Since last time, I've picked up a bit more knowledge, thanks in some part to various e-mail and forum discussions. Although the previous circuit will work if scaled up, the MOSFETs have to handle **all** of the reactive current. The power supply also needs to be stiffer, although total power consumption is only equal to the load, since the flyback currents regenerate the supply - which makes sense, since a pure inductance is entirely lossless. Here, I'll show you the physics:

Start with a variable dead time square wave. This is fed to a pair of MOSFETs (mind the transformer polarity). You don't have to use MOSFETs, that's just how I chose to draw it (but they ARE easier to use). At any rate, starting with the top MOSFET turning on:

- Inductor voltage (V
_{L}) rises to +V. - Inductor current starts increasing at the rate
*di/dt*= V/L. - When the MOSFET turns off, the current stored in the inductor looks for a way out and the voltage freewheels.
- Inductor voltage is clamped by the lower MOSFET, which has an intrinsic body diode. The same equation applies, and since voltage is negative (from the inductor's point of view),
*di/dt*is negative as well. Current starts falling. - Depending on the dead time, the flyback pulse may completely discharge the inductor, in which case a certain amount of dead time may appear where the inductor has little energy in it (undamped, parasitic capacitance will cause it to ring, excited by the
*dV/dt*of the fall to zero). This is shown on V_{L}as a partial line, and, due to the low duty cycle, very wide on the oscillograph shown below.

After the initial charging peak, the MOSFET releases the inductor, which flies back to the opposing MOSFET whose body diode clamps the pulse to the supply rail (which due to poor regulation is a bit higher in voltage, thus narrower in width than the charging pulse. The ringing is probably due to the solderless breadboard). Dampened ringing is seen after the pulse; without a snubber, it's downright ugly.

- If the MOSFET does turn on as its terminal voltage is reversing (remember, during the flyback pulse, the junction is reverse-biased, with the body diode sinking inductor current, which is dropping towards zero then reversing), the waveform is a continuous square wave.
- Inductor current I
_{L}ideally follows a triange wave, but loss including supply resistance, R_{ds}drop, induced/radiated power and others contribute to an RL exponential curve instead.

And lastly, to complete this discussion: with lossless devices, power consumption is zero because if you follow the amperes, you'll notice that, during a half cycle, one MOSFET first sinks current **back into** the power supply rail, before drawing it back out to charge the inductor so the opposing 'FET can do the same.

Although it does have full control of the inductor (and very simple ways to control power output: supply voltage, driven frequency or duty cycle), that control comes at the price of carrying **all** the reactive current. Just how much?

Well, if you consider the coupling factor K of the work coil to the work (imagine the work as a one-turn coil, loaded with resistance equal to the circumference's resistance) is roughly proportional to the ratio of their areas, then say you have a 2" dia. coil working into a 1" rod: the diameter is half, area is quarter and K ~= 0.25. That represents a circuit composed of roughly 3/4 of the work coil's (empty) inductance, which is lossless, plus a small (1/4) resistive component that actually dissipates power. As such, Q is around 3 or 4 (again, very roughly...I'm not doing mutual inductance or phased vectors here). And, as such, that means for every ampere of current sent to the inductor that's causing the work to heat, three amps go along just to make the inductor bounce! (Contrast this with a power transformer, where due to the high permeability iron core, inductance is high and reactive current is low, a mere fraction of total current capacity. This is the advantage of cored channel induction furnaces, which are unfortunately impractical on this scale.)

In a perfect world, these extra amps go along for the ride and come back to the power supply on the next cycle, and indeed it does, but we need to swap this current through real-world devices, which tend to get very expensive and power-consuming when you push a hundred or two amperes through them.

So let's cancel that nasty little reactive inductance. How do we do that? The exact opposite of inductance, capacitance. Now, there's two ways, series and parallel - series doesn't really work, because in a series circuit, current must be equal. (Voltage is canceled instead.) So we need a parallel resonant circuit to cancel the current. Voltage isn't an issue since devices which can handle upwards of 800V are easy to find.

Since the work coil is small, to get a reasonable frequency (usually in the range of 300Hz to 50kHz), a large capacitance is necessary, say 1 to 100uF. The low L/C ratio means large currents, which though bypassed away from our precious transistors, are now having a free-for-all at the capacitor. This needs stiff capacitors. Aside from industrial grade units manufactured specifically for this service, your best bet is probably polypropylene - anything rated for rough use; high pulse or continuous currents and so forth.

Now that we've got the reactive currents in check, we need to drive it efficiently. The most efficient choice by far is class D, that is, switching. If you operate your transistors in only two states, fully OFF (zero current) or fully ON (zero voltage), they never dissipate any power! Reality sticks us with leakage current, switching time and forward voltage, but we can still get 85% easily - that means 175W dissipated for a cool kilowatt output. But there's a catch: that capacitance doesn't like to be switched. The harmonics from shifting just a few volts in a quarter microsecond will draw tens of amperes! One easy way to allow things to change voltage rapidly is to add inductance, so let's add another circuit element.

I've split the tank capacitance into two elements, C_{match} and C_{T}, because that's how it actually works. Since C_{match} must vary with L_{match} for a given frequency, varying just L varies the resonant frequency of the tank (if you don't provide a seperate C_{match}). This actually means we are series resonant with L_{match} and C_{match}, while the tank is still parallel resonant. Viola!

More detailed info on the subject at Richie Burnett's Induction Heater.

One final blurb in this page before I say goodnight: I updated my driver schematic. It's a bit faster and a bit more powerful now.

I knew from day one the previous double-buffered drive was slow, but I didn't bother to fix it because I already had the SG3524 tacked between +V and GND on the breadboard. I got off my ass today and rearranged it, updated the follower with some faster transistors (f_{T} typ 8MHz instead of 3MHz) and wound a somewhat larger driver transformer (with 24AWG wire) as well. The G.P. PNP needs to be rated for supply voltage and at least 100mA; at +/-17V, 2N4403 is a close call but will work. I've got a 2SA1015 in there right now (complement to the oft-seen C1815); the last version had a pair of A970's. It pushes a 15 ohm resistor cleanly, so it's got plenty of current capacity. The oscillographs below are with 47 ohm > 2.7nF loads on each winding.

Medium duty cycle, 90kHz. Slight overshoot. "Gate" waveform (load capacitor voltage); winding side waveform is only slightly sharper.

Full duty cycle, no shoot-through that I can tell. A welcome change from the TIP31/32 follower I had previously which couldn't go much over 35% duty cycle!

8kHz waveform. You can start to see transformer inductance cutting into the flat areas. Much lower and flyback current upsets the null areas. I don't know what the step is on the positive pulse.

Oh yes, the induction heater! Next test I'm planning involves a half bridge of my six remaining STW11NB80's and whatever voltage I feel like bringing it up to. I need the tank capacitors first.

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