Download An inductance is a short circuit to Steady State DC.

ECE 3301
General Electrical Engineering
Section 23
Inductance
1
2
Inductance
• An electrical current in any conductor
causes a magnetic field to exist around the
conductor.
• The magnetic field forms a closed loop
around the conductor as illustrated below.
H
I
H
3
Inductance
• The magnetic field obeys the right-hand
rule.
• If the current is in the direction of the thumb
of the right hand, the magnetic field is
pointed in the direction the fingers form
around the conductor.
H
I
H
4
Inductance
• If the conductor is formed into a coil, the
magnetic field is reinforced by adjacent
conductors.
I
H
I
5
Inductance
• The magnetic field is said to “link” the turns
of the coil.
• Energy is stored in the magnetic field.
I
H
I
6
Inductance
• The effect of the magnetic field is to
maintain the current in the conductor.
• This effect is called inductance.
I
H
I
7
Inductance
• An inductance in a circuit is represented by
the symbol shown below.
i

v
L

8
Inductance
• An inductance in a circuit is represented by
the symbol shown below.
• Inductance is measured in henries (H).
i

v
L

9
Inductance
• The current-voltage relationship for
inductance is given by the equation
di
v = L dt
i

v
L

10
Inductance
• The voltage across an inductance is
proportional to the time-rate-of-change of
the current through the inductance.
• The constant of proportionality is called the
inductance.
i

v
L

di
v = L dt
11
i(t)

v(t) L

di
v = L dt
Current (A)
The current
source
Drives a timevarying current
through the inductor
1
0
-1
0
1
2
Time (S)
3
4
5
12

v(t) L
i(t)

Current (A)
The voltage
across the
inductance
di
v = L dt
Is proportional to the
time rate-of-change
of the current
1
0
-1
0
1
2
Time (S)
3
4
5
13
i(t)

v(t) L

Current (A)
The voltage
across the
inductance
di
v = L dt
Is proportional to the
slope of this
waveform
1
0
-1
0
1
2
Time (S)
3
4
5
14
i(t)

v(t) L
di
v = L dt
Current (A)

Slope = 0
1
0
-1
0
1
2
Time (S)
3
4
5
2
3
4
5
Voltage = 0
Voltage (V)
L
-1
0
1
-L
Time (S)
15
i(t)

v(t) L
di
v = L dt
Current (A)

1
Slope = 1
0
-1
0
1
2
Time (S)
3
2
3
4
5
4
5
Voltage = L(1)
Voltage (V)
L
-1
0
1
-L
Time (S)
16
i(t)

v(t) L
di
v = L dt
Current (A)

1
Slope = 0
0
-1
0
1
2
Time (S)
3
4
5
4
5
Voltage = 0
Voltage (V)
L
-1
0
1
2
3
-L
Time (S)
17
i(t)

v(t) L
di
v = L dt
Current (A)

1
Slope = 1
0
-1
0
Voltage (V)
L
-1
1
2
Time (S)
3
4
5
3
4
5
Voltage = L(1)
0
1
2
-L
Time (S)
18
i(t)

v(t) L
di
v = L dt
Current (A)

1
Slope = 0
0
-1
0
1
2
Time (S)
Voltage (V)
L
-1
3
4
5
4
5
Voltage = 0
0
1
2
3
-L
Time (S)
19
i(t)

v(t) L
di
v = L dt

The current
source
Drives a time-varying,
sinusoidal current
through the inductor
i(t) = 1 sin(2 t)
1.5
Current (A)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
20

v(t) L
i(t)
di
v = L dt

Period = T0
1.5
Current (A)
1
0.5
Amplitude
-0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
i(t) = A sin(2 f t)
Frequency = f =
1
T0
21

v(t) L
i(t)
di
v = L dt

The voltage
across the
inductance
Is proportional to the
time rate-of-change
of the current
i(t) = 1 sin(2 t)
1.5
Current (A)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
22
i(t)

v(t) L
di
v = L dt

i(t) = 1 sin(2 t)
d
v(t) = L sin(2 t) = 2 L cos(2 t)
dt
1.5
Current (A)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
23

v(t) L
i(t)
di
v = L dt

i(t) = 1 sin(2 t)
1.5
Current (A)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
Time (S)
-1.5
v(t) = 2 L cos(2 t)
Voltage (V)
2 L
-0.5
0
0.5
1
1.5
2
2.5
-2 L
Time (S)
24

v(t) L
i(t)
di
v = L dt

Current (A) and
Voltage (V)
 = 90 Deg.
-0.5
i
0
v
0.5
1
1.5
2
2.5
Time (S)
The current and voltage are 90 degrees “out of phase”
with each other.
The voltage leads the current by 90 degrees.
25
i(t)

v(t) L

di
v = L dt
Current (A)
The current
source
Drives a timevarying current
through the inductor
1
0.5
0
-1
0
1
2
Time (S)
3
4
5
26
i(t)

v(t) L

Current (A)
The voltage
across the
inductance
di
v = L dt
Is proportional to the
slope of this
waveform
1
0.5
0
-1
0
1
2
Time (S)
3
4
5
27
i(t)

v(t) L
di
v = L dt
Current (A)

The instantaneous
change in current
1
0.5
0
Voltage (V)
-1
-1
0
1
2
Time (S)
3
0
1
2
3
Time (S)
4
5
4
5
Causes an infinite
voltage pulse across
the inductor
28
i(t)

v(t) L
di
v = L dt
Current (A)

1
0.5
0
Voltage (V)
-1
-1
0
1
2
Time (S)
3
0
1
2
3
Time (S)
An infinite pulse
4
5
cannot
be achieved
!
4
5
29
Rule 1 of Inductance
• Since this infinite voltage impulse cannot be
physically realized, we conclude the first
rule-of-thumb about inductance.
• One cannot instantaneously change the
current through an inductance.
30
Rule 2 of Inductance
• Since the voltage across the inductance is
proportional to the time rate-of-change of the current
through the inductance, when the current is constant
(DC), the voltage is zero.
• A voltage of zero across a circuit element is the
definition of a short circuit. This leads to the second
rule-of-thumb about inductances.
• An inductance is a short circuit to Steady State
DC.
31
Inductance
i(t)

v(t)

i0
L
• Consider an inductance driven by a voltage
source.
• The inductance has an initial current of i0
amps.
32
Inductance
i(t)

v(t)

i0
L
• The voltage-current relationship is given by:
di
v=L
dt
33
Inductance
i(t)

v(t)

i0
L
• Performing a bit of calculus:
di
v=L
dt
v dt = L di
1
di = v dt
L
34
Inductance
i(t)

v(t)

L
i0
• Integrating both sides:
i(t)
t
i(t0)
t0
1
 dy =  v d

L

t
 i(t)
1
y 
= 
v d

L
i(t0)
t0
35
Inductance
i(t)

v(t)

L
i0
• Completing the integration:
t
1
i(t) – i(t0) = 
v d
L
t0
36
Inductance
i(t)

v(t)

i0
L
• Solving for i(t):
t
1
i(t) = 
v d + i(t0)
L
t0
37
Inductance
i(t)

v(t)

i0
L
• The current through an inductance is
proportional to the integral of the voltage
across the inductance plus the initial current
through the inductance.
t
1
i(t) = 
v d + i(t0)
L
t0
38
i(t)

v(t)

t
i0
L
1
i(t) = 
v d + i(t0)
L
t0
Voltage (V)
The voltage
source
Places a timevarying voltage
across the
inductance
1
0
-1
0
1
2
Time (S)
3
4
5
39
i(t)

v(t)

Voltage (V)
The current
through the
inductance
t
i0
L
1
i(t) = 
v d + i(t0)
L
t0
Is proportional to the
area under the
voltage waveform
1
0
-1
0
1
2
Time (S)
3
4
5
40
i(t)
t
1
i(t) = 
v d + i(t0)
L
L
i0
t0
Voltage (V)

v(t)

Accumulated
area
0
Current (A)
-1
Initial
current
-1
Final
current
1
0
1
2
Time (S)
3
4
5
2
Time (S)
3
4
5
I 0 + 2/L
I0
0
1
41
i(t)
t

v(t)
1
i(t) = 
v d + i(t0)
L
L

t0
The voltage
source
Places a sinusoidal
voltage across the
inductance
v(t) = sin(2 t)
1.5
Voltage (V)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
42
i(t)
t

v(t)
1
i(t) = 
v d + i(t0)
L
L

t0
The current
Is proportional to the
integral of the voltage
v(t) = sin(2 t)
1.5
Voltage (V)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
43
i(t)
t

v(t)
1
i(t) = 
v d + i(t0)
L
L

t0
t
1
i(t) = 
sin(2 ) d
L
t0
i(t) = –
1
2 L
cos(2t)
v(t) = sin(2 t)
1.5
Voltage (V)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
-1
-1.5
Time (S)
44
i(t)
t

v(t)
1
i(t) = 
v d + i(t0)
L
L

t0
v(t) = sin(2 t)
1.5
Voltage (V)
1
-0.5
0.5
0
-0.5 0
0.5
1
1.5
2
2.5
1
cos(2 t)
2
2.5
-1
Current (A)
-1.5
Time (S)
i(t) = –
1/2
L
-0.5
0
0.5
1
1.5
2 L
-1/2 L
Time (S)
45
i(t)
t

v(t)
1
i(t) = 
v d + i(t0)
L
L

t0
 = 90 Deg.
i
Voltage (V) and
Current (A)
v
-0.5
0
0.5
1
1.5
2
2.5
Time (S)
The current and voltage are 90 degrees “out of phase”
with each other.
The voltage leads the current by 90 degrees.
46
Power and Energy in an
Inductance
• The instantaneous power absorbed by any
circuit element is given by:
p = vi
• Using the voltage-current relationship for an
inductance:
di
v=L
dt
47
Power and Energy in an
Inductance
• Leads to the instantaneous power in an
inductance:
di
p=Li
dt
• By definition, power is given by:
dw
p=
dt
48
Power and Energy in an
Inductance
• Consequently :
dw
di
=Li
dt
dt
• and :
dw = L i di
49
Power and Energy in an
Inductance
• The energy stored in an inductance may be found
by integration.
• Assume an initial energy of zero and an initial
current of zero.
w
i
 dx = L  y dy




w

x 
1 2
= L y
2
0



i
0
50
Power and Energy in an
Inductance
• The energy stored in an inductance is:
1 2
w–0= Li –0
2
1 2
w= Li
2
51

v(t) L
i(t)
di
v = L dt

Current (A)
IM
-1
0
1
2
3
4
5
6
7
8
9
5
6
7
8
9
-I M
Time (S)
Voltage (V)
LI M
-1
0
1
2
3
4
-L I M
Time (S)
52

v(t) L
i(t)
p = vi

Voltage
(V)
Current (A)
LII MM
-1
0
1
2
3
4
5
6
7
8
9
5
6
7
8
9
-L-IIMM
Time
Time (S)
(S)
Power (W)
LI M2
-1
0
1
2
3
4
-L I M2
Time (S)
53

v(t) L
i(t)
1
2
w= Li
2

Current (A)
IM
-1
0
1
2
3
4
5
6
7
8
9
5
6
7
8
9
-I M
Time (S)
Energy (J)
LI M2/2
-1
0
1
2
3
4
Time (S)
54
Inductors in Series
a

vab = v1 + v2 + ··· + vk

v1
L1
di
di
di
vab = L1 + L2 + ··· + Lk
dt
dt
dt
L2
di
vab = (L1 + L2 + ··· + Lk )
dt


i
v2
vab

··· ··· ···

vk

b

di
vab = Leq
dt
Lk
Leq = L1 + L2 + ··· + Lk
55
Inductors in Parallel
Node
i
a

v

i1
···
i2 ···
ik
L1
L2 ···
Lk
···
b
t
1
 v dt + im(t0)
im(t) =
Lm 
t0
i = i1 + i2 + ··· + ik
56
Inductors in Parallel
Node
i
a

v

b
i1
···
i2 ···
ik
L1
L2 ···
Lk
···
t
t
t
t0
t0
t0
1
1
1


i =  v dt + i1(t0) +  v dt + i2(t0) + ··· + 
v dt + ik(t0)
L1
L2
Lk 
57
Inductors in Parallel
Node
i
a

v

b
i1
···
i2 ···
ik
L1
L2 ···
Lk
···
t
1 
1 1
i =  + + ··· +  
v dt + i1(t0) + i2(t0) + ··· + ik(t0)

L
L
L
2
k
 1
t0
58
Inductors in Parallel
Node
i
a

v

i1
···
i2 ···
ik
L1
L2 ···
Lk
b
···
1
1
1
1
= + + ··· +
Leq L1 L2
Lk
i(t0) = i1(t0) + i2(t0) + ··· + ik(t0)
59