Induction Heating Basics

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“ELECTROHEAT describes any heating process where electricity is the primary energy

source”. I.e. an electromagnetic field (or electromagnetic radiation) interacts with the part to

be heated (otherwise known as “the work-piece”), and causes it to be heated. Generally, the

frequency of the electromagnetic field interacting with the work piece characterises the

heating process. The frequency of the electromagnetic radiation can stretch from DC to

beyond daylight.

DC

Daylight

Resistance (DC)

Mains-frequency Induction (50Hz)

Medium-frequency Induction (1-20kHz)

Radio-frequency Induction (50kHz-10MHz) Dielectric (27-48MHz) Microwave (GHz) Infrared heating

Through

Surface

Freq

uenc

y

Hea

t Pen

etra

tion

shallow

surface

Deep

through

shallow

surface

through

I2 R H

eatin

g

This figure shows most of the different categories of electro-heat, arraigned in an ascending

order of frequency. At DC (or low frequency AC), direct resistance heating heats

conductive work pieces by attaching an electrode to each end of the material to be heated. A

large current is then passed through the work piece. This method features high power

densities and relatively high efficiencies. A typical use for this heating method is for

resistance welding of car bodies, etc.

Direct resistance welding is limited to relatively few materials and heating geometries,

however, as the current will flow through the shortest path between the electrodes, and the

highest resistance part of the circuit will get hottest, so if the work piece is a lower resistivity

than the electrodes, the electrodes will be heated more!

Billet heating by direct resistance heating is rather wasteful for this reason, as the part of the

material under and beyond the electrodes is not heated, and would be scrapped in a forging

application.

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A further limitation on the application of direct resistance heating is the contact pressure

required between the electrode and the work piece. This can be considerable, and can

damage some more fragile materials.

Direct resistance heating is a DC process. As the electro-heat frequency increases the

heating is done via a time-changing magnetic field inducing a current in a conducting work

piece. This current produces I2R heating in the work piece, and the process is known as

induction heating.

It can be seen that induction heating is split into three categories: Mains frequency, medium

frequency and radio frequency. This is for two reasons. The first is that the equipment for

generating the excitation current for the coil (or work head) that generates the time-varying

magnetic field has historically been very different.

MAINS FREQUENCY induction heating systems are excited by direct connection to the

utility supply. The current penetrates deeply into the work-piece and they tend to be used in

very high power systems.

MEDUIM FREQUENCY induction heating systems (500Hz-10kHz) have suffered a long

evolution. They are used for smaller work-pieces and the current penetrates about 1cm into

the work-piece. In their inception, in the 1920’s, the excitation was provided by the use of a

motor-generator set. As suggested by their name, these work by mechanically coupling a

high-frequency generator to the shaft of a mains-powered motor. They are primitive, noisy

and inefficient. They are also simple, and easy to service, and are hence still found in some

less technologically advanced countries.

The advent of the THYRISTOR in the 1960’s saw most MG sets being replaced by various

different topology inverters. Thyristor-based inverters are still in production, and modern

thyristors can carry several thousand amps and can block several thousand volts, whilst

offering a very low forward conducting voltage drop (1.4V). This capability means that

medium-frequency induction heaters with powers in excess of 500kW are often best

manufactured with IGBT-based inverters. Across the power range 50kW upwards, thyristor

based inverters are still being manufactured, and are characterised by their reliability and

simplicity, and their manufacturing expense and their inflexibility.

Recently (within the last 5 or so years), design emphasis has moved from thyristor bases

inverters to IGBT based inverters. IGBTs offer three principal advantages over thyristors:

The first advantage is that, unlike thyristors, IGBTs turn off when their gate signal is

removed. Once turned on, a thyristor will remain conducting until it is reverse biased. This

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means that the power-electronic topology of an IGBT-based inverter can be considerably

simpler. The second major advantage of an IGBT over a thyristor is the packaging

technology is considerably easier to use. These two improvements make the manufacturing

costs of the inverter considerably lower. The final improvement that IGBTs offer is that

they can have considerably faster switching speeds, so the operating frequency range of an

IGBT based system will be extended.

The operation of these inverters will be discussed in more detail at the end of this course

section.

Finally, we have RADIO-FREQUENCY induction heating, which covers the range 50kHz –

10MHz +, although it gets difficult to make work-head coils that work at the highest

frequencies. From the 1920’s, the excitation has been provided by thermionic valve-based

oscillators, originally made from modified radio transmitters. Themionic valves are

characterised by very fast switching times (ns) and very large forward conduction voltage

drops (100V). To make a workable induction heating system, a very high voltage DC power

supply is used (up to 12kV). Even with this high supply voltage, the efficiencies of these

“generators” is only in the region of 60%, so, with powers of 2½kW to 2MW, it can be seen

that a considerable amount of power is wasted.

This inefficiency, coupled with the safety and reliability concerns associated with a high

voltage DC link, compounded with the high unit cost for a power triode, have led, since the

1980’s to the increasing use of MOSFET based inverters. Currently, commercially available

inverters can deliver 1MW at 800kHz, and the frequency range is expanding.

So, to summarise, for medium frequency systems, the THYRISTOR is being replaced by the

IGBT, and for radio frequency systems, the TRIODE is almost completely superseded by

the MOSFET.

Having briefly explored how the different frequencies are generated, we should now look at

why the different frequencies are generated. This is all related to current penetration. At

low frequencies, the magnetic field penetrates deep into the work-piece, which means that

the induced current will flow deep into the work-piece. If the field can penetrate further

than half way through the material, then the field coming from the opposite direction will

cancel it out, as it will be acting in the opposite direction. Therefore, for a given piece of

material, there is a minimum frequency that can be used to heat it. It is this effect that

allows the use of a laminated steel magnetic core on a transformer.

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Conversely, if the frequency is too high, only the surface of the material is heated, which

may mean that the applied power has to be limited such that the outside is not overheated

(melted or burnt) before the inside is heated. It may be, of course, that you only want to heat

the outside!

Therefore, the frequency controls the depth of heating.

Moving up through the electromagnetic spectrum, we come to DIELECTRIC HEATING.

Dielectric heating uses the application of a high frequency electric field to align and move

the molecules of a non-conductive material. This movement causes friction within the

material, which causes it to heat. This process is not affected by penetration depth, but as

the intensity of the electric field is inversely proportional to the distance between the

electrodes, the thickness of the heated part is restricted. The most common use for dielectric

heating is therefore plastic welding, which is use to weld thin plastic sheet in many

applications, such as pencil cases, plastic sacks, etc.

Moving further up the electromagnetic spectrum, we get to possibly the best-known use of

electro-heat: the microwave. Microwaves are used in industrial heating as well as domestic,

for example drying paper as it is produced. As a far-field (i.e. wave) electromagnetic source

is used, care has to be taken when using it, as if any leakage occurs, operators can become

injured.

Microwave heating systems work by causing certain molecules to vibrate at a resonant

frequency (in the case of a domestic microwave oven, water molecules at 2.4GHz). Once

again, as in dielectric heating, this vibration then causes heating essentially by friction.

Microwaves are a shallow heating mechanism, with most of the incident energy being

absorbed within the first ~4cm of the target thickness (dependent on the material and the

frequency).

Moving further up the electromagnetic spectrum, we get to infrared heating. Infrared

heating describes the use of using the infrared radiation from a hot body to heat a colder

body. As radiation is a surface phenomenon, the surface of the emitter is cooled by the

radiation, and the surface of the target is heated, and the thermal conductivity of the heated

material carries the heat into the work-piece. There are three sub-categories of IR heating

long wave, medium wave and short wave, which correspond to increasing surface

temperatures of the emitting body. As the wavelengths increase, the heat tends to be able to

pass further through various materials. The heat rate is governed by the equation

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( )44TsAP θθε −=

where epsilon is a material constant called the emissivity.

Having defined electro heat in terms of operation, its use must be justified. The most

common criticism of electro heat in most forms is it’s cost, both operational and in terms of

capital plant cost. An example of this can be seen in the small company of R.S. Hall

Engineering Ltd, the small company that I worked in between finishing my degree and

starting my PhD. Seven years ago, they invested in an induction heater to replace a gas

furnace, for heat-treating steel. The induction heater heats steel bar from ambient to 1200C.

As the bar that they were heating could be a minimum of 9.5mm in diameter, they had to use

a 30kHz induction heater (we’ll derive this frequency later), and, to get the throughput they

needed, they chose a 30kW power unit. This cost them £22000. It was estimated that, if

they had rebuilt their gas-powered furnace from scratch, it would have only cost them

£8000. Clearly, for them, the initial outlay was offset by other factors.

One major factor is the energy cost. This will be shown (on the next slide) to be reduced.

From a process point of view, one of the most attractive features of electro heat in general is

the quality of the heat control. Most electro heat processes allow an instantaneous power

control with continual variability. Compare this with a coal-powered furnace that can take

several hours to warm up. Further, the power put into the process can be accurately

controlled with an electro heat system, whereas most combustion-based methods are subject

to fuel quality variations.

A further advantage of electro heat is that the power density within the work piece can be

very high. This is because the heat is developed within the work piece and therefore no

conduction process is required. The fact that the heat is developed in the work piece means

that it, rather than a surrounding furnace, is the hottest part in the system. These factors

combine to reduce the heating time, which reduces the material losses due to oxidisation and

chemical (metallurgical) changes such as decarburisation, which traditionally account for a

high proportion of the lost material in combustion-based heating processes.

As well as being able to modulate the power in the time-domain, the special control of the

heat available with induction, dielectric, and, to a lesser extent direct conduction heating, is

considerably better than any other means (such as an oxy-acetylene flame). This has its

advantages: the profiles of heated parts can be accurately controlled, such as the seam of a

plastic weld, or the part of a pop-rivet that bends.

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A further tangible improvement that electro heat offers over traditional heating processes it

the elimination of soot and smoke from the workplace. Most electro heat methods are also

much quieter than combustion processes, and therefore the replacement of combustion

heating processes with electro-heat processes results in a workplace that is considerably

more pleasant.

Again, the main criticism of electro heat is the cost. To make a true comparison of each

energy source, it is important to consider it’s calorific value and to compare the cost on an

equal plane.

Oil=20p/40MJ=0.5p/MJ

Gas=34p/106MJ=0.32p/MJ

Coal=4800p/27500MJ=0.17p/MJ

Electricity=4p/3.6MJ=1.1p/MJ

It is therefore obvious that coal is the cheapest form of energy calorifically, and electricity is

the most expensive, by a factor of about 6.

However, when we compare process efficiencies, a more realistic picture emerges.

Although gas has a higher calorific value per unit cost than electricity, the burner efficiency

is limited as gas is mixed with air rather than oxygen, so some of the energy goes into

heating the nitrogen. This is translated into hot exhaust gasses rather than hot material.

Further, getting the perfect mix between air and fuel is impossible, leaving some of the gas

partially un-burnt (carbon monoxide or soot). An efficiency of 60% for combustion is

therefore rather optimistic.

All we have done so far is to heat a flame – the flame then has to heat up a billet, generally

an inefficient process, with half of the heat going up the chimney. Finally, the equipment

utilisation has to be taken account of. With a gas furnace, a warm-up time of over an hour is

unexceptional, which means that the furnace will be left on over coffee breaks etc, wasting

energy.

The overall efficiency can be worked out

60/100*50/100*60/100=18%

…and the cost of the gas per useful MJ can then be found

0.32/0.18=1.77p/MJ

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A similar analysis can be carried out on the same process if it were to be done using direct

resistance heating. It can be seen that the system is not 100% efficient either. The “burner”

(i.e. energy conversion) efficiency is not 100%. This is due to I2R heating of the contact

electrodes etc, and is a function of the relative resistivity of the work piece and the

electrodes.

The heat transfer, though, is 100% efficient, as the heat is generated in the work piece.

The heat utilisation is not 100% because of the contact area. The metal under the contact

area is not heated uniformly, and may have to be scrapped, so the energy that has gone into

that part of the work-piece is wasted.

This gives an overall efficiency of 72%, and a cost per useful MJ of 1.52p, considerably

cheaper than the cost of gas! The electro-heat process suddenly seem more economically

viable!

Unfortunately, with direct resistance heating, the ends of the work-piece don’t get hot. In a

forging application, this can mean that 10% of the material can be wasted.

Alternatively, in a zone-heating application, this is a good thing.

For the forging case, the unheated part of the billet is generally scrap metal, which

introduces further costs into the equation. However, metal loss as scale (oxides), which can

be significant in a gas-fired system, is almost eliminated.

With induction heating, the heated length can be as long as the work-piece. The inverter can

be up to 98% efficient, and the coil can be over 90% efficient. The near-instantaneous

power means that the heat utilisation can be 100%, but there is loss due to radiation etc.

from the work piece. It is therefore possible to get a total energy efficiency of 80% from an

induction heating system with only minimal material wastage.

Both combustion furnaces and direct resistance heaters can give rise to a significant degree

of material wastage: 3% due to oxidation and metallurgical change in the case of

combustion furnaces, and up to 10% due to poor current distribution in the case of resistance

heaters. This increased material wastage can mean that the direct resistance heater is not an

economical choice for a process heating for billet forging applications. With its lack of

contact and increased power density, coupled with it’s ability to heat 100% of the billet

length with good heat distribution, induction heating can be a more economical choice.

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This picture shows a work head connected to a 7½ kW 150kHz (i.e. RF) induction heater,

heating an M10 bolt to well over 1000oC. As we shall see over the next few weeks, the

efficiency is a function of the material being heated, along with the heating profile, and, as

such, is variable. With billet heating, a heating efficiency of 70% is reasonable.

With induction heating, as with direct resistance heating, the heat is directly generated

within the material, so the heat transfer efficiency is 100%.

The heat utilisation, like the efficiency, is a function of the required heating pattern. It is

easy to heat 100% of the length of a billet; you simply put it in a long work-head coil. It is

more difficult to heat a narrow band, as focussing the magnetic flux can only be done to a

certain degree, with fringing over the (required) air-gap between the work-head and work-

piece spreading the heating somewhat. This means that although it is possible to make a hot

spot, it is difficult to avoid a warm area around it.

As induction heaters tend to give instantaneous heating control and can draw no quiescent

power, the time utilisation of energy can be 100% (i.e. when it’s not heating, it’s not

drawing power)

Therefore, for billet heating, we can get a 70% total energy efficiency, coupled with a sub-

1% total material wastage, if scale/decarburisation wastage is accounted for.

So, how does induction heating actually work? As briefly discussed in the last lecture,

induction heating works by inducing a current into a conducting object. It does it like this:

First, a copper coil (often a solenoid, but not exclusively), has a large, time varying current

set up in it by the imposition of a time-varying voltage across it (generally in the form of a

sine wave).

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This current then creates a time-varying magnetic field (for solenoid l

NIH = ), which will

cause a time varying flux ( HB µ= ).

E

I

If a conducting object is placed in the field, then a voltage will be induced around it

( BAdtd

E =ΦΦ= , ).

If the conducting object is a closed ring, then, like a shorted turn in a transformer, then the

voltage will cause a current to flow around the outside of the object.

( )jXRIV += ….jXR

VI

+=

The allowance for an impedance has to be made as this is an AC system:- If it were Dc, then

the rate of change of flux with time (dtdΦ

) would be zero, so no current would be induced.

Finally, this induced current causes I2R losses in the work piece, which makes this method

of heating effectively a resistance heating method, albeit with the current flowing at right

angles to that of direct resistance heating (i.e. around the billet rather than along it).

Having shown in some more detail how the basic principle of induction heating works by

basically considering the current flow in a thin tubular foil, we shall now look at what

happens to the induced current when induction-heating a solid work piece.

The answer to this question is fairly mathematically heavy, and to go too deeply into it

would be a bit of a waste of time. Therefore, I will simply give you a “hand-waving”

account of how the field, and therefore the current, is distributed throughout the materials,

followed by the analytical solution. This way we avoid vector integration, Bessel functions

etc.

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x

y

z

Ho

H1

H2

H3

To avoid having to discuss flux return paths and end effects, we consider a semi-infinite slab

of material being heated by an infinite 2-diamentional sheet of current just above it. This

diagram shows a finite part of a cross-section of the infinite set-up. The sheet of current that

represents the work head stretches infinitely to the right and left (in the x direction), and

infinitely forward and backward, or into and out of the page (the z direction). It occupies no

space a all in the y direction.

The semi-infinite slab that represents the work-piece also extends infinitely into and out of

the page, and infinitely to the left and right, but it goes from y=0 to y=-∞.

To see where the current goes, we can turn the homogeneous slab into a series of thin slices.

Consider the top slice. It has a time varying magnetic field, ( )tH ωcosˆ0 acting on it. In the

same way as the last slide, this will have a current density induced in it, ( )θω +tJ cos0 . The

phase shift (lag) is due to the inductance of the slab causing a lag between the EMF induced

in the slice and the current flowing through it.

This current density in the slab creates an opposing magnetic field in the slab, marked as H1.

The resistivity and the inductance of the slab reduce the magnitude of the current, and hence

the field, so H1 is smaller than H0.

Now consider the strip below. It sees a field that is equal to the vector sum of H0+H1 so,

this strip sees an attenuated field, as H0 opposes H1. This strip will therefore have an

attenuated current density induced in it ( )11 cos θω +tJ . This attenuated current density

creates a magnetic field H2.

The third strip down will see a field made up of the vector sum of H0,H1 and H2, i.e. a

further attenuated field, which will induce a smaller still current density, making the

resultant total field smaller and smaller as you go down through the y axis.

This effect, known as the ‘skin effect’, means that the field, and hence the loss, or heating

effect, is concentrated on the surface of the work-piece.

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It can therefore be shown, by letting the thickness of the slices tend to zero, and solving the

resultant differential equation, that the field in the x direction, the current in the z direction,

and the flux in the x direction all follow this form:

( )( )

( )

−

Φ=Φ==

−

δωδ yte

y

JyJ

HyHy

xx

zz

xx

cos

0)(

0)(

0)(

i.e. they are all of the form

−−

δωδ y

tey

cos

which is an exponential decay multiplied by an oscillating term with a variable phase shift.

0 1 2 3 4 5 0

0.2

0.4

0.6

0.8

1

y/δ

| J/J

0|

1/e

δ

Magnitude attenuation = δy

e−

0.3679

− δω ytcos → The oscillating term (note shifting phase)

This assumes that the current flowing through the coil is sinusoidal. Generally, this is the

case, with only a small distortion. The reason for this will be made more clear when we

have discussed how the coil is connected to the excitation circuit.

These terms are only true for a semi-infinite slab, so they have no direct use. However, they

are simple, and most of induction heating theory is based on them.

The most important part of the equations is the term δ. This is the skin depth, or the

penetration depth, and is the depth into the slab at which the current etc. has fallen to 1/e

(i.e. 1/natral number) of it’s surface value.

ωµρδ 2=

On inspection of the formula for the penetration depth, it can be seen that the heating depth

is a function of the resistivity, the permeability and the frequency. As the resistivity and the

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permeability of the work-piece are fixed by its material, the only way to actively control the

penetration depth of the current into the material is to alter the frequency. This is why

induction heating systems are divided into the three different frequency bands:

Mains – used for through heating large pieces of metal (gas cylinders etc)

MF – used for through heating smaller billets and strips – down to 15mm

R.F – used for surface heating or for heating very small pieces.

Although the current (and therefore the heat) is induced into the very surface of the material

under an RF magnetic field, RF induction heaters can be used for through heating by

allowing the heat to be conducted in through the material. This limits the rate at which the

material can be heated, as too high a power will result in the material melting on the surface

before the inside is even warm!

One other thing to note about the equations is the phase shift in the oscillating term. As the

y position decreases, the current, H-field and flux become more retarded.

The total current (per unit length) in the slab can be found by integrating the current density

from the surface to –infinity, with respect to the semi-infinite axis, y.

The current density has been defined as:

( )

−=

−

δωδ yteJyJy

zz cos0)(

JZ(0) is the surface current density. The phase shift on the current with depth has an effect

on the integration, which evaluates to:

−=4

cos2

)0( πωtJ

I z

This total current I can be considered to be flowing in 1 skin depth δ. Therefore, as far as

our semi-infinite slab goes, we have made the y-direction into a sheet of thickness δ that

carries all of the current in a uniform manner.

This is only a definition, remember, but it is extremely useful.

The most important bit is that it allows the surface power density to be defined, and from

that, an equivalent circuit.

So far, we have got as far as finding the total current in the work piece, and stating (although

proof can be found in the “induction heating handbook” by John Davies and Peter Simpson),

that the total current

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0452

)0(∠zJ

can be considered to be uniformly concentrated in the outer skin depth of the material

ωµρδ 2=

If this is true, then the surface power density can be found, by the use of

ρAl

R =

So, for a surface area of 1m×1m,

1m δ

1m

A=δ×1m, l=1m

So, using P=I2R,

yresistivitmA

mlJP z =×

×==×

= ρδ 1

1

2

)0(2

↑=I2

…in Wm-2

Now, it’s all very well knowing the power density in the work piece as a function of the

surface current in the work piece, but this has no real connection with induction heating: the

power is provided by a magnetic field, not by direct connection.

To relate the power to a coil, consider a long solenoid.

The total current density in the coil is equal to the coil current multiplied by the number of

turns divided by the length.

lNI

J c=0 Am-1

We can substitute this value for J0 into the surface power density to get

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δρ

δρ

22

/22

=

=lNIlNI

P cc Wm-2

From this, we can find the power density due to the coil current in a long solenoid. If we

now consider the field due to a solenoid

lNI

H =0 Am-1

this is equal to the surface current density. Therefore substituting H0 into the surface power

density equation,

δρ

22RMSHP =

The RMS symbol is inserted as a reminder that the rms value of the field must be used =

2H

To show the effect of the work piece properties on the surface power density, three

examples are worked through, each with the same field magnitude (100kAm-1) and

frequency (50Hz), but with varying permeabilities and resistivities.

Material Mild Steel Mild Steel Copper

Temperature θθθθw 20oC 800oC 20oC

Resistivity, ρρρρ 200nΩΩΩΩm 1.1µµµµΩΩΩΩm 17nΩΩΩΩm

Relative

permeability µµµµr

50 (<Curie) 1 (>Curie) 1 (non

magnetic)

Applied Field H0 100kAm-1 100kAm-1 100kAm-1

Frequency f 50Hz 50Hz 50Hz

Skin depth δδδδ 0.0045 m 0.0747 0.0093

Surface Power

densityPs

222kWm-2 73kWm-2 9.16kWm-2

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The first case is for mild steel at room temperature. The steel is below the Curie point

(~720oC), at which the ferromagnetic properties of the steel stop, and therefore a magnet

will not stick to it. This means that the relative permeability of the steel is relatively high.

Electrical engineers amongst you will notice that the permeability of the steel is

considerably lower than you would expect, with µr being well over 1000 normally.

This discrepancy can be explained by considering the given µr as a ‘large signal’ quantity,

and the expected µr being a small signal quantity. If the expected µr of 1000 is substituted

into

HB µ=

then you get

B=1000×4×π×10-7×100×103=124T

At the surface of the steel, but steel saturates at about 2T. The effect of the saturation is to

reduce the incremental permeability.

The effective relative permeability is therefore reduced to take this saturation into account –

i.e. the large signal incremental permeability is used. Effective values of 20 to 50 are often

used for steel under lower frequency induction heating fields.

Anyway, to find the surface power density, we first need to find the current penetration

depth

ππµωρδ

250104501020022

7

9

××××××== −

−

=0.0045m=4.5mm

Substituting the skin depth into the surface power density equation

( )0045.02

10200101002

9232

××××==

−

δρ

rmsHP =222kWm-2

Can you work out the other two situations?

Note, then, that the hot steel has a much deeper current penetration than the cold, due both

the (relatively slow) increase in the resistivity with temperature, and to the rapid reduction in

the permeability at curie.

Although the resistivity of the hot steel is over 5 times greater than the resistivity of the cold

steel, the losses in the steel are lower, as the current flows in an area that is over 16 times

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greater. Remember that the total current in the material is only a function of the H-field, so

the current is the same. The resistance is equal to the resistivity multiplied by the length

divided by the area, so if the penetration depth is smaller, the resistance, and hence the loss,

is higher.

From this, we can see that the steel is heated much faster below the Curie point than above it

for a fixed field strength.

This is a general conclusion, and is something to consider when dimensioning an induction

heating system.

Moving on now to the copper load, we can see that the penetration depth is just over twice

that for the cold steel load. Although the conducting area is similar, the per-unit surface

area effective resistance of the copper is much smaller that that of the steel. It can be seen

that the losses in copper, under the same field strength as the steel, are considerably smaller.

This can cause problems when induction heating copper!

On the other hand, if the H-field source (i.e. the work head) is also considered to be a semi-

infinite slab, starting where the work piece semi-infinite slab ends, then the analysis is valid

for the work head as well as the work piece. This allows the work head losses to be

estimated.

As the H-field at the very surface of the load slab is equal to the H-field at the very surface

of the source slab, all of the previous calculations can be used.

H0(workhead)=H0(workpiece)

Almost invariably, the work-head is made out of copper, although above 10MHz, the copper

is silver plaited to increase the surface conductivity.

As all of the losses worked out on the previous slide were for the same field intensity, the

relative efficiencies can be found for heating steel above and below Curie, and also for

heating copper.

Eff=Heat into load÷total heat

Total heat= Heat into load + Heat into work head

The theoretical efficiencies can be found

First for cool steel

%10016.9222

222 ×+

=E =96%

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similarly,

Hot steel=89%

Cold copper=50%

This calculation makes the assumption that the length and the width of the work piece and

the work head are equal. Whilst the width of a practical work piece may be equal to the

width of a practical work head, the length of the work head has to be longer, as it is wrapped

around the work piece. The efficiency is therefore always worse than this calculated value.

By substituting the skin depth equation into the surface power density equation, and

substituting this into the efficiency equation, both in terms of the work-head and the work

piece, then the best-case efficiency can be shown to be

rww

c

Eff

µρρ

+=

1

1

It is easy to draw the conclusion that for efficient heating, you need a highly conductive coil

and a highly resistive work piece, and if the work piece has a high permeability, then the

efficiency is improved.

Of course, many assumptions have been made here, not least the semi-infinite slab model:

this result suggests that you could induction heat air (resistivity infinite) with 100%

efficiency. This is clearly not the case.

Also, the analysis assumes that the work head is an homogeneous sheet of current, rather

that a series of discrete turns separated by layers of insulation. This means that the area that

the current flows in the work head is reduced, meaning that the local current density is

higher, increasing the losses. This is compounded by the fact that the length of the current

path around a practical work head is longer than the length of the current path around most

practical work pieces.

To make a better model of the work head / work piece pairing; which includes the reactive

as well as the resistive losses, we can create an equivalent circuit by considering the

magnetic flux.

We are first going to derive the EMF induced in the work piece by considering the magnetic

flux in the work piece.

If we start with the flux density. This follows the general form:

of 18

−=−

δωδ y

teBBy

cos0

where we have the surface value modified in amplitude and phase as you pass through the

material.

If we assume that the material is linear, and that

HB µ=

holds, then we can substitute µH0 for B0 and relate the flux density to the applied H-field.

We can then find the flux in the work piece by noting that the flux density is the rate of

change of flux with respect to space, so flux is the integral of flux density, with respect to

space.

Therefore, the total flux in the work piece can be found by integrating the flux density over

whole depth of the semi-infinite slab.

−=Φ4

cos2

00

πωµδµtH r

x

(note the similarity with the total current)

From the basic electromagnetic definitions, the EMF in a coil is equal to the number of turns

multiplied by the rate of change of the flux contained in the coil with respect to time.

dtd

NEΦ=

Of course, with the work piece, N=1, but we can refer this voltage back to the coil by

substituting the number of turns into N.

So, differentiating the total flux with respect to time, and multiplying by the number of coil

turns, we get the coil EMF due to the flux in the work piece.

+−=4

sin2

0 πωδµω tH

NE Vm-1

or, in cosine form

+=4

cos2

0 πωδµω tH

NE Vm-1

i.e. The voltage leads the H-field by 45o. If we consider that in the case of a long solenoid

of 19

lNI

H =0 Am-1

we can substitute for H to get

+=4

cos2

20 πωωδµ

tl

NIE Vm-1.

I.e. the voltage leads the current by 45o, or, more conventionally, the current lags the voltage

by 45o.

If we use the form

P=VI cos(φ)

where cos(φ) is the power factor, it can be seen that the power factor of the work piece is

cos(45o), or 1/√2

Alternatively, if we re-write the EMF equation so that E and I are complex AC

Signals rather than the time-domain signals used so far, we get

+=2

1

2

1

2

2

jl

NIE

ωδµ Vm-1

If we consider this to be in the form of Ohm’s law, then we can see that the impedance of

the work piece, referred to the coil terminals, is

+=2

1

2

1

2

2

jl

NZ

ωδµ Ωm-1

allowing equivalent values for the resistance and the reactance to be found: l

NR

2

2ωδµ=

Ωm-1,l

NX

2

2ωδµ= Ωm-1

Further, if we consider the reactance of an inductor

XL=ωL

The reactance can be expressed as an equivalent inductor

lN

L2

2δµ= Hm-1

of 20

This per-unit equivalent circuit can be used to analyse a very big (d>20δ) cylinder with a

solenoid wrapped tightly around it.

Dl

NRload πωδµ

2

2

= Ω

where πD is the length of the electrical path around the work-piece.

Similarly

Dl

NX load πωδµ

2

2

= Ω

Using the concept that the coil can be approximated to a semi-infinite slab, providing that

the conductor is over 10 times thicker than the skin depth of the current, it is possible to

define the equivalent circuit components in a similar manner.

Dl

NkR ccoil πωδµ

2

2

= , Dl

NkX ccoil πωδµ

2

2

=

Inspection of the equation shows that a coil factor, kc, has been included. This is to account

for the fact that the coil is made up of discrete turns rather than a homogeneous mass.

kc is normally of the range 1.1 to 2, depending on the shape and spacing of the coil

conductors. The lowest end of the range, being the most efficient, is for rectangular

conductors with minimal turn-on-turn insulation.

If round tube is used, then the efficiency is compromised by the reduction in the conducting

area, where only the bottom of the tube caries the full skin-depth of current, so a fair bit of

the area facing the work piece does not carry any current.

of 21

Once again, this analysis assumes that the coil is touching the work piece, and that the

diameter is much larger than the penetration depth.

If the coil is not touching the work piece, then there will be some ‘stray’ flux in the air gap

between the coil and the work piece. This stray flux will cause an EMF to be induced in the

coil, just like the flux in the work piece and the flux in the coil.

We can consider the field H0 to be constant across the air-gap by considering the penetration

depth.

ωµρδ 2=

the resistivity of air is infinite when it is not ionised (i.e. a spark or plasma). Therefore the

penetration depth of magnetic field into air is infinite, which means the rate of change of the

H-field with space is zero across the airgap. This means that the identity

H0(coil)=H0(workpiece)

Remains true in the presence of an air gap. Further, the equations for Rc, Xc, Rw and Xw also

remain correct. However, we now have to consider the flux in the air gap, which will

contribute to the coil terminal voltage.

The flux in the air-gap (Φgap) can be found from B=µH

ggap AH 00µ=Φ

Where Ag, the area of the air-gap, is equal to the area enclosed by the coil minus the area of

the work piece.

( )22

4 wokrpiececoilg ddA −= π

The coil voltage due to the air-gap can be found by finding the rate of change of the flux in

the air-gap with respect to time.

( )

=Φ

= ggap

g Atl

NIdtd

Ndt

dNV 0cos µω

( ) gg Atl

INV 0

2

sin ωµω=

Alternatively, in orthogonal complex AC form

of 22

l

AINjV g

g0

2 ωµ=

i.e. the voltage in the coil due to the air-gap is totally reactive, which makes sense as the air

does not get hot! The equivalent reactance of the air-gap can be found by dividing the

voltage by the current.

l

ANX g

g0

2ωµ=

Although the equivalent component for the air-gap is totally reactive, it does indirectly

increase the losses in the coil, as it increases the length of the current path in the work head.

Therefore, induction heating systems with a large air-gap are less efficient than those with a

small air-gap.

As the total flux within a coil gives rise to the EMF induced in the coil, it appears as if all of

the EMFs, and therefore all of the equivalent components, are in series. Therefore we have

derived a full equivalent circuit for a coil heating an electrically large load.

Xc Rc Xg

Lw

Rw

Example

This example is for a long thick bar of steel (300mm*2m) being heated from room

temperature to 1200oC under a field being excited at 50Hz. The coil is 500mm in diameter

and has 100 turns. It is the same length as the steel bar. Similar bars of steel are heated at

chesterfield cylinders at a power of 1MW prior to being forged to make gas cylinders.

Initially, we will find the equivalent circuit of the system with the steel at room temperature.

If we compare the skin depth (4.5mm at room temp), with the diameter (300mm), then it can

be seen that the condition given for the work piece to be electrically large is true (20δ<d).

of 23

Following the formula, Rcoil can be found:

ccc

c kdlN

R πωµδ2

20=

225.11050050210000104103.9 373

×××××××××××=

−−− πππcR

=0.0216Ω

Xc=0.0216Ω also

Lc=Xc/ω=69µH.

We have now found the equivalent component values for the coil. Moving on to the work

piece:

www

w dlN

R πωµδ2

2

=

22103005021000010450105.4 373

×××××××××××=

−−− πππ

=0.209Ω

Xw=0.209Ω also

Lw=Xw/ω=666µH.

Finally, the inductance due to the air gap. This is due to the area inside the coil that is not

accounted for by the work piece.

gg ANX ωµ02=

( )22

4 wcwcg ddAAA −=−= π

( )22 3.05.04

−= πgA =0.126m

126.050210410000 7 ×××××= − ππgX =0.25Ω

Lg=Xg/ω=790µH

So, if we were to apply 1000V to the terminals of the coil, what power would we get?

of 24

First find the coil current

ZV

I =

( ) ( )2222gcwcw XXXRRXRZ ++++=+=

( ) ( )22 25.0022.0209.0022.0209.0 ++++=Z =0.53Ω

Ω=∴

53.01000V

I =1874A

then find the power by using P=I2R

Power in work piece=I2Rw=734kW

Power in work head=I2Rc=77kW

The efficiency can be found by finding the useful power divided by the total power.

%100×+

=cw

w

PPP

E =734/(734+77)×100%=90.5%

The semi-infinite slab example, which does not account for the air gap, gives an efficiency

of 96%. It can be seen therefore that the air gap does indeed increase the losses in the coil,

despite dissipating no power itself.

It is all well and good being able to calculate the equivalent circuit values for an electrically

large work-piece in a long solenoid: this covers about 35% of all induction heating cases. A

further 35% of induction heating cases are for systems that are electrically small, and the

rest are for cases where the work head is short – it can be down to 1 turn.

One of the best conditions for rapid through heating is to have the skin-depth similar in

magnitude to the diameter, such that the majority of the load is being directly induction

heated, rather than being indirectly heated by thermal conduction from the heated surface.

Having an increased heating penetration gives a smaller temperature gradient through the

material for a given surface power density, allowing the middle to be heated at a much

higher rate than would be the case with an electrically large work piece, without the surface

melting or becoming degraded. As well as increasing the throughput, this is useful, as a

rapidly heated product will spend less time at an elevated temperature, so less material will

be lost through oxidation or metallurgical change.

of 25

To enable analysis of magnetically smaller parts, the equations for the resistance and

reactance due to the load are modified to include a correction variable.

c

ww l

pANR

2ωµ= ,

c

ww l

qANX

2ωµ= ,

Where Aw is the area of the workpiece

2

4 ww dAπ= .

Substituting 2

4 ww dAπ= into the new equations for Rw and Xw, and then comparing these

equations with those for electrically large cylinders, it can be seen that 2wd

p has replaced δ

as the effective width of the current flow in terms of the losses, and 2wd

q has replaced δ as

the effective width of the current flow in terms of the magnetic stored energy.

These modifiers (p & q) are due to the work-piece not being adequately approximated by a

semi-infinite slab. Their values have been derived from the solution of the Poncare integrals

used to analyse the magnetic field distribution in a finite geometric shape. As such, they are

a function of component shape and relative electrical size.

For cylinders with wdδ being below 8, a graphical lookup of the p and q values is often

used. For values of d/δ>8, the solution approximates to δd

q2= ,

δdp

+=

23.12

, and for

values of d/δ>20, the solution approximates to the semi-infinite slab assumption, so

δdqp

2== .

of 26

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8

p an

d q

valu

es

d/δ

p

q

The induction heating handbook, amongst other books, has the curves in. These curves can

also b found in some of the more advanced electromagnetic field textbooks, although in

some cases, p and q are reversed in meaning.

Example

This example is the same set up as before, except that the steel is at 800oC. The steel will be

past the Curie point, so the current penetration will be much deeper.

First, the relative current penetration depth must be found. The current penetration depth,

given by

ωµρδ 2= ,

has already been found to be 0.0747m or 74.7mm. The diameter of the steel bar is 300mm,

so the steel is not electrically large in diameter.

The ratio of the skin depth to the diameter has to be found: 300/75=4. This is lower than the

simple approximation, so the graphical lookup approach is used. First, we shall find the

equivalent resistance, Rw.

of 27

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8

p an

d q

valu

es

d/δ

p

q

Using the graphical lookup curve, we can find a value for p=0.38. This can be substituted

into the equivalent resistance formula

c

ww l

pANR

2ωµ=

The area of the work piece is also required

2

4 ww dAπ= =0.0707m2.

Remember that N is the number of turns in the work head coil, as the voltage is referred to

the work head coil terminals

20707.038.010000104502 7 ××××××=

−ππwR =0.053Ω

Moving on to Lw, the equivalent inductance of the work piece.

Both d/δ and Aw are common, but q is different from p. Therefore, for an electrically small

work piece, the reactance is different from the resistance. Inspection of the p and q curves

reveals that the q curve is always higher than the p curve, with the two asymptotically

converging. This means that the work-piece is always more reactive than resistive,

especially with small electrical diameter parts.

First, q needs to be found. Using the lookup curve, q is 0.54.

of 28

Substituting all of the relevant data into the Xw formula

20707.054.010000104502 72 ××××××==

−ππωµc

ww l

qANX =0.075Ω

Using L=X/ω, the equivalent inductor can be found

=0.075/(2×π×50)=240µH

As the frequency and the coil and work-piece dimensions are the same as in the previous

example, then the equivalent circuit values for the coil and the air-gap will be the same.

Therefore the power and the efficiency can be found, again with a 1000Vrms excitation

voltage.

First finding the current

I=V/Z

( ) ( )2222gcwcw XXXRRXRZ ++++=+=

( ) ( )22 25.0022.0075.0022.0053.0 ++++=Z =0.354Ω

Ω=∴

354.01000V

I =2820A

Using P=I2R, the losses in the work-piece and in the coil can be found:

Pw=421kW

Pc=172kW

The efficiency can be found by dividing the work-head power by the total power

cw

w

PPP

E+

= =421/(421+172)=71% - compare with 89% for the semi-infinite slab

This reduction in efficiency is due to the airgap, although the efficiency actually peaks when

the work piece is not electrically large, by concentrating the current. The peak efficiency

and therefore the best heating occur around d/δ=3.5, at the peak in the p-curve. This is very

important – it does not occur at d/δ=2, which is predicted by assuming that all of the current

flows in the first skin depth.

of 29

The reason for the lack of power for skin-depths of less than 1/3 of the diameter is a

phenomenon known as ‘flux cancellation’, where flux from one side cancels out flux from

the other side, causing the current flow around the inner circuits of the work-piece to be

reduced. This makes the work-piece a net reactive element in the equivalent circuit.

If the equations for the coil and the air-gap are compared to the new equations for the work-

piece, bearing in mind that2wd

p can replace δ as the effective width of the current flow in

terms of the losses it can be seen that they all share a number of common terms. They are

all in the same coil, so the number of turns and the coil length are common, and they are

excited at the same frequency. Further, the permeability of free space can be taken out.

This can massively simplify the calculation by pre-calculating a common constant:

c

c

lNf

K2

02 µπ=

wrw pAKR µ= , wrw qAKX µ=

ccrcc dKkXR δπ21==

gg KAX =

So far, although we have been calculating values for the equivalent reactance of the work-

head system, we have not really considered its impact, which is to increase the apparent

power (Vrms×Irms), without increasing the actual power (I2rms×R), for a given excitation

current. This means that the power factor is reduced by the reactance (p.f.=P/(VI)).

Given that the real power=I2R

And the apparent power=I2Z,

Power factor=tot

tot

ZR

Example

Find the power factor of the example coil equivalent circuit before and after curie.

Before:

22tottottot XRZ +=

Rtot=0.209+0.0216=0.231Ω

of 30

Xtot=0.209+0.0216+0.25=0.481Ω

Ztot=0.533Ω (already found to get the coil current)

p.f.=Rtot/Ztot=0.231/0.533=0.433

After:

Rtot=0.053+0.022=0.075Ω

Ztot=0.354Ω(already found to get the coil current)

p.f.=0.21

So, after curie, ≈1/5 of the terminal VA of the coil is used in the heating, so the coil would

need five times more VA to heat the work-piece than direct resistance heating would,

despite comparable efficiencies.

As the air-gap adds only to the reactance, the power factor of a real coil is always less than

1/√2, and decreases as the air-gap grows. If we examine the resistance and reactance due to

a large cylindrical work piece (or coil), substituting the skin depth into the equations

22.

2,

22 ρµωπωπµωµ

ρl

dNl

dNXR ==

we can see that the resistance and reactance due to the work piece and the coil are

proportional to the square-root of the frequency, whereas examination of the equation for the

reactance due to the air-gap shows that Xg is proportional to the frequency. This suggests

that as the frequency increases, the power factor will fall. It is not uncommon for RF

induction heating coils to have power factors of less than 1/15.

At the time of writing these notes, I was designing a 100kW induction heater to heat chain

from 720oC to over 1200oC in a coil ~1m long with 20 turns. This machine was to harden

chain for use in anchoring for ships. Due to the non-uniform shape of the coil, and its high

temperature, the area of the air-gap between the steel and the coil was considerably larger

than the cross-sectional area of the chain segments. I measured the power factor to be 0.063.

If this coil was to be directly connected to the 100kVA inverter, a power of only 6.3kW

would be developed. This would be an incredibly expensive heating method, in terms of

plant cost (£4000/kW).

This poor power factor was corrected by the standard method: A capacitor was inserted into

the circuit to supply negative kVARs to cancel out the positive reactive power provided by

the coil. The capacitor and inductor form a resonant circuit, and if this circuit is excited at

of 31

its resonant frequency, the power factor at its terminals will be 1. The actual form of the

circuit is dictated by the nature of the inverter used to supply it. There are two main types of

inverter: voltage-source and current-source. Voltage-source inverters produce a voltage

square-wave across their output terminals and the nature of the coil circuit determines the

current. Current-source inverters try to output a square-wave of current through their output

terminals, and the output circuit determines their output voltage.

If you were to connect a capacitor across the output terminals of a voltage-source inverter,

then the rapid rate of change of voltage with respect to time at the edge of each square-wave

half cycle would cause a large current spike through the capacitor (dtdv

CI = ). This spike

would create large losses in the switching devices within the inverter, and the transistors

would probably rapidly fail. Therefore, for a voltage source inverter, the power factor

correction capacitor is often connected in series with the induction-heating coil. The work-

head circuit and the capacitor form a resonant circuit, and tend to filter out all but the

fundamental of the excitation voltage. This means that although the output voltage of the

inverter is a square wave, the output current is a fairly pure sinusoid. The disadvantage of

this connection is that the inverter has to provide all of the coil current, which can be very

high if few turns are used. In this case, a transformer is used to scale the output voltage and

current appropriately.

D1 SW1 Q1

WORK HEAD

TANK CAPACITOR

D2 SW2 Q1 D3

SW3 Q2

D4 SW4 Q2

Cfilt

L2+RL CT

0 60 120 180 240 300 360 -100 -50

0 50

100

normalised time (deg.)

Cur

rent

(A)

Tank current Link current

0 60 120 180 240 300 360 -400 -200

0 200 400

normalised time (deg.)

Vol

tage

(V)

Conversely, the current-source inverter provides an essentially square-wave of current to the

load circuit. If a series-compensated circuit were used, the high di/dt at the edges of the

of 32

square-wave would cause large voltage spikes across the inductance of the work-head

circuit, which would be carried through the capacitor and the resistance to the inverter

terminals, and would probably rapidly destroy the switching elements in the inverter. A

parallel compensation circuit is therefore used for current-source induction heating inverters.

Once again, this circuit forms a resonant circuit that filters out all but the fundamental of the

excitation waveform, making the voltage across the inverter output terminals a fairly pure

sine-wave.

LSupply D1

SW1 Q

D2

SW2 Q

D3

SW3 Q

D4

SW4 Q

WORK HEAD

TANK CAPACITOR

0 60 120 180 240 300 360 -1000 -500

0 500

1000

Normalised time (deg.)

Vol

tage

(V)

Tank voltage Link voltage

0 60 120 180 240 300 360 -100 -50

0 50

100

Normalised time (deg.) C

urre

nt (A

)

In both cases, for MF and RF systems, the reactive elements in the circuit dominate the

resistive. This makes the resultant behaviour of both circuits highly resonant. As the phase

difference between the fundamental of the square-wave quantity (i.e. voltage for a voltage

source) and the sine wave (i.e. current for a voltage source) is zero when the switching

frequency of the inverter is at the resonant frequency of the network, the power factor is at a

maximum.

As the waveforms of the current and the voltage are different, the power factor cannot be 1.

In fact, for a sinusoidal current and a square wave voltage, the in-phase power factor is

4/(π√2) = 0.9. The square-wave component of the output is often scaled by this value to

give an equivalent sinusoidal voltage or current to make the maximum power factor be equal

to 1 when the current is in phase with the voltage.

Often, the system is run at a slightly inductive power factor to allow the inverter to

commutate correctly, but this is beyond the scope of this course. We will analyse the circuit

at it’s resonant frequency, as it is much simpler and yields sufficiently accurate results.

Consider the series resonant (voltage source) case. At resonance, the voltage across the

capacitor is equal in magnitude, but 180o out of phase, to the voltage across the total

inductive component of the equivalent work-head circuit. This means that the voltage

of 33

across Ccomp and Ltot cancel out, which means that the voltage across the resonant circuit

terminals is equal to the voltage across the equivalent series resistance. Therefore, the

resonant impedance of the circuit is equal to Rtot.

At resonance:

VC=-VL

∴VR=Vin

tot

in

tot

RRin R

VRV

II ===

∴ZT=Rtot

The resonant frequency can be found by equating the reactance of the capacitor with the

negative reactance of the inductance and rearranging. It can be seen that this frequency is

the undamped resonant frequency no matter what the loading is:

from VC=-VL

Solve for ω

LC1

0 =ω

The parallel case is a little more complex. Due to the circuit topology, the voltage across the

capacitor is equal to the voltage across the work-head.

At resonance, the current through the capacitor is equal in magnitude and 180o out pf phase

with the reactive current flowing through the work-head equivalent circuit – this means that

the reactive currents will cancel and the phase shift will be zero. Solving this equality for ω

yields the resonant frequency.

( )222 LRLjRV

I inwork ω

ω+−=

At resonance:

IC=VinjωC=-Im(Iwork)

222 LRLjV

CjV inin ω

ωω+

=

Solve for ω

of 34

CLCRL2

2

0−=ω

If the resonant frequency (in radians s-1) is substituted into the equation for Iwork, then the

terminal current will be equal to the real component of Iwork, as the imaginary component

will be cancelled out by the parallel capacitor’s current at resonance.

LCR

ZT =

In the case of the parallel-tuned circuit, the tuning capacitor and the inductance of the work-

head circuit affect the impedance.

The definition of resonant impedance brings us on to the last part of this course: Matching.

Although an inverter will have a nominal power specification, it will also have a maximum

voltage and a maximum current rating, and will only be able to deliver the rated power at the

maximum voltage and current. The maximum inverter voltage divided by the maximum

inverter current gives the inverter’s nominal output impedance. If the load circuit’s

impedance is above the inverter’s nominal impedance then the output current will not be

high enough to give the full power even when the voltage is turned up to the maximum

level.

Alternatively, if the load circuit’s impedance is lower than the inverter’s nominal

impedance, then the output current will equal the maximum inverter output current before

the voltage is turned up to full, again limiting the power.

If the output circuit’s impedance is wildly wrong, then the inverter may be damaged,

although most inverters have limiting circuits to prevent over currents.

Although it is sometimes possible to choose the number of coil turns etc to make the coil

impedance match the inverter, geometrical, mechanical or metallurgical constrains may

force the coil dimensions to be such that the required impedance is not directly attainable.

In this case, a transformer is often used. There are 2 circuit positions to insert the

transformer, the first being between the work-head and the capacitor (i.e. inside the resonant

circuit):

of 35

Placing a transformer inside the resonant circuit is useful if the required work-head current

or voltage can’t be matched to a standard capacitor. These are often used for single-turn

coils, where currents in excess of 20,000A are required at only 10-100V. MF induction

heating capacitors are generally limited by their terminal current, with 200A/stud being a

common limit, and 8 studs being used to terminate the capacitor. These capacitors often

cost in excess of £1000, and are rated in voltage at 600 to 1200V, depending on capacitance.

Generating the required compensation reactive current by the direct use of such capacitors

would clearly be expensive, as well as technically difficult, so a transformer is often used,

with a ratio of as much as 30:1, to scale the current down and the voltage up, such that the

capacitors and the inverter are better utilised. A transformer inserted between the capacitor

and the work head coil does have to carry the full kVA of the coil, and with power factors

being as low as 0.05, this can be up to 20 times the nominal heating power. This circuit

topology is not very efficient due to the high stress on the transformer, so the transformers

tend to be large and an efficiency of 60% or less can be expected from the resonant circuit

alone.

N 1

Where practical, therefore, a transformer is placed between the inverter terminals and the

resonant circuit. This topology is much more efficient, as the transformer only carries the

heating power, with the reactive power being contained within the resonant circuit.

The ratio of the transformer can be simply dimensioned. If ZT is the terminal impedance of

the resonant circuit, and ZI is the nominal impedance of the inverter, then the ratio can be

found:

T

I

ZZ

N =

This can be easily proven:

VI and II represent the voltage and the current at the inverter terminals

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VT and IT represent the voltage and current at the resonant circuit terminals

First take the voltage across the resonant circuit

NV

V IT =

Then find the current into the resonant circuit

NZV

ZV

IT

I

T

TT ==

The current into the transformer is the current into the resonant circuit divided by the ratio

2NZV

NI

IT

ITI ==

Re-arraigning

II

IT Z

IV

NZ ==2

Solving for N

T

I

ZZ

N =

So, we have found the equivalent circuit for a work-piece in a multi-turn coil, found out how

to chose a value of compensation capacitor to tune the coil to a specific frequency, and

chosen a transformer ratio to match the coil to the inverter. This is the end of the theoretical

part of the course.

Now for some examples:

Example:

Chain Heating

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What power inverter is required to heat the work-piece?

What rating capacitor is required? What transformer ratio is required to match to a 600Vrms Voltage-source output?

Solution:

First find the common constant:

c

c

lNf

K2

02 µπ= =28.42

Assume chain can be represented by k long round bars of same diameter as the wire used to make the links (k=2 –i.e. 2 bars!)

find skin depth in bar

714330261.122

−×××−×==

eeeππωµ

ρδ = 3.0476mm.

find the ratio diameter: skin-depth

3.3=δd

Therefore the graphical look-up table should be used

p=0.38, q=0.67

Resistance of chain = k×resistance of bar

4

2DAw

π= =7.9e-5m2

Freq 30kHz Chain Bar ∅ 10mm Link length 40mm Link width 20mm Chain Material Mild Steel θ Chain (start) 750oC

θ Chain (end) 950oC Power to chain 40kW ρ Chain 1.1×10-6 µ Chain 1 d coil 70mm l coil 1.2m ρ coil 0.017×10-6 N coil 12

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wrw pAKR µ= =28.42×1×0.38×7.9e-5=0.848mΩ

Rchain=1.69mΩ

Reactance of chain = k×reactance of bar

wrw qAKX µ= =28.42×1×0.67×7.9e-5=1.50mΩ

Xchain=2.99mΩ

Airgap area= Acoil-k(Abar)

4

2DAw

π= =7.9e-5m2

4

2DAcoil

π= =3.845e-3m2

Agap=3.69845e-3m2

gg KAX = =104.9mΩ

Resistance and reactance of coil standard (assume kc=1.2)

0

2ωµ

ρδ =c =0.379mm

ccrcc dKkXR δπ21== =0.5×28.42×1.2×π×70e-3×0.379e-3=1.42mΩ

It can be seen that the air-gap dominates the behaviour, and that the coil is not very efficient at all.

Power-

To find the current to achieve the specified power, use Pload=I2Rload.

369.1340−

==ee

RP

Iload

loadcoil =4.87kA

Use Ptot=I2Rtot to find the inverter power

Rtot=1.69e-3+1.42e-3=3.11mΩ

Ptot= I2Rtot=73.6kW

Capacitor-

re-arrange LC1

0 =ω to find C

20

1ωL

C =

L must be found

ωtot

tot

XL = =(2.99e-3+104.9e-3+1.42e-3)/(2π×30e3)= 0.58µH

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C=48µF

Assume voltage-source inverter

Vcap=Icap×1/(2πfC)

Icap=Icoil=4.87kA

Vcap=532V

Inverter Matching-

at resonance, Xterm=Rtot=1.69mΩ+1.42mΩ=3.11mΩ

Inverter required impedence=Xreq=V2/P

Xreq=6002/73600=4.9Ω

to find N, use N2=Xterm/Xreq

N2=3.11e-3/4.9=6.35e-4

N=0.025 (or 40:1)

Induction Heating Basics

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