Download 10.3 The ideal transformer

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10. Magnetically coupled networks
EMLAB
Transformer, Inductor
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1. Transformer
① Used for changing AC voltage levels.
② Transmission line : high voltage levels are used to decrease power loss due to
resistance of copper wires. The smaller magnitude of current, the less power loss,
when transmitting the same power.
③ Used for Impedance matching. Transformers are used to change magnitudes of
impedances to achieve maximum power transfer condition.
2. Inductors or transformers are difficult to integrate in an IC. (occupies large areas)
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Two important laws on magnetic field
Current
B-field
Current generates magnetic field
(Biot-Savart Law)
Current
Time-varying magnetic field
generates induced electric field that
opposes the variation. (Faraday’s
law)
Vinduced
B-field
Top view
Electric field
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Magnetic flux
Current

N
B
i
llength

B
llength
Magnetic flux :
 
NS
   B  da  BS  
i
S
llength
  Ni
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Self inductance
Current
Vinduced
Vinduced
d
 N
dt

B, 
Vinduced
d
N 2 S di
 N
 
dt
llength dt
Vinduced   L
di
dt
N 2S
L  
llength
L  N2
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Inductor circuit
(S : cross-section area of a coil, μ : permeability)

Magnetic field
   Ni  S
Total magnetic flux
linked by N-turn coil
Ampere’s Law
(linear model)
N
d
dt
Faraday’s
Induction Law
  N
d
S
di
di
  N2
L
dt
l
dt
dt
Assumes constant L
and linear models!
  L
Ideal Inductor
di
dt
The current flowing through a circuit induces magnetic field (Ampere’s law). A
sudden change of a magnetic field induces electric field that opposes the change
of a magnetic field (Faraday’s law), which appear as voltage drops across an
inductor terminals.
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Mutual Inductance
(1) When the secondary circuit is open

2  N 2
d
d  L  N
di
di
 N 2  1 i1   2 L1 1  L21 1
dt
dt  N1  N1
dt
dt
The current flowing through the primary circuit generates magnetic flux,
which influences the secondary circuit. Due to the magnetic flux, a
repulsive voltage is induced on the secondary circuit.
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Nomenclatures
Primary circuit
Secondary circuit
Primary coil
Secondary coil
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Secondary voltage and current with different coil winding directions
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Two-coil system
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(both currents contribute to flux)
(2) Current flowing in secondary circuit
d 1
di1
di2
1 
 L1
 L12
dt
dt
dt
d 2
di2
di1
2 
 L2
 L21
dt
dt
dt
Self-inductance
Self-inductance
Mutual-inductance
L12  L21  M
Mutual-inductance
(From reciprocity)
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The ‘DOT’ Convention
1 
d 1
di
di
 L1 1  M 2
dt
dt
dt
2 
d 2
di
di
 L2 2  M 1
dt
dt
dt
Dots mark reference polarity for
voltages induced by each flux
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Example 10.2
Mesh 1
Voltage terms
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Mesh 2
Voltage Terms
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Example 10.4
FIND THE VOLTAGE V0
 I2
I1
1. Coupled inductors. Define
their
voltages and currents
 
V1 V2
 
KVL : 2430  2 I1  V1
VS
KVL : - V2  j 2 I 2  2 I 2  0
Mutual inductance circuit
V1  j 4 I1  j 2( I 2 )
V0  2I 2
V2  j 2 I1  j 6( I 2 )
 / j2
0   j 2 I1  (2  j 2  j 6) I 2  / 2  j 4
VS  (2  j 4) I1  j 2 I 2


j 2VS  4  (2  j 4)2 I 2
j 2VS
j
2VS
I2 


 8  j16  j 16  8 j
V
2430
 5.373.42
V0  2 I 2  S 
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
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Example E10.3
WRITE THE KVL EQUATIONS
 I2
I1
 Va 
 Vb 
R1  R2  jL1 I1  R2  jM I 2  V1
 R2  jM I1  R2  R3  jL2 I 2  V1
1. Define variables for coupled inductors
2. Loop equations in terms of inductor
voltages
Va  R2 ( I1  I 2 )  V1  R1 I1  0
Vb  R3 I 2  V1  R2 ( I 2  I1 )  0
3. Equations for coupled inductors
Va  jL1 I1  jM (  I 2 )
Vb  jMI 1  jL2 (  I 2 )
4. Replace into loop equations and
rearrange
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Example 10.6
Z S  3  j1()
I1
DETERMINE IMPEDANCE SEEN BY THE SOURCE
Z L  1  j1()


V1
V
 2

Zi 
 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MI1  jL2 ( I 2 )
VS
?
I1
 Z S  jL1 I1  ( jM ) I 2  VS
 /( 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
Zi  4.3035.54()
4. Replace and do the algebra
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10.2 Energy analysis
di (t ) 
 di (t )
p1 (t )  1 (t )i1 (t )   L1 1  M 2  i1 (t )
dt
dt 

t1
t1
0
0
di (t ) 
 di (t )
p2 (t )   2 (t )i2 (t )   L2 2  M 1  i2 (t )
dt
dt 

W    p1 (t )  p2 (t )dt   1 (t )i1 (t )   2 (t )i2 (t )dt
 di (t )

di (t ) 
di (t ) 
 di (t )
   L1 1  M 2  i1 (t )   L2 2  M 1 i2 (t )dt
dt
dt 
dt
dt 


0 
t1

t1

d 1

 
L1i12 (t )  L2i22 (t )  2M i1 (t )i2 (t ) dt
dt 2

0
1
 L1 I12 (t1 )  L2 I 22 (t1 )  2 M I1 (t1 ) I 2 (t1 )  0
2


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Coupling coefficient
L I
2
1 1

(t1 )  L2 I 22 (t1 )  2 M I1 (t1 ) I 2 (t1 )  0
2

M  
M2  2
 I 2  0
L1  I1 
I 2    L2 
L1  
L1 

 M  L1 L2
M  k L1 L2
k 
M
L1 L2
; Coefficient of coupling
(0  k  1)
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Compute the energy stored in the mutually coupled inductors
Example 10.7
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 )  L2 i22 (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

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
Circuit in frequency domain
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10.3 The ideal transformer
1  N1
d 
v1 (t )  N1 (t )  v
N
dt
 1  1 First ideal transformer

d
v2 (t )  N 2
(t ) v2 N 2 equation
dt 
v1 (t )i1 (t )  v2 (t )i2 (t )  0 Ideal transformer is lossless
2  N 2
Insures that ‘no magnetic flux
goes astray’
Since the equations
are algebraic, they
are unchanged for
Phasors. Just be
careful with signs
i1
N
  2 Second ideal transformer
i2
N1 equations
Circuit Representations
v1 N1

;
v2 N 2
i1 N 2

i 2 N1
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Reflecting Impedances
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For future reference
*
 N  N 
S1  V1I1*  V2 1  I 2 2   V2 I 2*  S2
 N 2  N1 
N
n  2  turns ratio
N1
V1 N1

(both  signs at dots)
V2 N 2
Phasor equations for ideal transformer
I1
N
  2 (Current I 2 leaving transform er)
( I 2 )
N1
V2  Z L I 2 (Ohm' s Law)
N
N
V1 2  Z L I1 1
N1
N2
V2
n
I1  nI 2
V1 
ZL
n2
S1  S 2
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
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Non-ideal transformer
M  k L1 L2
V1  jL1 I1  jMI 2
Z L I 2  V2  0, (V2   jL2 I 2  jMI1 )
 ( Z L  jL2 ) I 2  jMI1  I 2 
j M
I1
Z L  jL2
(M ) 2
 V1  jL1 I1 
I1
Z L  jL2
V1
(M ) 2
(M ) 2   2 L1 L2  jL1Z L (1  k 2 ) 2 L1 L2  jL1Z L
 Z in   jL1 


I1
Z L  jL2
Z L  jL2
Z L  jL2
(1) k  1
V  jL1 
 Z L
 Z in  1  
I1  Z L  jL2 
(2) Z L  jL2
2
L 
N 
 Z in   1  Z L   1  Z L
 L2 
 N2 
To build ideal transformers,
following two conditions are
needed.
(1) k=1;
(2) ZL<<jωL;
EMLAB
Example 10.8
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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
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Thevenin’s equivalents 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’
EMLAB
Thevenin’s equivalents from primary
ZTH 
Z2
n2
In open circuit I1  0 and I 2  0
25
Equivalent circuit reflecting
into primary
VOC 
VS 2
n
Thevenin impedance will be the the
secondary mpedance reflected into
the primary circuit
Equivalent circuit reflecting
into secondary
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Example 10.9
Draw the two equivalent circuits
n2
Equivalent circuit reflecting
into secondary
Equivalent circuit reflecting
into primary
EMLAB
Find I1
Example E10.8
27
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
EMLAB
Find Vo
Example E10.9
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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
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Safety considerations
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
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