Chapter 18 Powerpoint

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Counter EMF (also known as Back EMF)
When the current through an inductor changes, the magnetic field also
changes. This changing magnetic field causes there to be an induced
voltage across the terminals of the inductor. This induced voltage’s
polarity is in a direction opposite to that of the original voltage applied.
This is basically Lenz’s law in a nutshell.
Current through inductors in a DC
circuit when switch closes
What do you think voltage across
an inductor looks like in a DC
circuit when the switch is closed?
Voltage through inductors in a DC
circuit when switch closes
Inductance has it greatest effect only when
there is a change in current.
What would happen in this circuit?
Inductance Summary
Inductors react against changes in current by consuming
or providing voltage in a polarity necessary to oppose the
change
When an inductor is faced with an increasing current, it
acts as a load, by consuming voltage. (series opposing)
When an inductor is faced with a decreasing current, it
acts as a source, by sourcing voltage. (series aiding)
The ability of an inductor to store energy in the form of a
magnetic field (and consequently opposing changes in
current) is called inductance.
The unit of inductance is the Henry – H
(Denoted as the letter L on a schematic)
• Random Video of the Day
Another RVOTD
Inductor Schematic Symbol
Typically, the range of inductors are from micro-Henry’s to milli-Henry’s
(H to mH)
• In the picture of the inductor below, since
current takes the path of least resistance,
why wouldn’t the current just bypass all
the turning and jump straight across the
tops of all the coils?
Because there is a clear enamel coating on the
wires that acts like an insulator
18.3 Inductance
Factors Affecting Coil Inductance
Factors Affecting Coil Inductance
N – Number of turns
• The design of the inductor is
most important in determining
the value of inductance.
• 1. The amount of turns in
creating the coil affect how
much voltage can be linked
between the turns. In a later
equation, this figure is denoted
by (N).
Coil Inductance Factors (cont.)
A – Area of coil
• 2. The diameter of the coil
affects inductance. A larger
diameter core results in more
magnetic lines of force than
compared to a smaller
diameter coil. Denoted by (A)
for area=πr2.
r
Coil Inductance Factors (cont.)
l – length of coil
• 3. As the length (l) of
the inductor grows,
the distance between
turns increases,
which causes the
magnetic field to get
weaker.
• Denoted by l, for
length in meters,
• not I for current!
l
Coil Inductance Factors (cont.)
µ - permeability of core
• 4. And finally, inductance is
affected by the core material in
which the coil is wrapped
around. Higher permeability
enables more flux to form.
More flux means more
inductance. This is denoted by
(µ) pronounced “mju”, not
known as micro in this case.
Table 16.2 from book
Material
Permeability, µ
(H/m)
Relative Permeability, µr (No
Units)
Air or Vacuum
1.26 x 10-6
1
Nickel
6.28 x 10-5
50
Cobalt
7.56 x 10-5
60
Cast Iron
1.1 x 10-4
90
Machine Steel
5.65 x 10-4
450
Transformer iron
core
6.9 x 10-3
5500
Silicon Iron
8.8 x 10-3
7000
Permalloy
0.126
100,000
Supermalloy
1.26
1,000,000
L=
Inductance equation
18.3.1
2
µN A
l
L= Inductance in Henrys, H
µ=permeability of core, µ0µr
µ0= 1.26 x 10-6 (The absolute
permeability of air)
µr= relative permeability of core
material, refer to Table 16.2 
N= number of coil turns
A= area of cross-section of coil
wire, πr2.
l= length of core material (m)
Table 16.2 from book
Material
Permeability, µ
(H/m)
Relative Permeability, µr
(No Units)
Air or Vacuum
1.26 x 10-6
1
Nickel
6.28 x 10-5
50
Cobalt
7.56 x 10-5
60
Cast Iron
1.1 x 10-4
90
Machine Steel
5.65 x 10-4
450
Transformer iron
core
6.9 x 10-3
5500
Silicon Iron
8.8 x 10-3
7000
Permalloy
0.126
100,000
Supermalloy
1.26
1,000,000
Inductance problem worked out
• Given: l=10cm r=1.5cm N=200 turns
l= .1m r=.015m N=200 turns
It’s an air core that we’re dealing with. So relative permeability is µr =1.
L= µN2A
l
L= (µ) (200)2 [π(.0152)]
.1
L= [(1.26 x 10-6)] (40000) (.000707)
.1
L= .000035632
.1
L= .00035632 H
or
356.3 µH
Another Inductance problem!
• Given:
l=15cm
r=.02m
N=500 turns
Machine Steel Core
• Find Inductance Value
Another Inductance problem!
• Given:
l = 15cm
r = .02m
L = 3mH
Nickel Core
• Find the number of turns
Another Inductance problem!
• What is the inductance if you take a wire and wrap it around a
pencil 15 times?
Types of Inductors
Air Core
Iron-Core
Types of Inductors
Low value L
High frequency applications
Large value L
Used in transformers
Mid value L
Used to reduce losses such
as eddy currents. Used for
higher DC current
applications
Ferrite Core
Low to High value L
Great Magnetic Conductor,
Low Electrical Conductor.
Suitable for high and low freq
applications
Toroidal core
Mid to High value L
Very low flux leakage loss
because of donut shape.
Variable Core
Variable L
Used for tuning circuits like
radios
Very low L
Coil printed Directly on circuit
board. For Freq above
500Mhz (Very high
Powdered iron-core
Printed Circuit board core
PCB Core Inductors
Measuring the induced voltage
across an inductor
• Equation 18.1:
This equation says that the higher the change in current, and the faster
that change occurs, the higher the voltage across the inductor will be.
• How much self-induced voltage occurs
across a 4H inductor when the current
going through it changes by 10Amps in 1
second?
• How much self-induced voltage occurs
across a 4H inductor when the current
going through it changes by 10Amps in
1ms?
1ms)
= 400V
Calculating total inductance LT
• What is the total inductance of this circuit?
62H
Calculating the total inductance in a circuit is done
the same way as calculating the total resistance
Calculating total inductance LT
• What is the total inductance of this circuit?
6H
Calculating the total inductance in a circuit is done
the same way as calculating the total resistance
Calculating total inductance LT
• What is the total inductance of this circuit?
LT = 15H
Calculating the total inductance in a circuit is done
the same way as calculating the total resistance
However! There is a subtle difference
in calculating LT that you don’t have to
consider when calculating RT
• What if the magnetic field from one
inductor interferes with the magnetic field
from the inductor next to it?
• In other words, if the magnetic field from
one inductor cuts the across the coils of
the inductor located physically next to it,
then extra inductance will be introduced,
(or some inductance will be cancelled out.)
THIS IS CALLED MUTUAL INDUCTANCE
MUTUAL INDUCTANCE
Mutual Inductance
Calculating total inductance if there
is mutual inductance involved
• In the circuit above, if the inductor’s
magnetic fields didn’t interfere with each
other, aka have no mutual inductance,
then the total inductance is just 13H.
• However, suppose the inductors magnetic
fields were completely overlapping each
other, then we would have to consider this.
• k is the coefficient of coupling and it is a
number ranging from 0% to 100%
• If the coils of 2 inductors were wrapped
around each other, then the magnetic
fields of each have no choice but to couple
themselves to the other inductor.
• In this case the coefficient of coupling, k, is
100%.
• If the inductors’ magnetic fields are not
coupled, then k = 0.
MUTUAL INDUCTANCE
k = .9
k = .3
k=0
Calculating total inductance if there
is mutual inductance involved
• Suppose the inductors were connected in
a series aiding arrangement with a
coefficient of coupling of 90%.
Calculating total inductance if there
is mutual inductance involved
• Suppose the inductors were connected in
a series opposing arrangement with a
coefficient of coupling of 90%.
2 coils have inductances of 8mH and
4.7mH. If the coefficient of coupling between
them is .82, what is the mutual inductance?
Now find the total inductance if they are series aiding and if they
are series opposing
LT = 8mH + 4.7 mH + 2·5.03mH = 22.76mH (If in series aiding arrangement)
LT = 8mH + 4.7 mH - 2·5.03mH = 2.64mH (If in series opposing arrangement)
Stray Inductance
• Talk about stray inductance
Go over Cap/Inductor Handout

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