Induction Heater Tutorial 10kw and 3kw.pdf

7/28/2019 Induction Heater Tutorial 10kw and 3kw.pdf

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http://inductionheatertutorial.com/inductionheater/index.html

Induction Heater Tutorial 10kw and 3kwDisclaimer: The topics discussed use high voltage and heat. They can cause property

damage as well as hurt and kill. This site and author have made this information public for

educational purposes only. Anyone who reads this and attempts to make a device based onany part of it does so at his/her own risk. This is disavows any responsibility, and does notencourage anyone to do this.***This may seem a bit random, but if you want to see me doing a really cool dance routinewith my 13 year old daughter, click here. The video gets interesting in 50 seconds.

An induction heater is an interesting device, allowing one to rapidly heat a metalobject. With enough power, one can even melt metal. The induction heater works without theneed for fossil fuels, and can anneal and heat objects of various shapes. I set out to make aninduction heater that could melt steel and aluminum. So far I have been able to feed an input

power of over 3 kilowatts! Now that I have done this I would like to share how it works, andhow you can build one. At the end of the tutorial I will discuss and show you how to build a

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 inverter. My

initial goal was to rapidly heat metals. My next goal was to levitate metals. I succeeded, butrealized that I could not levitate solid copper and steel. Their density was too great for themagnetic field. This was my final goal: to levitate and suspend molten copper and steel. Atthe end of this tutorial I will go into the development of a 10kw unit that realized this goal. Iwill also elaborate on the problems that had to be overcome in order to achieve this.

Let's start.My induction heater is an inverter. An inverter takes a DC power source and converts

it into AC power. The AC power drives a transformer which is coupled to a series LC tank.The inverter frequency is set to the tank's resonant frequency, allowing the generation ofvery high currents within the tank's coil. The coil is coupled to the workpiece and sets upeddy currents. These currents, traveling through a conductive, but slightly resistiveworkpiece, heat the piece. Remember, Power = Heat = R*I^2. The workpiece is like a one-turn coil; the work coils has several turns. Thus, we have a step-down transformer, so evenhigher currents are generated in the workpiece.

I would like to acknowledge the invaluable help from John Dearmond, Tim Williams,Richie Burnett and other members of the 4hv forum for helping me understand this topic.

Now, before we talk more, let's see some pictures of what it can do:

Later, I will give a link to a video showing it running. Here is the inverter:
http://inductionheatertutorial.com/inductionheater/index.htmlhttp://inductionheatertutorial.com/inductionheater/index.htmlhttp://inductionheatertutorial.com/inductionheater/index.html


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What I will now do is go over each part. Then, I will give the schematics, go over them andhow you can build this device.

Induction Heater Components

We will talk about each component making up the induction heater. First, there is theworkcoil. This is what heats the workpiece. The workcoil will get very hot from the highcurrent going through it and the radiation of heat from the workpiece.

The workcoil is attached to the LC tank. This can either be a series or parallelresonant tank. The tank and coil need to be cool, so I implemented a plumbing-type designthat 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. Thetransformer is connected to the inverter.

The inverter chops the DC power source at a particular frequency. This is theresonant frequency of the tank. Now, as the workpiece heats and goes through its curie point- the temperature when the metal is no longer ferromagnetic - the resonant frequencychanges. The inverter needs to stay locked on as closely as possible to the current resonantfrequency to achieve the fullest power. Some will do this manually, using an oscilloscope to

monitor the waveforms, or using a voltmeter on the tank and tuning the frequency to thehighest tank voltage. Another method is using a phase locked loop (PLL) to monitor thephase relationship of the inverter voltage and tank voltage. This is the method I use and I willdiscuss this in detail later on.

Let's start with how to easily make a workcoil. We will be using frequencies in the10s to 100s of kilohertz (kHz), so metals will conduct the current only slightly below thesurface. This is the skin effect. The current depth in mm is

Depth (mm) = 76/(F)

So, the wider the tubing, the lower the resistance. We also want to use tubing so we

can water-cool the coil. I purchased some refrigerator 3/8" copper tubing from Home Depot.You will also need some 1/2" copper pipe and the necessary fittings so you can feed waterthrough one end, have it circulate through the coil, and come out the other end. I have brassfittings with nipples so I can attach some tubing to my fountain pump, and a return tube tomy ice water bath.

This is the tubing I got from Home Depot.


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I want to mention a few points about the workcoil:

More turns allows you to heat a bigger piece of metal. The coil should allow you toeasily heat your workpiece, or to do so with small movements in and out of the field. Themore turns, the less induced voltage, and less induced current in the workpiece. If theinduced current is too low you may never achieve a high enough temperature to get beyondthe Curie point, where you will then get a significant boost in heating. I believe this occurs,

because of the change in the workpiece molecular arrangement, reducing the quenchingeffect on the coil.

You will also have a lower Fres for the same tank capacitance. This results in deepercurrent penetration into the workpiece, which may or may not be desired depending on yourapplication. All this means it will take longer to heat the metal for the same input power. Tocompensate you will need a higher voltage going to the workcoil if you want to maintain thesame rate of heating. You can compensate for more turns on your workcoil with fewer turnson your coupling transformer. However, you will still be faced with the issue of needingmore input power to achieve the higher excitation voltage on the workpiece. You can getmore input power by having a higher input voltage or drawing more current.

LC Tank: Polypropylene Film Capacitor BankFor my first capacitor bank I purchased my caps from Illinois Capacitor. You canalso purchase them from Newark Electronics.

The induction heater uses a workcoil as a step-down transformer. This transformersteps the voltage down, but increases the available current to the workpiece, which is theone-turn coil that completes the transformer. The magnetic flux is coupled to our workpiece.The better the coupling, the more efficient is our workcoil. The closer the workpiece is to thecoil the better the energy transfer.

This is the workcoil and tank. The capacitors are high voltage metallized power filmsnubbers.


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The workcoil is made from shaping the 3/8" copper tubing. I use brass compression

fittings to attach it to the LC tank. The tank is made from two 1" x 3/16" thick copper bars. Idrill holes in the bars to accommodate the capacitors. We need a capacitor that can handleseveral hundreds of amps of current. I purchased some pulse capacitors with current ratingsof 14A, 3000vdc, 750vac. With 20 capacitors this is close to 300A average current. Thecoupling transformer fits over the copper tubing. If you look closely, you will see the

fountain pump submerged in water. This pumps ice water through the tank and back out intothe bucket. Water flows in from the bottom left, through the copper pipe soldered to the busbar, through the coil, over the bank to the upper left, and through the tubing connected to theother bus bar, and out on the upper right. You should also take note where the workcoilconnects with the capacitor bank. It does not connect both leads at the front end; instead, thecoil connects to opposite ends. This ensures that the capacitors share an equal current load.Otherwise, if both end connected to the front, the capacitors closest to the coil would handlethe brunt of current because the resistance would be the least. When you are dealing withhundreds of amps, small changes in R are significant.


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These are the bars with the holes drilled in them. The tank uses 20 capacitors, but you

can use any number that gives you the capacitance and current handling capacity that yourequire.


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First, you need to determine what operating frequency you will use. Higher

frequencies have greater skin effect (less penetration) and are good for smaller objects.Lower frequencies are better for larger objects and have greater penetration. Higherfrequencies have greater switching losses, but there is less current going through the tank. Ichoose a frequency near 70khz and wound up with about 66khz. My capacitor bank is 4.4ufand can handle over 300A. My coil is near 1uH. The capacitors are from Illinois Capacitors.Mine are 0.22uf/3000vdc. The model number is 224PPA302KS.

Fres = 1/2(LC)

Once you wind your coil you can get an idea of its value by making a simple RLCcircuit with it and connect it to a function generator and scope. I used a 1R resistor and a500pf capacitor. I increased my function generator sine wave and measured the voltageacross R. At resonance the LC impedance drops and the voltage drop across R peaks. Thisgave me a ballpark figure, but you can just go by the calculation.

Now, as far as the workcoil goes you can form the workcoil by driving a piece ofPVC tubing into the ground. I used a 1" pipe (1.5" OD). Take the copper tubing and fill it

with sand or salt. Make sure it is completely filled. This way it will act like a solid tube andwill not collapse when you bend it. Fix one end with something like a heavy vice and work


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the tubing around your PVC tube until you have your desired number of turns. Four to fiveturns at 1.5-2" will give you a coil with an inductance between 0.8 - 1.3 uH.

You can see how nicely the coil forms around the pipe. Once you are happy with theturns and shape you can blow the sand out with an air compressor.

Power Supply: Voltage double and regulated source

I need to talk about two power supplies for the unit. One is the high voltage DC that

the inverter converts to AC for feeding the tank. You need an unregulated, smoothed source.You can use 110vac through a rectifier and smoothed with a 1000uf-1500uf capacitor for asupply of 170vdc. I used a voltage multiplier to convert it to 320vdc. Below are some basicschematics for a voltage doubler. I used the third variation for mine. Make sure you haveyour rectifier on a large heat sink because it will be conducting a lot of amperes. My rectifieris rated for 25A/500vac.

When I transitioned to my 10kw unit I increased the size of my high voltage supply.Each capacitor is rated for 450vdc, so I can go up to 900vdc between both ends. I use two

50A rectifiers giving me 100A.The second power supply you will need will

be a 15vdc regulated source. It is imperitive that it is

regulated because the PLL has a voltage controlledoscillator. The VCO determines the output frequencybased on input voltage it receives. The frequencyrange it can generate is based on its supply voltage,Vss. If the supply voltage wanders, the oscillatorfrequency will wander and this will definitely throwyou out of resonance.

Now, when I made the 10kw unit it uses fourmosfets instead of two. This is twice the amount ofgate charging. You need to make sure your 15vdcsupply can supply the amps to rapidly charge the

gates. It should also have a robust transformer andcapacitor on the end to make sure there is plenty ofcharge available. I plan on adding an outboard passtransistor. One problem that plagued me for thelongest time was a jittery inverter current when Ireached modest power levels. The current would jump

back and forth when compared to the inverter voltage.It appeared as if two currents were competing. At firstI thought this was EMI affecting my gate drive and I

spent the longest time trying to fix it. I noticed that when I disconnected any one of the fourmosfets the current tracing was perfect. This led me to believe that I was falling short oncharging all the gates rapidly enough, and the mosfets were not all conducting identically.Remember, you need to fully turn the mosfets on in the shortest time possible. I put a scope


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on the gates and noticed that the slopes changed when I added the fourth gate. I solved theproblem by adding a 39000uf capacitor to my power supply. This was with a 1.6Atransformer and a 2A 15v regulator. The tracing was perfect. I plan on changing thetransformer to 3A and adding adding the outboard pass transistor just to play it safe.

Ferrite Toroids, RCL Theory and Transformer CouplingI guess the best way to understand what is going on is to start with the workcoil and

work backwards. Remember from earlier I said that the workcoil is the primary end of astep-down transformer. We have hundreds of amps flowing through here and this creates avoltage in the workpiece. We achieve these high currents because the RCL tank is atresonance. This means that the inductive reactance and capacitive reactance cancel out, andall we are left with is the small, real resistance.

Below we have a RCL circuit with a resistance of 4R, Zl = 4ohms and Cl = 3ohms.The reactive impedance cancels to 1ohm, giving us a phase shift of 18degrees leading. Theinductor wins and the inductor voltage leads the current. There is only one current runningthrough the series circuit. You can also say the inductor voltage leads the voltage across the

resistor, because the voltage and current of the resistor are in phase. Remember, the voltagedrop across an inductor is a reaction against a change in current through it. The instantaneousvoltage is zero when the current is at a peak because the change in current is zero, manifested

by the zero slope.

If the inductive and capacitive reactance cancel out the phase shift is zero. Thecurrent in the circuit is in phase with the voltage.

Now here is an important point. The maximum power transfer will occur when thecurrent is in phase with the voltage.

So, at resonance, the current in the series circuit is in phase with the voltage source. Ifwe are out of resonance the current phase is shifted from zero with respect to the voltage. Ifthere is more inductive reactance the current lags the voltage; if there is more capacitive

reactance the current leads the voltage. You can also say the capacitor voltage lags thecurrent.


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What is the voltage source for the series tank? It is our coupling transformer. I amexperimenting with different materials and turns, but right now I am using an iron poweredcore from Amidon Corp made from Type 3 material. This material is good from frequencies

between 0.05Mhz and 0.5Mhz. I used two toroids. Each is 2.25" in diameter and 0.565"thick. I wound 14g wire around for 20-26 turns. I am still trying to figure out the optimum

turns and the best material. The lower the turns the greater greater the exciting voltage to thetank. However, magnetization current goes up as does the load on the inverter.Below are the two toroids. I use two to prevent saturation. I wonder how three would

do?Here I have wound 14g wire around. The

transformer does not impart a phase shift if place on thetank correctly. If you flip it around you will introduce a180 degree shift which will prevent the PLL fromlocking onto the frequency. Just turn it around. Whichway is the right way? Use the right-hand rule.

Here is a solenoid with the current flowing inthe direction shown. Put your right thumb in thedirection of the current and your fingers curl in thedirection of the B field. The field outside of the coil isnot important to us; the field inside the solenoid sumsto one large field going from right to left. If we had ametal bar or part of the toroid's arc inside, the fieldwould travel through it.

So here is a mock-up of the coupling transformer. The current travels to the positiveterminal of our toroid transformer output. Using the right hand rule we can realize thedirection of the B field for each turn. The black arrow on the toroid shows the direction thefield travels in the core. Using the right-hand rule again we see that the current travelsthrough the copper tubing from left to right towards the positive terminal of our RLC tank.We will use this as the positive lead for monitoring our tank capacitor voltage later. If youare unsure which end is which you can wind a few turns of wire as a secondary and scope theends. The voltages in and out should be in phase.

Ferrite Transformer

When I started this project I didn't understand how one determined the number ofturns to put on the primary coupling transformer. There are several factors to consider. First,


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the wire needs to be able to handle the current. If you are dealing with high frequencies, themajority of the current is conducted on the surface. This is the skin effect. You will need tohave several insulated strands to increase the surface area; these strands will need to betwisted in order to reduce eddy currents. As you pack more wire into the space, heating

becomes more significant. If your wire is not robust enough you might need a cooling

system.The power to your system has a voltage and a current. If you have the means to runhigh voltages, you can adjust your windings to keep the primary current low enough toreduce the heating of your transformer and switches. If I want to keep the primary currentlow I need more turns on the primary. As long as I have enough voltage, the same primarycurrent will yield a much larger secondary current. Let's go over an example:

My transformer has 10 turns on the primary and one on the secondary (this is theresonant tank). Let's assume that the load across the secondary is 1 ohm. If I have 100v onthe primary, a 10:1 transformer gives us 10v on the secondary. 10A of secondary currentrequires 1A of primary current. The power draw is 100W. If I want to draw less current I canwind a 20:1 transformer. Now, 200v on the primary results in 10v on the secondary. The

current is still 10A on the secondary, but it is 0.5A on the primary. This means that as long asI have a higher voltage supply, I can reduce the current my inverter requires, and stillmaintain the same power to my workpiece. If I have 400v available, I can draw the same 1Aon the primary, but have 20A available on the secondary.

When heating small pieces of metal with small coils, the current demand will go upquickly as there is little material to quench the tank. You want a lot of turns on the primary inorder to keep the current draw low while still supplying a lot of current to the tank. If you

plan on heating large pieces of metal, the tank gets quenched and the current draw will be toolow for effective heating. You need less turns on the primary in order to provide a higherexcitation voltage to the tank.

Let's look at another example where the workpiece is quenching the tank. In this caseyou don't have enough voltage to get an adequate current to flow in the tank. If you have200v on a 20:1 transformer you will have 10v on the secondary. If the load is 1R you willhave 10A on the secondary and 0.5A in the primary. If our maximum voltage is 200v weneed to draw more current, making sure our switches can handle this of course. By changingto a 10:1 transformer we get 20v @ 20A on the secondary; the primary we have 200v @ 2A.We are drawing more power and we have doubled the output current at the expense ofneeding to deal with four times the primary current. As long as the primary circuit canhandle this we have solved the problem. As you go lower on the turns you need to make sureyou do not saturate the core. Also remember that a small amount of the total primary currentis magnetization current.

Oscilloscope TracingsThe inverter outputs a drive voltage to the coupling transformer. The current in is in

phase with the current out. When the tank is at resonance, the tank current is in phase withthe drive current of the coupling transformer, and is in phase with the inverter input voltage.If anything, you want the current to slightly lag the voltage because the mosfets behave

better when facing an inductive, rather than a capacitive, load. This has to do with themosfets conducting in the reverse direction. The tracing below is clean and allows me toreach very high power levels while maintaining relatively cool mosfets.


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Now, if the tanks are above resonance we have more inductive reactance. The tank'snet current will lag the driving voltage from the coupling transformer. Since the input andoutput current of the coupling transformer are in phase, the tank's current is lagging theinverter driving voltage. Below you can see the dominating inductive reactance results in theinverter current (triangle-looking wave) is almost 90 degrees lagging the inverter voltage(square wave).

If we are below the resonant frequency capacitive reactance results in the currentleading the inverter voltage. Also, there is ringing in the current waveform and at the invertervoltage transitions. This noise gets worse with higher power levels and can result in mosfetfailure.


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Below is another example of ringing. You can see ringing on the voltage at thetransition and on the current waveform. I have positioned them apart for easier viewing. Thisis due to high inductance on the gate. Heavy current on the gate causes a large Ldi/dt. The

problem can usually be solved by either increasing the gate resistance (increase the resistorvalue), or decrease the stray inductance by shortening the gate lead. I was able to almosteliminate the ringing by shortening the gate lead, but then I did not have enough length forconnecting two in parallel. So I changed the value of R from 5 ohms to 10 ohms. The firstimage is with the 5 ohm gate resistor. I was still able to charge the gate with 10 ohms in asufficiently short amount of time at 15v.

These images show the waveform after the fix: shortening the gate leads as much aspossible to still allow room for paralleling two of them and increasing the gate resistor from5R to 10R. Notice the clean voltage square wave and the smooth current curve. The secondimage is a blow-up of the first.


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Below is a basic sketch of the half-bridge inverter. The coil in the middle is the couplingtransformer to the resonant tank. The arrows show the paths the current takes as the switchesalternate between closed-open and open-closed.

Below are two sketches. Sketch I shows ringing if there is too long of a delay duringswitching. If the next switch does not close in time, the inductive kick will drive the voltagetoo high, causing an overshoot, followed by a large dip when it finally closes. Sketch IIshows profound voltage sagging in the middle of the waveform. I had this happen when thedecoupling capacitors went bad, shorting the current path. The capacitors are needed toremove any DC component from the pulse.

I would like just mention that the inductive waveforms is really an exponential curve. If wecan approximate the tank above resonane as a RL circuit responding to a step response


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The solution to

is

Analysis of a capacitor dominant RC circuit will yield something similar. When dealing witha RCL step response one has

and the 2nd order differential equation is

and the general solution is

If the system is underdamped the solution has the form:

V(t) = e-t(Bcos(t) + Bsin(t))

Oscilloscope Tracings II

Let's continue our discussion of oscilloscope tracings so we can better understandhow the inverter is going to work and lock onto resonance. From the last page I mentionedthat voltage across the tank capacitor lags the current by 90 degrees. At resonance, the tankcurrent has a zero phase shift with respect to source (inverter) tank voltage because theinductive and capacitive reactance cancel out. If you display the inverter voltage andcapacitor voltage together, you can see the sinusoidal capacitor voltage lags the invertervoltage by 90 degrees. The square wave is the inverter voltage, but you would get the same

relationship if you scoped the voltage output of the toroid transformer.


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We will monitor this relationship. We are at resonance when our PLL chip keeps Vcninety degrees lagging behind Vinverter. Now, we can easily exceed the chips maximuminput voltage, so we need to clip the top and bottom the the capacitor voltage, and keep it toa maximum of 15v. We do this with some clamping diodes yeilding this waveform, whichwill be the signalin input on pin 14 of the HEF4046.

Below is a diagram of the scoped voltages. Using a differential probe, the positivelead goes to the positive inverter lead going to the toroid and the negative to the negativelead. Using a second differential probe we scope the + and - ends of the capacitor tank. Vcwill lag Vinv or Vtank. We will have to invert the Vc waveform, that is shift it 180 degrees,in order for the PLL to work, which I will discuss shortly.


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Now, we are ready to talk about the phase locked loop chip - the HEF4046. After thisdiscussion, we will have enough information to understand the workings of the inverter andhow it maintains a lock on the resonance.

Oscilloscope Tracings IIII have to share some bad waveforms I got one day. I hadn't used my heater all

summer and wanted to try it out before giving it to a friend. Below are the voltage/currentwaveform. Underneath is a tracing of the gate drive signal and the inverter voltage fromanother run. Notice how the current is no longer a nice sinusoid. The negative current

prematurely starts to rise and then go back down before resuming its normal cycle.

Here is another image. The waveform is different from above, but still bizarre and nota good sinusoid.

This is the inverter voltage (yellow) and gate drive (blue). Notice how the voltageheavily sags and the gate signal is no longer a clean square wave.


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

As you can see, I was getting strange waveforms and I did not know why.At first I thought it was the mosfets so I swapped them out. When that failed to fix

the problem I redid the gate resistors and shielding. Then, I pulled out the inverter capacitorsand replaced them. Still no good. Frustrated, I took out the board and replaced the gate drivecapacitors. When this failed I redid the entire circuit board thinking I was getting some typeof cross-talk or a failed component. I saved myself from buying another tank capacitor byconnecting the coupling transformer to another LC tank. Again, I had the same problem.

I thought I checked everything and I couldn't understand how the waveform haddeteriorated. Sometimes, the current appeared to go at twice the frequency of the invertervoltage. Then, I had a final thought. I started looking at my high voltage DC supply. I musthave reconnected the HV wires to the inverter early in the summer. Notice in the picture howthey are not together.


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

Amazingly, after days of racking my brain, this simple solution was all that wasneeded. I twisted the HV wires close (as I had done in the past) and made sure they wereclose on my inverter board before splitting to each of the HV rails. At the frequency I amdriving my coil, stray inductance and capacitance on the HV lines is significant and clearlyaffected my waveforms. Not only did this affect the LC tank, but it affected the gate signaland the voltage supply signal to the circuitry, making things even worse. Hopefully, myexperience will make someone's life easier if these symptoms appear.

Below are the waveforms for the inverter voltage and current immediately after thisrepair. I have the frequency deliberately higher than resonance to prevent reverse currents.


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Below is the gate signal after this repair.

Phase Locked Loop (PLL) BasicsWhen you read about induction heaters and inverters you will probably come across

the term phase locked loop. The people writing the tutorials will assume you know all aboutthese. I will make the opposite assumption and give you a brief understanding of the conceptso you can understand how this will help maintain resonance with our induction heater.

A PLL consists of three parts: a voltage controlled oscillator (VCO), a loop filter anda phase detector. The VCOout drives the device, or inverter gate in our case. It also closesthe loop by feeding itself back into the phase detector so it can get compared with a referencesignal.

The VCO generates a 50% duty cycle square wave; the frequency depends on the inputvoltage to the VCO. The higher the VCOinput (pin 9) voltage the higher the VCOoutputfrequency; the lower the voltage the lower the frequency. The PLL phase detector comparesthe phases of two inputs: the reference signal on pin 14 and the VCOout frequency. The

phase detector has two options for outputs: PCA1 and PCA2. We use the former, which is aXOR gate.


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Figure 2

The logic is high if one of the two inputs is high; otherwise it is low. It will generate asquare wave whose width is based on the phase difference of the two signals. If the twowaves are 90 degrees out of phase the average value of Vphi is Vdd/2. The loop filter takesthe phase detector output and converts this to the input voltage to the VCO. The simplestfilter is a RC low-pass filter. The cut-off frequency will determine how sensitive the PLL isto phase changes, and how well it stays locked on the reference signal.

So what happens? At resonance the tank current is real and in phase with the couplertransformer voltage, which is in phase with the inverter voltage. The tank capacitor voltagelags the tank current by 90 degrees; therefore, it lags the inverter voltage by 90 degrees. Nowas the workpiece heats its ferromagnetic properties change. The workcoil becomes a variableinductor and affects the resonant frequency of the tank. If the effective resonance goes down,it seems to the circuit that we increased on drive frequency to the tank. This makes the tankmore inductive. Inductance causes the source voltage lead the tank current. That is, the tankcurrent is forced to lag the inverter voltage. The capacitor voltage initially lagged the current

by 90 degrees. This means the capacitor voltage lags the inverter voltage even more asshown below.

Figure 3

Below we can see the relationships with Vinv, Vcap and Vphi. Vphi is high Vinv or Vcap ishigh, but not both.


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Figure 4

The top shows Vinv and Vc. An increase in inductive reactance is the same as if weincreased our inverter drive frequency. We lower it by decreasing the voltage to VCOin. Wesee in the top pair that as Vc shifts more to the right of Vinv the XOR region increases.However, we need it to decrease in order to yield a lower voltage for VCO. We achieve this

by inverting Vc to Vc_inverted. Now as Vc_inverted shifts to the right, Vphi decreases. Weintegrate this to a voltage value and use this for VCOin. A smaller VCOin results in a lowerfrequency and we stay in resonance. The frequency range is determined by resistors on pins11 and 12 of the PLL. When, VCOin is at ground the frequency is at the low-end of therange; when it is at the supply voltage it is at the high end.

When we are at resonance - inverter voltage and current are in phase - the invertervoltage leads the tank capacitor voltage by 90 degrees. Vphi is half of half a pulse width (seeFigure 2 and 4). The average voltage is Vdd/2, or 7.5v if our supply is 15v. So, 7.5v atVCOin will keep us close to resonance if our center frequency is Fres. The problem is that

Fres changes with different workpieces and during heating. However, the PLL will adjustitselft to maintain a lock on the phase relationship.The scope images below show these waveforms. The first picture is at a lower

frequency than the bottom picture. Shown are Vcap_inverted, Vinv, and Vphi. The capacitorvoltage is clipped to protect the PLL chip.

The capacitor voltage is a clean signal, and was distorted when I tried to show threesignals. Below is just the inverter and tank capacitor voltage.


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We need to discuss a few more things about the PLL next.

Phase Locked Loop (PLL) Basics IIIf you will recall, here is a block diagram of the PLL device.

There are several filters one can use for the feedback loop. The simplest is the passivelow-pass RC filter. I used the active integrator, which use a R and C element. To ensure aDC bias does not work itself into the capacitor, I put a discharge resistor in parallel with C.

The active filter has more gain than the passive filter. The phase shift in the beginning is -90.I don't know if this helps keep our signals at -90 or not. I scoped both the passive and activefilter action by monitoring the relationship of the inverter voltage and current, and I can saythat the latter maintained a tighter lock on a -90 phase difference during changes in the tank'sresonant frequency. Below is a table of some filters. I used one similar to the second. I add avariable voltage input to V- on the op-amp, which allows me to fine tune the frequency. Iusually tune it slightly above resonance, using a voltage monitor on the tank voltage for thenear-high point. One other thing: you need a gain of -1 after the active filter because itinverts the signal. The -1 gain op-amp will restore the proper polarity.


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Let's talk about how we set the free-running PLL frequency and the range it can capture. Ifthe resonant frequency falls with the PLL capture range, the PLL will be able to find the

frequency that maintains the 90 degree shift that we want, and maintain this phase lock as thefrequency required for this phase difference changes over a wider range of frequencies.Here is the chip

These formulas can be off and require constants as shown below:


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reverse. I go up the left hand side to the 60khz row and across to the 15v supply line. I gostraight down and get a C1 of 300pf. This will be my starting point for my equations. UsingC1 = 330pf, I will pick some R values and measure the actual frequency in order todetermine the values of the constants K1 and K2.

We want to have the center frequency, (Fmin + Fmax)/2, equal our resonant

frequency, and we want about 10-15kHz on either side. Now, the chips can vary from theequation by a factor of 4, so you need to multiply each equation by a constant. Take a 100kresistor for R2 and R1. Ground pin 9 and measure Fmin. Next, connect pin 9 to Vdd andmeasure Fmax. This will give you K1 and K2. I measured 50kHz for Fmin, giving me a K1of 1.81. I then connected pin 9 to Vdd and got 154khz. Subtracing Fmin, 50khz, I was able todedue that K2 equals 3.78. My frequency is 65kHz, so I want something between 50-80kHz.I will use a 330pf capacitor, as determined from the graph, and values of K1 = 1.81 and K2 =3.78. I now use these values to determine the true values of R1 and R2 that I need, which is100k and 348k. The calculations are below. Of course, you need to verify this with yourscope.

Fmin = 50khz = 1.81/(100,000 x 362e-12)

Fmax = 80khz = 50khz + 3.78/(348,000 x 362e-12)

On my circuit I add a trim pot and another resistor in parallel to R2 with an optionaljumper. This gives me a selection of resonant frequency ranges.

So, how does our circuit come together? Let's see.

Induction Heater Inverter Schematic

Most of the electronics components on the schematic are from Digikey Corp andMouser Electronics.

PLL OVERVIEWThe PLL receives two inputs through pins 14 and 9. Pin 14 is the clamped capacitor

tank voltage. It is inverter (shifted 180 degrees) in order for the feedback to work properly.The high voltages are kept down with R1. All inverter grounds are isolated from earthground. C7 and resistors on pins 11 and 12 set the capture range. Jumper JP1 converts pin12's resistor from 100k to 60k. R5 affords you the ability to vary the capture range even morefor tuning the center frequency to the tank resonant frequency. We will discuss this at theend.

PCAout goes through the active integrator filter, which is made up of a quad op-amp.The integrator output then goes through a filter with a gain of -1 to restore the polarity of thesignal. During use, jumper JP2 is open and JP3 is closed to allow the feedback to get to pin9. The drive frequency leaves pin 4 and drives a non-inverting and inverting gate drive.These chips drive the primary of a 1:1:1 gate drive transformer, T1. C1 removes DC bias.Diodes D5 and D6 offer some delay so both mosfets are not on at the same time. Theseseries diodes have nothing to do with reverse currents, like the one's you are used to seeingacross the DS juntion. Again, they are for timing. Your tracings should be short to the gatedrive on the mosfet. I have connectors on my board going to wires which run to the chips onlarge heat sinks. The wire acts like an antenna and you can get noise which will induce wildoscillations in your mosfets, destroying them. I put a ferrite bead that attenutes frequencies

above 300khz right before the lead to the gate drive. This works perfectly. Below are the


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tracings going to the gates and then showing the tracing from one of the gates and theinverter output.

TIMING DELAY FOR MOSFETSThese are the gates drive signals going to the mosfets. The signals are superimposed.

The small slope is part of the delay imparted by the series diodes D5 and D6.

This is the inverter tracing on top, and one of the gate drive signals on the bottom.With this mosfet, when the gate is high the DS junction grounds the power, so the voltagedrops to zero.

Below is the timing showing ZVS. The voltage goes through zero volts exactly whenthe current is zero.

Below, one of the series diodes is shorted, so we can compare the timing of the signalgoing to both gates. The bottom tracing has the transformer gate drive going directly to thegate. The tracing above it goes through the diode The temporal difference between the gatedrive with and without the series diode is close to 100ns.


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Below are the gate drive waveforms with both series diodes working.

Below, both series diodes are shorted, and we can see the time to reach the samevoltage is delayed by about 200ns.

OP-AMP INTEGRATORThe op-amp is centered around Vdd/2. R10 moves the center point on the integrator

allowing you to fine tune PLL frequency. You can force it to stay a little above resonance byadjusting it. When connecting it to the circuit, set it up so clockwise motion increases thePLL frequency.

MOSFETS, ZERO VOLT SWITCHING, and CURRENT MONITORINGMosfets U1 and U2 have ultra-fast diodes across the source and drain to protect the

slower acting intrinsic drain diodes. There are no series isolation diodes with the mosfets fortwo reasons. We are doing zero volt/current switching which is guarenteed when the circuitis in tune by the PLL. When we switch the mosfet there is no current or voltage on thedevice. Secondly, the present day mosfets have very fast intrinsic diodes, rated for flywheelservice. Capacitors C1 and C2 set a point half-way above ground, which gets charged when

U1 is open, and discharges to ground when U1 closes and U2 opens. The current transformer


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T2 uses a 1:100T ratio to monitor the inverter current. The 100R resistor means that every1V on the oscilloscope is 1A of current going to the coupling transformer.

Now, let's look at the tank circuit schematic. The inverter output is coupled to thetank through T3, which is a 20:1 toroid transformer. The 20 turn primary is connected to theinverter output. The coper tubing which form the connects for workcoil and capacitor serves

as a one-turn primary. You can experiment with different toroid materials and turn-ratios.The resonant frequency will changes as the material goes through its curie point.Below is a picture of the current conduction through the inverter during different

phases of the power transfer cycle. It shows how the free-wheeling diodes come into play todivert the reverse current around the mosfet.

During Mode 1, the upper mosfet is conducting and transfering power to the resonanttank through the coupling transformer in our circuit. In Mode 2, the mosfets aretransitioning, and the upper mosfet turns off slightly before the bottom one turns on. Here,current is conducted through the free-wheeling diode of the lower mosfet. In Mode 3, thelower mosfet turns on, and the resonant tank throws the power back through mosfet. In Mode3, both mosfets are off during the transition, and the upper mosfet's free-wheeling diodeconducts the current.

TUNINGYou will have to tune the PLL to your tank's resonant frequency. To do this just

connect jumper JP2. Leave jumper JP3 open, which goes to the integrator. With a volt meter,measure the voltage at pin 9 and the inverter ground. Trim R6 until you have one half ofyour supply voltage. Accounting for the diode voltage drop on the regulated 15vdc supply,this should be around 7.2v. You will need a differential set of oscilloscope probes to do thisnext part right. Put one probe pair across the current transformer, which would be acrossR15. Put another probe pair across the inverter output at J2. This will monitor invertervoltage and current. Using a variac, set the voltage input to your inverter high voltage supplyto a low value like 30-40vac. Trim R5 until you have the current and voltage in phase. Acruder method uses an unregulated rectifier with a smoothing capacitor with the voltageinput being the tank capacitor. Monitor the voltage for a maximum.


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Once you are confident that the PLL's center frequency is close to the resonantfrequency, open jumper JP2 and close jumper JP3. Turn on the inverter first and then turn onthe variac to the voltage doubler, which provides the high voltage for the inverter. Slowlyincrease the voltage while monitoring the inverter voltage and current waveforms. After 20or 30v you should see it lock onto Fres. The inverter output will be a nice square wave and

the current will be close to a smooth sinusoidal tracing. If all you see is a triangle-lookingwave for the current you probably have the polarity wrong on your capacitor voltage input topin 14. The quickest fix is to swap the connections going to the coupling transformer. Try itagain and it should work.

PLEASE NOTE THERE ARE MORE SCHEMATICS AT THE END OF THIS TUTORIAL(YOU HAVE MORE WEB PAGES TO GO)


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You can watch a video of it working here.

If you have read this much, you might want to read a bit more. These two other sites are wellworth the reading and explain things from their perspective with regards to the theory andconstruction of an induction heater.

Neon-John's Induction Heatinghttp://inductionheatertutorial.com/inductionheater/www.neon-john.com/Induction

Tim Williams Index Link

http://webpages.charter.net/dawill/tmoranwmsComments should go to jonathan_at_houseofficer.com

The next page shows images of the actual circuit.

PC Professor Computer Training, Service & Supporthttp://www.pcprofessor.com/

Pictures of the Actual Circuit

Making a circuit board and components

Everything starts with making my own circuit board. I got positive photoresist boardsand made my own with a fluorescent light. You need a positive developer, such as sodiummetasilicate pentahydrate. Then, I used Ammonium Persulfate in an etching tank andagitator. The result is below. If you need electronic parts, Digikey and Mouser are goodsources. There is always Ebay.
http://inductionheatertutorial.com/inductionheater/www.neon-john.com/Inductionhttp://inductionheatertutorial.com/inductionheater/www.neon-john.com/Inductionhttp://webpages.charter.net/dawill/tmoranwmshttp://webpages.charter.net/dawill/tmoranwmshttp://www.pcprofessor.com/http://www.pcprofessor.com/http://www.pcprofessor.com/http://webpages.charter.net/dawill/tmoranwmshttp://inductionheatertutorial.com/inductionheater/www.neon-john.com/Induction


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Induction Heater Levitation Tutorial

Induction heating and levitation is pretty cool. Using a levitation coil, you levitate aconductive object in the magnetic field and heat within that field. Depending on the metaland power setting you can even boil it mid-air. Aluminum will levitate and melt easily at 1-1.5kw of input power. You can levitate copper and steel balls. You can even melt them;

however, solid balls were too dense at my 2.5kw power level. In order to melt solid copperand steel you need near 8kw of power; suspending molten copper and steel requires over10kw of power. Component heating can be an issue so you need to make sure you have arobust cooling system for the mosfets, igbts, diodes, transformer and coil.

Current going through the coil sets up a magnetic field. This field, according toLenz's law, sets up an opposite magnetic field in the workpiece. This magnetic field opposesthe one inducing it, and repels the object upwards. The picture below shows a snapshot intime. The field alternates. The coil also increases in diameter as one moves upwards. Thisresults in there being a magnetic force underneath the object, but nothing directly above it.This results in an upwards force. The object moves up until the distance of the workpiece tothe inner surface of the coil is such that the magnetic field is too weak to drive it up anymore. The bucking plate at the top turns in the opposite direction. The two fields cancel outso there is no upward driving force at this point. It is a null zone.


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Now, the magnetic field created in the workpiece creates circulating eddy currents.These currents heat the workpiece. The closer the workpiece gets to the coil the better thecoupling, which creates more heating. You will find if you gently push the object down witha quartz rod it will heat up very quickly.

Below are some diagrams showing what was just discussed.

Below are the pictures of the levitation coil. The turns are tight, so you will need touse sand or salt so you can bend it without deforming the tubular shape. The coil is a conicalhelix. The bottom has a smaller inner diameter than the top. Make a bow to reverse the

direction and turn 1-2.5 coils in the opposite direction for the bucking plate. Keep the coiltight, but make sure the coils don't short. You will need a quartz rod to hold the object inplace until it levitates, or while it is heating. This is one levitation coil that I made. I madeanother one that is slightly larger. When I am levitating dense metals I keep the bucking

plate further from the main coil to minimize the downward forces on the workpiece.


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IXYS Power Mosfets and microprocessor trackingThe mosfets are from IXYS Corp, which can be purchased from Digikey

Corporation. The microcontroller is from Arduino.Ten kilowatts is a lot of power, and this is what is needed to levitate molten copper

and steel. At this power level high currents cause oscillations on the gate and the PLL

topology is not good enough to maintain a tight resonant lock. I wanted something that couldfind the resonant frequency with any coil and lock onto it without any manual adjustment.Precise resonant locking and tracking was accomplished with a microprocessor-PLL

circuit. I use the PLL to find the phase difference between the inverter and tank capacitor.Ninety degrees is the correct phase difference. I use the microprocessor to monitor the PLLoutput and develop a DC voltage that corresponds to the phase difference. I use this DCvoltage as the input to the PLL's VCO in order to maintain the correct frequency. Here is animportant point: tune to a slightly higher frequency so the current slightly lags the voltage. Ifyou are too close, or the current leads the voltage, the mosfets will heat up. I got my mosfetsto get hot with small currents if I had the tuning too close to ZVS. When I tuned the currentto slightly lag I had 10x the current going through them and they still remained cool (with

forced air convection).Next, I had to move to a 240 vac line @ 30-50A. At this power level heating becomes

a very real issue. I have two 100 cfm fans blowing on each 5"x5" heatsink for the mosfets. Iswitched to mosfets because they work a better at the 100 khz frequency range, and theyhave less switching losses. I am currently using the IXYS Polar HiPerFET seriesIXFN56N90P mosfets. I am using two in parallel for each leg of the half-bridge. At 25c eachmosfet can handle 56A. I am figuring that I should keep each under 30A, which is why Ihave two of them. They are rated for 900v.

Running mosfets in parallel can be tricky. First, you want to make sure they all comefrom the same lot. The problem with running more than one device is unequal heating andoscillations on the gate. Fortunately, because these mosfets have positive thermalcoefficients, the hotter they get the less current they conduct. This way, one mosfet does notrun away and carry more and more current as it gets hotter than its partner. Make sure themosfets for each leg are on the same heat sink. Second, you need the have enough resistanceon the gate to prevent oscillations. Five ohms is enough. I used 10 ohms because the Ldi/dtwas too high with the former, resulting in ringing during the transitions. I originally hadferrite beads on the gate leads, but I eliminated them in order to shorten the lead length. Thisresulted in even less ringing to the point of it almost being non-existant. The gate resistorwas sufficient to prevent oscillations on the parallel devices.

ATMega Microprocessor and the Arduino

The key to having the driver being able to find, trackand maintain a perfect resonance lock is the ATMega328microprocessor. Of course, one can choose anothermicroprocessor. You will need to be able to easily program it.I used the Arduino Duemilanove. This board allowed me toeasily interface with my computer and upload code. A USBcable connects from the computer to the board. The DIP chipfits into the socket and that is all. Arduino offers somesolutions to mount the chip with all the necessary hardwareinto your project; I found it very easy to just put in a socketwith the minimal connections on my own to get it working.


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Litz wire, magnet wire, ferrite toroidsI then found that the coupling transformer was getting hot. It got so hot the wire

started melting though the insulation. I switched the toroid from the powered iron to a 3C90ferrite toroid core. I use 4 1" thick x 2.5" OD x 1.5" ID toroids. I got them from Ferroxcube.The part number for the 3C90 material is TC25151. The high permeability ferrite and the

number of them assures me there will be no core heating or saturation. The wire is #10 and Iuse about 20 turns. I found that at lower turns the primary current was too high. By addingmore turns, I needed more primary voltage, but I used less current. I then mounted adedicated 100cfm fan for cooling the transformer.

Right now I am trying ferrite toroids that are larger to accomodate a thicker gauge ofwire. I am using the ZP48613TC toroids from Magnetics. They are about 4" x 3" x 1/2". I amusing seven of them to reduce the flux density. They are coated so I don't need to worryabout shorting the wire. The wire is made from 23g magnet wire. I have 64 strands (good for46A of DC current) twisted and braided. This took a little time, but I got it down to make 25'cord in a short period. 100khz should use 26g, but I think the 23g will work out well enough.I don't want to braid several hundred strands. I will go over some flux calculations on the

next page.The original primary used 10g wire. This was unable to handle the current, and

despite forced convection cooling, overheated and melted the vinyl wire coating.

Below is the braided litz wire that I made from magnet wire

Below is the ferrite toroid wrapped with 26 turns of the magnet wire braid


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drive circuitry from the high powered inverter section. I now have two boards. One boardtracks and generates the drive signal. All of the critical signals are shielded with a grounding

plane. The signal exits in a shielded wire that is connected to ground. This goes to the secondboard. On this board the gate drive lead is shielded with the shield connected to ground. Themosfet modules are electrically isolated from the heat sink. If they are connected the heat

sink will broadcast the inverter waveform and interfere with the feedback signals. The gatedrive signal goes through a damping resistor. I originally had a ferrite bead, but found thiswas not necessary. I have copper shielding completely covering the gate drive lead. Insidethe shielding I have the lead going to a 10 ohm resistor. It is important that each switch hasits own resistor as close to the gate as possible to reduce oscillations. So far this has workedout well. I have seen what happens with poor shielding, and it isn't pretty.

Below is a picture of the inverter with the shielded leads.

I also had to modify the feedback control loops. I use RC networks to filter the pulse-waves from the PLL to create a DC feedback signal. I then use RC networks to take thePWM output and convert it into a steady DC signal for the PLL's VCO input. The better myfiltering for a smooth signal, the longer the delay. If the delay is too long the system

oscillates. If the filtering is not sufficient, the drive signal is course. The trick was to find theright balance.

The tank capacitor was another issue. A rough calculation showed that I was alreadymaxing out the capability of my capacitor bank. I had 20 capacitors that were each rated for amaximum of 14A. I must have had between 200-300A going through them with my 2kwunit. I switched to a water-cooled Celem capacitor that was rated for 1000A. Theconnections were made so that every contact point shared current equally. If you make theconnections incorrectly, all the current will go through the closest point and rapidly overheatthe copper. I witnessed this and it was not pretty.

I will start elaborating on the modified drive circuit and the inverter on the followingpages (to come shortly).Here is the 10kw unit in action


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Here are some links to the levitation coil in action for the 2kw unitLevitating and melting aluminum http://www.youtube.com/watch?v=Q6Zrnv4OtbULevitating and melting copper scraphttp://www.youtube.com/watch?v=PVtcp4JZ8FA&feature=relatedLevitating copper tubing http://www.youtube.com/watch?v=SZx2IeNB1Ac&feature=related

10kw Induction Heater Inverter SchematicI got most of my components from Digikey Corp and Mouser Electronics. The

microprocessor is from ArduinoMicroprocessor/PLL controlled driver

10kw Inverter for use with levitation and iron forging
http://www.youtube.com/watch?v=Q6Zrnv4OtbUhttp://www.youtube.com/watch?v=Q6Zrnv4OtbUhttp://www.youtube.com/watch?v=PVtcp4JZ8FA&feature=relatedhttp://www.youtube.com/watch?v=PVtcp4JZ8FA&feature=relatedhttp://www.youtube.com/watch?v=SZx2IeNB1Ac&feature=relatedhttp://www.youtube.com/watch?v=SZx2IeNB1Ac&feature=relatedhttp://www.youtube.com/watch?v=SZx2IeNB1Ac&feature=relatedhttp://www.youtube.com/watch?v=PVtcp4JZ8FA&feature=relatedhttp://www.youtube.com/watch?v=Q6Zrnv4OtbU


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Home http://inductionheatertutorial.com/inductionheater/index.html

Electronic Supplies

Things that I found useful for the project:1. oscilloscope2. differential probes3. current meter4. multimeter5. circuit breadboard for prototyping6. circuit board fabrication supplies: circuit board, ammonium persulfate, positive

developer, etching tank7. solder station8. Digikey9. Mouser10.Newark11.Mapp Gas torch12.Variac
http://inductionheatertutorial.com/inductionheater/index.htmlhttp://inductionheatertutorial.com/inductionheater/index.htmlhttp://inductionheatertutorial.com/inductionheater/index.htmlhttp://inductionheatertutorial.com/inductionheater/index.html

Induction Heater Tutorial 10kw and 3kw.pdf

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