoltages and currents
2. Write loop equationsin terms of coupledinductor voltages
3. Write equations forcoupled inductors
4. Replace into loop equationsand do the algebra
LEARNING EXAMPLEWrite the mesh equations
1V
21 II
2V
32 II
1. Define variables for coupled inductors
2. Write loop equations in terms of coupled inductor voltages
1
21111 Cj
IIVIRV
0)(1
123232221
Cj
IIIIRVIRV
0)( 233342
32 IIRIR
Cj
IV
)()(
)()(
322212
322111
IILjIIMjV
IIMjIILjV
3. Write equations for coupled inductors
4. Replace into loop equations and rearrange terms
321
1
11
11
1
1
MIjICj
MjLj
ICj
LjRV
332
21
3222
11
1
1
10
IRLjMj
ICj
RLjMjRMjLj
ICj
MjLj
3342
2
2321
1
0
IRRCj
Lj
IRMjLjMIj
LEARNING EXAMPLE
)(13 jZS )(11 jZL
)(21 jLj )(22 jLj)(1 jMj
DETERMINE IMPEDANCE SEEN BY THE SOURCE1I
VZ Si
1I 2I
1V
2V
SS VVIZ 11
1. Variables for coupled inductors
2. Loop equations in terms of coupled inductors voltages
022 IZV L
3. Equations for coupled inductors
)( 2111 IMjILjV )( 2212 ILjMIjV
4. Replace and do the algebra
0)()(
)(
221
211
ILjZIMj
VIMjILjZ
L
SS
Mj
LjZL
/
)/( 2
SL
LS
VLjZ
IMjLjZLjZ
)(
)())((
2
12
21
2
2
11
)()(
LjZ
MjLjZ
I
VZ
LS
Si
11
)1(33
2
j
jjZ i
j
j
1
133
j
j
1
1
)(5.25.32
133
j
jjZ i
)(54.3530.4 iZ
THE IDEAL TRANSFORMER
Insures that ‘no magnetic fluxgoes astray’
11 N 22 N
2
1
2
1
22
11
)()(
)()(
N
N
v
v
tdt
dNtv
tdt
dNtv
First ideal transformerequation
0)()()()( 2211 titvtitv Ideal transformer is lossless
1
2
2
1
N
N
i
i Second ideal transformer
equations
1
2
2
1
2
1
2
1 ;N
N
i
i
N
N
v
vCircuit Representations
REFLECTING IMPEDANCES
dots)at signs (both 2
1
2
1
N
N
V
V
r)transforme leaving I(Current 21
2
2
1
N
N
I
I
Law) s(Ohm' 22 IZV L
2
11
1
21 N
NIZ
N
NV L 1
2
2
11 IZ
N
NV L
LZN
NZ
I
V2
2
11
1
1
sideprimary the into
reflected , impedance, LZZ 1
For future reference
2*22
*
1
22
2
12
*111 SIV
N
NI
N
NVIVS
ratio turns 1
2
N
Nn
21
21
21
21
SSn
ZZ
nIIn
VV
L
Phasor equations for ideal transformer
LEARNING EXAMPLEDetermine all indicated voltages and currents
25.04/1 n
Strategy: reflect impedance into theprimary side and make transformer“transparent to user.”
21n
ZZ L
LZ
16321 jZ
5.1333.25.1342.51
0120
1250
01201 jI
1 1 1 (32 16) 2.33 13.5
83.36 13.07
V Z I j
12 14 (current into dot)
II I
n
dot) to opposite is ( 112 25.0 VnVV
CAREFUL WITH POLARITIES ANDCURRENT DIRECTIONS!
USING THEVENIN’S THEOREM TO SIMPLIFY CIRCUITS WITH IDEAL TRANSFORMERS
Replace this circuit with its Theveninequivalent
00
121
2
InII
I11 SVV
112
11SOC
S nVVnVV
VV
To determine the Thevenin impedance...
THZReflect impedance intosecondary
12ZnZTH
Equivalent circuit with transformer“made transparent.”
One can also determine the Theveninequivalent at 1 - 1’
USING THEVENIN’S THEOREM: REFLECTING INTO THE PRIMARY
Find the Thevenin equivalent ofthis part
00 21 II and circuit open Inn
VV SOC
2
Thevenin impedance will be the thesecondary mpedance reflected intothe primary circuit
22
n
ZZTH Equivalent circuit reflecting
into primary
Equivalent circuit reflecting into secondary
LEARNING EXAMPLEDraw the two equivalent circuits
2n
Equivalent circuit reflecting into secondary
Equivalent circuit reflectinginto primary
LEARNING EXAMPLE oV Find
secondary intoreflect tobetter is compute To oV
But before doing that it is better to simplify the primary using Thevenin’s Theorem
Thevenin equivalent of this part
dV
904dOC VV
02444
4
j
jVd
)4||4(2 jZTH
14.42 33.69 ( )OCV V
)(24 jZTH44
1688
44
162
j
jj
j
jZTH
2
48
1
1
1
62 j
j
j
j
jZTH
j
1
9024
This equivalent circuit is now transferred tothe secondary
LEARNING EXAMPLE (continued…)
Thevenin equivalent of primary side
2n
69.3384.28520
2
jVo
04.1462.20
69.3384.282
Equivalent circuit reflecting into secondary
Circuit with primary transferred to secondary
LEARNING EXAMPLE 2121 ,,, VVIIFind
Nothing can be transferred. Use transformer equations and circuitanalysis tools
21
21
nIIn
VV
Phasor equations for ideal transformer
022
0101
211 I
VVV :1 Node @
022 2212
I
j
VVV :2 @Node
4 equations in 4 unknowns!
2n21
12
221
121
2
2
02)1(
01022
II
VV
IVjV
IVV
051I
05.22I
05)2)(1( 11 VjV
43.6324.2
05
21
051 jV 43.635
43.63522V