Post on 20-May-2018
EE 70Winter Quarter 2008
Magnetic Circuits and Transformers(Chapter 15)
EE 70Clifford Hwang
February 28, 2008
EE 70Winter Quarter 2008
Agenda• Physical model for resistors• Physical model for capacitors• Physical model for inductors• Mutual inductance• Transformers• Diodes
EE 70Winter Quarter 2008
Physical Model of Resistors
• ρ: resistivity (fundamental property of the material)
• Resistance depends on how long the object is andits cross-sectional area
• See p. 30-31
AL
Rρ
=
EE 70Winter Quarter 2008
Capacitance of a Parallel-Plate Capacitor
WLAdA
C == ε
mF 1085.8 120
−×=ε
0εεε r=
(derived using Gauss’ Law)
(vacuum permittivity)
Dielectric constant (value is material dependent)
EE 70Winter Quarter 2008
Sample Calculation of Capacitance
( ) nF77.1F10x1770m10x1
m02.0m
F10x854.8dA
C
m10x1mm1.0d
m02.0)m10x20)(m10x10()cm20)(cm10(WLA
124
212
0
4
222
==
==
==
====
−−
−
−
−−
ε
Fairly large
EE 70Winter Quarter 2008
Inductance
• Wire coiled around a magnetic core• Current flow àMagnetic field / flux• Core improves magnetic flux and can
re-direct it
B
B
I
I
EE 70Winter Quarter 2008
Flux Linkages
φλ
φ
N
BA
dA
=
=
⋅= ∫ AB
Magnetic flux passing through a surface area A:
For a constant magnetic flux density perpendicular to the surface:
The flux linking a coil with N turns:
EE 70Winter Quarter 2008
Definition of Inductance
icurrentlinkagesFlux
Lλ
==
φλ N=Substitute for the flux linkages using
iN
Lφ
=
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Reluctance
RNi
=φ
• The magnetic flux can also be defined in terms of a quantity called reluctance– Magnetic equivalent of resistance
RN
iN
L2
==φ
Permeability (material dependent)
(constant, time independent)
EE 70Winter Quarter 2008
Faraday’s Law
Faraday’s law of magnetic induction:
The voltage induced in a coil whenever its flux linkages are changing. Changes occur from:
• Magnetic field changing in time
• Coil moving relative to magnetic field
dtdi
Ldt
)Li(ddtd
V ===λ
EE 70Winter Quarter 2008
Mutual Inductance• Current flowing through one coil can produce a magnetic
field that can induce a current in a different coil
1
11
1
111
i
iL
λ
λ
=
= ←
2
22
2
222
i
iL
λ
λ
=
= ←
2
12
2
21
1
21
1
12
iiiiM
λλλλ==== ←←
Self inductance for coil 1
Self inductance for coil 2
Mutual inductance between coils 1 and 2:
i1 i2
v1 v2
+ +
––
EE 70Winter Quarter 2008
Circuit Equations for Mutual Inductance
dtdi
Ldtdi
Mdt
dv
dtdi
Mdtdi
Ldt
dv
iLMiMiiL
22
122
211
11
2212
2111
+±==
±==
+±=±=
λ
λλλ
• The mutual inductance can increase or decrease the voltage• Depends on the direction of the field produced
EE 70Winter Quarter 2008
Mutual Inductance• To determine whether or not the mutual inductance increases
or decreases the voltage, one of the terminals is “dotted”• Currents entering the dotted terminals produce aiding fluxes (based
on right hand rule)
Fields are aiding Fields are opposing
EE 70Winter Quarter 2008
Example 15.8
mHN
L
mHN
L
410
200
110
100
7
222
2
7
221
1
===
===
R
R
WeberturnsampereR /)(107 −=
Self inductance:
EE 70Winter Quarter 2008
Example 15.8
15
7111
1 1010
100i
iiN −===R
φ
mH2i
M
i10200N
1
21
15
1221
==
×== −
λ
φλ
Mutual inductance:
(Magnetic flux from left coil) (Flux linkage on right coil, due to flux from left coil)
EE 70Winter Quarter 2008
Example 15.8
dtdi
Mdtdi
Le
dtdi
Mdtdi
Le
1222
2111
−=
−=
• Flux produced by i2 opposes the flux produced by i1• Proof: use right hand rule• If there’s a dot on the lower terminal of the left coil, then
add a dot to the upper terminal of the right coil
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Transformers• An application of mutual inductance• Used to step up or step down AC voltages
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Ideal Transformers
01
m10
1
m101
1
1m11
)tsin(NV
dt)tcos(NV
dt)t(vN1
(t)
Law sFaraday'by dtd
N)tcos(V)t(v
φωω
φωφφ
φω
+=+=+=
==
∫ ∫
EE 70Winter Quarter 2008
Ideal Transformers
[ ]
)t(vNN
)tcos(VNN
)tsin(dtd
NV
Ndtd
N)t(v
11
2m1
1
2
o1
m1222
==
+==
ω
φωω
φ
Step up voltage: N2 > N1Step down: N2 < N1
EE 70Winter Quarter 2008
Lenz’s Law• Why is i2 going from left to right?
• Unlike Example 15.8, there is no applied voltage / current on the right side
• Polarity of the induced voltage is such that the voltage would produce a current (through an external resistance) that opposes the original change in flux linkages
Top view
EE 70Winter Quarter 2008
Ideal Transformers
)()(
0 since0
12
12
2211
2211
tiNN
ti
iNiNiNiN
=
=≈≈=−= RRF φ
The magneto-motive force (mmf) applied to the core:
If the voltage is stepped upthe current is stepped down
EE 70Winter Quarter 2008
Ideal Transformers
)t(i)t(v)t(p
)t(i)t(vNN
)t(i)t(v)t(p
111
211
2222
=
==
( ) ( )tptp 12 =
• Since an ideal transformer is a purely inductive circuit (no resistance), there should be no power lost• This means that the AC current will be stepped down if the AC
voltage is stepped up• Net power is neither generated nor consumed by an ideal transformer
)()( 12
12 ti
NN
ti =
EE 70Winter Quarter 2008
Impedance Transformations
'L
2
1
2'L
21
12
121
112
2
2L
ZNN
Z)N/N()N/N(
)N/N()N/N(
Z
==
==IV
IV
• Goal: transform the circuit into one where the load impedance (ZL) and transformer are merged into a single impedance (ZL’)• Circuit elements on the secondary side are reflected to the primary side
’
1
1LZ
IV
=′
L
2
2
1'L Z
NN
Z
=
EE 70Winter Quarter 2008
Example 15.11
45282845220002000j2000
2000j10001000ZRZ
2000j1000)20j10()10(ZNN
Z
20j10Z
'L1s
2L
2
2
1'L
L
∠=∠=+=
++=+=
+=+=
=
+=
• Impedance transformations are used to find the voltages across the transformer
Zs
EE 70Winter Quarter 2008
o
oo
o
oo
o
43.186.790
)43.632236)(453536.0(
)2000j1000)(453536.0(Z
A453536.045282801000
Z'L11
s
s1
∠=
∠−∠=
+−∠==
−∠=∠∠
==
IV
VI
Example 15.11
EE 70Winter Quarter 2008
Example 15.11
oo
oo
43.1860.79)43.186.790(101
45536.3)453536.0(10
11
22
12
12
∠=∠==
−∠=−∠==
VV
II
NNNN
We can now calculate the current and voltage phasors on the secondary side using the turns ratio:
EE 70Winter Quarter 2008
Example 15.12
Ω=
=
=
∠==
10)1000(101
01000101
2
1
2
1
2'1
1
2'
RNN
R
VNN
V sso
• Circuit elements and sources can also be reflected from the primary side to the secondary side of the transformer
Reciprocal of ZL’ equation
Same as ideal voltage transformation
EE 70Winter Quarter 2008
Example 15.12
ooo
o
oo
o
43.1806.79)0100(45220
43.63510
)0100(45220
)10/20(400100
)0100(20202010
2010102010
1
'2
∠=∠∠
∠=
∠∠
∠+=
∠++
=++
+=
−Tan
jj
Vj
jV s
Same result as Example 15.11
EE 70Winter Quarter 2008
• Resistivity is in between a conductor and an insulator– Conductor: ρ < 10-2 Ω-cm (ex.
aluminum, copper)– Insulator: ρ > 105 Ω-cm (ex.
silicon dioxide)
• Resistivity can be controlled precisely and even electrically modified
• Resistivity, doping concentration, and carrier mobility can be obtained from graphs
What is a Semiconductor?
pqnq11
pn µµσρ
+==
EE 70Winter Quarter 2008
Why do different materials have different ρ?
• Different materials have different energy band diagrams
An electron in a single atom can only exist at certain energy levels
Energy levels split as atoms are brought together, which may result in gaps
Ec
Ev
EE 70Winter Quarter 2008
• The allowed energy levels are so close to one another that they can be modeled as a continuous band
• Gaps can occur between continuous bands– The size of the energy gap between the filled and unfilled levels
determines whether you have a conductor, insulator, or semiconductor
Band Diagrams
Conductor
Insulator (large Eg) or Semiconductor (Eg ~ 1eV)
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EE 70Winter Quarter 2008
• To adjust the number of carriers in a semiconductor, impurities can be added– Use a nearby element in the periodic table
• Modeled in the band diagram as an Ed (donor/electron) or Ea (acceptor/hole) level
Doping
Want small gap
EE 70Winter Quarter 2008
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