Transcript No Slide Title

MAGNETICALLY COUPLED NETWORKS
LEARNING GOALS
Mutual Inductance
Behavior of inductors sharing a common magnetic field
Energy Analysis
Used to establish relationship between mutual reluctance and
self-inductance
The ideal transformer
Device modeling components used to change voltage and/or
current levels
Safety Considerations
Important issues for the safe operation of circuits with transformers
BASIC CONCEPTS – A REVIEW
Magnetic field
Total magnetic
flux linked by Nturn coil
Ampere’s Law
(linear model)
Faraday’s
Induction Law
Assumes constant L
and linear models!
Ideal Inductor
MUTUAL INDUCTANCE
Overview of Induction Laws Induced links
on second
Magnetic coil (2 )
flux
Total flux linkage
  N (webers)
If linkage is created by a current flowing
through the coils…
  Li (Ampere’s Law)
The voltage created at the terminals of
the components is
vL
di
dt
(Faraday’s Induction Law)
What happens if the flux created by the
current links to another coil?
One has the effect of mutual inductance
TWO-COIL SYSTEM
Self-induced
(both currents contribute to flux)
Mutual-induced
Linear model simplifying
notation
THE ‘DOT’ CONVENTION
COUPLED COILS WITH DIFFERENT WINDING CONFIGURATION
Dots mark reference
polarity for voltages
induced by each flux
A GENERALIZATION
Assume n circuits interacting
n
i   ij
j 1
i  Total flux linking circuit i
ij  Flux linking circuit i caused by a current
in circuit j
For linear inductor models
ij  Lij i j
Lii  " self inductance " of circuit i
Lij  L ji  Mutual inductance between circuits
i and j
Special case n=2
1  L11i1  L12i2 Linear Model :
2  L21i1  L22i2 L12  L21
THE DOT CONVENTION REVIEW
LEARNING EXAMPLE
Currents and voltages follow
passive sign convention

( v 2 ( t ))

 i2 ( t )
 i1 ( t )
Flux 2 induced
voltage has + at dot
di1
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )  M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1
For other cases change polarities or
current directions to convert to this
basic case
 di 
 di 
v1 ( t )  L1   1   M   2 
 dt 
 dt 
 di 
 di 
 v 2 (t )  M   1   L2   2 
 dt 
 dt 
di1
di
M 2
dt
dt
di
di
v2  M 1  L2 2
dt
dt
v1   L1
LEARNING EXAMPLE
Mesh 1
Voltage terms
LEARNING EXAMPLE - CONTINUED
Mesh 2
Voltage Terms
More on the dot convention
di1
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )  M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1

1v 
di1
di
M 2
dt
dt
di
di
v2   M 1  L2 2
dt
dt
 v1   L1

 i1

 v2

 i2
di1
di
M 2
dt
dt
di
di
 v2  M 1  L2 2
dt
dt
v1  L1
Equivalent to a
negative mutual
inductance
LEARNING EXTENSION
Write the equations for v1 (t ), v2 (t )


Convert to
basic case
 v1 (t )   L1
v2 ( t )   M
di
di
(t )  M 2 (t )
dt
dt
di1
di
(t )  L2 2 (t )
dt
dt
di
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )   M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1
PHASORS AND MUTUAL INDUCTANCE
di1
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )  M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1
Assuming complex
exponential sources
V1  jL1I1  jMI 2
V2  jMI 1  jL2 I 2
Phasor model for mutually
coupled linear inductors
LEARNING EXAMPLE
CASE I
I
The coupled inductors can be connected in four different ways.
Find the model for each case
Currents into dots
I
 V1 
CASE 2
 V2 
Currents into dots
 V1 
V2  jMI  jL2 I
V  j ( L1  L2  2 M )  jLeq I
V  V1  V2
V1  jL1  jMI
I
I
V  V1  V2
V1  jL1I  jMI
 V2 
V2  jMI  jL2 I
V  j ( L1  2 M  L2 ) I
Leq
Leq  0 imposes a physical constraint
on the value of M
CASE 3
I1
Currents into dots
I2
V 
V 
I  I1  I 2
 I 2  I  I1
V  jL1I1  jMI 2
V  jMI 1  jL2 I 2
V  jL1I1  jM ( I  I1 )
V  jMI 1  jL2 ( I  I1 )
V  j ( L1  M ) I1  jMI
 /( L2  M )
V   j ( L2  M ) I1  jL2 I  /( L1  M )
L1L2  M 2
V  j
I
L1  L2  2 M
CASE 4
Currents into dots
I1
 I2
I1
V 
( L1  L2  2 M )V  j  M ( L2  M )  L2 ( L1  M ) I
I2
I  I1  I 2
 ( V ) 
V  jL1I1  jMI 2
 V  jMI 1  jL2 I 2
L1L2  M 2
V
I
L1  L2  2 M
LEARNING EXAMPLE
FIND THE VOLTAGEV0
 I2
1. Coupled inductors. Define their
voltages and currents
I1
 
V1 V2
 
KVL : 2430  2 I1  V1
VS
KVL : - V2  j 2 I 2  2 I 2  0
MUTUAL INDUCTANCECIRCUIT
V1  j 4 I1  j 2( I 2 )
V0  2I 2
V2  j 2 I1  j 6( I 2 )
VS  (2  j 4) I1  j 2 I 2
 / j2
0   j 2 I1  (2  j 2  j 6) I 2  / 2  j 4


j 2VS  4  (2  j 4)2 I 2
j 2VS
2VS
j
I2 


 8  j16  j 16  8 j
VS
2430
V0  2 I 2 
 5.373.42

4  2 j 4.4726.57
2. Write loop equations
in terms of coupled
inductor voltages
3. Write equations for
coupled inductors
4. Replace into loop equations
and do the algebra
LEARNING EXAMPLE
I1  I 2
Write the mesh equations
3. Write equations for coupled inductors
I 2  I3

V1

V2


V1  jL1 ( I1  I 2 )  jM ( I 2  I 3 )
V2  jM ( I1  I 2 )  jL2 ( I 2  I 3 )
4. Replace into loop equations and
rearrange terms
1. Define variables for coupled inductors

1 
 I1
V   R1  jL1 
j

C

1

1 
 I 2  jMI 3
  jL1  jM 
j

C

1
2. Write loop equations in terms of coupled
inductor voltages

V  R1 I1  V1 
I1  I 2
jC1
 V1  R2 I 2  V2  R3 ( I 2  I 3 ) 
 V2 
1 
 I1
0   jL1  jM 
j

C

1
I 2  I1
0
jC1
I3
 R4 I 3  R3 ( I 3  I 2 )  0
jC 2

1 
 I 2
  jL2  jM  R2  jM  jL2  R3 
j

C

1
  jM  jL2  R3 I 3
0   jMI 1   jL2  jM  R3 I 2


1
  jL2 
 R4  R3  I 3
j C 2


LEARNING EXTENSION
FIND I1 , I 2 ,V0
VS
V0   j 4I 2
 
V1 V2
 
1. Define variables for coupled inductors
Voltages in Volts
2. Loop equations
Impedances in Ohms
VS  V1  4 I1  0 Currents in ____
V2  (2  j 4) I 2  0
3. Coupled inductors equations
V1  j 4 I1  jI 2
V2  jI1  j8 I 2
4. Replace and rearrange
(4  j 4) I1  jI 2  VS  /  j
 /( 4  j 4)
jI1  (2  j 4) I 2  0
(1  8(1  j )(1  2 j )) I 2  jV S
I2 
jV S
 j 240
240



 7  24 j  j 24  7 j 2516.26
I 2  0.96  16.26( A)
jI1  (2  j 4) I 2  0 / j  I1  j (2  j 4) I 2
I1  190  4.4763.43  0.96  16.26
I1  4.29137.17( A)
V0   j 4 I 2  1  90  4  0.96  16.26
V0  3.84  106.26(V )
LEARNING EXTENSION
WRITE THE KVL EQUATIONS
 I2
I1
 Va 
 Vb 
1. Define variables for coupled inductors
2. Loop equations in terms of inductor
voltages
Va  R2 ( I1  I 2 )  V1  R1I1  0
 Vb  R3 I 2  V1  R2 ( I 2  I1 )  0
3. Equations for coupled inductors
Va  jL1I1  jM ( I 2 )
Vb  jMI 1  jL2 ( I 2 )
4. Replace into loop equations and
rearrange
 R1  R2  jL1 I1   R2  jM I 2  V1
  R2  jM I1   R2  R3  jL2 I 2  V1
LEARNING EXAMPLE
Z S  3  j1()
I1
DETERMINE IMPEDANCE SEEN BY THE SOURCE Z i 
Z L  1  j1()

V1 
V
 2

 I2
jL1  j 2()
jL2  j 2()
jM  j1()
1. Variables for coupled inductors
2. Loop equations in terms of coupled
inductors voltages
Z S I1  V1  VS
 V2  Z L I 2  0
3. Equations for coupled inductors
V1  jL1I1  jM ( I 2 )
V2  jMI 1  jL2 ( I 2 )
4. Replace and do the algebra
 Z S  jL1 I1  ( jM ) I 2  VS
VS
I1
 /( Z L  jL2 )
 ( jM ) I1  ( Z L  jL2 ) I 2  0  / jM
( Z
S

 jL1 )( Z L  jL2 )  ( jM ) 2 I1
 ( Z L  jL2 )VS
VS
( jM )2
Zi 
 ( Z S  jL1 ) 
I1
Z L  jL2
( j1) 2
1
1 j
Z i  3  j3 
 3  j3 

1  j1
1 j 1 j
Z i  3  j3 
1 j
 3.5  j 2.5()
2
Z i  4.3035.54()
WARNING: This is NOT a phasor
LEARNING EXTENSION
DETERMINE IMPEDANCE SEEN BY THE SOURCE
Z L  2  ( j 2 || j1)
Z S  2  j1()
ZL  2 

 V2
VS
I1
V1
2
 2  2 j ( )
j
I2

1. Variables for coupled inductors
2. Loop equations V1  Z S I1  VS
V2  Z L I 2  0
4. Replace and do the algebra
3. Equations for coupled inductors
V1  jL1I1  jMI 2
V2  jMI 1  jL2 I 2
VS
( jM )2
Zi 
 ( Z S  jL1 ) 
I1
Z L  jL2
One can choose directions
for currents.
If I2 is reversed one gets
1
(1  2 j )
( j1) 2
 (2  j ) 

Z i  (2  j1)  j 2 
the same equations than
2(1  2 j ) (1  2 j )
(2  2 j )  2 j
in previous example.
Solution for I1 must be the
1 2 j
same and expression for
Zi  2  j 
 2.1  0.8 j ()
2
2
(
1

2
)
impedance must be the same
Z i  2.2520.85()
ENERGY ANALYSIS
We determine the total energy stored in a coupled network
This development is different from the one in the book. But the final
results is obviously the same
t
EQUATIONSFOR COUPLED INDUCTORS
d 1 2
1

pT (t )   L1i1 (t )  Mi1 (t )i2 (t )  L2 i22 (t ) 
di
di
dt  2
2
 / i1 (t )

v1 (t )  L1 1 (t )  M 2 (t )

dt
dt
1 2
1
w
(
t
)

L
i
(
t
)

Mi
(
t
)
i
(
t
)

L2i22 (t )
di1
di2
11
1
2
2
2
v2 (t )   M
(t )  L2
( t )  / i2 ( t )
2
2
dt
dt
1M 2
1M 2

i1 (t ) 
i1 (t )
TOTALPOWER SUPPLIED TO NETWORK
2 L2
2 L2
2
2
pT (t )  v1 (t )i1 (t )  v2 (t )i2 (t )




1
M 2
1
M



w
(
t
)

L

i
(
t
)

L
i
(
t
)

i1 (t ) 
1
1
2 2



di1
di2
2
L2 
2
L2
pT (t )  L1i1 (t ) (t )  Mi1 (t )
(t )

dt
dt
M2
M  L1L2
w (t )  L1 
0
di
di
 M 1 (t )i2 (t )  L2 i2 (t ) 2 (t )
L2
dt
dt
M
2
Coefficient of
1 di12
1
di
k

di
i
2
1
2
(t ) M
(t )
(t )
L1L2 coupling
2 dt
2 dt
dt

LEARNING EXAMPLE
Compute the energy stored in the mutually coupled inductors
t  5ms
Assume steady state operation
We can use frequency domain techniques

Merge the writing of the loop and coupled
inductor equations in one step
k 1
L1  2.653mH , L2  10.61mH
1
1
w (t )  L1i12 (t )  Mi1 (t )i2 (t )  L2i22 (t )
2
2
MUST COMPUTE M , i1 (t ), i2 (t )
L1, L2 , k  M  k L1L2
M  5.31mH
L1  377  2.653 103  1
L2  4, M  2
I1

 I2

Circuit in frequency domain
2 I1  ( j1I1  j 2 I 2 )  240
4 I 2  ( j 2 I1  j 4 I 2 )  0
SOLVE TO GET
I1  9.41  11.31( A), I 2  3.3333.69( A)
 i1 (t )  9.41cos(377t  11.31)( A)
i2 (t )  3.33 cos(377t  33.69)( A)
WARNING: The term 377t is in radians!
t  0.005s  377t  1.885(rad )  108
i1 (0.005)  1.10( A), i2 (0.005)  2.61( A)
w (0.005)  0.5  2.653  103 (1.10) 2
 5.31 103 (1.10)  (2.61)
 0.5  101.61 103  (2.61) 2 ( J )
w(0.005)  22.5mJ
LEARNING EXTENSION
DETERMINE ENERGY STORED AT t  10ms
f  60 Hz    378.9( s 1 )
L1  2  L1  0.00528( H )  L2
M  0.00264( H )
f  60 Hz


1
1
w (t )  L1i12 (t )  Mi1 (t )i2 (t )  L2i22 (t )
2
2
2 I1  ( j 2 I1  j1I 2 )  1230
( j1I1  j 2 I 2 )  (2  j 2) I 2  0
i1 (t )  3.75 cos(378.9t  8.66)( A)
i2 (t )  1.875 cos(378.9t  98.66)( A)
378.9(rad / sec)  0.010(sec)  3.789(rad )  217.1
(2  j 2) I1  jI 2  1230
 I 2  0.5 jI1
jI1  2 I 2  0
(2  j 2  0.5) I1  1230
1230
1230

 3.75  8.66( A)
2.5  j 2 3.2038.66
I 2  0.5 jI1  0.5  90  3.75  8.66
I1 
 1.875  98.66
Go back to time domain
i1 (0.010)  3.3( A)
i2 (0.010)  0.91( A)
w (0.010)  0.5 * 0.00528 * (3.3) 2
 0.00264 * (3.3)( 0.91)
 0.5 * 0.00528 * (0.91) 2 ( J )
w(0.010)  0.00264 * (3.32  (3.3)(0.91)  0.912 )
w(0.010)  0.030J  30mJ
THE IDEAL TRANSFORMER
1  N1
2  N 2
Insures that ‘no magnetic flux
d 
goes astray’
v1 (t )  N1 (t )  v
N
dt
 1  1 First ideal transformer

d
v2 (t )  N 2
(t ) v2 N 2 equation
Since the equations
dt 
are algebraic, they
v1 (t )i1 (t )  v2 (t )i2 (t )  0 Ideal transformer is lossless
i1
N
  2 Second ideal transformer
i2
N1 equations
Circuit Representations
v1 N1

;
v2 N 2
are unchanged for
Phasors. Just be
careful with signs
i1 N 2

i2 N1
REFLECTING IMPEDANCES
For future reference
*
 N  N 
S1  V1I1*  V2 1  I 2 2   V2 I 2*  S2
 N 2  N1 
n
V1 N1

(both  signs at dots)
V2 N 2
Phasor equations for ideal transformer
I1 N 2

(Current I2 leaving transformer)
I 2 N1
V2  Z L I 2 (Ohm's Law)
N
N
V1 2  Z L I1 1
N1
N2
N2
 turns ratio
N1
V2
n
I1  nI 2
V1 
ZL
n2
S1  S2
Z1 
2
N 
V1   1  Z L I1
 N2 
2
N 
V1
 Z1   1  Z L
I1
 N2 
Z1  impedance, Z L , reflected
into the primary side
LEARNING EXAMPLE
Determine all indicated voltages and currents
n  1/ 4  0.25
I1 
1200
1200

 2.33  13.5
50  j12 51.4213.5
V1  Z1I1 
Z1
1200
Z1  Z 2
Z1I1  (32  j16)  2.33  13.5
Strategy: reflect impedance into the
primary side and make transformer
“transparent to user.”
ZL
Z1 
SAME
COMPLEXITY
Z1
32  j16
1200 
 120
Z1  Z 2
51.4213.5
V1  35.7826.57  2.33  13.5  83.3613.07
n2
ZL
CAREFUL WITH POLARITIES AND
CURRENT DIRECTIONS!
I2 
I1
 4I1 (current into dot)
n
V2  nV1  0.25V1 ( is opposite to dot)
Z1  32  j16
LEARNING EXTENSION Find the current I1
n2
I2 
I1
n
Strategy: reflect impedance into the
primary side and make transformer
“transparent to user.”
ZL
Z1 
Z1 
Z1
n2
4  j2
 1  j 0.5
4
Z i  3  j 2.5 Voltage in Volts
120
120
I1 

 3.0739.81( A)
3  j 2.5 3.91  39.81
LEARNING EXTENSION
Impedance in Ohms
...Current in Amps
Find Vo
Strategy: Find current in secondary and then use Ohm’s Law
I2 
I1
I
 V0  2  1  3.0739.81(V )
2
2
USING THEVENIN’S THEOREM TO SIMPLIFY CIRCUITS WITH IDEAL TRANSFORMERS
Replace this circuit with its Thevenin
equivalent
I2  0 
  I1  0  V1  VS1
I1  nI 2 
V1  VS1 
  VOC  nVS1
V2  nV1 
To determine the Thevenin impedance...
Reflect impedance into
secondary
ZTH
ZTH  n2 Z1
Equivalent circuit with transformer
“made transparent.”
One can also determine the Thevenin
equivalent at 1 - 1’
USING THEVENIN’S THEOREM: REFLECTING INTO THE PRIMARY
ZTH 
Z2
n2
Find the Thevenin equivalent of
this part
In open circuit I1  0 and I 2  0
VOC 
Equivalent circuit reflecting
into primary
VS 2
n
Thevenin impedance will be the the
secondary mpedance reflected into
the primary circuit
Equivalent circuit reflecting
into secondary
LEARNING EXAMPLE
Draw the two equivalent circuits
n2
Equivalent circuit reflecting
into secondary
Equivalent circuit reflecting
into primary
LEARNING EXAMPLE
Find Vo
Thevenin equivalent of this part
To compute Vo is better to reflect into secondary
But before doing that it is better to simplify the primary using Thevenin’s Theorem

Vd

ZTH  2 
ZTH 
 j16 8  j8  j16

4  j4
4  j4
2  j6 1  j 8  j 4


1 j 1 j
2
VOC  Vd  4  90
Vd 
 j4
24  90
240 
4  j4
1 j
VOC  14.47  33.69(V )
ZTH  2  (4 ||  j 4)
ZTH  4  j 2()
This equivalent circuit is now transferred to
the secondary
LEARNING EXAMPLE (continued…)
n2
Thevenin equivalent of primary side
Transfer to
secondary
Circuit with primary transferred to secondary
Vo 
2
2  28.84  33.69
28.84  33.69 
20  j 5
20.62  14.04
LEARNING EXTENSION
Find I1
n2
2  j 2 ( )
I1
Equivalent circuit reflecting
into primary
0.5

360
120
2
Notice the position of the
dot marks
I1 
360  60
2  j1.5
I1 
360
2.5  36.86
LEARNING EXTENSION
Find Vo
n2
8  j8
240
Transfer to
secondary
 VO 
2
40
VO 
2
200
(8  j8)  2
VO 
400
12.81  38.66
LEARNING EXAMPLE
Find I1, I 2 ,V1,V2
2V1  V2  2 I1  100
 I1  50
 V1  (1  j )V2  2 I 2  0
n2
V2  2V1
I1  2 I 2
I 2  2.50
 V1  (1  j )( 2V1 )  50
V1 
Nothing can be transferred. Use
transformer equations and circuit
analysis tools
Phasor equations for ideal transformer
V2
n
I1  nI 2
V  100 V1  V2
@ Node 1 : 1

 I1  0
2
2
V V V
@Node 2 : 2 1  2  I 2  0
2
j2
V1 
4 equations in 4 unknowns!
50
50

 563.43
1  j 2 2.24  63.43
V2  2 563.43
SAFETY CONSIDERATIONS: AN EXAMPLE
Houses fed from different distribution
transformers
Braker X-Y opens, house B
is powered down
When technician resets the
braker he finds 7200V between
points X-Z
when he did not expect to find any
Good neighbor runs an extension
and powers house B
LEARNING BY APPLICATION
Why high voltage transmission lines?
CASE STUDY: Transmit 24MW over 100miles with 95% efficiency A. AT 240V
B. AT 240kV
l
Given : Conductor resistance , R 
A
  resistivity of material (e.g., copper  8  108 /m)
l  length of conductor  160 .9 Km  1.609  105 m
A  cross section   r 2
Required : Maximum losses, Ploss  24 MW  0.05  1.2 MW
Known : Line losses : Ploss  RI 2
24  106W
A. At 240V one needs a current I l 
 105 A
240V
8 108 1.609 105
10
1.2 10 W  2 

10
 r  8.624m
 r2
6
24  106W
2
B. At 240kV one needs a current I l 

10
A
3
240  10 V
8 108 1.609 105
4
1.2 10 W  2 

10
 r  0.8624cm
 r2
6
LEARNING EXAMPLE
Rating a distribution transformer
Households per transformer  10
Maximum current per household  200 A
Determining ratio
V2
n
V1
n
240
1

13800 57.5
Determining power rating
Max secondary current  2000 A
n
I1
 34.78 A
 I1  2000
57.6
I2
 P  13,800  34.78  480kVA
Also: P  240(V )  2000( A)
LEARNING EXAMPLE
Battery charger using mutual inductance
0.5 ARMS
100mARMS
Smaller inductor
Assume currents in phase
BATTERY CHARGER WITH HIGH FREQUENCY SWITCH
APPLICATION EXAMPLE
INDUCED ELECTRIC NOISE
LAC  LDC  10nH
“NOISE”
 0.1
CASE 1: AC MOTOR (f=60Hz)
VNOISE  1.88V
CASE 2: FM RADIO TRANSMITTER
VNOISE  3.14V
!!!
LEARNING EXAMPLE
LINEAR VARIABLE DIFFERENTIAL TRANSFORMER (LVDT)
NO LOAD!
V  10V
IN
RMS
f  2kHz
| k  k | 0.8
12
13
| V || V |
O
IN
L  1 
 

L  0.8 
S
P
2
LVDT - CONTINUED
L
 1.25  1.5625
L
S
2
P
Complete Analysis
Assuming zero- phase input
Design equations for
inductances
FOR EXAMPLE
I  10mA 

V  10V

f  2000 Hz 
P
IN
RMS
RMS
LEARNING BY DESIGN
Use a 120V - 12V transformer to build a 108V autotransformer
Conventional transformer
Use the subtractive connection
on the 120V - 12V transformer
Auto transformer connections
Vxz  Vxy  V yz
Vxz  Vxy  Vxz
Circuit representations
DESIGN EXAMPLE
DESIGN OF AN “ADAPTOR” OR “WALL TRANSFORMER”
Design constrains and requirements
P  2W
DC
| V | 12
2
  0.6
Transformer turn ratio
 25 : 2
Notice safety margin
Specify 100mA rating for extra
margin!
Transformers