| Electronics > Projects, Designs, and Technical Stuff |
| Small radius heat induction coil connected via coaxial cable? |
| << < (3/3) |
| Plasmateur:
--- Quote from: T3sl4co1l on January 04, 2019, 05:37:01 pm ---What kind of coil geometries would be acceptable to your application, and what L and R parameters would they have? That is the question, as far as being able to tune it properly. The two-capacitor network acts as an impedance divider, which when the capacitive reactance is canceled out, also acts as an impedance transformer. Usually there is an inductor in there, as well. --- End quote --- Yes. I removed the inductor awhile ago. Originally this was used for plasma generation, but I could never get the plasma to ignite with the additional inductor as well as the auxiliary capacitors. --- Quote ---The easiest way to demonstrate this working (given that familiarity with reactances is a prerequisite), is to picture a resonant tank: a capacitor and inductor in parallel. At resonance, reactances cancel, and the parallel equivalent impedance is infinite (like an open circuit), and the series equivalent impedance is zero (like a short circuit). Suppose we connect a resistor in series with the loop: the series equivalent is then just that resistor (at resonance). Or suppose we connect one in parallel: same thing with the parallel equivalent. What if we do both series and parallel resistors? We have some choice of which places to put them, which gives the four permutations of L-match networks. Suppose we measure the series equivalent impedance, with a resistor in parallel with the capacitor. That is, a circuit of (port)--L--(C || R)--GND. What is the resistance at resonance? (But first, what is resonance in this circuit?) As it turns out, the resonant frequency is slightly different, by an amount on the order of -1 / (R^2 C^2). For non-precision cases, and Q > 5 or so, this is negligible, but it is worth noting that the frequency changes. (This is more obvious with exaggerated values, say R < sqrt(L/C), and the transient response rather goes "thump" instead of "dinggg"!) So that's easy to approximate out, and we're left with the resistance question. As it happens, the R value is inverted relative to the resonant impedance sqrt(L/C). That is, the equivalent series resistance measured is around Req = L / RC. Consider what happens at resonance: the loop current rises, and the voltage (across the inductor or capacitor) rises. Both rise proportionally, the ratio being the resonant impedance, sqrt(L/C). We can start with a low voltage, apply it in series to an LC tank, and get a large voltage on the resonant node -- "Q multiplication". If we load that resonant node, the Q is spoiled and the voltage drops relatively rapidly, which is to say, it is a high impedance node -- we have transformed the source impedance. It works inversely, starting from a high impedance and making a low impedance, just the same. This is the basic, qualitative way in which the tuning network operates. To match any impedance, high or low, two L-match networks are glued together back to back; which puts two inductors in series so only one physical component is needed. That leaves two variable capacitors, in effect, one to match the source and the other to match the load. :) L-match calculator to play around with: https://www.daycounter.com/Calculators/L-Matching-Network-Calculator.phtml Tim --- End quote --- Tim. Thank you for taking the time to explain this. You are awesome. What I am having a hard time understanding is that I don't have a load resistor, but an inductor in place of it. I understand how to solve for Q when there is a load resistor, but not when it is replaced by an inductor. |
| T3sl4co1l:
--- Quote from: Plasmateur on January 04, 2019, 08:09:40 pm ---What I am having a hard time understanding is that I don't have a load resistor, but an inductor in place of it. I understand how to solve for Q when there is a load resistor, but not when it is replaced by an inductor. --- End quote --- Simple -- your inductor isn't an inductance, it has resistance too! The Q factor or ESR or EPR of it can be estimated from the geometry and materials, or measured in operation. (Noteworthy, tungsten's resistivity changes wildly with temperature so the Q will change as well. The figure measured when cold will not be the same at final temperature.) --- Quote from: Plasmateur on January 04, 2019, 08:02:38 pm ---I might drop the on hand 13.56MHz source I have and just build one based on this thing I'll have to contact them to see what frequencies this can get up to without overheating the transistors. --- End quote --- Would guess a MHz is pushing it... Hmm weird, there's some sort of driver chip in the middle. Did they-- is it actually a class D driver, with analog feedback? An attempt at a boosted "Royer" (actually Baxandall) oscillator as is commonly seen? That's pretty awful... the problem is this: for the oscillator to start, it must start from a small-signal condition, amplifying noise until the dominant mode(s) take over, at which point the transistors begin to saturate, the amplitude limits, and stable (periodic) and efficient operation is had. You can't just go from one to the other -- attempting to feed that small noise signal into some kind of comparator (if it's a gate driver IC, the typical input characteristic is a logic-level threshold with hysteresis), will result in the tallest (amplitude) signal taking over, which typically means oscillating near the maximum toggle rate of the driver itself. Thus cooking the driver and its transistor. This assumes it is actually a gate driver; even if it's not, the fact that there's no TVS diodes protecting the transistors is damning enough to me... Tim |
| coppercone2:
the calculator on the website estimates resistance and q |
| Navigation |
| Message Index |
| Previous page |