Module 5

Inductance5.1 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.

SN

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.

I

Magnetic Fluxlines

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 =FA (Teslas) (1)

where F = number of flux lines (Webers)A = Area (m2)

1 Tesla = 1 Weber/metre

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.

m r =m

m o(2)

Materials fall into the following categories

Relative Permeability Name Examplem < 1 diamagnetic nitrogen1 < m < 100 paramagnetic copper, aluminiumm > 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 (At)

f = NImA

B =fA

= NIm

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 = Ndf

dt (V) (3)

where u = induced voltageN = number of turnsf = magnetic flux

This is Faraday’s law.

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

L =

N2 mAl

(Henries) (4)

where N = number of turnsm = magnetic permeabilityA = cross sectional area of corel = length of core

l

A

m

Figure 5.5

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 dfdi (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 = Ndf

dt= N

df

diÊ Ë Á ˆ

¯ ˜

didt

Ê Ë Á ˆ

¯ ˜ (6)

Substituting in equation (5) we see that the voltage across the inductor produced by a particular change in current will be

u L = L didt (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 5.1A 10 mH coil has a signal with the current shown below passed through it. What will be the voltage waveform across the coil?

8

4

0

i L

1086420

milliseconds

SolutionVL = L di

dt= L Di

Dt

DiDt

for 2 - 7 ms period = 10 A5ms

= 2000 A / s

uL = 10 x 10-3 H . 2000 A/s = 20 v

DiDt

for 7 - 8 ms period = -10 A1ms

= -10, 000 A / s

uL = 10 x 10-3 H . -10,000 A/s = -100 v

-80

-40

0v L

1086420

milliseconds

5.6 Symbols

Aircore

Ironcore

Ferritecore

Adjustable Alternatesymbol

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

L1 L2 L3 LNVb

Figure 5.7

Q. What happens if we place inductors in series?

A. This is fairly obvious. Two similar coils in series is just like one coil twice as long, therefore

L T = L1 + L 2 + L 3... + LN (8)

This is the same as for resistors.

The parallel combination of inductors is also the same as for resistors.

L1 L2 L3 LNVb

Figure 5.8VT

L T=

V1

L1+

V2

L2+ ... VN

L N

L T =1

1L1

+1

L2+

1L3

... +1

LN(9)

5.8 Step Response

Let us repeat the step response analysis that we did for capacitors now with inductors.

R

L nL

iL

Vb

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

L nL

iL

Vb

• 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

R(1- e- t /( L / R) ) (10)

1.0

0.8

0.6

0.4

0.2

0.0

i L

543210

time constants

Storage

Figure 5.10

Note that the time constant for this circuit is

t =LR (11)

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 = Vbe -t / t (12)

1.0

0.8

0.6

0.4

0.2

0.0

n L

543210

time constants

Storage

Figure 5.11

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

Vb L nL

iL

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 = -Vie-t /t (13)

where Vi = Amplitude of reverse voltage

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

n L

543210

time constants

Decay

Figure 5.13

• this voltage causes a decaying current to continue to flow in the same direction

i L = I me -t / t =Vb

Re -t / t (14)

1.0

0.8

0.6

0.4

0.2

0.0

i L

543210

time constants

Decay

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 5.2Lets use a circuit that will not open circuit the inductor thus causing very high voltages.

2 kW

50 V 4 H

iL

3 kW

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.

SolutionVoltage 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

iL = Im(1-e-t / t) = VbR (1-e-t /(L/R)) = 50

2000(1-e-t /0.002)

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(R1 + R2 ) =502k (2k + 3k) =125 volts

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

i L = I m e-t / t =50

2000e- t / 0.0008

-120

-80

-40

0

40v

L volt

s

20151050

ms

20

10

0

i L mA

20151050

ms

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

u L = L didt (15)

Take the integral of both sides and multiply by L and we have an expression for the current.

i L =1L

uL dtÚ (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.

R

LVmsinwt

i

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

u L = L didt

= L d sin wtdt

= wL coswt (17)

The voltage across the inductor is a cos function whilst the current through it is a sin function.

2

1

0

-1

-2

350300250200150100500degrees

Inductor

Voltage

Current

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°

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

5mF1 volt100 Hz Z

200mH

Calculate the input impedance for the series circuit above.

Solution Zt = Zr + Zc + Zl = R + (-jXc) + jXl

R = 250W

X c =1

wC=

12 p100 Hz 5x 10-6 F

=1

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 volt100 Hz Z 250W 200mH5mF

Calculate the input impedance for the parallel circuit above.

Solution

R = 250W

X C =1

wC=

12 p 100 Hz 5 x10 -6 = 318.3W

XL = wL = 2 p 100 Hz 200 x 10-3 = 125.66W

Z = R //(- jXc ) //(jXL )

=1

1R

+1

-jX c+

1jX L

=1

1250

+1

-j318.3+

1j125.66

=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)

|Z| = 315.6W

R = 250W

XC = 318.3W

f = -37.6˚

XL = 125.6W

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 =VZ

=1

315.6= 3.16 mA

Vsupply = 1 V

VR = 792 mV

VC = 1V

f = -37.6˚i = 3.17 mA

VL = 398.0mV

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.

Vsupply = 1 V iR = 4.0 mA

iC = 3.14 mA

f = -50.3˚

i = 6.26 mA

iL = 7.95 mA

Figure 5.19

5.10 Filters using inductors

R

LInput 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

LInput Output

High Pass

Vout = VinjX L

R + jXL (20)

VoutVin

=1

RjX L

+ 1(21)

VoutVin

=1

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

wc Pass bandStop band

gain

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 =RL into equation (22).

VoutVin

=1

1 - j1 (23)

1

-j1

Mag = 1.414

f = 45˚

Figure 5.22

VoutVin

=1–0°

1.414 – - 45°VoutVin

= 0.707 –45°

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 =RL

(24)

Example 5.5Plot 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 HzVoutVin

=1

1 - j100W / (2p 1Hz 15.9x10 -3H)

VoutVin

=1

1 - j1x103 ª 0 –90°

Gain = 0 Phase = 90°

At 1000 HzVoutVin

=1

1 - j100W / (2p 1000 Hz 15.9x 10-3 H)

VoutVin

=1

1 - j1= 0.707 –45°

Gain = 0.707 Phase = 45°

At 1 MHzVoutVin

=1

1 - j100W / (2p 1MHz 15.9 x10-3 H)

VoutVin

=1

1 - j1x 10-3 ª 1.0 –0°

Gain = 1.0 Phase = 0°

10.707

1kHz1Hz 1 MHz

1kHz1Hz 1 MHz

Phase

Gain

0̊

45˚

90˚

Freq

Freq

Low Pass

R

LInput Output

Low Pass

Figure 5.23Calculate the gain for this filter.

Vout = VinR

R + jXL(25)

VoutVin

=1

1 +jXLR

=1

1 +jwLR

(26)

Q. What does this equation mean?

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

wcPass band Stop band

gain

w

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 =RL into equation (26).

VoutVin

=1

1 + j1ª 0.707 – - 45°

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°

10.707

1kHz1Hz 1 MHz

1kHz1Hz 1 MHz

Phase

Gain

0̊

-45˚

-90˚

Freq

Freq

5.11 RLC circuits

LC

Figure 5.25

Lets now look at a circuit with L and C in parallel. The impedance of the circuit will be

Z =1

1jX L

+1

- jX C

=1

1jwL

+1

- j / wC

What happens when XC = XL= X.

Z =1

1jX

+1

-jX

=1

- j 1X

+ j 1X

=10

= •(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 , 12pfC

= 2pfL

f r2 =

12p2pLC

f r =1

2p LC

We may make a filter out of this LC element

LC

R

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.

10.707

Gain

Freqfr

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.

†

Qp =R

2pfrL= 2pfrRC =

frBW

†

BW = fh - fl

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.

L

C

R

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.

10.707

Gain

Freqfr

Figure 5.29The quality factor Q for a series resonant circuit is the ratio of the reactance to the resistance.

†

Qs =2pfrL

R=

12pfrRC

=fr

BWBW =

frQ

= fh - fl

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.

LR

We often need to include this resistance in our calculations if it is significant compared to the resistance in the circuit.

R

L

Input Output

High Pass

RL

If R is not >> RL then

Vout = Vin

R L + jX L

R L + jX L + R=

11+ R /(R L + jX L )

Vout

Vin

=1

1+ R /(R L + jwL)Compare for ideal inductorVoutVin

=1

1 - jR / wL This resistance will also affect the time constant.

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 = N pdf pdt

(27)

However because the secondary has no current applied to it, it uses the magnetic flux produced by the primary.

us = Nsdf m

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.

k =fmfp

(29)

Therefore

u s = kNsdfpdt (30)

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 = Nsdfmdi p

(H) (31)

M = N pdfpdis

(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 = Mdi pdt (34)

u p = Mdi sdt (35)

Let us now see how a voltage on the primary will be transformed at the secondary.

From (34) us = Mdi pdt

substitute for M from (31)

us = Nsdf mdi p

di pdt

cancel dip and substitute for fm from (29)

u s = Nsdkfp

dt

let us assume coupling is ideal (k=1) and substitute for df pdt

from (30)

us =NsN p

u p (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.

i s =NpNs

i p (37)

Also the impedance (Zp) looking through the primary to a load (ZL) connected to the secondary is transformed.

Z p =NpNs

Ê

Ë Á Á

ˆ

¯ ˜ ˜

2

Z L (38)

Example 5.6We 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.

Np Ns240 VAC 12 VAC 10W

SolutionThe turns ratio to get 12 volts from 240 volts would be

N sN p

=12240

= 0.05

For example if the primary had 1000 turns the secondary would have to have 50 turns.

Secondary current would be

I s =PsVs

=10W12V = 833mA

Primary current would therefore be

I p =NsNp

I s = 0.05 x 833mA = 41 mA

5.13 Transformer design

Transformers take many forms dependant on their application