induction heater tutorial

5/24/2018 Induction Heater Tutorial

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Induction Heater Tutorial

10kw and 3kw

Disclaimer: The topics discussed use high voltage and heat. They can cauproperty damage as well as hurt and kill. This site and author have made thinformation public for educational purposes only. Anyone who reads this aattempts to make a device based on any part of it does so at his/her own ris

This is disavows any responsibility, and does not encourage anyone to do this.

An induction heater is an interesting device, allowing one to rapidly heat

metal object !ith enough power, one can even melt metal The induction heat

works without the need "or "ossil "uels, and can anneal and heat objects

various shapes I set out to make an induction heater that could melt steel a

aluminum #o "ar I have been able to "eed an input power o" over 3 kilowatts$ %

that I have done this I would like to share how it works, and how you can bui

one At the end o" the tutorial I will discuss and show you how to build

levitation coil that will allow you to boil metals while suspended in mid air$

The first part of this tutorial will go through my development of a 3kw invertey initial goal was to rapidly heat metals. y ne!t goal was to levitate metal" succeeded, but reali#ed that " could not levitate solid copper and steel. Thedensity was too great for the magnetic field. This was my final goal: to levitaand suspend molten copper and steel. At the end of this tutorial " will go inthe development of a $%kw unit that reali#ed this goal. " will also elaborate the problems that had to be overcome in order to achieve this.

&et's start.

y induction heater is an inverter. An inverter takes a D( power source aconverts it into A( power. The A( power drives a transformer which is coupled a series &( tank. The inverter fre)uency is set to the tank's resonant fre)uencallowing the generation of very high currents within the tank's coil. The coil coupled to the workpiece and sets up eddy currents. These currents, travelithrough a conductive, but slightly resistive workpiece, heat the piece. *emembe+ower -eat *"0. The workpiece is like a one1turn coil2 the work coils hseveral turns. Thus, we have a step1down transformer, so even higher currents agenerated in the workpiece.

" would like to acknowledge the invaluable help from ohn Dearmond, Tim 4illiam

*ichie 5urnett and other members of the 6hv forum for helping me understand thtopic. 7ow, before we talk more, let's see some pictures of what it can do:


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&ater, " will give a link to a video showing it running. -ere is the inverter:


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4hat " will now do is go over each part. Then, " will give the schematics, over them and how you can build this device.

Induction Heater &omponents

4e will talk about each component making up the induction heater. 8irst, theis the workcoil. This is what heats the workpiece. The workcoil will get very hfrom the high current going through it and the radiation of heat from tworkpiece.

The workcoil is attached to the &( tank. This can either be a series or parallresonant tank. The tank and coil need to be cool, so " implemented a plumbin

type design that allows me to pump water through the coil using a fountain pump

The resonant tank is coupled to the power source with a coupling transformer. Ttransformer is connected to the inverter.

The inverter chops the D( power source at a particular fre)uency. This is tresonant fre)uency of the tank. 7ow, as the workpiece heats and goes through icurie point 1 the temperature when the metal is no longer ferromagnetic 1 tresonant fre)uency changes. The inverter needs to stay locked on as closely possible to the current resonant fre)uency to achieve the fullest power. 9owill do this manually, using an oscilloscope to monitor the waveforms, or using

voltmeter on the tank and tuning the fre)uency to the highest tank voltagAnother method is using a phase locked loop +&&; to monitor the pharelationship of the inverter voltage and tank voltage. This is the method " uand " will discuss this in detail later on.

&et's start with how to easily make a workcoil. 4e will be using fre)uencies the $%s to $%%s of kilohert# k-#;, so metals will conduct the current onslightly below the surface. This is the skin effect. The current depth in mm is

Depth mm; 8;


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9o, the wider the tubing, the lower the resistance. 4e also want to use tubing we can water1cool the coil. " purchased some refrigerator 3/?@ copper tubing fr-ome Depot. ou will also need some $/0@ copper pipe and the necessary fittinso you can feed water through one end, have it circulate through the coil, acome out the other end. " have brass fittings with nipples so " can attach sotubing to my fountain pump, and a return tube to my ice water bath.

This is the tubing " got from -ome Depot.

" want to mention a few points about the workcoil:

ore turns allows you to heat a bigger piece of metal. The coil should allow yto easily heat your workpiece, or to do so with small movements in and out of tfield. The more turns, the less induced voltage, and less induced current in tworkpiece. "f the induced current is too low you may never achieve a high enoutemperature to get beyond the (urie point, where you will then get a significaboost in heating. " believe this occurs, because of the change in the workpiemolecular arrangement, reducing the )uenching effect on the coil.

ou will also have a lower 8res for the same tank capacitance. This results deeper current penetration into the workpiece, which may or may not be desirdepending on your application. All this means it will take longer to heat tmetal for the same input power. To compensate you will need a higher voltagoing to the workcoil if you want to maintain the same rate of heating. ou ccompensate for more turns on your workcoil with fewer turns on your couplitransformer. -owever, you will still be faced with the issue of needing moinput power to achieve the higher e!citation voltage on the workpiece. ou cget more input power by having a higher input voltage or drawing more current

'& Tank( )olypropylene *ilm &apacitor +ank

8or my first capacitor bank " purchased my caps from "llinois (apacitor. can also purchase them from 7ewark Blectronics.

The induction heater uses a workcoil as a step1down transformer. This transformsteps the voltage down, but increases the available current to the workpiecwhich is the one1turn coil that completes the transformer. The magnetic flu! coupled to our workpiece. The better the coupling, the more efficient is oworkcoil. The closer the workpiece is to the coil the better the energy transfer


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This is the workcoil and tank. The capacitors are high voltage metalli#ed powfilm snubbers.

The workcoil is made from shaping the 3/?@ copper tubing. " use brass compressifittings to attach it to the &( tank. The tank is made from two $@ ! 3/$=@ thicopper bars. " drill holes in the bars to accommodate the capacitors. 4e needcapacitor that can handle several hundreds of amps of current. " purchased sopulse capacitors with current ratings of $6A, 3%%%vdc,


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These are the bars with the holes drilled in them. The tank uses capacitors, but you can use any number that gives you the capacitance and currehandling capacity that you re)uire.


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8irst, you need to determine what operating fre)uency you will use. -ighfre)uencies have greater skin effect less penetration; and are good for smallobEects. &ower fre)uencies are better for larger obEects and have greatpenetration. -igher fre)uencies have greater switching losses, but there is lecurrent going through the tank. " choose a fre)uency near


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on a large heat sink because it will be conducting a lot of amperes. y rectifiis rated for 0CA/C%%vac.

4hen " transitioned to my $%kw unit " increased the si#e of my high voltasupply. Bach capacitor is rated for 6C%vdc, so " can go up to I%%vdc between boends. " use two C%A rectifiers giving me $%%A.

The second power supply you will need will be a $Cvdc regulated source. "t

imperitive that it is regulated because the +&& has a voltage controlloscillator. The H( determines the output fre)uency based on input voltage receives. The fre)uency range it can generate is based on its supply voltagHss. "f the supply voltage wanders, the oscillator fre)uency will wander and thwill definitely throw you out of resonance.

7ow, when " made the $%kw unit it uses four mosfets instead of two. This is twithe amount of gate charging. ou need to make sure your $Cvdc supply can suppthe amps to rapidly charge the gates. "t should also have a robust transformand capacitor on the end to make sure there is plenty of charge available. " plon adding an outboard pass transistor. ne problem that plagued me for t

longest time was a Eittery inverter current when " reached modest power levelThe current would Eump back and forth when compared to the inverter voltage. appeared as if two currents were competing. At first " thought this was Baffecting my gate drive and " spent the longest time trying to fi! it. " noticthat when " disconnected any one of the four mosfets the current tracing wperfect. This led me to believe that " was falling short on charging all tgates rapidly enough, and the mosfets were not all conducting identicall*emember, you need to fully turn the mosfets on in the shortest time possible.put a scope on the gates and noticed that the slopes changed when " added tfourth gate. " solved the problem by adding a 3I%%%uf capacitor to my pow


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supply. This was with a $.=A transformer and a 0A $Cv regulator. The tracing wperfect. " plan on changing the transformer to 3A and adding adding the outboapass transistor Eust to play it safe.

*errite Toroids, -&' Theory and Trans"ormer &oupling

" guess the best way to understand what is going on is to start with tworkcoil and work backwards. *emember from earlier " said that the workcoil

the primary end of a step1down transformer. 4e have hundreds of amps flowithrough here and this creates a voltage in the workpiece. 4e achieve these hicurrents because the *(& tank is at resonance. This means that the inductireactance and capacitive reactance cancel out, and all we are left with is tsmall, real resistance.

5elow we have a *(& circuit with a resistance of 6*, Jl 6ohms and (l 3ohmThe reactive impedance cancels to $ohm, giving us a phase shift of $?degreleading. The inductor wins and the inductor voltage leads the current. There only one current running through the series circuit. ou can also say tinductor voltage leads the voltage across the resistor, because the voltage a

current of the resistor are in phase. *emember, the voltage drop across inductor is a reaction against a change in current through it. The instantaneovoltage is #ero when the current is at a peak because the change in current #ero, manifested by the #ero slope.

"f the inductive and capacitive reactance cancel out the phase shift is #ero. Tcurrent in the circuit is in phase with the voltage.

7ow here is an important point. The ma!imum power transfer will occur when tcurrent is in phase with the voltage.


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9o, at resonance, the current in the series circuit is in phase with tvoltage source. "f we are out of resonance the current phase is shifted from #ewith respect to the voltage. "f there is more inductive reactance the currelags the voltage2 if there is more capacitive reactance the current leads tvoltage. ou can also say the capacitor voltage lags the current.

4hat is the voltage source for the series tankK "t is our coupling transformer.am e!perimenting with different materials and turns, but right now " am using

iron powered core from Amidon (orp made from Type 3 material. This material good from fre)uencies between %.%Ch# and %.Ch#. " used two toroids. Bach 0.0C@ in diameter and %.C=C@ thick. " wound $6g wire around for 0%10= turns. " still trying to figure out the optimum turns and the best material. The lower tturns the greater greater the e!citing voltage to the tank. -owevemagneti#ation current goes up as does the load on the inverter.

5elow are the two toroids. " use two to prevent saturation. " wonder how thrwould doK

-ere " have wound $6g wire around. The transformer does not impart a phase shiif place on the tank correctly. "f you flip it around you will introduce a $degree shift which will prevent the +&& from locking onto the fre)uency. uturn it around. 4hich way is the right wayK Lse the right1hand rule.


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-ere is a solenoid with the current flowing in the direction shown. +ut yoright thumb in the direction of the current and your fingers curl in t

direction of the 5 field. The field outside of the coil is not important to uthe field inside the solenoid sums to one large field going from right to lef"f we had a metal bar or part of the toroid's arc inside, the field would travthrough it.

9o here is a mock1up of the coupling transformer. The current travels to tpositive terminal of our toroid transformer output. Lsing the right hand rule can reali#e the direction of the 5 field for each turn. The black arrow on ttoroid shows the direction the field travels in the core. Lsing the right1harule again we see that the current travels through the copper tubing from left right towards the positive terminal of our *&( tank. 4e will use this as tpositive lead for monitoring our tank capacitor voltage later. "f you are unsuwhich end is which you can wind a few turns of wire as a secondary and scope tends. The voltages in and out should be in phase.


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*errite Trans"ormer

4hen " started this proEect " didn't understand how one determined the numbof turns to put on the primary coupling transformer. There are several factors consider. 8irst, the wire needs to be able to handle the current. "f you adealing with high fre)uencies, the maEority of the current is conducted on tsurface. This is the skin effect. ou will need to have several insulated stranto increase the surface area2 these strands will need to be twisted in order reduce eddy currents. As you pack more wire into the space, heating becomes mosignificant. "f your wire is not robust enough you might need a cooling system.

The power to your system has a voltage and a current. "f you have the means run high voltages, you can adEust your windings to keep the primary current lenough to reduce the heating of your transformer and switches. "f " want to kethe primary current low " need more turns on the primary. As long as " haenough voltage, the same primary current will yield a much larger secondacurrent. &et's go over an e!ample:

y transformer has $% turns on the primary and one on the secondary this is tresonant tank;. &et's assume that the load across the secondary is $ ohm. "fhave $%%v on the primary, a $%:$ transformer gives us $%v on the secondary. $of secondary current re)uires $A of primary current. The power draw is $%%4. "fwant to draw less current " can wind a 0%:$ transformer. 7ow, 0%%v on the primaresults in $%v on the secondary. The current is still $%A on the secondary, b

it is %.CA on the primary. This means that as long as " have a higher voltasupply, " can reduce the current my inverter re)uires, and still maintain tsame power to my workpiece. "f " have 6%%v available, " can draw the same $A the primary, but have 0%A available on the secondary.

4hen heating small pieces of metal with small coils, the current demand will up )uickly as there is little material to )uench the tank. ou want a lot turns on the primary in order to keep the current draw low while still supplyia lot of current to the tank. "f you plan on heating large pieces of metal, ttank gets )uenched and the current draw will be too low for effective heatin


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ou need less turns on the primary in order to provide a higher e!citativoltage to the tank.

&et's look at another e!ample where the workpiece is )uenching the tank. "n thcase you don't have enough voltage to get an ade)uate current to flow in ttank. "f you have 0%%v on a 0%:$ transformer you will have $%v on the secondar"f the load is $* you will have $%A on the secondary and %.CA in the primary. our ma!imum voltage is 0%%v we need to draw more current, making sure o

switches can handle this of course. 5y changing to a $%:$ transformer we get 0M 0%A on the secondary2 the primary we have 0%%v M 0A. 4e are drawing more powand we have doubled the output current at the e!pense of needing to deal wifour times the primary current. As long as the primary circuit can handle this have solved the problem. As you go lower on the turns you need to make sure ydo not saturate the core. Also remember that a small amount of the total primacurrent is magneti#ation current.

.scilloscope Tracings

The inverter outputs a drive voltage to the coupling transformer. The curre

in is in phase with the current out. 4hen the tank is at resonance, the tacurrent is in phase with the drive current of the coupling transformer, and is phase with the inverter input voltage. "f anything, you want the current slightly lag the voltage because the mosfets behave better when facing inductive, rather than a capacitive, load. This has to do with the mosfeconducting in the reverse direction. The tracing below is clean and allows me reach very high power levels while maintaining relatively cool mosfets.


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7ow, if the tanks is above resonance we have more inductive reactance. The tanknet current will lag the driving voltage from the coupling transformer. 9ince tinput and output current of the coupling transformer are in phase, the tankcurrent is lagging the inverter driving voltage. 5elow you can see the dominatiinductive reactance results in the inverter current triangle1looking wave; almost I% degrees lagging the inverter voltage s)uare wave;.


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"f we are below the resonant fre)uency capacitive reactance results in tcurrent leading the inverter voltage. Also, there is ringing in the currewaveform and at the inverter voltage transitions. This noise gets worse wihigher power levels and can result in mosfet failure.


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5elow is another e!ample of ringing. ou can see ringing on the voltage at ttransition and on the current waveform. " have positioned them apart for easiviewing. This is due to high inductance on the gate. -eavy current on the gacauses a large &di/dt. The problem can usually be solved by either increasing tgate resistance increase the resistor value;, or decrease the stray inductanby shortening the gate lead. " was able to almost eliminate the ringing shortening the gate lead, but then " did not have enough length for connectitwo in parallel. 9o " changed the value of * from C ohms to $% ohms. The firimage is with the C ohm gate resistor. " was still able to charge the gate wi

$% ohms in a sufficiently short amount of time at $Cv.


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These images show the waveform after the fi!: shortening the gate leads as muas possible to still allow room for paralleling two of them and increasing tgate resistor from C* to $%*. 7otice the clean voltage s)uare wave and the smoocurrent curve. The second image is a blow1up of the first.


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5elow is a basic sketch of the half1bridge inverter. The coil in the middle the coupling transformer to the resonant tank. The arrows show the paths tcurrent takes as the switches alternate between closed1open and open1closed.


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5elow are two sketches. 9ketch " shows ringing if there is too long of a delduring switching. "f the ne!t switch does not close in time, the inductive kiwill drive the voltage too high, causing an overshoot, followed by a large dwhen it finally closes. 9ketch "" shows profound voltage sagging in the middle the waveform. " had this happen when the decoupling capacitors went bad, shortithe current path. The capacitors are needed to remove any D( component from tpulse.

" would like Eust mention that the inductive waveforms is really an e!ponenticurve. "f we can appro!imate the tank above resonane as a *& circuit respondito a step response

The solution to


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is

Analysis of a capacitor dominant *( circuit will yield something similar. 4hdealing with a *(& step response one has

and the 0nd order differential e)uation is

and the general solution is

"f the system is underdamped the solution has the form:

Ht; e1Nt5cos t ; O 5sin t ;;

.scilloscope Tracings II

&et's continue our discussion of oscilloscope tracings so we can bett

understand how the inverter is going to work and lock onto resonance. 8rom tlast page " mentioned that voltage across the tank capacitor lags the current I% degrees. At resonance, the tank current has a #ero phase shift with respect source inverter; tank voltage because the inductive and capacitive reactancancel out. "f you display the inverter voltage and capacitor voltage togetheyou can see the sinusoidal capacitor voltage lags the inverter voltage by degrees. The s)uare wave is the inverter voltage, but you would get the sarelationship if you scoped the voltage output of the toroid transformer.


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4e will monitor this relationship. 4e are at resonance when our +&& chip keeps ninety degrees lagging behind Hinverter. 7ow, we can easily e!ceed the chima!imum input voltage, so we need to clip the top and bottom the the capacitvoltage, and keep it to a ma!imum of $Cv. 4e do this with some clamping diodyeilding this waveform, which will be the signalin input on pin $6 of t-B86%6=.


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5elow is a diagram of the scoped voltages. Lsing a differential probe, tpositive lead goes to the positive inverter lead going to the toroid and tnegative to the negative lead. Lsing a second differential probe we scope theand 1 ends of the capacitor tank. Hc will lag Hinv or Htank. 4e will have invert the Hc waveform, that is shift it $?% degrees, in order for the +&& work, which " will discuss shortly.


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7ow, we are ready to talk about the phase locked loop chip 1 the -B86%6=. Aftthis discussion, we will have enough information to understand the workings the inverter and how it maintains a lock on the resonance.

.scilloscope Tracings III

" have to share some bad waveforms " got one day. " hadn't used my heater asummer and wanted to try it out before giving it to a friend. 5elow are tvoltage/current waveform. Lnderneath is a tracing of the gate drive signal athe inverter voltage from another run. 7otice how the current is no longer a nisinusoid. The negative current prematurely starts to rise and then go back dobefore resuming its normal cycle.


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-ere is another image. The waveform is different from above, but still bi#ar

and not a good sinusoid.


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This is the inverter voltage yellow; and gate drive blue;. 7otice how tvoltage heavily sags and the gate signal is no longer a clean s)uare wave.

-ere is another gate wave that is abnormal taken at a different time.


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As you can see, " was getting strange waveforms and " did not know why.

At first " thought it was the mosfets so " swapped them out. 4hen that failed fi! the problem " redid the gate resistors and shielding. Then, " pulled out tinverter capacitors and replaced them. 9till no good. 8rustrated, " took out tboard and replaced the gate drive capacitors. 4hen this failed " redid the enticircuit board thinking " was getting some type of cross1talk or a failcomponent. " saved myself from buying another tank capacitor by connecting t

coupling transformer to another &( tank. Again, " had the same problem.

" thought " checked everything and " couldn't understand how the waveform hdeteriorated. 9ometimes, the current appeared to go at twice the fre)uency of tinverter voltage. Then, " had a final thought. " started looking at my hivoltage D( supply. " must have reconnected the -H wires to the inverter early the summer. 7otice in the picture how they are not together.


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They should be together to cancel out any stray inductance as shown below.


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Ama#ingly, after days of racking my brain, this simple solution was all that wneeded. " twisted the -H wires close as " had done in the past; and made suthey were close on my inverter board before splitting to each of the -H rails. the fre)uency " am driving my coil, stray inductance and capacitance on the lines is significant and clearly affected my waveforms. 7ot only did this affethe &( tank, but it affected the gate signal and the voltage supply signal to tcircuitry, making things even worse. -opefully, my e!perience will make someonelife easier if these symptoms appear.

5elow are the waveforms for the inverter voltage and current immediately aftthis repair. " have the fre)uency deliberately higher than resonance to preve

reverse currents.


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5elow is the gate signal after this repair.


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)hase 'ocked 'oop /)'' +asics

4hen you read about induction heaters and inverters you will probably coacross the term phase locked loop. The people writing the tutorials will assuyou know all about these. " will make the opposite assumption and give youbrief understanding of the concept so you can understand how this will hemaintain resonance with our induction heater.

A +&& consists of three parts: a voltage controlled oscillator H(;, a lofilter and a phase detector. The H(out drives the device, or inverter gate our case. "t also closes the loop by feeding itself back into the phase detect

so it can get compared with a reference signal.


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The H( generates a C%P duty cycle s)uare wave2 the fre)uency depends on t

input voltage to the H(. The higher the H(input pin I; voltage the higher tH(output fre)uency2 the lower the voltage the lower the fre)uency. The +&& phadetector compares the phases of two inputs: the reference signal on pin $6 athe H(out fre)uency. The phase detector has two options for outputs: +(A$ a+(A0. 4e use the former, which is a Q* gate.

8igure 0


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The logic is high if one of the two inputs is high2 otherwise it is low. "t wigenerate a s)uare wave whose width is based on the phase difference of the tsignals. "f the two waves are I% degrees out of phase the average value of Hpis Hdd/0. The loop filter takes the phase detector output and converts this the input voltage to the H(. The simplest filter is a *( low1pass filter. Tcut1off fre)uency will determine how sensitive the +&& is to phase changes, ahow well it stays locked on the reference signal.

9o what happensK At resonance the tank current is real and in phase with tcoupler transformer voltage, which is in phase with the inverter voltage. Ttank capacitor voltage lags the tank current by I% degrees2 therefore, it lathe inverter voltage by I% degrees. 7ow as the workpiece heats its ferromagnetproperties change. The workcoil becomes a variable inductor and affects tresonant fre)uency of the tank. "f the effective resonance goes down, it seems the circuit that we increased on drive fre)uency to the tank. This makes the tamore inductive. "nductance causes the source voltage lead the tank current. This, the tank current is forced to lag the inverter voltage. The capacitor voltainitially lagged the current by I% degrees. This means the capacitor voltage lathe inverter voltage even more as shown below.

8igure 3

5elow we can see the relationships with Hinv, Hcap and Hphi. Hphi is high Hinv Hcap is high, but not both.


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8igure 6

The top shows Hinv and Hc. An increase in inductive reactance is the same as we increased our inverter drive fre)uency. 4e lower it by decreasing the voltato H(in. 4e see in the top pair that as Hc shifts more to the right of Hinv tQ* region increases. -owever, we need it to decrease in order to yield a lowvoltage for H(. 4e achieve this by inverting Hc to HcRinverted. 7ow HcRinverted shifts to the right, Hphi decreases. 4e integrate this to a voltavalue and use this for H(in. A smaller H(in results in a lower fre)uency and stay in resonance. The fre)uency range is determined by resistors on pins $$ a$0 of the +&&. 4hen, H(in is at ground the fre)uency is at the low1end of trange2 when it is at the supply voltage it is at the high end.

4hen we are at resonance 1 inverter voltage and current are in phase 1 tinverter voltage leads the tank capacitor voltage by I% degrees. Hphi is half half a pulse width see 8igure 0 and 6;. The average voltage is Hdd/0, or


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The capacitor voltage is a clean signal, and was distorted when " tried to shthree signals. 5elow is Eust the inverter and tank capacitor voltage.


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4e need to discuss a few more things about the +&& ne!t.

)hase 'ocked 'oop /)'' +asics II

"f you will recall, here is a block diagram of the +&& device.

There are several filters one can use for the feedback loop. The simplest is tpassive low1pass *( filter. " used the active integrator, which use a * and


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element. To ensure a D( bias does not work itself into the capacitor, " putdischarge resistor in parallel with (. The active filter has more gain than tpassive filter. The phase shift in the beginning is 1I%. " don't know if thhelps keep our signals at 1I% or not. " scoped both the passive and active filtaction by monitoring the relationship of the inverter voltage and current, andcan say that the latter maintained a tighter lock on a 1I% phase differenduring changes in the tank's resonant fre)uency. 5elow is a table of sofilters. " used one similar to the second. " add a variable voltge input to H1

the op1amp, which allows me to fine tune the fre)uency. " usually tune slightly above resonance, using a voltage monitor on the tank voltage for tnear1high point. ne other thing: you need a gain of 1$ after the active filtbecause it inverts the signal. The 1$ gain op1amp will restore the proppolarity.

&et's talk about how we set the free1running +&& fre)uency and the range it ccapture. "f the resonant fre)uency falls with the +&& capture range, the +&& wibe able to find the fre)uency that maintains the I% degree shift that we wanand maintain this phase lock as the fre)uency re)uired for thisphase differenchanges over a wider range of fre)uencies.

-ere is the chip


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These forumulas can be off and re)uire constants as shown below:

ou can also use graphs on the manufacturer datasheets to get you in the ballpafor the values you need. The first step is to determine the capacitor value th

will get you near your 8res at a given Hdd voltage. Determine the * value yneed for 8min, and then determine the * you need for 8ma!.


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&et's do a )uick e!ample. y 8res is =Ck-# and my supply is $Cv. Actually, supply is $6.6v, because " have a diode to protect from hooking up the pos aneg in reverse. " go up the left hand side to the =%kh# row and across to the $supply line. " go straigtht down and get a ($ of 3%%pf. This will be my startipoint for my e)uations. Lsing ($ 33%pf, " will pick some * values and measuthe actual fre)uency in order to determine the values of the constants F$ and F0

4e want to have the center fre)uency, 8min O 8ma!;/0, e)ual our resona

fre)uency, and we want about $%1$Ck-# on either side. 7ow, the chips can vafrom the e)uation by a factor of 6, so you need to multiply each e)uation byconstant. Take a $%%k resistor for *0 and *$. Sround pin I and measure 8mi7e!t, connect pin I to Hdd and measure 8ma!. This will give you F$ and F0.measured C%k-# for 8min, giving me a F$ of $.?$. " then connected pin I to Hand got $C6kh#. 9ubtracing 8min, C%kh#, " was able to dedue that F0 e)uals 3.

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