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Designing an Auto Match Unit for RF supply

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GeordieBoy

Sat Dec 11 2010, 11:53AM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

It looks like you are confusing the imaginary (reactive) part of impedance with the resistive (real) impedance. You can't match an RF source into a lossless tank circuit by driving it where the impedance has a magnitude of 50 ohms...

A lossless LC circuit by definition has nowhere to dissipate the power! Whilst V and I will be in the ratio of 50:1 when the magnitude of the impedance is 50 ohms, the current will be out of phase. This represents total reflected power and the RF source will either shutdown, foldback the output power or go up in smoke.

Impedance matching is all about converting the real part of the load impedance to appear as 50 ohms of resistance at the RF source _AND_ cancelling any remaining reactive (imaginary) component.

In order to do this you need to know (or make a guess) what your load is. Then work out the relevant component values in your pi-match, L-match or whatever circuit to transformer the impedance to 50+j0 ohms as seen at the RF generator's output port.

In the case of a plasma discharge for something like PECVD, there will be a severe mismatch initially before a discharge is struck in the chamber. Initially the RF source will see a lot of reflected power, and likely with V and I not even in the ratio of 50:1 Depending on how the RF source tolerates this condition (and the type of matching circuit used) the voltages and currents in the load network may ring up to sufficient levels to cause ionisation and a discharge to strike in the chamber. If this happens the discharge will then stabilise at a point where the energy in equals the energy out and it will neither grow nor diminish. It is the energy transferred from the tank circuit to the plasma discharge that sets the real part of the tank circuit's impedance, and this is what you want to match into in order to deliver maximum power.

I hope this helps,

-Richie,

Steve Conner

Sat Dec 11 2010, 03:17PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Yes, like we told you already. You can't design a matching circuit until you know what the real part of your load is. And with plasma you never know what the real part of your load is until you've built the thing and tried it out.

When doing Tesla coil modelling, I used to use the "hungry streamer" model, where I assumed the plasma grew until its size was limited by the maximum power transfer condition. In other words, if you can only get the plasma lit, it'll adjust itself to match your RF generator, IF (a subtle point) your generator's maximum power point is also its proper operating point.

The hungry streamer model didn't work too well, but neither do any of the other Tesla coil streamer models.

Ash Small

Sat Dec 11 2010, 05:11PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Thanks, Geordie Boy and Steve McConnor. I had assumed, from what I'd read in an induction heater article, that the load due to the plasma would effectively increase the inductance of the coil (In the same way that a cored coil has a greater inductance than one with an air core.

Do I take it that I merely have to place a resistance in series with the coil in the QuickSmith simulation?

Would I also be correct in assuming that a 600 Watt RF generator (assuming that 600 Watts is the power drawn from the mains, and an efficiency of around, say, 75%), would supply 150 Volts at 3 Amps into a 50 Ohm load? (450 Watts).

How would I go about guesstimating the load? will this not also be dependant on the operating pressure?

I assume the output from the RF generator is variable up to 600 Watts, but as it is controlled via a 15 pin 'D' connector I'll have to contact the manufacturer to find out. I don't want to do this until I've done as much research as possible and I know exactly what questions I need to ask them)

Ash Small

Sun Dec 12 2010, 12:17PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Ash Small wrote ...

Thanks, Geordie Boy and Steve McConnor. I had assumed, from what I'd read in an induction heater article, that the load due to the plasma would effectively increase the inductance of the coil (In the same way that a cored coil has a greater inductance than one with an air core.

Do I take it that I merely have to place a resistance in series with the coil in the QuickSmith simulation?

.

Having slept on this, I realize that the load on the coil can be compared to the losses in a transformer once the core is saturated, and it gets hot. The energy is no longer being stored but is converted to heat, although each time the field collapses you still have a current flow in the plasma (electrons moving in one direction, positive ions moving in the other) which will induce a current in the coil so some energy is effectively stored, over and above that which is due to the inductance of the coil alone.

The resistive load will be equal to the heat and light produced, including losses in the tank circuit etc. the energy in the tank circuit will stabilise at some point, assuming all other factors are constant, including the temperature of the system, so the full output from the RF generator will be converted to heat and light, assuming the impedance is matched.

I know that commercial matching circuits are only around 95% efficient at best, so with losses in the generator itself, I can only expect around 450 Watts of plasma from a 600 Watt generator at best. I'll have to think about this a bit more.......

Ash Small

Tue Dec 14 2010, 06:24PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

I've input various different loads into QuickSmith, using the circuits below, with an inductance of 3 microhenries at a frequency of 14 MHz, but I think there is something wrong with my method.

I've input the load as a resistance in series with the coil, and I don't think the results I'm getting are correct, for one thing, I'm not able to get any results for loads over 50 Ohms in the second circuit, which is supposed to match loads greater than 50 Ohms.

What am I doing wrong? the results I obtained are plotted below.

Mattski

Tue Dec 14 2010, 07:44PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

The circuit on the right can only match to 50 ohms loads which are inductive *and* outside the unit impedance and admittance circles. Meaning that the load resistance must be less than 50ohms (or else the load starts in the unit impedance circle and there is no way to use just capacitors to match it, you need an inductor) and the inductance must be high enough that the conductance (the real part of 1/(jwL+R)) is less than 1/50 siemens.

This stuff is all easier if you understand exactly how smith charts work, because then you can sketch stuff out on paper or in your head to see if it's feasible before fine tuning it on the computer. One introduction can be found here, though it uses an impedance-only smith chart while I prefer to use an impedance-admittance chart personally.

Edit: I almost forgot about this site which has some excellent smith chart applets, although I think they're all for transmission lines, not LC networks.

Ash Small

Tue Dec 14 2010, 07:58PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

So the basic method I'm using is correct?

Adding the load on the coil as a series resistance with the coil?

I'm assuming QuickSmith works out everything else for me. (I've not yet been able to download the tutorial, due to browser problems, I think.)

Mattski

Tue Dec 14 2010, 09:05PM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

If your system has reached a somewhat steady state and operates at a single frequency then it is probably valid to model the loss as a series resistance, much like the radiation resistance of an antenna. The tricky part is estimating what R values you expect, I really don't know how to estimate from theory but you should be able to measure it in a similar way to how you measured inductance. The difference is you need to find not only the gain but the phase shift to find R.

It needs to be measured while a load is in place, and probably as a function of temperature too since I think it will change a lot as the load heats up.

Ash Small

Tue Dec 14 2010, 09:57PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Thanks for your input, Mattski. I'll sleep on it and then probably post again tomorrow.

Ash Small

Wed Dec 15 2010, 09:20AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Well, I've added a 4 microhenry inductance in series with the input and I can now tune both circuits to 1500 Ohms and beyond.

I don't recall seeing any inductors in the commercial AMU's that I've seen, though, but I may have not noticed them. (there certainly weren't any water cooled inductors, but they probably wouldn't need cooling anyway)

I imagine one circuit will be more efficient than the other, but I've not really looked into this yet.

Just puzzling over the points raised in this article:

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