cont
Outline - Part 2
EE 411: Circuit Theory
Sinusoidal Steady-State Analysis - II
© Copyright by Margarida Jacome
EE411: SL12
! Series, Parallel, and !-to-Y Simplifications
! Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
! Phasor Diagrams
! Mutual Inductance
! The Transformer
! The Ideal Transformer
1
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Problem 1
Ig
j3#
-j3#
Ig
Problem 1 (cont)
Ig
Use the mesh-current method to find
the phasor current Ig.
5#
j3#
-j3#
Ig
(5 + j2) Ia– 5Ig = j15
5#
o
5"0 = 5
5"0o
5"0o
j2#
Ia
o
+ 5"-90
–
o
o
o
5"-90 = 5 (cos (-90 ) + j sin (-90 )) = -5j
5"0o
5"0
o
Ig= ?
– 5 Ia + 5 Ig= -j15
j2#
Ia
(Cramer’s Method)
o
+ 5"-90
–
!=
-5
5
(5 + j2)
-5
-5
-j15
(5 + j2)
j15
= 25 - 5(5+ j2) = - j10
2 equations:
j3Ig + 5 (Ig– Ia) – j3 (Ig– 5) = 0
5 (Ia– Ig) – 5j + j2 (Ia– 5) = 0
Ng =
5 Ig– 5 Ia + j15 = 0
$
!
(5 + j2) Ia– 5Ig – 15j = 0
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Ig =-j3 = 3 "-90o A
3
= -j75 + j15 (5 + j2) = -30
Ig = Ng / ! = -30/(- j10)
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4
Problem 2
200k#
Problem 2 (cont)
200k#
Find the steady state expression for
vo(t) when vg(t) = 2.5 cos 107t V
1.25pF
b
a
o
40k#
10k#
–
10pF
+
–
+
-5V
10k#
KCL on (b):
1.25pF $ 1/(j (107)(1.25%10-12)) = -j80k#
10
-Va
40
–
+
Va
-j10
Vo
-j80
+
Va – Vo
200
+
Va
40
=0
=0
(1)
' Va = - j0.5 Vo (2)
Substituting (2) in (1):
(26 + j20)(- j0.5) Vo – Vo = 50 $ -j13 Vo + 10 Vo – Vo = 50
Vo =
50
= 1.8 + j2.6 = 3.16"55.30o
9 – j13
Back to time domain:
(& = 107 rad/s)
Va – 2.5
104
+
Va – Vo
Va
+
2%106
-j104
+
Vo 50k#
–
+
Va
=0
4%104
-Va
Vo = 0
–
4%104 -j8%104
6
Outline - Part 2
$ 20 Va – 50 + j20 Va + Va – Vo + 5 Va = 0
' (26 + j20 ) Va – Vo = 50
+
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5
Problem 2 (cont)
Va – 2.5
5V
-5V
KCL on (a):
10pF $ 1/(j (107)(10-11)) = -j10k#
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-j10k#
+
–
Impedance of capacitors for & = 107 rad/s:
40k#
–
2.5"0o
+
vo(t) 50k#
–
cont
vg(t)
Vg = 2.5"0
5V
Vb = 0 (assuming that OPAMP
in linear region)
-j80k#
! Series, Parallel, and !-to-Y Simplifications
! Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
! Phasor Diagrams
! Mutual Inductance
! The Transformer
! The Ideal Transformer
vo(t) = 3.16 cos (107t + 55.30o)V
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8
Phasor Diagram
Problem 3
What is the relationship between Vs and VL
(load voltage) graphically?
Sketch in the complex plane showing the relationships of the phasor voltages
and phasor currents.
Assume I = I "0o. Sketch V .
m
j50#
I
10#
-j50#
VL
I
VS
VL = j50 I = (50 "90 ) I .
VR = 10 I.
Re
L2
(5)
Ib
Ia
VR1
(6)
Vs
VL
I
L2
VL
VR1
Ia
VL
VL1
(5)
Ib
I
! I = Ia + Ib
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Ib ?
Ia
VR1
VL
I
! VL1 leads I by 90o
! |VL1| = |I| &L1
! VR1 is in phase with I
10
! |VR1| = |I| R
1
Outline - Part 2
Vs = VL + VR1 + VL1
VL1
R2
VL
! Ia is in phase with VL
! Magnitude: |Ia| = |VL|/R2
Ib
I?
9
cont
R2
Ib
(4)
! Ib lags behind VL by 90o
! |Ib| = |VL|/&L2
What is the relationship between Vs and VL (load voltage) graphically?
L1
R1
I
–
VL
Ib
Problem 3 (cont)
Ib
Ia
(3)
VL + VR + VC
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Ia
+
VL
Ia
Ia
VL1?
VS = VL + VR + VC
VC = -j50 I = (50 "-90o) I .
+
VL
(2)
Ia ?
VL
–
o
Vs +
–
R1
Vs +
–
VL + VR
+
VC
–
L1
I
Im
+ V – + V –
L
R
Vs +–
s
(1)
VL1
! Series, Parallel, and !-to-Y Simplifications
! Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
! Phasor Diagrams
! Mutual Inductance
! The Transformer
! The Ideal Transformer
! VL1 leads I by 90o
! |VL1| = |I| &L1
! VR1 is in phase with I
! |VR1| = |I| R1
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12
Time
TimeDomain
Domain
Mutual Inductance
Mutual Inductance
! The magnetic field considered in our study of inductors
was restricted to a single circuit
R1
# inductance (or self-inductance) is the parameter that relates
voltage to a time-varying current in the same circuit
vs(t) +–
i1(t)
#Use mesh currents, i.e., write the circuit
equations in terms of the coil currents
M
L1
i2(t)
L2
(1) Choose reference directions for each
current (arbitrary directions, as before)
R2
(2) Add voltages around each closed path.
! We now consider the situation in which two circuits are
linked by a magnetic field
# in this case, the voltage induced in the second circuit can be
related to a time-varying current in the first circuit by as parameter
known as mutual inductance
Two magnetically coupled coils
Because of the mutual inductance M, there
will be two voltages across each coil:
! (Self-)Inductances: L1, L2
" A self induced voltage: L1(di1(t)/dt)
! Mutual Inductance: M
" A mutually induced voltage: M(di2(t)/dt)
M = k L1 L2
Polarity??
0"k"1
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13
Polarity of the Mutually Induced
Voltage
Depends on the way the coils are wound in relation to the reference direction
of the currents.
(A
iA
iC
A
C
B
"
D
(D
core material
Arbitrarily select one terminal of the second coil (e.g., terminal C) and assign a current into it.
Use the right-hand rule to determine the magnetic field established by the current (iC).
If the fluxes have the same direction, place a dot on the terminal where the test
current iC enters. (Else, place
the dot
on the
terminal where the current leaves).
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perfect
coupling
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14
Polarity of the Mutually Induced
Voltage
Depends on the way the coils are wound in relation to the reference direction
of the currents.
i induces (
(A
iA
iC
A
C
!( induces v
!t
B
Procedure:
1. Arbitrarily select one terminal of one coil (e.g., terminal A) and mark it with a dot. Assign a
current into the dotted terminal. Use the right-hand rule to determine the magnetic field
established by the current (iA).
2.
coupling coefficient
"
D
(D
core material
If the fluxes have the same direction, place a dot on the terminal where the test
current iC enters. (Else, place the dot on the terminal where the current leaves).
When
Whenthe
thereference
referencedirection
directionfor
foraacurrent
currententers
entersthe
thedotted
dottedterminal
terminalofofaacoil,
coil,
the
thereference
referencepolarity
polarityofofthe
thevoltage
voltagethat
thatititinduces
inducesininthe
theother
othercoil
coilisispositive
positive
atatits
itsdotted
dottedterminal.
terminal.
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EE411
16
Time
TimeDomain
Domain
Time
TimeDomain
Domain
Dot Convention - Schematics
Example 2
# A dot is placed on one terminal of each winding. These dots carry the polarity
information.
i1(t)
M
i1(t)
M
+
+
+
+
vL
2
_
vL
1
_
vL
2
_
L1
vL = L1
1
vL = L2
L2
R1
M
i2(t)
vL
1
_
Self induced voltages
2
i2(t)
L1
L2
+
+
vs(t) +–
i1(t)
L1
Mutually induced voltages:
di1(t)
+M
dt
di2(t)
di2(t)
vL = L1
1
dt
+M
dt
di1(t)
vL = L2
2
dt
–M
di2(t)
dt
dt
di2(t)
–M
di1(t)
di1(t)
dt
-vs(t) + R1 i1(t) + L1
L2
i2(t)
dt
R2 i2(t) + L2
dt
– M
dt
– M
di2(t)
dt
di1(t)
dt
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M
(1)
L1
i2
Leq
=0
=0
18
L2
i2
L1
di1(t)
dt
L2
–
+
M
(b)
d(i2(t))
vab = L1 d (i1(t) – i2(t)) + M
dt
dt
19
vab = Leq
–
KVL:
=0
di2
dt
M
i1
(2)
di2(t)
dt
=0
+
R2
i1
di1(t)
di1(t)
+M
Give Leq
(a)
-vs(t) + R1 i1(t) + L1
dt
Problem 4
M
L1
dt
Time
TimeDomain
Domain
Example 3
i1(t)
di2(t)
+M
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17
Time
TimeDomain
Domain
vs(t) +–
di1(t)
R2
dt
When the reference direction for a current
enters the dotted terminal of a coil, the reference
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polarity of the voltage that it induces in the otherEE411
coil is positive at its dotted terminal.
R1
di2(t)
R2 i2(t) + L2
i2(t)
L2
di2
dt
–
L1
i2 – i 1
i2
L2
M
d(i2 – i1)
dt
+
(1)
di (t)
d(i2(t))
0 = L1 d (i2(t) – i1(t)) – M
– M d (i2(t) – i1(t)) + L2 2
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dt
dt
dt
EE411
(2)
20
Time
TimeDomain
Domain
Time
TimeDomain
Domain
Problem 4 (cont)
Problem 4 (cont)
d(i2(t))
vab = L1 d (i1(t) – i2(t)) + M
dt
dt
d(i2(t))
0 = L1 d (i2(t) – i1(t)) – M
dt
dt
vab = L1
di (t)
– M d (i2(t) – i1(t)) + L2 2
dt
dt
di1(t)
dt
0 = (M – L1)
+ (M – L1)
di1(t)
dt
di2(t)
(1)
dt
di1(t)
solving for
di2(t)
+ (L1 + L2 – 2M)
dt
…
dt
(2)
rewrite:
vab = L1
di1(t)
dt
di1(t)
0 = (M – L1)
di2(t)
+ (M – L1)
dt
dt
+ (L1 + L2 – 2M)
di2(t)
di1(t)
dt
dt
vab
(M – L1)
0
(L1 + L2 – 2M)
=
L1
(L1 + L2 – 2M)
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21
Time
TimeDomain
Domain
dt
=
=
=
vab
i1
(L1 + L2 – 2M)
2+
L1
2
L1 L2 – 2M L1 – M + 2M L1 – L1
(L1 + L2 – 2M)
i1
2
L1
vab
Leq
(a)
L2
i2
Leq
vab = Leq
di1(t)
dt
(L1 + L2 – 2M)
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23
di1(t)
dt
dt
L1
i2
L1
i2
L1
i2
L2
+
L2
–
(2)
vab
L1 L2 – M 2
M
(1)
di2
M
i1
i1
(b)
vab = Leq
–
Give Leq
M
L1(L1 + L2 – 2M) – (M – L1)2
L1 L2 – M 2
Leq =
Problem 5
(a)
(L1 + L2 – 2M)
22
Time
TimeDomain
Domain
Problem 4 (cont)
di1(t)
vab
L1(L1 + L2 – 2M) – (M – L1)2
(M – L1)
(M – L1)
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(L1 + L2 – 2M)
=
M
(b)
+
di2
dt
+
i2 – i 1
M
L2
d(i2 – i1)
dt
–
KVL:
d(i2(t))
vab = L1 d (i1(t) – i2(t)) – M
dt
dt
(1)
di (t)
d(i2(t))
0 = L1 d (i2(t) – i1(t)) + M
+ M d (i2(t) – i1(t)) + L2 2
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dt
dt
dt
EE411
(2)
24
Time
TimeDomain
Domain
Time
TimeDomain
Domain
Problem 5 (cont)
d(i2(t))
vab = L1 d (i1(t) – i2(t)) – M
dt
dt
d(i2(t))
0 = L1 d (i2(t) – i1(t)) + M
dt
dt
Problem 5 (cont)
(1)
di1(t)
vab = L1
di (t)
+ M d (i2(t) – i1(t)) + L2 2
dt
dt
(2)
di1(t)
0 = -(M + L1)
di2(t)
– (M + L1)
dt
dt
(1)
dt
solving for
+ (L1 + L2 + 2M)
di2(t)
di1(t)
…
dt
(2)
dt
rewrite:
vab = L1
di1(t)
dt
0 = -(M + L1)
di2(t)
– (M + L1)
di1(t)
dt
dt
+ (L1 + L2 + 2M)
di2(t)
di1(t)
dt
dt
vab
-(M + L1)
0
(L1 + L2 + 2M)
=
=
L1
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(L1 + L2 + 2M)
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25
26
Time
TimeDomain
Domain
Time
TimeDomain
Domain
Problem 5 (cont)
di1(t)
dt
=
=
=
vab
(L1 + L2 + 2M)
L12 + L1 L2 + 2M L1 – M 2 – 2M L1 – L12
(L1 + L2 + 2M)
L1 L2 + M 2
(L1 + L2 – 2M)
4H
M
L1(L1 + L2 + 2M) – (M + L1)2
L1 L2 – M 2
Leq =
Problem 6
(a)
(L1 + L2 + 2M)
i1
i1
L1
vab
Leq
5#
L2
i2
ig(t)
(b)
ig
vab
for the configuration of
Problem 4, sign was a
minus.
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vab
L1(L1 + L2 + 2M) – (M + L1)2
-(M + L1)
-(M + L1)
(L1 + L2 + 2M)
vab = Leq
di1(t)
dt
4
di1(t)
dt
8H
Write the set of mesh-current equations that
describe the circuit in terms of i1 and i2.
20# i1
16H
i2
60#
voltage drop caused
by mutual inductance
+ 20 (i1(t) – i2(t)) + 8 d (ig(t) – i2(t)) + 5 (i (t) – i (t)) = 0
1
g
dt
20 (i2(t) – i1(t)) + 60 i2(t) + 16 d (i2(t) – ig(t)) – 8 di1(t) = 0
dt
dt
27
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voltage drop caused
by mutual inductance
28
Time
TimeDomain
Domain
Problem 7
Give vx (t) in terms of i1(t).
i1(t)
+
M
' v1(t) = L1
L2 v2(t)
_
_
di1(t)
= -L2
di1(t)
di1(t)
+ M
dt
dt
+ L2
di1(t)
+ M
– M
d(-i1(t))
– M
dt
d(i1(t))
dt
dt
di1(t)
– M
d(i1(t))
dt
di2(t)
v2(t) = L2
_
' vx(t) = L1
4H
KCL: i1(t) = -i2(t)
+
v1(t) L1
vx(t)
KVL: vx(t) = v1(t) – v2(t)
i2(t)
+
Q1
dt
ig(t)
8H
20# i1
16H
ig
60#
i2
dt
d(i1(t))
dt
dt
di1(t)
vx(t) = (L1 + L2 + 2M)
5#
Write the set of mesh-current equations that
describe the circuit in terms of i1 and i2.
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29
dt
Time
TimeDomain
Domain
Time
TimeDomain
Domain
Power Calculation
i1(t)
+
v1(t)
_
L1
Energy Calculation
i2(t)
M
L2
i1(t)
di1(t)
dt
+
+
+
v1(t)
v2(t)
i1(t) + M
but :
_
_
p = L1
di2(t)
dt
di2(t)
dt
p = L1
di1(t)
dt
i2(t)
M
v2(t)
p = v1(t) i1(t) + v2(t) i2(t)
= L1
i1(t) + L2
i1(t) + L2
i1(t) +
di2(t)
di1(t)
dt
dt
i2(t) + M
di1(t)
dt
i2(t)
#
i2(t) = d (i1(t)i2(t))
dt
EE411
L1
di1(t)
dt
L2
i1(t) + L2
_
di2(t)
dt
i2(t) + M d (i1(t)i2(t))
dt
t
p dt’ = (1/2) L1 i12(t) + (1/2) L2 i22(t) + M i1(t)i2(t)
0
w(t) = (1/2) L1 i12(t) + (1/2) L2 i22(t) + M i1(t)i2(t)
di2(t)
i2(t) + M d (i1(t)i2(t))
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dt
dt
30
instantaneous energy stored in the two coils
31
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32
Time
TimeDomain
Domain
Energy Calculation (cont)
i1(t)
Mutual Inductance in the Frequency Domain
i1(t)
i2(t)
M
+
+
+
v1(t)
_
p = L1
L1
di1(t)
dt
L2
v1(t) = L1
di2(t)
dt
i2(t) – M d (i1(t)i2(t))
dt
dt
di2(t)
v2(t) = L2
dt
I1
w(t) = (1/2) L1 i12(t) + (1/2) L2 i22(t) – M i1(t)i2(t)
j&M
+ M
v1(t) = L1
+ M
dt
d(i1(t))
_
I1
I2
j&L2
33
ig(t) 10#
2mH
10#
70"0o
vg(t) = 70cos 5000t V.
vg(t) +–
+
–
frequency
domain
KVL:
& = 5000 rad/s
Ig
L1 = 2mH $ j&L1 = j(5000)(2%10-3) = j10#
L2 = 8mH $ j&L2 =
j(5000)(8%10-3)
M = 2mH $ j&M =
j(5000)(2%10-3)
= j40#
= j10#
70"0o
+
–
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dt
dt
j&M
j&L1
_
_
I2
j&L2
V2
_
V1 = j& L1 I1 – j& M I2
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Austin
2=
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10#
j10#
IL
j10#
30#
j40#
35
j10#
Ig
j& L2 I2 – j& M I1
34
& = 5000 rad/s
IL
j40#
j10#
30#
8mH
– M
dt
d(i1(t))
Problem 8 (cont)
(a) Find the steady-state expressions for the
currents ig(t) and iL(t) when
iL(t)
2mH
dt
di2(t)
d(i1(t))
+
V1
V2
V2 = j& L2 I2 + j& M I1
Problem 8
– M
+
V1 = j& L1 I1 + j& M I2
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_
di1(t)
v2(t) = L2
dt
+
j&L1
L2 v2(t)
_
d(i1(t))
+
V1
+
v1(t) L1
_
di1(t)
i2(t)
M
+
L2 v2(t)
_
_
i1(t)
+
v1(t) L1
v2(t)
i1(t) + L2
i2(t)
M
30#
ig(t) = 5cos (5000t – 37.87o) A
!
iL(t) = cos (5000t – 180o) A
70 = (10 + j10) Ig + j10 IL
Ig = 4 – j3 = 5"-37.87o A
0 = j10 Ig +(30 + j40) IL
IL = -1 A = 1"-180o
70
j10
0
30 + j40
Ig =
= 4 – j3 A
10 + j10
j10
j10
30 + j40
10 + j10
70
j10
0
10 + j10
j10
j10
30 + j40
IL =
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= -1 A
36
Problem 8 (cont)
Time Domain:
ig(t) 10#
vg(t) +–
Problem 8 (cont)
(b) Find the coefficient of coupling
iL(t)
2mH
2mH
k=
M
=
L1L2
2
= 0.5
ig(t) 10#
vg(t) +–
16
2mH
(c) Find the energy stored in the magnetically
coupled coils at t = 100)µs and t = 200)µs
iL(t)
2mH
8mH
30#
30#
8mH
t = 100)µs:
$ 5000t =(5000)(100)µ)(10-6) = 0.5) = 90o
iL(100)µs) = cos (-90o) = 0
o
iL(t) = cos (5000t – 180 ) A
ig(100)µs) = 5cos (52.13o) = 3A
o
ig(t) = 5cos (5000t – 37.87 ) A
w(100)µs) = (1/2)(2%10-3)(32) + 0 + 0
= 9mJ
w(t) = (1/2) L1 ig2(t) + (1/2) L2 iL2(t) + M ig(t)iL(t)
(see slide 32 -- note polarity of current i2(t))
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EE411
37
Problem 8 (cont)
ig(t) 10#
vg(t) +–
2mH
8mH
30#
t = 200)µs:
$ 5000t =(5000)(200)µ)(10-6) = ) = 180o
iL(200)µs) = cos (0o) = 1A
o
iL(t) = cos (5000t – 180 ) A
ig(200)µs) = 5cos (143.13o) = -4A
o
ig(t) = 5cos (5000t – 37.87 ) A
!
w(200)µs) = (1/2)(2%10-3)(42) + (1/2)(8%10-3)(12) + (2%10-3)(-4%1) = 12mJ
w(t) = (1/2) L1 ig2(t) + (1/2) L2 iL2(t) + M ig(t)iL(t)
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EE411
38
Outline - Part 2
iL(t)
2mH
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EE411
39
! Series, Parallel, and !-to-Y Simplifications
! Source Transformations and Thévenin-Norton Equivalent
Circuits
! Node-Voltage Method and Mesh-Current Methods
! Phasor Diagrams
! Mutual Inductance
! The Transformer
! The Ideal Transformer
contMargarida Jacome - UT Austin
EE411
40
!
The Transformer
Linear Transformer Circuit
! Device based on magnetic coupling, used in both
communication and power circuits
! A simple transformer is formed when two coils are wound on a single
core to ensure magnetic coupling
# Communications: used to match impedances and eliminate dc
signals from portions of the system
(a)
ZS
" Linear transformer
+s
V
–
# Power circuits: used to establish voltage levels that facilitate
transmission, distribution and consumption of electrical power
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EE411
Vs V
+s
–
R1
I1
(b)
j&L1
j&L2
I2
(a)
Load impedance
R1 = resistance of primary winding
R2 = resistance of secondary winding
L1 = self-inductance of primary winding
L2 = self-inductance of secondary winding
M = mutual inductance
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EE411
Analysis of the circuit aims at
determining:
I1 = primary current of the
transformer
I2 = secondary current of the
transformer
Zab = impedance seen looking
into the transformer from
43
terminals a,b
42
Primary and Secondary Currents
ZS
ZL
secondary winding
of the transformer
(connected to the
load)
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EE411
I1
(b)
KVL:
R1
R2
j&M
I1 =
+ s Vs
V
–
(d)
Zab
(d)
41
R2 (c)
j&M
ZL
j&L2
primary winding
of the transformer
(connected to the
source)
Parameters of Linear Transformer
Circuit
(a)
j&M
(b)
! Understanding of the sinusoidal steady-state behavior of
the transformer is required in the analysis of
communication and power circuits
Frequency domain circuit
model for a transformer
used to connect a load to
source
R2 (c)
j&L1
" Ideal transformer
ZS
R1
j&L1
Z11
j&L2
I2
ZL
self impedance of the mesh
containing the primary winding
I2 =
Z22
Z11 Z22 + &2M2
j&M
Z11 Z22 + &2M2
Vs
Vs
Vs = (ZS + R1 + j&L1) I1 – j&M I2
0 = – j&M I1 + (j&L2 + R2 + ZL) I2
Z22
Vs = Z11 I1 – j&M I2
0 = – j&M I1 + Z22 I2 # I2 =
j&M
2 2
Vs = Z11 I1 + & M I1
Z22
I1
Z22 Jacome - UT Austin
Margarida
Z22: self impedance of the mesh 44
EE411
containing the secondary winding
Zab: Impedance “seen” by the
Source
Zint
(a)
ZS
R1
R2
j&M
(a)
Z22
I1 =
I1
+ s Vs
V
–
Reflected Impedance Zr
j&L1
(b)
Zab ?
Zint =
Z11 Z22 + &2M2
ZL
I2
j&L2
Vs
I1
Z11 Z22 + &2M2
=
Z22
= Z11 +
Vs
I1
+ s Vs
V
–
Zab= R1 + j&L1 +
&2M 2
j&L2 + R2 + ZL
Reflected impedance is due to the existence of mutual inductance. If the two coils
are decoupled, M is zero, and thus Zr is zero.
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EE411
45
j&L1
j&L2
ZL
&2M 2
j&L2 + R2 + ZL
(b)
ZS
+ s Vs
V
–
R1
R2
j&M
j&L1
ZL
j&L2
(b)
# Load impedance: ZL = RL + jXL
Reactance XL is positive if the load is inductive and negative if the load is capacitive
2
2
# Reflected impedance: Zr = & M [R2 + RL – j (&L2 + XL)]
|Z22|2
# Writing reflected impedance in rectangular form:
&2M 2
Zr =
R2 + RL + j (&L2 + XL)
46
Reflected Impedance Zr
(a)
Zr =
+ s Vs
V
–
ZL
Impedance of the second coil & load impedance
transmitted (or reflected) to the primary side
of the transformer.
Zab
&2M2
Z22
R2
j&M
I2
&2M2
Z22
Reflected impedance Zr
Reflected Impedance Zr (cont)
R1
j&L2
Zab = R1 + j&L1 +
!
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(a)
j&L1
Z22
&2M2 – Z
&2M2
S = R1 + j&L1 +
Z22
Z22
Z11 = ZS + R1 + j&L1
ZS
R2
j&M
(b)
&2M2
Zr =
Zab= Zint – ZS = Z11 +
R1
ZS
=
&2M2 [R2 + RL – j (&L2 + XL)]
(R2 + RL )2 + (&L2 + XL)2
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EE411
=
&2M 2
|Z22|2
scaling
factor
[R2 + RL – j (&L2 + XL)]
magnitude of self impedance of47the mesh
containing the secondary winding
sign of the reactive component
(&L2 + XL) is reversed.
Thus,
Thus,the
thelinear
lineartransformer
transformerreflects
reflectsthe
theconjugate
conjugateofofthe
theself-impedance
self-impedanceofofthe
thesecondary
secondary
circuit
scaled
by
some
Margarida
Jacome
- UT
Austinfactor.
circuit(Z
(Z22*)*)into
intothe
theprimary
primarywinding
winding
scaled
by
some
factor.
48
22
EE411
Thevenin Equivalent
(a)
ZS
R1
R2
j&M
+ s Vs
V
–
j&L1
Problem 9
(a)
ZTh = Zab = R1 + j&L1 +
j(5000)(16%10-3)
= j80#
j&L2 = j(5000)(64%10-3) = j320#
-j/(&CL) = -j160#
j&M = j(160k)#
&2M2
Z22
&2M2
Zab = R1 + j&L1 +
= 40 + j80 +
Z22
(160k)2
120 + j320 – j160
= 40 + j80 + 76.8k2 – j102.4k2 = 40 + 76.8k2 = 40+60 = 100 #
(b)
Purely resistive means: 102.4k2 = 80 $ k = 0.88
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EE411
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49
Thevenin Equivalent w.r.t. to
Terminals c,d
ZS
R1
+ s Vs
V
–
R2
j&M
j&L1
R1
(c)
ZS
I1
ZL
j&L2
R2
j&M
+ s Vs
V
–
I1
R2
j&M
j&L1
RTh = R2 + j&L2 + Z’r
VTh = j&M I1 = j&M
(d)
(c)
j&L2
(d)
(c)
j&L2
!
50
RTh
j&L1
(d)
R1
Zab= 100 #
Thevenin Equivalent w.r.t. to
Terminals c,d
VTh = ?
ZS
L1 L2
= k (16)(64)10-6 = k32mH
Moving to
frequency domain:
Zr
ZTh
+ s Vs
V
–
M =k
(b)
j&L1 =
(a)
1.25µF
64mH
(b)
ZS
The value of k is adjusted until Zab is purely
resistive when & = 5k rad/s. Find Zab.
120#
16mH
ZL
j&L2
k
40#
Vs
ZS + R1 + j&L1
Zr =
(VOC $ I2 = 0)
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EE411
where Z’r is the impedance reflected from the secondary winding
when the voltage source is substituted by a short-circuit.
51
&2M 2
j&L2 + R2 + ZL
$
Z’r=
&2M 2
j&L1 + R1 + ZS
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EE411
(‘symmetry’)
52