http://inductionheatertutorial.com/inductionheater/index.html Induction Heater Tutorial 10kw and 3kw Disclaimer: 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 on any part of it does so at his/her own risk. This is disavows any responsibility, and does not encourage anyone to do this. ***This may seem a bit random, but if you want to see me doing a really cool dance routine with 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 metal object. With enough power, one can even melt metal. The induction heater works without the need for fossil fuels, and can anneal and heat objects of various shapes. I set out to make an induction 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, and how 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, but realized that I could not levitate solid copper and steel. Their density was too great for the magnetic field. This was my final goal: to levitate and suspend molten copper and steel. At the end of this tutorial I will go into the development of a 10kw unit that realized this goal. I will 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 of very high currents within the tank's coil. The coil is coupled to the workpiece and sets up eddy currents. These currents, traveling through a conductive, but slightly resistive workpiece, heat the piece. Remember, Power = Heat = R*I^2. The workpiece is like a oneturn coil; the work coils has several turns. Thus, we have a step-down transformer, so even higher 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: What I will now do is go over each part. Then, I will give the schematics, go over them and how you can build this device. Induction Heater Components We will talk about each component making up the induction heater. First, there is the workcoil. This is what heats the workpiece. The workcoil will get very hot from the high current 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 parallel resonant tank. The tank and coil need to be cool, so I implemented a plumbing-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. The transformer is connected to the inverter. The inverter chops the DC power source at a particular frequency. This is the resonant 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 frequency changes. The inverter needs to stay locked on as closely as possible to the current resonant frequency 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 the highest tank voltage. Another method is using a phase locked loop (PLL) to monitor the phase relationship of the inverter voltage and tank voltage. This is the method I use and I will discuss this in detail later on. Let's start with how to easily make a workcoil. We will be using frequencies in the 10s to 100s of kilohertz (kHz), so metals will conduct the current only slightly below the surface. 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 water through one end, have it circulate through the coil, and come out the other end. I have brass fittings with nipples so I can attach some tubing to my fountain pump, and a return tube to my ice water bath. This is the tubing I got from Home Depot. 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 to easily heat your workpiece, or to do so with small movements in and out of the field. The more turns, the less induced voltage, and less induced current in the workpiece. If the induced current is too low you may never achieve a high enough temperature to get beyond the 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 quenching effect on the coil. You will also have a lower Fres for the same tank capacitance. This results in deeper current penetration into the workpiece, which may or may not be desired depending on your application. All this means it will take longer to heat the metal for the same input power. To compensate you will need a higher voltage going to the workcoil if you want to maintain the same rate of heating. You can compensate for more turns on your workcoil with fewer turns on your coupling transformer. However, you will still be faced with the issue of needing more input power to achieve the higher excitation voltage on the workpiece. You can get more input power by having a higher input voltage or drawing more current. LC Tank: Polypropylene Film Capacitor Bank For my first capacitor bank I purchased my caps from Illinois Capacitor. You can also purchase them from Newark Electronics. The induction heater uses a workcoil as a step-down transformer. This transformer steps the voltage down, but increases the available current to the workpiece, which is the one-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 the coil the better the energy transfer. This is the workcoil and tank. The capacitors are high voltage metallized power film snubbers. through the copper pipe soldered to the bus bar. instead. you will see the fountain pump submerged in water. the coil connects to opposite ends. It does not connect both leads at the front end. The tank is made from two 1" x 3/16" thick copper bars. if both end connected to the front. I purchased some pulse capacitors with current ratings of 14A. over the bank to the upper left. I use brass compression fittings to attach it to the LC tank. This ensures that the capacitors share an equal current load. 750vac. through the coil. When you are dealing with hundreds of amps. I drill holes in the bars to accommodate the capacitors. . With 20 capacitors this is close to 300A average current. We need a capacitor that can handle several hundreds of amps of current. This pumps ice water through the tank and back out into the bucket. and through the tubing connected to the other bus bar.The workcoil is made from shaping the 3/8" copper tubing. small changes in R are significant. and out on the upper right. 3000vdc. the capacitors closest to the coil would handle the brunt of current because the resistance would be the least. You should also take note where the workcoil connects with the capacitor bank. If you look closely. The coupling transformer fits over the copper tubing. Water flows in from the bottom left. Otherwise. The tank uses 20 capacitors.These are the bars with the holes drilled in them. . but you can use any number that gives you the capacitance and current handling capacity that you require. Higher frequencies have greater skin effect (less penetration) and are good for smaller objects. At resonance the LC impedance drops and the voltage drop across R peaks. I increased my function generator sine wave and measured the voltage across R. Mine are 0.5" OD). but you can just go by the calculation.4uf and can handle over 300A. Take the copper tubing and fill it with sand or salt. Fres = 1/2π√(LC) Once you wind your coil you can get an idea of its value by making a simple RLC circuit with it and connect it to a function generator and scope. My capacitor bank is 4. My coil is near 1uH. you need to determine what operating frequency you will use.22uf/3000vdc. I used a 1R resistor and a 500pf capacitor. Fix one end with something like a heavy vice and work . Make sure it is completely filled. The model number is 224PPA302KS. I choose a frequency near 70khz and wound up with about 66khz.First. The capacitors are from Illinois Capacitors. Higher frequencies have greater switching losses. Now. as far as the workcoil goes you can form the workcoil by driving a piece of PVC tubing into the ground. but there is less current going through the tank. This way it will act like a solid tube and will not collapse when you bend it. Lower frequencies are better for larger objects and have greater penetration. This gave me a ballpark figure. I used a 1" pipe (1. This is twice the amount of gate charging. One problem that plagued me for the longest time was a jittery inverter current when I reached modest power levels. I noticed that when I disconnected any one of the four mosfets the current tracing was perfect. When I transitioned to my 10kw unit I increased the size of my high voltage supply. Once you are happy with the turns and shape you can blow the sand out with an air compressor.the tubing around your PVC tube until you have your desired number of turns. smoothed source. The frequency range it can generate is based on its supply voltage. I used the third variation for mine. At first I thought this was EMI affecting my gate drive and I spent the longest time trying to fix it. and the mosfets were not all conducting identically. Make sure you have your rectifier on a large heat sink because it will be conducting a lot of amperes. I use two 50A rectifiers giving me 100A. Power Supply: Voltage double and regulated source I need to talk about two power supplies for the unit.8 . The second power supply you will need will be a 15vdc regulated source. You can use 110vac through a rectifier and smoothed with a 1000uf-1500uf capacitor for a supply of 170vdc.1. The VCO determines the output frequency based on input voltage it receives. Vss. Four to five turns at 1. The current would jump back and forth when compared to the inverter voltage. It appeared as if two currents were competing. You need to make sure your 15vdc supply can supply the amps to rapidly charge the gates. My rectifier is rated for 25A/500vac. If the supply voltage wanders. the oscillator frequency will wander and this will definitely throw you out of resonance. You need an unregulated. Below are some basic schematics for a voltage doubler. Each capacitor is rated for 450vdc. you need to fully turn the mosfets on in the shortest time possible. I put a scope . I used a voltage multiplier to convert it to 320vdc. I plan on adding an outboard pass transistor. Remember. You can see how nicely the coil forms around the pipe. This led me to believe that I was falling short on charging all the gates rapidly enough. It should also have a robust transformer and capacitor on the end to make sure there is plenty of charge available.3 uH.5-2" will give you a coil with an inductance between 0. Now. It is imperitive that it is regulated because the PLL has a voltage controlled oscillator. when I made the 10kw unit it uses four mosfets instead of two. so I can go up to 900vdc between both ends. One is the high voltage DC that the inverter converts to AC for feeding the tank. on the gates and noticed that the slopes changed when I added the fourth gate. RCL Theory and Transformer Coupling I guess the best way to understand what is going on is to start with the workcoil and work backwards. giving us a phase shift of 18degrees leading. if there is more capacitive reactance the current leads the voltage. Below we have a RCL circuit with a resistance of 4R. the voltage drop across an inductor is a reaction against a change in current through it. There is only one current running through the series circuit. The instantaneous voltage is zero when the current is at a peak because the change in current is zero. The tracing was perfect. the current in the series circuit is in phase with the voltage source. We achieve these high currents because the RCL tank is at resonance. Ferrite Toroids. I solved the problem by adding a 39000uf capacitor to my power supply. Remember. The current in the circuit is in phase with the voltage. . real resistance. This was with a 1. because the voltage and current of the resistor are in phase. I plan on changing the transformer to 3A and adding adding the outboard pass transistor just to play it safe. If there is more inductive reactance the current lags the voltage. You can also say the capacitor voltage lags the current. manifested by the zero slope. Remember from earlier I said that the workcoil is the primary end of a step-down transformer.6A transformer and a 2A 15v regulator. The inductor wins and the inductor voltage leads the current. The reactive impedance cancels to 1ohm. Now here is an important point. If the inductive and capacitive reactance cancel out the phase shift is zero. The maximum power transfer will occur when the current is in phase with the voltage. If we are out of resonance the current phase is shifted from zero with respect to the voltage. Zl = 4ohms and Cl = 3ohms. This means that the inductive reactance and capacitive reactance cancel out. You can also say the inductor voltage leads the voltage across the resistor. We have hundreds of amps flowing through here and this creates a voltage in the workpiece. at resonance. So. and all we are left with is the small. 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 of turns to put on the primary coupling transformer.25" in diameter and 0. If you flip it around you will introduce a 180 degree shift which will prevent the PLL from locking onto the frequency. I use two to prevent saturation. I wonder how three would do? Here I have wound 14g wire around. We will use this as the positive lead for monitoring our tank capacitor voltage later.5Mhz. The transformer does not impart a phase shift if place on the tank correctly. The lower the turns the greater greater the exciting voltage to the tank. . Each is 2. I am experimenting with different materials and turns. the field would travel through it. First. The current travels to the positive terminal of our toroid transformer output. The black arrow on the toroid shows the direction the field travels in the core. Using the right hand rule we can realize the direction of the B field for each turn. This material is good from frequencies between 0. If you are unsure which end is which you can wind a few turns of wire as a secondary and scope the ends.What is the voltage source for the series tank? It is our coupling transformer. Which way is the right way? Use the right-hand rule. the field inside the solenoid sums to one large field going from right to left. Using the right-hand rule again we see that the current travels through the copper tubing from left to right towards the positive terminal of our RLC tank. If we had a metal bar or part of the toroid's arc inside. Below are the two toroids. Here is a solenoid with the current flowing in the direction shown. magnetization current goes up as does the load on the inverter. Put your right thumb in the direction of the current and your fingers curl in the direction of the B field. but right now I am using an iron powered core from Amidon Corp made from Type 3 material. However. I am still trying to figure out the optimum turns and the best material.565" thick. So here is a mock-up of the coupling transformer. There are several factors to consider. The field outside of the coil is not important to us. Just turn it around. I used two toroids.05Mhz and 0. I wound 14g wire around for 20-26 turns. If the load is 1R you will have 10A on the secondary and 0. If your wire is not robust enough you might need a cooling system. Let's assume that the load across the secondary is 1 ohm. The power draw is 100W. When heating small pieces of metal with small coils. If our maximum voltage is 200v we need to draw more current. If anything. You want a lot of turns on the primary in order to keep the current draw low while still supplying a lot of current to the tank. these strands will need to be twisted in order to reduce eddy currents. This has to do with the mosfets conducting in the reverse direction. and is in phase with the inverter input voltage. and still maintain the same power to my workpiece. rather than a capacitive. Oscilloscope Tracings The inverter outputs a drive voltage to the coupling transformer. the majority of the current is conducted on the surface. The current in is in phase with the current out. If you are dealing with high frequencies. but it is 0. You need less turns on the primary in order to provide a higher excitation voltage to the tank. As long as I have enough voltage. The current is still 10A on the secondary. load. you want the current to slightly lag the voltage because the mosfets behave better when facing an inductive. If you have the means to run high voltages. Now. As you pack more wire into the space. the same primary current will yield a much larger secondary current. 200v on the primary results in 10v on the secondary. As you go lower on the turns you need to make sure you do not saturate the core. This means that as long as I have a higher voltage supply. This is the skin effect. If I want to keep the primary current low I need more turns on the primary.the wire needs to be able to handle the current. the primary we have 200v @ 2A. When the tank is at resonance. 10A of secondary current requires 1A of primary current. . If I have 100v on the primary. In this case you don't have enough voltage to get an adequate current to flow in the tank. heating becomes more significant. You will need to have several insulated strands to increase the surface area. As long as the primary circuit can handle this we have solved the problem. Let's look at another example where the workpiece is quenching the tank. The power to your system has a voltage and a current. If I have 400v available. If you have 200v on a 20:1 transformer you will have 10v on the secondary. making sure our switches can handle this of course. the tank gets quenched and the current draw will be too low for effective heating. Let's go over an example: My transformer has 10 turns on the primary and one on the secondary (this is the resonant tank). you can adjust your windings to keep the primary current low enough to reduce the heating of your transformer and switches. We are drawing more power and we have doubled the output current at the expense of needing to deal with four times the primary current. If I want to draw less current I can wind a 20:1 transformer. a 10:1 transformer gives us 10v on the secondary. I can reduce the current my inverter requires. By changing to a 10:1 transformer we get 20v @ 20A on the secondary. If you plan on heating large pieces of metal.5A on the primary.5A in the primary. the tank current is in phase with the drive current of the coupling transformer. I can draw the same 1A on the primary. the current demand will go up quickly as there is little material to quench the tank. but have 20A available on the secondary. The tracing below is clean and allows me to reach very high power levels while maintaining relatively cool mosfets. Also remember that a small amount of the total primary current is magnetization current. Also. if the tanks are above resonance we have more inductive reactance. This noise gets worse with higher power levels and can result in mosfet failure. . there is ringing in the current waveform and at the inverter voltage transitions. the tank's current is lagging the inverter driving voltage. The tank's net current will lag the driving voltage from the coupling transformer. Since the input and output current of the coupling transformer are in phase. Below you can see the dominating inductive reactance results in the inverter 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 current leading the inverter voltage.Now. Notice the clean voltage square wave and the smooth current curve. I have positioned them apart for easier viewing. These images show the waveform after the fix: shortening the gate leads as much as possible to still allow room for paralleling two of them and increasing the gate resistor from 5R to 10R. You can see ringing on the voltage at the transition and on the current waveform. I was still able to charge the gate with 10 ohms in a sufficiently short amount of time at 15v. . The problem can usually be solved by either increasing the gate resistance (increase the resistor value). I was able to almost eliminate the ringing by shortening the gate lead. Heavy current on the gate causes a large Ldi/dt. The first image is with the 5 ohm gate resistor. or decrease the stray inductance by shortening the gate lead.Below is another example of ringing. This is due to high inductance on the gate. So I changed the value of R from 5 ohms to 10 ohms. but then I did not have enough length for connecting two in parallel. The second image is a blow-up of the first. The coil in the middle is the coupling transformer to the resonant tank. I would like just mention that the inductive waveforms is really an exponential curve. shorting the current path. If we can approximate the tank above resonane as a RL circuit responding to a step response . causing an overshoot. followed by a large dip when it finally closes. Sketch II shows profound voltage sagging in the middle of the waveform.Below is a basic sketch of the half-bridge inverter. the inductive kick will drive the voltage too high. Sketch I shows ringing if there is too long of a delay during switching. I had this happen when the decoupling capacitors went bad. Below are two sketches. If the next switch does not close in time. The capacitors are needed to remove any DC component from the pulse. The arrows show the paths the current takes as the switches alternate between closed-open and open-closed. . the tank current has a zero phase shift with respect to source (inverter) tank voltage because the inductive and capacitive reactance cancel out. The square wave is the inverter voltage. At resonance. When dealing with a 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-αt(Bcos(ѡt) + Bsin(ѡt)) Oscilloscope Tracings II Let's continue our discussion of oscilloscope tracings so we can better understand how the inverter is going to work and lock onto resonance. you can see the sinusoidal capacitor voltage lags the inverter voltage by 90 degrees. but you would get the same relationship if you scoped the voltage output of the toroid transformer. If you display the inverter voltage and capacitor voltage together. From the last page I mentioned that voltage across the tank capacitor lags the current by 90 degrees.The solution to is Analysis of a capacitor dominant RC circuit will yield something similar. Now. and keep it to a maximum of 15v. we can easily exceed the chips maximum input voltage. Using a differential probe. that is shift it 180 degrees. in order for the PLL to work. We will have to invert the Vc waveform. the positive lead goes to the positive inverter lead going to the toroid and the negative to the negative lead. Below is a diagram of the scoped voltages. Using a second differential probe we scope the + and . We do this with some clamping diodes yeilding this waveform. We are at resonance when our PLL chip keeps Vc ninety degrees lagging behind Vinverter.We will monitor this relationship. which I will discuss shortly.ends of the capacitor tank. which will be the signalin input on pin 14 of the HEF4046. . Vc will lag Vinv or Vtank. so we need to clip the top and bottom the the capacitor voltage. The waveform is different from above. but still bizarre and not a good sinusoid. Notice how the voltage heavily sags and the gate signal is no longer a clean square wave.Now. Underneath is a tracing of the gate drive signal and the inverter voltage from another run. After this discussion. Below are the voltage/current waveform. This is the inverter voltage (yellow) and gate drive (blue).the HEF4046. The negative current prematurely starts to rise and then go back down before resuming its normal cycle. Oscilloscope Tracings III I have to share some bad waveforms I got one day. we will have enough information to understand the workings of the inverter and how it maintains a lock on the resonance. Here is another image. Notice how the current is no longer a nice sinusoid. I hadn't used my heater all summer and wanted to try it out before giving it to a friend. we are ready to talk about the phase locked loop chip . . Still no good. When this failed I redid the entire circuit board thinking I was getting some type of cross-talk or a failed component. I pulled out the inverter capacitors and replaced them. I took out the board and replaced the gate drive capacitors. Frustrated. Then. I thought I checked everything and I couldn't understand how the waveform had deteriorated. I was getting strange waveforms and I did not know why. As you can see. Then. . I saved myself from buying another tank capacitor by connecting the coupling transformer to another LC tank. When that failed to fix the problem I redid the gate resistors and shielding. the current appeared to go at twice the frequency of the inverter voltage. Again. I had the same problem. Sometimes. I started looking at my high voltage DC supply. I must have reconnected the HV wires to the inverter early in the summer. Notice in the picture how they are not together. At first I thought it was the mosfets so I swapped them out. I had a final thought.Here is another gate wave that is abnormal taken at a different time. after days of racking my brain. Below are the waveforms for the inverter voltage and current immediately after this repair. this simple solution was all that was needed. Amazingly. I have the frequency deliberately higher than resonance to prevent reverse currents. I twisted the HV wires close (as I had done in the past) and made sure they were close on my inverter board before splitting to each of the HV rails. but it affected the gate signal and the voltage supply signal to the circuitry. .They should be together to cancel out any stray inductance as shown below. making things even worse. my experience will make someone's life easier if these symptoms appear. At the frequency I am driving my coil. stray inductance and capacitance on the HV lines is significant and clearly affected my waveforms. Not only did this affect the LC tank. Hopefully. The VCOout drives the device. The people writing the tutorials will assume you know all about these. The VCO generates a 50% duty cycle square wave. It also closes the loop by feeding itself back into the phase detector so it can get compared with a reference signal.Below is the gate signal after this repair. the lower the voltage the lower the frequency. A PLL consists of three parts: a voltage controlled oscillator (VCO). I will make the opposite assumption and give you a brief understanding of the concept so you can understand how this will help maintain resonance with our induction heater. Phase Locked Loop (PLL) Basics When you read about induction heaters and inverters you will probably come across the term phase locked loop. the frequency depends on the input voltage to the VCO. The phase detector has two options for outputs: PCA1 and PCA2. or inverter gate in our case. The higher the VCOinput (pin 9) voltage the higher the VCOoutput frequency. We use the former. which is a XOR gate. a loop filter and a phase detector. . The PLL phase detector compares the phases of two inputs: the reference signal on pin 14 and the VCOout frequency. The tank capacitor voltage lags the tank current by 90 degrees. Vcap and Vphi. The workcoil becomes a variable inductor and affects the resonant frequency of the tank. So what happens? At resonance the tank current is real and in phase with the coupler transformer voltage. it seems to the circuit that we increased on drive frequency to the tank. but not both. This means the capacitor voltage lags the inverter voltage even more as shown below. therefore. which is in phase with the inverter voltage. If the effective resonance goes down. Vphi is high Vinv or Vcap is high. The simplest filter is a RC low-pass filter. This makes the tank more inductive. If the two waves are 90 degrees out of phase the average value of Vphi is Vdd/2. The capacitor voltage initially lagged the current by 90 degrees. Now as the workpiece heats its ferromagnetic properties change.Figure 2 The logic is high if one of the two inputs is high. otherwise it is low. The loop filter takes the phase detector output and converts this to the input voltage to the VCO. Inductance causes the source voltage lead the tank current. Figure 3 Below we can see the relationships with Vinv. That is. it lags the inverter voltage by 90 degrees. . and how well it stays locked on the reference signal. The cut-off frequency will determine how sensitive the PLL is to phase changes. the tank current is forced to lag the inverter voltage. It will generate a square wave whose width is based on the phase difference of the two signals. and was distorted when I tried to show three signals. The average voltage is Vdd/2. or 7.inverter voltage and current are in phase . Vinv. A smaller VCOin results in a lower frequency and we stay in resonance.5v at VCOin will keep us close to resonance if our center frequency is Fres. When we are at resonance . The frequency range is determined by resistors on pins 11 and 12 of the PLL. The problem is that Fres changes with different workpieces and during heating. However. An increase in inductive reactance is the same as if we increased our inverter drive frequency. we need it to decrease in order to yield a lower voltage for VCO. Shown are Vcap_inverted. The capacitor voltage is clipped to protect the PLL chip. the PLL will adjust itselft to maintain a lock on the phase relationship. We integrate this to a voltage value and use this for VCOin. Vphi decreases. 7. Below is just the inverter and tank capacitor voltage.5v if our supply is 15v. The first picture is at a lower frequency than the bottom picture. when it is at the supply voltage it is at the high end. The capacitor voltage is a clean signal.the inverter voltage leads the tank capacitor voltage by 90 degrees. Vphi is half of half a pulse width (see Figure 2 and 4). VCOin is at ground the frequency is at the low-end of the range. . So. We lower it by decreasing the voltage to VCOin. The scope images below show these waveforms. However. When. We see in the top pair that as Vc shifts more to the right of Vinv the XOR region increases.Figure 4 The top shows Vinv and Vc. We achieve this by inverting Vc to Vc_inverted. and Vphi. Now as Vc_inverted shifts to the right. I used one similar to the second. I used the active integrator. The active filter has more gain than the passive filter. I scoped both the passive and active filter action by monitoring the relationship of the inverter voltage and current. The phase shift in the beginning is -90. The -1 gain op-amp will restore the proper polarity. using a voltage monitor on the tank voltage for the near-high point. The simplest is the passive low-pass RC filter. Phase Locked Loop (PLL) Basics II If you will recall. There are several filters one can use for the feedback loop. I add a variable voltage input to V. which allows me to fine tune the frequency.We need to discuss a few more things about the PLL next. Below is a table of some filters. and I can say that the latter maintained a tighter lock on a -90 phase difference during changes in the tank's resonant frequency. One other thing: you need a gain of -1 after the active filter because it inverts the signal. I usually tune it slightly above resonance.on the op-amp. I don't know if this helps keep our signals at -90 or not. which use a R and C element. To ensure a DC bias does not work itself into the capacitor. here is a block diagram of the PLL device. . I put a discharge resistor in parallel with C. Let's talk about how we set the free-running PLL frequency and the range it can capture. the PLL will be able to find the frequency that maintains the 90 degree shift that we want. If the resonant frequency falls with the PLL capture range. and maintain this phase lock as the frequency 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: . my supply is 14.You can also use graphs on the manufacturer datasheets to get you in the ballpark for the values you need. Determine the R value you need for Fmin.4v. Let's do a quick example. and then determine the R you need for Fmax. Actually. The first step is to determine the capacitor value that will get you near your Fres at a given Vdd voltage. My Fres is 65kHz and my supply is 15v. because I have a diode to protect from hooking up the pos and neg in . Using C1 = 330pf. connect pin 9 to Vdd and measure Fmax.reverse. I measured 50kHz for Fmin. jumper JP2 is open and JP3 is closed to allow the feedback to get to pin 9. they are for timing. Ground pin 9 and measure Fmin. PLL OVERVIEW The PLL receives two inputs through pins 14 and 9. Pin 14 is the clamped capacitor tank voltage. The integrator output then goes through a filter with a gain of -1 to restore the polarity of the signal. you need to verify this with your scope. which is 100k and 348k. I have connectors on my board going to wires which run to the chips on large heat sinks. Induction Heater Inverter Schematic Most of the electronics components on the schematic are from Digikey Corp and Mouser Electronics. I put a ferrite bead that attenutes frequencies above 300khz right before the lead to the gate drive. I go straight down and get a C1 of 300pf. giving me a K1 of 1.000 x 362e-12) Fmax = 80khz = 50khz + 3. The high voltages are kept down with R1. (Fmin + Fmax)/2. This gives me a selection of resonant frequency ranges. Fmin = 50khz = 1. T1. These series diodes have nothing to do with reverse currents. The wire acts like an antenna and you can get noise which will induce wild oscillations in your mosfets. so I want something between 50-80kHz. which is made up of a quad op-amp. Your tracings should be short to the gate drive on the mosfet. I then connected pin 9 to Vdd and got 154khz. This works perfectly. These chips drive the primary of a 1:1:1 gate drive transformer. We will discuss this at the end. All inverter grounds are isolated from earth ground. Next. I will pick some R values and measure the actual frequency in order to determine the values of the constants K1 and K2. 50khz. like the one's you are used to seeing across the DS juntion. and we want about 10-15kHz on either side. as determined from the graph. We want to have the center frequency. Below are the .81.81/(100. equal our resonant frequency. This will be my starting point for my equations. It is inverter (shifted 180 degrees) in order for the feedback to work properly. Jumper JP1 converts pin 12's resistor from 100k to 60k.78/(348. This will give you K1 and K2. Of course. Take a 100k resistor for R2 and R1. I will use a 330pf capacitor. During use. the chips can vary from the equation by a factor of 4. My frequency is 65kHz. Now. I now use these values to determine the true values of R1 and R2 that I need. destroying them. PCAout goes through the active integrator filter.000 x 362e-12) On my circuit I add a trim pot and another resistor in parallel to R2 with an optional jumper. how does our circuit come together? Let's see. The drive frequency leaves pin 4 and drives a non-inverting and inverting gate drive. so you need to multiply each equation by a constant. The calculations are below. So. C1 removes DC bias. and values of K1 = 1. C7 and resistors on pins 11 and 12 set the capture range. I go up the left hand side to the 60khz row and across to the 15v supply line. Diodes D5 and D6 offer some delay so both mosfets are not on at the same time. Subtracing Fmin. I was able to dedue that K2 equals 3. R5 affords you the ability to vary the capture range even more for tuning the center frequency to the tank resonant frequency.81 and K2 = 3. Again.78.78. .tracings going to the gates and then showing the tracing from one of the gates and the inverter output. TIMING DELAY FOR MOSFETS These are the gates drive signals going to the mosfets. when the gate is high the DS junction grounds the power. With this mosfet. The small slope is part of the delay imparted by the series diodes D5 and D6. Below. one of the series diodes is shorted. This is the inverter tracing on top. The bottom tracing has the transformer gate drive going directly to the gate. The signals are superimposed. Below is the timing showing ZVS. and one of the gate drive signals on the bottom. so we can compare the timing of the signal going to both gates. The tracing above it goes through the diode The temporal difference between the gate drive with and without the series diode is close to 100ns. so the voltage drops to zero. The voltage goes through zero volts exactly when the current is zero. and discharges to ground when U1 closes and U2 opens. The current transformer . both series diodes are shorted. When connecting it to the circuit. Secondly.Below are the gate drive waveforms with both series diodes working. set it up so clockwise motion increases the PLL frequency. OP-AMP INTEGRATOR The op-amp is centered around Vdd/2. rated for flywheel service. MOSFETS. the present day mosfets have very fast intrinsic diodes. and CURRENT MONITORING Mosfets 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 for two reasons. which gets charged when U1 is open. Below. When we switch the mosfet there is no current or voltage on the device. Capacitors C1 and C2 set a point half-way above ground. R10 moves the center point on the integrator allowing you to fine tune PLL frequency. ZERO VOLT SWITCHING. We are doing zero volt/current switching which is guarenteed when the circuit is in tune by the PLL. You can force it to stay a little above resonance by adjusting it. and we can see the time to reach the same voltage is delayed by about 200ns. and the upper mosfet's free-wheeling diode conducts the current. To do this just connect jumper JP2. set the voltage input to your inverter high voltage supply to a low value like 30-40vac. Monitor the voltage for a maximum. Now. The inverter output is coupled to the tank through T3. both mosfets are off during the transition. Trim R6 until you have one half of your supply voltage. The coper tubing which form the connects for workcoil and capacitor serves as a one-turn primary. which is a 20:1 toroid transformer.T2 uses a 1:100T ratio to monitor the inverter current. and the upper mosfet turns off slightly before the bottom one turns on. Trim R5 until you have the current and voltage in phase. During Mode 1. You can experiment with different toroid materials and turn-ratios. this should be around 7. With a volt meter. TUNING You will have to tune the PLL to your tank's resonant frequency. The 100R resistor means that every 1V on the oscilloscope is 1A of current going to the coupling transformer. Accounting for the diode voltage drop on the regulated 15vdc supply. The resonant frequency will changes as the material goes through its curie point. This will monitor inverter voltage and current. Leave jumper JP3 open. It shows how the free-wheeling diodes come into play to divert the reverse current around the mosfet. The 20 turn primary is connected to the inverter output. Put one probe pair across the current transformer. measure the voltage at pin 9 and the inverter ground. the mosfets are transitioning. You will need a differential set of oscilloscope probes to do this next part right. Put another probe pair across the inverter output at J2. current is conducted through the free-wheeling diode of the lower mosfet. In Mode 3. let's look at the tank circuit schematic. Using a variac. A cruder method uses an unregulated rectifier with a smoothing capacitor with the voltage input being the tank capacitor. and the resonant tank throws the power back through mosfet. which goes to the integrator. which would be across R15. the lower mosfet turns on.2v. In Mode 3. In Mode 2. the upper mosfet is conducting and transfering power to the resonant tank through the coupling transformer in our circuit. Here. Below is a picture of the current conduction through the inverter during different phases of the power transfer cycle. . Slowly increase the voltage while monitoring the inverter voltage and current waveforms. The quickest fix is to swap the connections going to the coupling transformer. open jumper JP2 and close jumper JP3. PLEASE NOTE THERE ARE MORE SCHEMATICS AT THE END OF THIS TUTORIAL (YOU HAVE MORE WEB PAGES TO GO) . which provides the high voltage for the inverter. Turn on the inverter first and then turn on the variac to the voltage doubler. After 20 or 30v you should see it lock onto Fres. Try it again and it should work. If all you see is a triangle-looking wave for the current you probably have the polarity wrong on your capacitor voltage input to pin 14.Once you are confident that the PLL's center frequency is close to the resonant frequency. The inverter output will be a nice square wave and the current will be close to a smooth sinusoidal tracing. You need a positive developer. The result is below. If you have read this much. These two other sites are well worth the reading and explain things from their perspective with regards to the theory and construction of an induction heater. Digikey and Mouser are good sources.neon-john. There is always Ebay. I used Ammonium Persulfate in an etching tank and agitator. such as sodium metasilicate pentahydrate. .com/ Pictures of the Actual Circuit Making a circuit board and components Everything starts with making my own circuit board.com/inductionheater/www.pcprofessor. I got positive photoresist boards and made my own with a fluorescent light. PC Professor Computer Training.charter. Then. Neon-John's Induction Heating http://inductionheatertutorial.You can watch a video of it working here.com The next page shows images of the actual circuit.com/Induction Tim Williams Index Link http://webpages. If you need electronic parts. Service & Support http://www.net/dawill/tmoranwms Comments should go to jonathan_at_houseofficer. you might want to read a bit more. . you levitate a conductive object in the magnetic field and heat within that field. The picture below shows a snapshot in time. This field. The coil also increases in diameter as one moves upwards. but nothing directly above it. It is a null zone. sets up an opposite magnetic field in the workpiece. solid balls were too dense at my 2. transformer and coil. This magnetic field opposes the one inducing it. igbts. In order to melt solid copper and steel you need near 8kw of power. The object moves up until the distance of the workpiece to the inner surface of the coil is such that the magnetic field is too weak to drive it up any more. This results in an upwards force. The two fields cancel out so there is no upward driving force at this point. however.Induction Heater Levitation Tutorial Induction heating and levitation is pretty cool. Component heating can be an issue so you need to make sure you have a robust cooling system for the mosfets. suspending molten copper and steel requires over 10kw of power. You can even melt them. Depending on the metal and power setting you can even boil it mid-air. diodes.5kw of input power. You can levitate copper and steel balls. Aluminum will levitate and melt easily at 11. The field alternates. Using a levitation coil. and repels the object upwards. Current going through the coil sets up a magnetic field. This results in there being a magnetic force underneath the object. according to Lenz's law. . The bucking plate at the top turns in the opposite direction.5kw power level. The closer the workpiece gets to the coil the better the coupling. which creates more heating. Make a bow to reverse the direction and turn 1-2. The coil is a conical helix. The bottom has a smaller inner diameter than the top. . so you will need to use sand or salt so you can bend it without deforming the tubular shape. Below are some diagrams showing what was just discussed. This is one levitation coil that I made. You will find if you gently push the object down with a quartz rod it will heat up very quickly. the magnetic field created in the workpiece creates circulating eddy currents. When I am levitating dense metals I keep the bucking plate further from the main coil to minimize the downward forces on the workpiece. You will need a quartz rod to hold the object in place until it levitates. but make sure the coils don't short.5 coils in the opposite direction for the bucking plate.Now. These currents heat the workpiece. Below are the pictures of the levitation coil. The turns are tight. or while it is heating. I made another one that is slightly larger. Keep the coil tight. I got my mosfets to get hot with small currents if I had the tuning too close to ZVS. I am figuring that I should keep each under 30A. The microcontroller is from Arduino. I am using two in parallel for each leg of the half-bridge. I have two 100 cfm fans blowing on each 5"x5" heatsink for the mosfets. Here is an important point: tune to a slightly higher frequency so the current slightly lags the voltage. Of course. Next. They are rated for 900v. Running mosfets in parallel can be tricky. First. When I tuned the current to slightly lag I had 10x the current going through them and they still remained cool (with forced air convection). but I eliminated them in order to shorten the lead length. I use this DC voltage as the input to the PLL's VCO in order to maintain the correct frequency. the mosfets will heat up. I used 10 ohms because the Ldi/dt was too high with the former. The problem with running more than one device is unequal heating and oscillations on the gate. and this is what is needed to levitate molten copper and steel. one can choose another microprocessor. one mosfet does not run away and carry more and more current as it gets hotter than its partner. Second. because these mosfets have positive thermal coefficients. track and maintain a perfect resonance lock is the ATMega328 microprocessor. ATMega Microprocessor and the Arduino The key to having the driver being able to find. Make sure the mosfets for each leg are on the same heat sink. I found it very easy to just put in a socket with the minimal connections on my own to get it working. or the current leads the voltage. the hotter they get the less current they conduct. Five ohms is enough.IXYS Power Mosfets and microprocessor tracking The mosfets are from IXYS Corp. The gate resistor was sufficient to prevent oscillations on the parallel devices. I originally had ferrite beads on the gate leads. 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 use the microprocessor to monitor the PLL output and develop a DC voltage that corresponds to the phase difference. If you are too close. This resulted in even less ringing to the point of it almost being non-existant. Ten kilowatts is a lot of power. Arduino offers some solutions to mount the chip with all the necessary hardware into your project. you want to make sure they all come from the same lot. resulting in ringing during the transitions. . I used the Arduino Duemilanove. I wanted something that could find the resonant frequency with any coil and lock onto it without any manual adjustment. At this power level heating becomes a very real issue. which is why I have two of them. which can be purchased from Digikey Corporation. Precise resonant locking and tracking was accomplished with a microprocessor-PLL circuit. I had to move to a 240 vac line @ 30-50A. I am currently using the IXYS Polar HiPerFET series IXFN56N90P mosfets. At 25c each mosfet can handle 56A. I switched to mosfets because they work a better at the 100 khz frequency range. A USB cable connects from the computer to the board. You will need to be able to easily program it. This way. I use the PLL to find the phase difference between the inverter and tank capacitor. Ninety degrees is the correct phase difference. This board allowed me to easily interface with my computer and upload code. Fortunately. you need the have enough resistance on the gate to prevent oscillations. The DIP chip fits into the socket and that is all. and they have less switching losses. They are about 4" x 3" x 1/2". ferrite toroids I then found that the coupling transformer was getting hot. I have 64 strands (good for 46A of DC current) twisted and braided. overheated and melted the vinyl wire coating. The high permeability ferrite and the number of them assures me there will be no core heating or saturation. I found that at lower turns the primary current was too high. magnet wire. 100khz should use 26g. It got so hot the wire started melting though the insulation. but I think the 23g will work out well enough. I needed more primary voltage.Litz wire. The wire is #10 and I use about 20 turns. This took a little time. Right now I am trying ferrite toroids that are larger to accomodate a thicker gauge of wire. By adding more turns. I use 4 1" thick x 2. The original primary used 10g wire. I don't want to braid several hundred strands. They are coated so I don't need to worry about shorting the wire. This was unable to handle the current. I am using the ZP48613TC toroids from Magnetics. and despite forced convection cooling.5" OD x 1. I got them from Ferroxcube. 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 . but I got it down to make 25' cord in a short period. I switched the toroid from the powered iron to a 3C90 ferrite toroid core. but I used less current.5" ID toroids. I will go over some flux calculations on the next page. The part number for the 3C90 material is TC25151. I then mounted a dedicated 100cfm fan for cooling the transformer. I am using seven of them to reduce the flux density. The wire is made from 23g magnet wire. Now. Celem Capacitor Tinned copper braid. EMI. Depending on your leads you will need anywhere from 1/4 to 1/2 inch copper braid. and it gets worse at the power levels needed for levitation. This is a Magnetics. we would use 4. This envelops the gate leads and is connected to the ground. If one exceeds this value the permeability will go towards one (the permeability of air) and the transformer will no longer function as a transformer.4cm2. We have 8 of them for a total cross-sectional area of 14. in a worse-case scenario.44 for a sine wave. With use a multiplication factor of 4 for a square-wave. Inc P material core. So. The solution begins by separating the . According to this chart. Magnetic noise is a big problem.Ferrite Toroid One needs to make sure the maximum flux density of the core is not exceeded.8cm2. We'll use 20 for convenience. At 100C this number will be closer to 3000 gauss. and the RMS value is near 350V. it will cease transferring power. We are using between 19-24 turns depending on the heating application. The tinned copper resists corrosion. The formula is Bmax = E x 108 / (4 x N x F x A) This yields: 350 x 108 / (4 x 20 x 100. The cross-sectional area of the ZP48613TC is 1. but how it affects the PLL feedback loop on the driver. we still have a 10x safety margin. One not only needs to worry about how the emi affects the gate drive on the inverter. the maximum flux density is 5000 gauss at 25C.000 x 14. The voltage is close to a pure square wave with a slight ripple voltage. Tinned Copper Braided Shielding. That is. or copper braided shielding wire is essential to keep the emi out of the gate leads. Please refer to the chart below provided by this company.6) = 300 gauss So. let's calculate the flux density of our ferrite coupling transformer. we need to calculate the flux density of our core. I must have had between 200-300A going through them with my 2kw unit. I then use RC networks to take the PWM output and convert it into a steady DC signal for the PLL's VCO input. The connections were made so that every contact point shared current equally. If the delay is too long the system oscillates. The mosfet modules are electrically isolated from the heat sink. The tank capacitor was another issue. I originally had a ferrite bead. I now have two boards. but found this was not necessary. the drive signal is course. Here is the 10kw unit in action . The signal exits in a shielded wire that is connected to ground. and it isn't pretty. A rough calculation showed that I was already maxing out the capability of my capacitor bank. So far this has worked out well. It is important that each switch has its own resistor as close to the gate as possible to reduce oscillations. Below is a picture of the inverter with the shielded leads. On this board the gate drive lead is shielded with the shield connected to ground. If the filtering is not sufficient. I will start elaborating on the modified drive circuit and the inverter on the following pages (to come shortly). If they are connected the heat sink will broadcast the inverter waveform and interfere with the feedback signals. One board tracks and generates the drive signal. Inside the shielding I have the lead going to a 10 ohm resistor. I use RC networks to filter the pulsewaves from the PLL to create a DC feedback signal.drive circuitry from the high powered inverter section. The trick was to find the right balance. The better my filtering for a smooth signal. I had 20 capacitors that were each rated for a maximum of 14A. I witnessed this and it was not pretty. I also had to modify the feedback control loops. I have seen what happens with poor shielding. If you make the connections incorrectly. The gate drive signal goes through a damping resistor. I have copper shielding completely covering the gate drive lead. all the current will go through the closest point and rapidly overheat the copper. All of the critical signals are shielded with a grounding plane. This goes to the second board. I switched to a water-cooled Celem capacitor that was rated for 1000A. the longer the delay. com/watch?v=SZx2IeNB1Ac&feature=related 10kw Induction Heater Inverter Schematic I got most of my components from Digikey Corp and Mouser Electronics.youtube.youtube.Here are some links to the levitation coil in action for the 2kw unit Levitating and melting aluminum http://www.com/watch?v=PVtcp4JZ8FA&feature=related Levitating copper tubing http://www. The microprocessor is from Arduino Microprocessor/PLL controlled driver 10kw Inverter for use with levitation and iron forging .youtube.com/watch?v=Q6Zrnv4OtbU Levitating and melting copper scrap http://www. multimeter 5. positive developer. Digikey 9.html Electronic Supplies Things that I found useful for the project: 1. ammonium persulfate. solder station 8.Home http://inductionheatertutorial. Newark 11. etching tank 7. oscilloscope 2.com/inductionheater/index. Variac . differential probes 3. circuit board fabrication supplies: circuit board. circuit breadboard for prototyping 6. current meter 4. Mouser 10. Mapp Gas torch 12.
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