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Magnetically Coupled Networks
1
2
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
3
dt
diMv 1
2
dt
diMv 2
1
4
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
6
Mutually Coupled CoilsThe 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
dt
idL
dt
idMtv
dt
idM
dt
idLtv
22
12
2111
)(
)(
i1(t)
+
–
v1(t) L1
i2(t)
+
–
v2(t)L2
M
7
EXAMPLEWrite a set of mesh-current equations that describe the circuit shown in terms of the currents i1, i2, and i3
8
(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
The second mesh equation is
The third mesh equation is
dt
diii
dt
diivg
23121 5.498
1232312 865.440 iiiiii
dt
d
dt
di
3232
13 2065.490 iiidt
diii
dt
d
9
Mutually Coupled Coils (AC)
• The frequency domain model of the coupled circuit is essentially identical to that of the time domain
2212
2111
IIV
IIV
LjMj
MjLj
10
Source Input ImpedanceLinear 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
LPRPP
Si f ZZZZ
IV
Z
L1 L2
M
ZL+–
VS
Z
11
Transformer
source load
THE LINEAR TRANSFORMER
ZS
ZLI1
I2
R1 R2
jωL1jωL2
Vs
a
b
c
d
jωM
source Load
12
2221
2111
0 IZLjRMIj
MIjILjRZV
L
ss
L
s
ZLjRZ
LjRZZ
2222
1111
122
222211
2
222211
221
IZ
MjV
MZZ
MjI
VMZZ
ZI
s
s
13
22
22
1122
222211
int1 Z
MZ
Z
MZZZ
I
Vs
L
sab
ZLjR
MLjR
ZZ
MZZ
22
22
11
22
22
11
14
REFLECTED IMPEDANCE
LLL jXRZ
LL
LL
LL
LLr
XLjRRZ
M
XLRR
XLjRRM
XLjRR
MZ
222
22
22
22
22
2222
22
22
LL XLjRRZ 2222
15
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 transformercouples 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?
EXAMPLE
16
S 0 L U T I 0 N
a) frequency-domain equivalent circuit of the system
;3495.0
;16004400
;36009400
2
1
HM
jjLj
jjLj
.2500400
101
;120034006
jjCj
jjMj
17
3700700360020010050011 jjjZ
c). The self-impedance of the secondary circuit.
900900250080060010022 jjjZ
d). The impedance reflected into the primary winding.
e). The scaling factor by which Z22* is reflected is 8/9
800800900900
9
8 900900
900900
12002
jjjj
Z r
18
f). The impedance seen looking into the primary terminals of the transformer is the impedance of primary winding plus the reflected impedance, thus
440010008008003600200 jjjZab
g). Calculate I1 and I2
rmsmAj
jZZ
VI
abs
s
0
0
1
57.7125.636020
45001500
0300
rmsmAIj
jI 0
12 43.6363.59900900
1200
19
WIP
isloadthetodeliveredpoweraverageThei
rmsVIjIZV
rmsVIjV
arertransformeofterminalstheatvoltagesTheh
ab
84.2800
).
63.538.28544001000
82.852.1562500800
).
2
2
0111
022
20
WIPab 00.410002
1
j). The average power delivered to the transformer is
Therefore
%11.7110000.4
84.2 x
21
Energy Analysis
• An energy analysis of the mutually coupled inductors provides an expression for the instantaneous stored energy
• The sign is positive (+) if currents are both entering (or leaving) the dots; sign is negative (-) if currents are otherwise
)()()()()( 212
22212
1121 titiMtiLtiLtw
22
212222
12112
1 iMiiLiLtw
0212222
12112
1 iMiiLiL
022 2121
2
22
11
MLLiii
Li
L
MLL 21
10 21 kLLkM
23
Quantifying the Coupling• The mutual inductance, M, is in the range
• The coefficient of coupling (k) between two inductors is defined as
– for k > 0.5, inductors are said to be tightly coupled– for k 0.5, coils are considered to be loosely coupled
210 LLM
1021
LL
Mk
24
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.
DRILL EXERCISE