Most literature in relation to this either approaches it from the perspective of stepping up/down the voltage or matching impedances. What is confusing me is how are you meant to match the impedance of a circuit that theoretically has zero impedance (neglecting negligible elements like ESR).
Under what theory would a circuit have zero resistance?
A sorely incomplete one, I hope.
Since after all, without resistance, there can be no power transfer!
No, it is indeed precisely the ESR that remains, which is reflected through the transformer, and matched.
Where does the ESR come from? How to calculate it?
The work coil is a transformer, one with poor coupling. The work is a shorted turn, except it's not really a short, it has self-inductance (the inductance of one turn around the current path) and resistance (which is of course the material resistance, factored by skin depth or material thickness, size and shape, and so on). This sets the current due to the induced voltage; the current is then reflected back into the work coil as a load.
Because the transformer has poor coupling, it has a lot of leakage inductance. So a reasonably detailed model would look like an L1 || (L2 + R) circuit, where L1 is the coil's self inductance, L2 is the leakage inductance, and R is the work resistance.
This still doesn't help you much, because L2 is really hard to put a number on, and R is tricky to do any better than estimate, given simple conditions (like a bunch of cylinders).
I give this as background, but not as actually useful help. So what does help? Sqash it all together and call it an estimated Q factor (which at its simplest, reduces to an L + R or L || R equivalent, at a given frequency).
What good is that? Two things. One, Q factor varies widely with conditions anyway, so it's not very useful knowing it precisely. Second, because of this, the power supply needs to be able to work into the same range that your load is expected to cover.
A very practical example is a steel melting process, which might go from a Q of 5 (loaded with cold magnetic metal: the magnetic hysteresis loss about doubles the loss (ESR)!) to a Q of 10 or 20 (above Curie temperature, or maybe slumped into a small heel), to back down to 10 or so (with a full molten load).
The Q is proportional to the ratio of areas between work coil and load. You always want to be as close to the work as possible, to keep efficiency high; if you avoid designing for very low Q (say, 1-3), that's fine -- you can always increase Q by adding an inductor (hopefully high quality, to maintain efficiency) in series with the work coil. You can't reduce it, though!
The Q also varies with work resistance:
Lower resistivity materials, like aluminum, bronze and copper, tend to reflect magnetic fields rather than dissipate power. The Q is higher, maybe 2-3x that of hot steel. Also, the coil losses are much larger, because the coil itself is carrying about as many amp-turns as are induced in the work, and is made of the same resistivity!
Higher resistivity materials, like stainless, titanium, graphite and so on, tend to work very nicely -- as long as they are thicker than several skin depths. Simply put: as long as the field is being dissipated into the material (that's what skin effect is: the cancellation of magnetic field due to induced currents), it will be heated effectively.
Clearly, at some point, resistivity can be too high, and the Q will rise again. This is true of ceramics and composites, which don't have enough bulk conductivity to do much (yet they may be excellent absorbers at microwave frequencies, say -- SiC and ferrite are good examples). A perfect insulator has no induced current at all, and therefore does not load the coil.
Based on this, and given a ballpark Q range of perhaps 5 to 30, and an inverter range of 2:1, you can design the transformer and taps.
30/5 = 6. 6 is almost 3 octaves (which is
, so the transformer probably needs 3 or 4 taps.
That is, given some work coil inductance (say 1uH nominal), and a frequency (say 50kHz), you therefore need 10uF to resonate with it, and you get a resonant impedance of 0.32 ohms. For Q = 5 to 30, you get ESR = 10 to 63 mohm.
The inverter might deliver a 320Vpk square wave at 50kHz, at up to 30A RMS. That's 10kVA capacity, so you can drive a 5kW load anywhere from 1:2 to 2:1 mismatch (i.e., at 320V and 15A, or 160V and 30A -- you need some way to adjust the effective voltage, either by PWM*, or varying the supply voltage, or adjusting the driven frequency away from resonance**.
*This has to be phase shift PWM in an H bridge. (The reactive currents don't work out if you PWM anything else, like a half bridge -- try it.)
**This saves on power conversion stages, but has an ominous side effect: the control loop is almost impossible to stabilize. The "plant" has a complex pole pair, where the Q factor of that pole is (more or less) equal to the work coil Q (so, it's very peaky when the Q is high), and the pole frequency is equal to the difference between resonant frequency and driven frequency. Which in turn depends on the control state..... so, yeah.
You can guard band it, but you end up with a ponderously slow control, which can be dangerous for the inverter's health. (Say if the control doesn't respond quick enough, and it dips below setpoint, hits resonance, and current builds up 2-3x what it should be, and...) A DSP could be designed to solve for this and change the controller's time constant accordingly, but geez, that's a lot of work. (It might be simpler to apply a state space solution, simply cranking through a pile of linear algebra every cycle. You'd end up with something that doesn't quite look like PWM or PID control, and would have the advantage of being the most agile possible control.)
Anyway, the example inverter has load impedance range of 160/30 = 5.3 to 21.3 ohms (a factor of +/- 2, so to speak), which needs to be matched to a 10-60 mohm load. A ratio of 23:1 to 18:1 is minimal (i.e., 2 taps).
But that's only at one inductance, and one frequency. For each of these, you get a proportional spread in load impedance. A practical design might consider work coils 1-10uH, at frequencies 20-80kHz, matched with one capacitor of 5uF, or a selection of capacitors somewhere from 0.4uF to 63uF. This gives a huge spread of impedances, so you'll likely compromise by cutting off certain combinations: large and small inductances, at high and low frequencies, respectively. This gives a sort of tilted prism, in the L-Freq-Q operating space.
Anyway, whatever the compromise made, you'll likely want to settle with a transformer around, say, 4:1 tap range. Pile up the ferrite and you can manage this with only a few secondary turns -- convenient as they'll need to be water cooled.
Note that leakage inductance, in the transformer, needs to be small. The secondary should fit closely around the primary. It's fine to wrap the cores with the primary, then assemble the secondary [tubing] into it. Even better is using a single tube (as large as possible) down the center, and a huge tube around the outside, with the tubes joined at the top and bottom so you get one big shorted turn, except one end isn't actually shorted, but the inner and outer connect to separate plates that go to the capacitor and work coil. (For this one, you'd probably close and flood the center, and wrap the outer with cooling tubes, or flood the whole thing, primary and all, which is nice for cooling the primary and cores, but not so nice for insulation.)
(Can you tell I've done this before?
)
Tim
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