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Inductor
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
The Basic Inductor
When a length of wire is formed into a coil., it
becomes a basic inductor. When there is current in
the inductor, a three-dimensional magnetic field is
created.
A change in current
causes the magnetic
S
N
field to change. This in
turn induces a voltage
across the inductor that
opposes the original
change in current.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
The Basic Inductor
One henry is the inductance of a coil when a current,
changing at a rate of one ampere per second, induces one
volt across the coil. Most coils are much smaller than 1 H.
The effect of inductance is greatly
magnified by adding turns and winding
them on a magnetic material. Large
inductors and transformers are wound
on a core to increase the inductance.
Magnetic core
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Faraday’s law
Faraday’s law was introduced in Chapter 7 and repeated
here because of its importance to inductors.
The amount of voltage induced in a coil is directly
proportional to the rate of change of the magnetic field
with respect to the coil.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Lenz’s law
Lenz’s law was also introduced in Chapter 7 and is an
extension of Faraday’s law, defining the direction of the
induced voltage:
When the current through a coil changes and an
induced voltage is created as a result of the changing
magnetic field, the direction of the induced voltage is
such that it always opposes the change in the current.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Lenz’s law
A basic circuit to demonstrate Lenz’s law is shown.
Initially, the SW is open and there is a small
current in the circuit through L and R1.
L
VS
SW
+
R1
R2


Electronics Fundamentals 8th edition
Floyd/Buchla
+
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Lenz’s law
SW closes and immediately a voltage appears
across L that tends to oppose any change in current.
L

+
VS
+
SW
R1
R2


Electronics Fundamentals 8th edition
Floyd/Buchla
+
Initially, the meter
reads same current
as before the switch
was closed.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Lenz’s law
After a time, the current stabilizes at a higher level
(due to I2) as the voltage decays across the coil.
L
VS
SW
+
R1


R2
+
Later, the meter
reads a higher
current because of
the load change.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Practical inductors
In addition to inductance, actual inductors have
winding resistance (RW) due to the resistance of the
wire and winding capacitance (CW) between turns.
An equivalent circuit for a practical inductor
CW
including these effects is shown:
Notice that the winding resistance
is in series with the coil and the
winding capacitance is in parallel
with both.
Electronics Fundamentals 8th edition
Floyd/Buchla
RW
L
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Types of inductors
There are a variety of inductors, depending on the
amount of inductance required and the application.
Some, with fine wires, are encapsulated and may
appear like a resistor.
Common symbols for inductors (coils) are
Air core
Electronics Fundamentals 8th edition
Floyd/Buchla
Iron core
Ferrite core
Variable
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Factors affecting inductance
Four factors affect the amount of inductance for a
coil. The equation for the inductance of a coil is
N 2 A
L
l
where
L = inductance in henries
N = number of turns of wire
 = permeability in H/m (same as Wb/At-m)
l = coil length on meters
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
What is the inductance of a 2 cm long, 150
turn coil wrapped on an low carbon steel core that
is 0.5 cm diameter? The permeability of low
carbon steel is 2.5 x104 H/m (Wb/At-m).
A  πr 2  π  0.0025 m   7.85 105 m 2
2
N 2 A
L
l
2
150 t   2.5 104 Wb/At-m  7.85 105 m2 

0.02 m
 22 mH
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Practical inductors
Inductors come in a variety of sizes. A few
common ones are shown here.
Encapsulated
Electronics Fundamentals 8th edition
Floyd/Buchla
Torroid coil
Variable
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Series inductors
When inductors are connected in series, the total
inductance is the sum of the individual inductors.
The general equation for inductors in series is
LT  L1  L2  L3  ...Ln
If a 1.5 mH inductor is
connected in series with
an 680 H inductor, the
total inductance is 2.18 mH
Electronics Fundamentals 8th edition
Floyd/Buchla
L
1
L
2
1
.
5
m
H 6
8
0

H
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Parallel inductors
When inductors are connected in parallel, the total
inductance is smaller than the smallest one. The
general equation for inductors in parallel is
LT 
1
1 1 1
1
   ... 
L1 L2 L3
LT
The total inductance of two inductors is
LT 
1
1 1

L1 L2
…or you can use the product-over-sum rule.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Parallel inductors
If a 1.5 mH inductor is connected in
parallel with an 680 H inductor,
the total inductance is 468 H
L1
1.5m
H
Electronics Fundamentals 8th edition
Floyd/Buchla
L2
680
H
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Inductors in dc circuits
When an inductor is connected
in series with a resistor and dc
source, the current change is
exponential.
Vinitial
t
0
Inductor voltage after switch closure
Ifinal
R
L
0
Current after switch closure
Electronics Fundamentals 8th edition
Floyd/Buchla
t
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Inductors in dc circuits
The same shape curves are
seen if a square wave is
used for the source. Pulse
response is covered further
in Chapter 20.
VS
VL
R
VS
Electronics Fundamentals 8th edition
Floyd/Buchla
L
VR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Inductive reactance
Inductive reactance is the opposition to
ac by an inductor. The equation for
inductive reactance is
X L  2πfL
The reactance of a 33 H inductor when a
frequency of 550 kHz is applied is 114 W
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Inductive reactance
When inductors are in series, the total reactance is the sum of the
individual reactances. That is,
X L(tot )  X L1  X L2  X L3    X Ln
Assume three 220 H inductors are in series with a 455 kHz
ac source. What is the total reactance?
The reactance of each inductor is
X L  2 fL  2  455 kHz  220 μH   629 W
X L(tot )  X L1  X L2  X L3
 629 W  629 W  629 W  1.89 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Inductive reactance
When inductors are in parallel, the total reactance is the reciprocal of
the sum of the reciprocals of the individual reactances. That is,
X L(tot ) 
1
1
1
1
1


  
X L1 X L2 X L3
X Ln
If the three 220 H inductors from the last example are placed
in parallel with the 455 kHz ac source, what is the total
reactance?
The reactance of each inductor is 629 W
X L(tot ) 
Electronics Fundamentals 8th edition
Floyd/Buchla
1
1

 210 W
1
1
1
1
1
1


+
+
X L1 X L2 X L3 629 W 629 W 629 W
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Inductive phase shift
When a sine wave
is applied to an
inductor, there is a
phase shift between
voltage and current
such that voltage
always leads the
current by 90o.
Electronics Fundamentals 8th edition
Floyd/Buchla
VL 0
90
I 0
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Sinusoidal response of RL circuits
When both resistance and inductance are in a series
circuit, the phase angle between the applied voltage and
total current is between 0 and 90, depending on the
values of resistance and reactance.
VL
VR
V R lags V S
VL lead s VS
R
L
VS
I
I lags V S
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Impedance of series RL circuits
In a series RL circuit, the total impedance is the phasor
sum of R and XL.
R is plotted along the positive x-axis.
XL is plotted along the positive y-axis.
 XL 

 R 
  tan 1 
Z
Z is the diagonal
Z
XL
XL

R

R
It is convenient to reposition the
phasors into the impedance triangle.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Impedance of series RL circuits
Sketch the impedance triangle and show the
values for R = 1.2 kW and XL = 960 W.
Z
1.2 kW 
2
+  0.96 kW 
 1.33 kW
  tan 1
0.96 kW
1.2 kW
 39
Electronics Fundamentals 8th edition
Floyd/Buchla
2
Z = 1.33 kW

39o
XL =
960 W
R = 1.2 kW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Analysis of series RL circuits
Ohm’s law is applied to series RL circuits using
quantities of Z, V, and I.
V  IZ
V
I
Z
V
Z
I
Because I is the same everywhere in a series circuit,
you can obtain the voltage phasors by simply
multiplying the impedance phasors by the current.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Analysis of series RL circuits
Assume the current in the previous example is 10 mArms.
Sketch the voltage phasors. The impedance triangle from
the previous example is shown for reference.
The voltage phasors can be found from Ohm’s
law. Multiply each impedance phasor by 10 mA.
Z = 1.33 kW

39o
R = 1.2 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XL =
960 W
VS = 13.3 V

39o
VL =
9.6 V
VR = 12 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Variation of phase angle with frequency
Phasor diagrams that have reactance phasors can only
be drawn for a single frequency because X is a
function of frequency.
Increasing f
As frequency changes,
X
Z
the impedance triangle
for an RL circuit changes
Z
X
as illustrated here
because XL increases with
Z
X
increasing f. This



determines the frequency
R
response of RL circuits.
3
2
1
1
Electronics Fundamentals 8th edition
Floyd/Buchla
2
L3
L2
L1
3
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Phase shift
For a given frequency, a series RL circuit can be used to
produce a phase lead by a specific amount between an
input voltage and an output by taking the output across
the inductor. This circuit is also a basic high-pass filter, a
circuit that passes high frequencies and rejects all others.
R
Vout
Vin
Vout
f
Vin
L
Vout
(phase lead)
Vin
f

Electronics Fundamentals 8th edition
Floyd/Buchla
VR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Phase shift
Reversing the components in the previous circuit
produces a circuit that is a basic lag network. This circuit
is also a basic low-pass filter, a circuit that passes low
frequencies and rejects all others.
L
Vin
VL
R
Vin
Vin
Vout
f (phase lag)
Vout
f
Vout
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Sinusoidal response of parallel RL circuits
For parallel circuits, it is useful to review conductance,
susceptance and admittance, introduced in Chapter 10.
Conductance is the reciprocal of resistance.
Inductive susceptance is the reciprocal
of inductive reactance.
G
1
R
1
BL 
XL
1
Admittance is the reciprocal of impedance. Y 
Z
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Sinusoidal response of parallel RL circuits
In a parallel RL circuit, the admittance phasor is the sum
of the conductance and inductive susceptance phasors.
The magnitude of the susceptance is Y  G2 + BL 2
The magnitude of the phase angle is   tan 1 
G
VS
G
BL
BL
Electronics Fundamentals 8th edition
Floyd/Buchla
BL 

G 
Y
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Sinusoidal response of parallel RL circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BL is plotted along the negative y-axis.
 BL 

G 
  tan 1 
Y is the diagonal
G
VS
G
BL
BL
Electronics Fundamentals 8th edition
Floyd/Buchla
Y
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Sinusoidal response of parallel RL circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
1
1
1
B

 0.629 mS
G 
 1.0 mS
L
2 10 kHz  25.3 mH 
R 1.0 kW
Y  G 2 + BL 2 
1.0 mS +  0.629 mS  1.18 mS
2
2
G = 1.0 mS
VS
f = 10 kHz
Electronics Fundamentals 8th edition
Floyd/Buchla
R
1.0 kW
L
25.3 mH
BL =
0.629 mS
Y=
1.18 mS
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Analysis of parallel RL circuits
Ohm’s law is applied to parallel RL circuits using
quantities of Y, V, and I.
I
Y
V
I
V
Y
I  VY
Because V is the same across all components in a
parallel circuit, you can obtain the current in a given
component by simply multiplying the admittance of
the component by the voltage as illustrated in the
following example.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Analysis of parallel RL circuits
Assume the voltage in the previous example is 10 V.
Sketch the current phasors. The admittance diagram
from the previous example is shown for reference.
The current phasors can be found from Ohm’s
law. Multiply each admittance phasor by 10 V.
G = 1.0 mS
BL =
0.629 mS
Y=
1.18 mS
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 V
=
IL =
6.29 mA
IR = 10 mA
IS =
11.8 mA
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Phase angle of parallel RL circuits
Notice that the formula for inductive susceptance is
the reciprocal of inductive reactance. Thus BL and IL
are inversely proportional to f: BL  1
2 fL
As frequency increases, BL
and IL decrease, so the angle
between IR and IS must
decrease as well.

IL
Electronics Fundamentals 8th edition
Floyd/Buchla
IR
IS
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 11
1
Series-Parallel RL circuits
Series-parallel RL circuits are combinations of both series and
parallel elements. The solution of these circuits is similar to
resistive combinational circuits but you need to combine reactive
elements using phasors.
The components in the
R2
R1
yellow box are in series and
Z1
Z2
those in the green box are
L1
L2
also in series.
Z1  R12  X L21
and
Z 2  R22  X L22
Electronics Fundamentals 8th edition
Floyd/Buchla
The two boxes are in parallel. You can
find the branch currents by applying
Ohm’s law to the source voltage and
the branch impedance.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.