Download 5: Inductors transformers

• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Lecture 5
Inductors and transformers
•  Inductor
–  Hydraulic analogy
–  Duality with capacitor
–  Charging and discharging
•  Transformer
–  Physics
–  Fundamental equations
–  Dot convention
–  Comparison to voltage divider
Robert R. McLeod, University of Colorado
http://hilaroad.com/camp/projects/magnet.html
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• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
The Inductor
Packages
http://www.electronicsandyou.com/
electronics-basics/inductors.html
Physics
http://www.orble.com/userimages/user2207_1162388671.JPG
Symbol
Air core / generic
Ferrite core
Laminated iron core
Robert R. McLeod, University of Colorado
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• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Inductor hydraulics
• 
• 
• 
The hydraulic analogy to an inductor is a frictionless waterwheel with mass.
Kinetic energy is stored in the wheel.
The finite mass makes the wheel resist changes in current.
Pump is turned on at t=0.
Wheel is initially
stationary , that is,
I(0)=0.
At a later time…
…the wheel will be
turning with a finite
current.
If the pump belt was
removed, current
would continue to
flow.
Robert R. McLeod, University of Colorado
https://ece.uwaterloo.ca/~dwharder/Analogy/Inductors/
52
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Inductor+resistor hydraulics
“charging” phase
Valve (switch) is open
Wheel is initially
stationary , that is,
IINDUCTOR(0)=0.
Pump is turned on at t=0.
So all current initially goes
through resistor according
to Ohm’s law.
Pump is constant current
At a later time…
Valve (switch) is open
The wheel is turning,
offering very little
resistance.
Thus a large current will go
through the wheel. The
same amount is still going
through R (by Ohm’s law)
but this is now a small
fraction of the total current.
Robert R. McLeod, University of Colorado
https://ece.uwaterloo.ca/~dwharder/Analogy/Inductors/
53
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Inductor+resistor hydraulics
“discharge” phase
Valve (switch) is closed at t=0
Wheel is initially moving,
that is, IINDUCTOR(0)>0.
By conservation of current,
this current must run
through resistor, dissipating
energy.
Valve (switch) is closed
At a later time…
The kinetic energy in the
wheel is gone and it is
nearly stationary.
Approximately no current
is flowing.
Robert R. McLeod, University of Colorado
https://ece.uwaterloo.ca/~dwharder/Analogy/Inductors/
54
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
RL circuit
0.2 ms
1 KHz
Vsource
I
VL
+
VL
I
-
• 
• 
• 
• 
• 
Energy is stored in the magnetic field of the inductor.
Power is finite, so the energy does not change instantaneously.
The magnetic field is directly related to the current.
Thus inductors resist changes in the current.
Inductors act like short at DC, open at infinite frequency
dI
VL = L
dt
Robert R. McLeod, University of Colorado
L
τ L = = GL = 0.1[ms]
R
55
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
RL charge/discharge
Discharge
• 
• 
Charge
• 
At t=0, the voltage source is
replaced by a short.
What is the current versus time?
• 
Boundary conditions
At t=0, a 12 V voltage source
replaces the short.
What is the current versus time?
Boundary conditions
I 0 = VSource R = 1.2 [A ]
I0 = 0
I (t = ∞) = VSource R = 1.2 [A ]
I (t = ∞) = 0
Time constant
Time constant
Discharging exponential function
Charging exponential function
I (t ) = I 0e − t τ L
I (t ) = I 0 1 − e − t τ L
(
1.2
1.2
1.0
1.0
0.8
0.8
I [A]
I [A]
τ L = GL = 1 mH 10 Ω = 0.1 ms τ L = GL = 1 mH 10 Ω = 0.1 ms
0.6
0.4
)
0.6
0.4
0.2
0.2
0.1
0.2
t [mS]
0.3
0.4
Robert R. McLeod, University of Colorado
0.5
0.1
0.2
0.3
t [mS]
0.4
0.5
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• Lecture 5: Transformers
ECEN 1400 Introduction to Analog and Digital Electronics
The Transformer
Packages
http://
www.audiophilejournal.co
m/what-is-a-switchingpower-supply/
http://www.ehow.com/
how_7842346_buildsimple-electricaltransformer-school.html
http://
en.wikipedia.org/
wiki/Transformer
Physics
http://www.witricity.com/pages/technology.html
Symbol
Robert R. McLeod, University of Colorado
57
• Lecture 5: Transformers
ECEN 1400 Introduction to Analog and Digital Electronics
Hydraulic analogy
Primary
NP gear teeth
Secondary
NS gear teeth
• Two paddle-wheels connected by ideal gears with different #s of teeth.
• Rotation ratio of the two shafts is inverse to gear ratio:
ω P NS
=
ω S NP
IP NS
=
IS NP
T
N
• Torque ratio is proportional to gear ratio: P = P
TS N S
V
N
• Voltage (pressure) is proportional to torque, so: P = P
VS N S
⎛ NS ⎞ ⎛ NP ⎞
VS
= I SVS = PS
• Is power conserved? PP = I PVP = ⎜ I S
⎝ N P ⎟⎠ ⎜⎝ N S ⎟⎠
• Current is proportional to rotation rate, so
• Note gear connection could be flipped, reversing the secondary direction.
http://en.wikipedia.org/wiki/Talk%3AHydraulic_analogy
Robert R. McLeod, University of Colorado
http://pptcrafter.wordpress.com/2013/01/17/drawing-and-animating-gears-in-powerpoint/
58
• Lecture 5: Transformers
ECEN 1400 Introduction to Analog and Digital Electronics
Physics of the transformer
VP
-
Magnetic flux
(primary side)
A
Φ = NP
µ0 I P
2R
+
Φ
IP
NS
Magnetic flux
(secondary side)
A
Φ = NS
µ0 I S
2R
IS
-
Load
+
VS
Robert R. McLeod, University of Colorado
Faraday’s Law
(primary side)
dΦ
VP = N P
dt
NP
(1)
(2)
(3)
Faraday’s Law
(secondary side)
dΦ
VS = N S
dt
(4)
59
• Lecture 5: Transformers
ECEN 1400 Introduction to Analog and Digital Electronics
Engineering equations of
transformer
Divide equation 2 by equation 4
VP N P
=
VS N S
•  VS is proportional to VP
•  NP/NS = “turns ratio”
•  E.g. Turns ratio of 10:1 would step
primary voltage down by factor of 10.
VS = VP/10
Divide equation 1 by equation 3, rearrange
IP NS
=
IS NP
•  IS is proportional to IP
•  Current increases if voltage decreases
•  E.g. Turns ratio of 10:1 would step
primary current up by factor of 10.
IS = 10 IP
Multiply these two equations
VP I P PP [W ] N P N S
=
=
=1
VS I S PS [W ] N S N P
Robert R. McLeod, University of Colorado
Power is conserved.
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• Lecture 5: Transformers
ECEN 1400 Introduction to Analog and Digital Electronics
Dot convention
If either winding
is reversed, the
secondary
current and
voltage are
reversed.
http://en.wikipedia.org/wiki/Transformer
•  Dots indicates + voltage terminals.
•  Primary current in, secondary current out dot.
Robert R. McLeod, University of Colorado
http://en.wikipedia.org/wiki/Dot_convention
61
• Lecture 5: Transformers
ECEN 1400 Introduction to Analog and Digital Electronics
Compare voltage
divider to transformer
+
IP
RP
VP
+
R2
+
IS
+
RS
-
RS
VP
VS
VS
-
VS =
IS
IP
RS
VP
RP + RS
IS = IP
RS
PS = VS I S =
VP I P
RP + RS
• VS lower than VP by R ratio
• IS same as IP
• Voltage can only decrease
• Power lower by R ratio
• Circuits share ground
• Works at all frequencies
• Cheap, small
Robert R. McLeod, University of Colorado
-
-
VS =
1
VP
NP NS
N
IS = P IP
NS
PS = VS I S = VP I P
• VS lower than VP by N ratio
• IS greater than IP by N ratio
• Voltage can increase or decr.
• Power conserved
• Circuits isolated
• Works only for AC
• Expensive, heavy
62
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 5.1
Q: A constant (“DC”) voltage is placed
across an inductor on the left and a very
high frequency time-varying (“AC”)
voltage is placed across the inductor on the
right. Large current will flow in which
cases?
A: No in the DC and no in the AC
B: Yes in the DC and no in the AC
C: No in the DC and yes in the AC
D: Yes in the DC and yes in the AC
E: It depends on the AC frequency.
The inductor is a short at DC and an open at high frequency.
Robert R. McLeod, University of Colorado
63
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 5.2
Q: At t=0, a switch is closed and the
uncharged inductor L begins to
charge. What is the current through
the inductor at t=1 second?
L / R = 1ms
A: 0 A
One second is 1000x the L/R time
B: 1 mA
constant, so the inductor will be fully
charged and operating like a short.
C: 10 mA
The voltage across R is thus 10 V.
I = V / R = 10 mA
D: 100 mA
E: I have no idea how to find this.
Robert R. McLeod, University of Colorado
64
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Quiz 5.3
Q: A voltage V1 = 9 sin(omega t) is
applied to the primary (left) side of
this transformer. What is V2 on the
secondary side?
A: V2 = 3 sin(omega t)
B: V2 = 9 sin(omega t)
N
C: V2 = 27 sin(omega t) V = N V = 3V
D: V2 = 9 sin(3 omega t)
E: V2 = 9 sin(omega / 3 t)
S
S
P
P
P
Robert R. McLeod, University of Colorado
65
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Bonus quiz 5.1
Q: A function generator is hooked to
the series RL circuit above. The peak
current through the inductor depends
on
A: The peak voltage of the source
B: The frequency of the source
C: The order of R and L in series
The system is linear, so the
D: A and B but not C •  inductor
current is
to the source
E: A and C but not B proportional
voltage.
• 
• 
Robert R. McLeod, University of Colorado
The current through the
inductor will drop when
the frequency is > 1/(L/R)
The order makes no
difference
66
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Bonus quiz 5.2
Q: The current through an inductor of
inductance L=2 H is
I(t) = 1 + cos(500 t).
The voltage on the inductor is:
A:
B:
C:
D:
E:
V(t)= -1000 sin(500 t)
V(t)=2t + (1/250) sin(500 t)
V(t)=1/2 + (1/2) cos(500 t)
V(t)=2/(1 + cos(500 t))
V(t)=2 + 2 cos(500 t)
VL = L
dI
= 2[− 500 sin (500t )] = −1000 sin (500t )
dt
Robert R. McLeod, University of Colorado
67
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Bonus quiz 5.3
Q: As the frequency of the function
generator is increased from zero, the peak
voltage across the inductor stays near zero,
then begins to increase. The frequency
where this begins to occur is about
A: 5 mHz
L / R = 5 ms
B: 200 Hz
1
C: 500 Hz
f = = 200 Hz
T
D: 2 KHz
E: 200 MHz
Robert R. McLeod, University of Colorado
68
• Lecture 5: Inductors
ECEN 1400 Introduction to Analog and Digital Electronics
Bonus quiz 5.4
I (t ) = I 0e
−
t
L R
t
−
1
I 0 = I 0e L R
4
L ⎛1⎞
t = − ln⎜ ⎟
R ⎝4⎠
Q: A function generator is
hooked to the series RL
circuit above. The function
generator is set to a square
wave which alternates
between V0 and 0V (high/
low) with a period >> RL.
If a transition from V0 to
0V occurs at t=0, when
does the current through the
inductor drop to ¼ of its
value at t=0?
Robert R. McLeod, University of Colorado
L ⎛1⎞
A : t = − ln⎜ ⎟
R ⎝4⎠
B:t = −
L ⎛3⎞
ln⎜ ⎟
R ⎝4⎠
C:t = −
R ⎛1⎞
ln⎜ ⎟
L ⎝4⎠
D:t = −
R ⎛3⎞
ln⎜ ⎟
L ⎝4⎠
E:t = e
−
1
4
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