Download EEE 302 Lecture 11 - Universitas Udayana

Magnetically Coupled Networks
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Magnetically Coupled Networks
• A new four-terminal element, the transformer, is
introduced in this chapter
• A transformer is composed of two closely
spaced inductors, that is, two or more
magnetically coupled coils
– primary side is connected to the source
– secondary side is connected to the load
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di1
v2   M
dt
di 2
v1   M
dt
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Dot Convention
• dot convention: dots are placed beside each
coil (inductor) so that if the currents are
entering (or leaving) both dotted terminals,
then the fluxes add
• right hand rule says that curling the fingers (of
the right hand) around the coil in the direction
of the current gives the direction of the
magnetic flux based on the direction of the
thumb
• We need dots on the schematic to know how
the coils are physically oriented out one
another
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Mutually Coupled Coils
The following equations define the coupling between the two
inductors assuming that each respective current enters the dot
side which is also the positive voltage side where L1 and L2
are the self-inductances of the coils (inductors), and M is the
mutual inductance between the two coils
M
i1(t)
+
v1(t)
–
L1
i2(t)
L2
+
v2(t)
–
d i1
d i2
v1 (t )  L1
M
dt
dt
di
di
v2 (t )  M 1  L2 2
dt
dt
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EXAMPLE
Write a set of mesh-current equations that describe the
circuit shown in terms of the currents i1, i2, and i3
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(In the following set of mesh-current equations, voltage drops appear as
positive quantities on the right-hand side of each equation.)
Summing the voltages around the first mesh yields
d
di2
vg  8i1  i2   9 i1  i3   4.5
dt
dt
The second mesh equation is
di2
d
04
 4.5 i1  i3   6i2  i3   8i2  i1 
dt
dt
The third mesh equation is
d
di2
0  9 i3  i1   4.5
 6i3  i2   20i3
dt
dt
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Mutually Coupled Coils (AC)
• The frequency domain model of the coupled
circuit is essentially identical to that of the time
domain
V1  j L1 I1  j M I 2
V2  j M I1  j L2 I 2
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Source Input Impedance
Linear Transformer
The source sees an input impedance, Zi, that is the sum
of the primary impedance, and a reflected impedance,
ZR, due to the secondary (load) side
VS
Zi 
 Z P  Z R  Z P  f Z L 
IP
M
Z
VS
+
–
L1
L2
ZL
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THE LINEAR TRANSFORMER
Transformer
source
ZS
a
R2
jωM
c
I2
I1
jωL1
Vs
source
R1
load
b
jωL2
d
ZL
Load
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Vs  Z s  R1  jL1 I1  jMI 2
0   jMI1  R2  jL2  Z L I 2
Z11  Z s  R1  jL1
Z 22  R2  jL2  Z L
Z 22
I1 
V
2
2 s
Z11Z 22   M
j M
j M
I2 
V 
I1
2
2 s
Z11Z 22   M
Z 22
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Vs
Z11Z 22   2 M 2
 2M 2
 Z int 
 Z11 
I1
Z 22
Z 22
Z ab  Z11 
 M
2
Z 22
2
 Zs
 2M 2
 R1  jL1 
R2  jL2  Z L 
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REFLECTED IMPEDANCE
Z L  RL  jX L
 2M 2
Zr 
R2  RL  j L2  X L 
 2 M 2 R2  RL   j L2  X L 

R2  RL 2  L2  X L 2

 2M 2
Z 22
2
R2  RL   j L2  X L 
Z22  R2  RL  jL2  X L 
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EXAMPLE
The parameters of a certain linear transformer are R1 = 200 Ω,
R2 = 100 Ω, L1 = 9 H, L2 = 4 H, and k = 0.5. The transformer
couples an impedance consisting of an 800 Ω resistor in series with a
1 µF capacitor to a sinusoidal voltage source. The 300 V (rms) source
has an internal impedance of 500 + j100 Ω and a frequency of 400
rad/s.
a) Construct a frequency-domain equivalent circuit of the system.
b) Calculate the self-impedance of the primary circuit.
c) Calculate the self-impedance of the secondary circuit.
d) Calculate the impedance reflected into the primary winding.
e) Calculate the scaling factor for the reflected impedance.
f) Calculate the impedance seen looking into the primary terminals of the
transformer.
g) Calculate the rms value of the primary and secondary current.
h) Calculate the rms value of the voltage at the terminals of the load and
source.
i) Calculate the average power delivered to the 800 Ω resistor.
j) What percentage of the average power delivered to the transformer is
delivered to the load?
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S0LUTI0N
a) frequency-domain equivalent circuit of the system
jL1  j 4009  j 3600 ;
jL2  j 4004  j1600 ;
M  0.5 94  3H ;
jM  j 4003  j1200 ;
1
106

  j 2500 .
jC j 400
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Z11  500  j100  200  j3600  700  j3700 
c). The self-impedance of the secondary circuit.
Z22  100  j600  800  j 2500  900  j900 
d). The impedance reflected into the primary winding.
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 1200

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

900  j900  900  j900  800  j800 
Zr  

9
 900  j900 
e). The scaling factor by which Z22* is reflected is 8/9
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f). The impedance seen looking into the primary terminals of the
transformer is the impedance of primary winding plus the
reflected impedance, thus
Z ab  200  j3600  800  j800  1000  j 4400 
g). Calculate I1 and I2
Vs
30000
I1 

Z s  Z ab 1500  j 4500
 20  j 60  63.25  71.57 0 mA rms 
j 1200
0
I2 
I1  59.6363.43 mA rms 
900  j900
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h). The voltages at the terminals of transformer are
V2  800  j 2500I 2  156.52  8.820 V rms 
V1  Z ab I1  1000  j 4400I1  285.385.630 V rms 
i ). The average power delivered to the load is
P  800 I 2  2.84 W
2
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j). The average power delivered to the transformer is
Pab  1000 I1  4.00 W
2
Therefore
2.84

x100  71.11%
4.00
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Energy Analysis
• An energy analysis of the mutually coupled inductors
provides an expression for the instantaneous stored
energy
w(t )  L1i1 (t )  L2 i2 (t )  M i1 (t ) i2 (t )
1
2
2
1
2
2
• The sign is positive (+) if currents are both entering
(or leaving) the dots; sign is negative (-) if currents
are otherwise
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wt   L i  L i  Mi1i2
1
2
1
2
2
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1
2
2
2 2
L i  L i  Mi1i2  0
2
11
2
2 2
1
2
2
 L1
L2 

  i1i2
i

i
1
2
 2

2




L1L2  M  0
L1L2  M
M  k L1L2
0  k  1
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Quantifying the Coupling
• The mutual inductance, M, is in the range
0  M  L1 L2
• The coefficient of coupling (k) between two
inductors is defined as

0  k 


 1
L1 L2 
M
– for k > 0.5, inductors are said to be tightly
coupled
– for k  0.5, coils are considered to be loosely
coupled
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DRILL EXERCISE
The self-inductances of the coils shown are L1 = 5 mH and
L2 = 33.8 mH. If the coefficient of coupling is 0.96,
calculate the energy stored in the system in millijoules
when (a) i1 = 10 A, i2 = 5 A; (b) i1 = -10 A, i2 = -5 A; (c) i1 =
-10 A, i2 = 5 A; and (d) i1 = 10 A, i2 = -5 A.
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