Module 5
Inductance
5.1 Magnetic Fields
Remember back to the magnetic field surrounding a bar magnet. The field lines go from
the North to South pole.
Unlike electric field lines the magnetic field lines are in fact loops that never originate or
terminate.
N
S
Figure 5.1
A magnetic field may also be produced by passing a current through a conductor. In this
case the field line will encircle the conductor.
Magnetic Flux
lines
I
Figure 5.2
The conductor may be wound into a coil which will concentrate the flux lines through the
centre of the coil.
I
Figure 5.3
5.2 Flux Density
The number of flux lines per unit area determines the flux density (B). This has the unit of
Teslas.
B=
F
A
where
(Teslas)
F = number of flux lines (Webers)
A = Area (m2)
1 Tesla = 1 Weber/metre
(1)
5.3 Permeability
Q. What happens if we place a material other than air inside a coil?
A. It depends if the material is magnetic or not.
Magnetic materials are materials in which magnetic lines may be easily set up.
This is described as the permeability of the material.
The permeability of a vacuum is
m0 = 4p x 10-7
Wb/A.m
The permeability of other materials are usually referred to a vacuum. This gives materials a
relative permeability.
mr =
m
mo
(2)
Materials fall into the following categories
Relative Permeability
Name
Example
m<1
diamagnetic
nitrogen
1 < m < 100
paramagnetic
copper, aluminium
m > 100
ferromagnetic
iron ,nickel, steel
The placement of a magnetic material inside a coil will allow more flux lines to be created.
The material placed inside a coil is referred to as the ‘core’.
So far we have simply seen the static field case, ie. a steady magnetic field produced by a
constant current.
What happens if we vary the magnetic field?
F = NI
f = NImA
B=
f
= NIm
A
(At)
5.4 Induction
Changing Flux
Figure 5.4
If a conductor is placed in a varying magnetic field so that the changing flux lines cross the
conductor then a voltage is produced in the conductor. If the conductor is coiled more of
the conductor is cut by the flux lines and the voltage will be increased.
u =N
df
dt
where
This is Faraday’s law.
(V)
u = induced voltage
N = number of turns
f = magnetic flux
(3)
In figure 5.3 we saw that passing a current through a coil caused a magnetic field to be
produced.
Q. What happens if we use an alternating current to drive the coil?
A. An alternating magnetic field will be produced.
Q. Will this alternating (varying) magnetic field induce a voltage into the coil as described
by Faraday’s law.
A. Yes it will.
This is Lenz’s law
An induced effect is always such as to oppose the cause that produced it.
5.5 Self Inductance
The ability to oppose a change in current flow is called self inductance. When dealing with
single coils it is often just called inductance.
The inductance of a coil depends on many physical parameters related to its construction.
An approximation may be used
N 2 mA
L=
l
where
(Henries)
N = number of turns
m = magnetic permeability
A = cross sectional area of core
l = length of core
A
m
l
Figure 5.5
(4)
Self Learning: Read about types of inductors, Boylestad 12.5.
The inductance of a coil can also be calculated from the rate of change of flux with current.
L =N
df
di
(H)
(5)
We can see from this that a large inductance will produce a large flux change when the
current varies.
We can rewrite equation (3) as
u =N
df Ê df ˆ Ê di ˆ
= ÁN ˜Á ˜
dt Ë di ¯ Ë dt ¯
(6)
Substituting in equation (5) we see that the voltage across the inductor produced by a
particular change in current will be
uL = L
di
dt
(7)
It is important to note that it is not a large magnitude of current that produces a high
output voltage. It is a rapid change in current which produces a large voltage.
Example 5.1
A 10 mH coil has a signal with the current shown below passed through it. What will be
the voltage waveform across the coil?
iL
8
4
0
0
Solution
VL = L
Di
Dt
2
4
6
milliseconds
8
di
Di
=L
dt
Dt
10 A
= 2000 A / s
for 2 - 7 ms period =
5ms
uL = 10 x 10-3 H . 2000 A/s = 20 v
10
Di
Dt
-10 A
= -10, 000 A / s
for 7 - 8 ms period =
1 ms
uL = 10 x 10-3 H . -10,000 A/s = -100 v
vL
0
-40
-80
0
2
4
6
milliseconds
8
10
5.6 Symbols
Air
core
Iron
core
Ferrite
core
Adjustable
Alternate
symbol
Figure 5.6
Inductors are often very specialised and are constructed for a particular circuit. Some
common symbols are shown.
5.7 Series and Parallel Inductors
Vb
L1
L2
L3
Figure 5.7
Q. What happens if we place inductors in series?
LN
A. This is fairly obvious. Two similar coils in series is just like one coil twice as long,
therefore
(8)
L T = L 1 + L 2 + L 3 ... + L N
This is the same as for resistors.
The parallel combination of inductors is also the same as for resistors.
Vb
L1
L2
L3
LN
Figure 5.8
VT V1 V2
V
=
+
+ ... N
L T L1 L2
LN
1
LT =
1
1
1
1
+
+
... +
L1 L 2 L 3
LN
(9)
5.8 Step Response
Let us repeat the step response analysis that we did for capacitors now with inductors.
R
iL
Vb
L
nL
Figure 5.9
An inductor is placed in series with a resistor and connected to a DC voltage source.
• prior to connection to the source there is no current flow through the inductor
• there is therefore no flux lines around the inductor
R
iL
Vb
L
nL
• as current begins to flow a magnetic field will build
• this field is changing therefore it is producing a voltage across the coil
• this induced voltage opposes the supplied voltage therefore limiting the current flow
• the current slowly builds until it reaches the maximum allowed due to the resistor
(Vb/R)
The faster you try to change the current the more the inductor opposes it. (The fastest
change possible is a step function)
We can evaluate this mathematically
i L = I m (1 - e -t / t ) =
Vb
(1- e - t /( L / R) )
R
(10)
1.0
Storage
0.8
iL
0.6
0.4
0.2
0.0
0
1
2
3
4
time constants
Figure 5.10
Note that the time constant for this circuit is
t=
L
R
(11)
5
The voltage across the inductor will start at the supply voltage. Because it is opposing the
current it looks like an open circuit and thus has the full supply voltage across it.
The voltage slowly decreases to zero as the current finally builds up and reaches a steady
state.
VL = Vb e -t / t
(12)
1.0
Storage
0.8
n
L
0.6
0.4
0.2
0.0
0
1
2
3
time constants
Figure 5.11
4
5
I want to see it with water!!
If the supply is disconnected and the circuit shorted out as shown below the following will
occur.
R
iL
Vb
L
nL
Figure 5.12
• prior to disconnecting the supply a steady magnetic field existed around the inductor
• when the current stops this field collapses
• a collapsing field is a changing field so a voltage is induced across the inductor
• the voltage may be large due to the rapid change in current and will be negative
because it is a reduction in current (di/dt)
u L = -Vi e -t /t
where
(13)
Vi = Amplitude of reverse voltage
0.0
Decay
-0.2
n
L
-0.4
-0.6
-0.8
-1.0
0
1
2
3
time constants
Figure 5.13
4
5
• this voltage causes a decaying current to continue to flow in the same direction
i L = I m e -t / t =
Vb -t / t
e
R
(14)
1.0
Decay
0.8
iL
0.6
0.4
0.2
0.0
0
1
2
3
4
5
time constants
Figure 5.14
• the inductor tries to keep the same current flowing
The inductor will try very hard to maintain the same current flowing. If the inductor is
open circuited, as happens in figure 5.12 when the switch is changing over, then the
voltage will rise to a very high value.
Note: This high 'back EMF' can destroy electronic circuits.
Here comes the water again!!!
Example 5.2
Lets use a circuit that will not open circuit the inductor thus causing very high voltages.
2 kW
iL
50 V
3 kW
4H
Draw a graph of voltage on the inductor and current through the inductor if the switch is
closed at time equals zero and opened at time equals 10 ms.
Solution
Voltage connected
Current flows through L and R1
Current through 3 kW - not interested
t for 2 kW and 4 H = 4/2000 = 2 ms
In 10 ms the inductor will store for 5 t's
i L = Im (1- e -t / t) = Vb (1- e-t /(L / R) ) = 50 (1- e -t / 0.002 )
R
2000
VL = Vbe-t / t = 50e-t / 0.002
Voltage disconnected
Current now flows through L, R1 and R2
Voltage across L reverses to supply this current
Current prior to switching is Im
Voltage required to maintain current is
Vi = I m (R 1 + R 2 ) =
50
(2k + 3k ) = 125 volts
2k
t for 2 kW + 3 kW and 4 H = 4/5000 = 0.8 ms
u L = - Vi e -t / t = -125 e -t / 0.0008
50 -t / 0.0008
i L = I m e -t / t =
e
2000
20
0
i mA
L
v volts
L
40
-40
-80
-120
10
0
0
5
10
ms
15
20
0
5
10
ms
15
20
5.9 Inductors and AC Signals
The inductor is a device that produces a magnetic field. We have seen that the magnetic
field is dependant on the current through the inductor and the voltage produced in an
inductor is dependant on how quickly the current is changed. This is a derivative
uL = L
di
dt
(15)
Take the integral of both sides and multiply by L and we have an expression for the
current.
iL =
1
Ú u L dt
L
(16)
Let us apply a sinusoidal current to an RL circuit. the current through the resistor and
inductor must be the same, they are in series.
i
R
Vmsinwt
L
Figure 5.15
The voltage across the resistor will be in phase with the current through it as before. Thus
a sinusoidal current
The voltage across the inductor will be
uL = L
di
d sin wt
=L
= wL cos wt
dt
dt
(17)
The voltage across the inductor is a cos function whilst the current through it is a sin
function.
2
Inductor
Current
1
0
Voltage
-1
-2
0
50
100
150
200
degrees
250
300
Figure 5.16
There is a 90° phase shift between the current and voltage.
The current lags the voltage by 90°
or
The voltage leads the current by 90°
350
I can’t understand why this happens!!!!!!!
The maximum voltage induced into a coil occurs when the current is being changed most
rapidly. Look at a sine wave and see where the greatest change occurs. It is where the
wave crosses the zero axis. This is where the largest voltage is produced.
Q. Is there a law like Ohm’s law that applies to inductors for AC signals?
A. Yes but just like capacitors an inductor does not have resistance.
Look back to equation (17) and see that a current of sin wt produced a voltage of
wL coswt . The wL term is the effective AC resistance of the inductor. We call it the
inductive reactance.
X L = wL
(18)
This describes the magnitude of opposition to current but not the phase shift. To add this
information we make use of the j operator.
jX L = jwL
(19)
We can now work out the ‘AC resistance’ of a circuit containing resistors, capacitors and
inductors. This is called the impedance of the circuit, symbol Z. It is the sum of all the
resistances and reactances in a circuit.
Example 5.3
250W
1 volt
100 Hz
Z
5mF
200mH
Calculate the input impedance for the series circuit above.
Solution
Zt = Zr + Zc + Zl = R + (-jXc) + jXl
R = 250W
Xc =
1
1
1
=
=
wC 2 p100 Hz 5 x 10 -6 F 3.14 x10 -3
= 318.3W
XL = wL = 2 p 100 Hz 200 x 10-3 H = 125.66W
Z = R +(-jXC ) + jXL= 250 - j 318.3 + j125.66W
= 250 - j192.6W
= 315.6 W
– - 37.6°
Example 5.4
1 volt
100 Hz
Z
250W
5mF
Calculate the input impedance for the parallel circuit above.
Solution
R = 250W
XC =
1
1
=
= 318.3W
wC 2 p 100 Hz 5 x10 -6
XL = wL = 2 p 100 Hz 200 x 10-3 = 125.66W
200mH
Z = R //(- jX c ) //(jX L )
=
=
=
1
1
1
1
+
+
R -jX c jX L
=
1
1
1
1
+
+
250 -j318.3 j125.66
1
0.004 + j3.14 x10 -3 - j7.96 x10 -3
1
6.26 x10 -3
– - 50.3°
= 159.7 –50.3° = 102.1 + j122.9W
Remember this can be represented with phasor diagrams. Lets look at example 5.3. (Series
components)
X L = 125.6W
R = 250W
f = -37.6˚
|Z| = 315.6W
X C = 318.3W
Figure 5.17
Q. What about voltage and current?
A. Remember the three components are in series so each have the same current. Use this
as a reference.
i=
V
1
=
= 3.16 mA
Z 315.6
V L = 398.0mV
V R = 792 mV
i = 3.17 mA
f = -37.6˚
VC = 1V
Vsupply = 1 V
Figure 5.18
Parallel components may be represented the same way. Look at example 5.4.
The voltage across all three components is the same. Use this as a reference.
iC = 3.14 mA
iR = 4.0 mA
Vsupply = 1 V
f = -50.3˚
iL = 7.95 mA
i = 6.26 mA
Figure 5.19
5.10 Filters using inductors
R
Input
L
Output
High Pass
Figure 5.20
Because of the change in reactance of an inductor with frequency we may use them to
make filters.
The circuits in figure 5.20 and figure 5.23 are simple voltage dividers except we are using
a resistor and an inductor instead of two resistors.
The output will be a fraction of the input (as for any voltage divider) and is calculated in
the normal manner.
High Pass
R
Input
L
Output
High Pass
jX L
Vout = Vin
R + jX L
(20)
Vout
1
=
R
Vin
+1
jX L
(21)
Vout
1
=
Vin 1 - jR / wL
(22)
This is the transfer function or gain of the filter.
Q. What does this equation mean?
A. When the frequency of the input signal (w) gets low the gain gets small.
When the frequency of the input signal gets high the gain gets large.
This is a high pass filter. It only passes high frequencies.
1
0.707
gain
Stop band
wc
Pass band
w
Figure 5.21
How can the circuit do this?
Remember the inductor has a reactance X L = wL , as w (2pf) becomes larger the reactance
becomes larger. Thus the output from the voltage divider becomes closer to one.
Q. At what point do we say the filter has cutoff the signal?
A. When R = wL.
Substitute w =
R
into equation (22).
L
Vout
1
=
Vin 1 - j1
(23)
1
f = 45˚
-j1
Mag = 1.414
Figure 5.22
Vout
1 –0°
=
Vin 1.414 – - 45°
Vout
= 0.707 –45°
Vin
At this point the output signal is 0.707 of the input and it is 45° out of phase. This is the
cutoff frequency or breakpoint of the filter.
Thus the cutoff frequency occurs when
wc =
R
L
(24)
Example 5.5
Plot the gain and phase of the output signal for the high pass RL filter circuit when R =
100W and L = 15.9 mH.
Solution
At 1 Hz
Vout
1
=
Vin
1 - j100W / (2p 1 Hz 15.9x 10 -3 H)
Vout
1
=
ª 0 –90°
Vin 1 - j1x10 3
Gain = 0 Phase = 90°
At 1000 Hz
V out
1
=
Vin 1 - j100W / (2p 1000 Hz 15.9 x 10 -3 H)
Vout
1
=
= 0.707 –45°
Vin 1 - j1
Gain = 0.707 Phase = 45°
At 1 MHz
V out
1
=
Vin 1 - j100W / (2p 1MHz 15.9 x10 -3 H)
Vout
1
=
ª 1.0 –0°
-3
Vin 1 - j1x 10
Gain = 1.0
Phase = 0°
1
0.707
Gain
1Hz
1kHz
1 MHz
1Hz
1kHz
1 MHz
Freq
90˚
Phase45˚
0˚
Freq
Low Pass
L
Input
R
Output
Low Pass
Calculate the gain for this filter.
V out = Vin
Vout
=
Vin
R
R + jX L
1
1
=
jX
jwL
1+ L 1+
R
R
Q. What does this equation mean?
Figure 5.23
(25)
(26)
A. When the frequency of the input signal (w) gets low the gain gets large.
When the frequency of the input signal gets high the gain gets small.
This is a low pass filter. It only passes low frequencies.
1
0.707
gain
w
Pass band
wc
Stop band
Figure 5.24
The breakpoint (cutoff frequency) is calculated in a similar way as for the high pass filter.
That is when R = wL
Substitute w =
R
into equation (26).
L
Vout
1
=
ª 0.707 – - 45°
Vin 1 + j1
This is the half power point as demonstrated for the high pass.
Repeating Example 5.5 but with the R and L reversed in the low pass configuration we
find.
At 1 Hz
Gain = 1.0
Phase = 0°
At 1000 Hz
Gain = 0.707 Phase = -45°
At 1 MHz
Gain = 0.0
Phase = -90°
1
0.707
Gain
1Hz
1kHz
1 MHz
Freq
1Hz
1kHz
1 MHz
Freq
0˚
Phase-45˚
-90˚
5.11 RLC circuits
C
L
Figure 5.25
Lets now look at a circuit with L and C in parallel. The impedance of the circuit will be
Z=
1
1
1
+
jX L - jX C
=
1
1
1
+
jwL - j / wC
What happens when XC = XL= X.
1
1
1
Z=
=
= =•
1
1
1
1 0
+
-j + j
jX
-jX
X
X
(W)
This occurs at a particular frequency. We call this the resonant frequency. If we go above
this frequency we find that XL becomes bigger than XC or lower in frequency vice versa.
Either way the impedance becomes finite and a phase shift exists.
At resonance the capacitor and inductor cancel each other out. Above and below this
frequency they don't fully cancel and the parallel LC combination looks like a capacitor or
inductor.
The resonant frequency may be calculated
X C = XL
,
1
2p2pLC
1
fr =
2p LC
f r2 =
1
= 2pfL
2pfC
We may make a filter out of this LC element
R
C
L
Figure 5.26
This is simply a voltage divider with the LC circuit as the bottom arm. At the resonant
frequency the impedance of the LC is infinite. At other frequencies it is less.
1
0.707
Gain
fr
Freq
Figure 5.27
This is a bandpass filter. We would use this type of filter to take out a signal from a lot of
other signals, ie tune a radio station.
The "quality factor" of the filter is described by Q, which for a parallel resonant circuit is
the ratio of the resistance.to the reactance.
f
R
Q =
= 2pf RC = r
p 2pf L
r
BW
r
†
†
BW = f - f
h l
Similarly we may put the LC in series. At resonance the capacitor and inductor
will cancel but because they are in series the impedance will be zero at
resonance and non-zero at other frequencies.
R
C
L
Figure 5.28
This will give us a bandstop filter. This could be used to reject an unwanted signal from a
lot of other, ie remove hum from a music signal.
1
0.707
Gain
fr
Freq
Figure 5.29
The quality factor Q for a series resonant circuit is the ratio of the reactance to the
resistance.
2pf L
1
f
r =
Q =
= r
s
R
2pf RC BW
r
†
f
BW = r = f - f
Q h l
Non-Ideal Inductors
We have looked at inductors so far that have been ideal, ie. had no winding resistance.
Real inductors will have a finite resistance due to the wire they are made from. The
resultant component can therefore be thought of as an inductor in series with a resistor.
R
L
We often need to include this resistance in our calculations if it is significant compared to
the resistance in the circuit.
R
RL
Output
Input
L
High Pass
If R is not >> RL then
R L + jX L
1
Vout = Vin
=
R L + jX L + R 1 + R /(R L + jX L )
Vout
1
=
Vin 1 + R /( R L + jwL)
Compare for ideal inductor
Vout
1
=
This resistance will also affect the time constant.
Vin 1 - jR / wL
5.12 Mutual Inductance
We can also use inductors to isolate two circuits electrically and transform voltages and
currents. The multi-coil configuration is called a transformer.
L1
L2
Figure 5.30
One coil is the input or primary coil and the other is the output or secondary coil.
The equations for coils we have seen so far apply to both the primary and secondary coils.
u p = Np
df p
dt
(27)
However because the secondary has no current applied to it, it uses the magnetic flux
produced by the primary.
df m
u s = Ns
dt
(28)
For this reason we define the voltage produced on the secondary as a result of the
magnetic flux fm. This is the flux produced by the primary that cuts the secondary coil.
Q. How much of the primary flux cuts the secondary coil?
A. It depends on how close the coils are and how well coupled they are.
We define the coupling coefficient (k) that describes this.
fm
k=
fp
(29)
Therefore
u s = kN s
df p
(30)
dt
We saw previously that the self inductance of a single coil (L) related the rate of change of
current through the coil to the voltage across it. Similarly with two coupled coils we relate
the change in flux in one coil to the change in current in the other. This is called the mutual
inductance (M)
M = Ns
M = Np
df m
di p
df p
di s
(H)
(31)
(H)
(32)
This can be simplified in terms of L which is something we can measure.
M = k L pLs
(33)
How then do we relate the current in the primary to the voltage in the secondary?
us = M
di p
dt
di s
up = M
dt
(34)
(35)
Let us now see how a voltage on the primary will be transformed at the secondary.
us = M
From (34)
di p
dt
substitute for M from (31)
df m di p
u s = Ns
di p dt
cancel dip and substitute for fm from (29)
u s = Ns
dkf p
dt
let us assume coupling is ideal (k=1) and substitute for
us =
Ns
up
Np
df p
dt
from (30)
(36)
We see that the secondary voltage is equal to the primary voltage times the turns ratio for
an ideal transformer.
A similar relationship applies for current except the relationship is the inverse turns ratio.
is =
Np
Ns
ip
(37)
Also the impedance (Zp) looking through the primary to a load (ZL) connected to the
secondary is transformed.
2
ÊN ˆ
Z p = ÁÁ p ˜˜ Z L
Ë Ns ¯
(38)
Example 5.6
We need a transformer to provide a 12 volts AC signal to power a portable radio. The
radio consumes 10 Watts of power and it will be run from the 240 volt AC mains.
240 VAC
Np
Ns
12 VAC 10W
Solution
The turns ratio to get 12 volts from 240 volts would be
Ns
12
=
= 0.05
N p 240
For example if the primary had 1000 turns the secondary would have to have 50 turns.
Secondary current would be
Is =
Ps 10W
=
= 833mA
Vs 12V
Primary current would therefore be
Ip =
Ns
I s = 0.05 x 833mA = 41 mA
Np
5.13 Transformer design
Transformers take many forms dependant on their application