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Large DRSSTC: choosing secondary impedance

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Uspring

Fri Apr 25 2014, 09:14AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
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Posts: 711

I haven't though too hard (let alone done any modelling) about the applicability of TL theory to the secondary though - might be looking at this completely incorrectly.

Paul Nicholson has done extensive work around secondary modelling.

My impression from that is, that a transmission line approximation is roughly correct. His analysis is much ore detailed, though.. If you drive your secondary with frequencies a lot above the fundamental, you'll observe resonances corresponding to higher transmission line modes.

Wrt to the optimal secondary impedance:
The power transfer to the secondary depends on the arc (resistive) load. For a given primary current it has a max at a specific arc resistance. If primary and secondary resonate at the same frequency this max is reached approximately, when Qsec = 1/k. Qsec is the ratio of arc resistance and secondary impedance. Arc resistance depends very much on how much power you put into the arc and will decrease with more power. Typical values are maybe around a few hundred kohms, so the optimal secondary impedance would be k times a few hundred kohms, i.e. maybe 50k. I tend to believe, that a lower value might work somewhat better particularly for high power coils.

Since the arc detunes the secondary to a varying extent, the initial assumption, that primary and secondary resonate at the same frequency cannot really be upheld. Optimally this condition should be reached at the end of primary rampup.

Steve Conner

Fri Apr 25 2014, 12:21PM

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

Interesting. Since arc load is a function of power transfer, and power transfer is a function of arc load, it sounds as if the loaded Q of the secondary would tend towards 1/k as the drive level was increased.

It would follow that coupling would ultimately limit the power throughput of the system: no amount of extra primary current could make up for a lack of coupling.

How does primary Q enter into the equation? I always tried to design so that both primary and secondary would have a loaded Q of 10, but I don't have an intuitive feel for how the load on the secondary affects the Q of the primary.

Uspring

Fri Apr 25 2014, 01:55PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
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Whether there is an upper limit to power througput depends on how much the arc resistance changes with power. For no change of arc resistance, the power throughput would be proportional to Ip^2.

For primary Q I have 2 equations for the limits of high and low arc loads. They are in terms of the primary loss resistance, from which you can calculate the primary Q:

Rp = wÂ°2 * Lp * Ls / Rarc, for large Rarc

and

Rp = k^2 * Lp / Ls * Rarc for small Rarc.

In essence you always need to take the lower ones of the Rps. Or better, think of both Rps in parallel. The first equation depends sensitively on tuning. And these equations hold only for ZCS.

Goodchild

Fri Apr 25 2014, 03:19PM

Registered Member #2292 Joined: Fri Aug 14 2009, 05:33PM
Location: The Wild West AKA Arizona
Posts: 795

Wow you guys are smart! So if I have this straight assuming we treat this system as a transmission line, we are essentially dealing with a series circuit that includes Zsec, Zarc, and the impedance of the earth ground Zgnd.

Now assuming we have no ground strikes Zarc stays the same (with a constant power that is). As soon as we get a ground strike Zarc drops off significantly. This would then prompt a large change in current, but because current can't change instantaneously threw Zsec this current has to propagate across the secondary.

So with classic transmission line theory we want to match the terminating impedance Zgnd to the secondary circuit impedance Z sec. Using Udoâ€™s example 160K would be the terminating Z.

What Iâ€™m not fully understanding, why is 160K to large? Is this because of practical implementation? ie your earth ground would never be this large?

Hydron

Fri Apr 25 2014, 05:44PM

Registered Member #30656 Joined: Tue Jul 30 2013, 02:40AM
Location: UK
Posts: 208

160k is too large because any practical ground is way less (mine was ~1k). Even if you had really dry soil or something, you couldn't run a TC with a ground that bad, the amps of secondary current would produce 100s of kV across it.

Uspring: How did you go about analyzing your arc measurements? I've got some preliminary data for my coil, but the only analysis I've done so far was using the scope maths functions (not yet sure how to deal with the 2 megasample files!).

A quick look at what I did manage suggests arc impedance drops off precipitously with ground strikes (to sub 50K ohm, clamping toroid voltage to <100kV), but I'm not willing to put hard numbers to anything yet.

Tonight I'll try and improve my setup and get some better data, along with some simultaneous camera exposure. Hopefully I can get the scope to let me only capture the coil on-time, which should make the data more manageable and let me use higher sample rates. Will post it all (probably in a new thread) once I've got something I'm happy with.

Steve Conner

Fri Apr 25 2014, 07:05PM

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

160k is too large because it would damp the desired resonance of the secondary coil and eat up all the power that is supposed to go into sparks.

Uspring

Fri Apr 25 2014, 07:30PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
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I measured current between secondary top and toroid and between toroid and breakout point. From this you can derive top voltage and arc current. The DSO can dump txt files with the ADC readings, which I imported to a spreadsheet. Beyond voltage and current magnitudes also the phase between them is interesting. To extract the phase, I wrote a little program which calculated the times of zero crossing by interpolation.

Hydron

Fri Apr 25 2014, 10:47PM

Registered Member #30656 Joined: Tue Jul 30 2013, 02:40AM
Location: UK
Posts: 208

Uspring: sorry, I should have been a bit clearer.
I know what to do (find the difference between the two measured currents to get topload charging current. integrate to get topload voltage, use I and V and the phase to look at streamer impedance), but not the best way how to do the analysis.
I haven't touched matlab etc for years, and I can't import 2 megasamples (!) into a spreadsheet, so I was hoping you had a quick and easy answer I could steal.

I can actually use the scope software to do the integration, scaling, subtraction etc, but I can't get the instantaneous phase or magnitude out of it without passing the data off to matlab. I'd also like something easier to make the requisite pretty graphs with.

Edit: I had posted some scope captures here, but I no longer trust them - the only low value resistors (for the current shunts) I could find last night were wirewounds which turned out to be more inductive than I expected, so I'm gonna get some better data before sharing.

Intra

Mon Apr 28 2014, 08:50AM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

Kizmo wrote ...

I managed to have quite epic secondary failure last night and now im looking forward of winding new one. With topload the failed secondary was around 70kOhm impedance at resonance. From earlier conversations I recall several coilers stating that somewhat optimal secondary impedance for big coil is less than 50kOhm. Is this still true? Old secondary was 315x1270mm and I think i will make it a little larger for next coiling season.

I had same issue. That's not an impedance problem. It can be a wire's coat. In some cases, if source wire for secondary is old, epoxy coating may crack under the influence of time. Therefore reduced dielectric breakdown strength and produce winding electrical breakdown. Also, common secondary coating may have been not large enough for sufficient dielectric strength.

Uspring

Mon Apr 28 2014, 09:41AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Here is an equation, which describes primary Q within a simple TC model, that neglects any losses except loading by the arc.

Qpri = Qsec/k^2 * (1 - f^2/fs^2)^2 + 1/(k^2 * Qsec)

f is the frequency the coil is running on and fs the secondary resonance. The equation is exact for those f, for which there is zero current switching, e.g. at the poles.
Generally it is an advantage to have a low Qpri since it means a fast transfer of energy in the primary tank to the secondary. Otherwise you would build up a lot of energy in the primary tank by the driver without it getting to the secondary. Also, a large amount of energy in the primary tank will cause large losses in it due to copper and MMC resistance.

Qpri gets minimal with respect to f, if f=fs. If primary and secondary resonances are at the same frequency and the coil is run e.g. at the lower pole, then f = fs/sqrt(1+k). The above equation then simplifies to:

Qpri = Qsec * (1+k)^2 + 1/(k^2 * Qsec)

Detuning, i.e. choosing primary and secondary fres different, will in general increase the first term considerably. This can be an advantage for some coils, since it allows to match the drivers voltage and current capabilities to the primary tank.

The equations also show a minimum for certain Qsecs. This gives an idea on how to choose secondary impedance, since Qsec is the ratio of arc load resistance to secondary impedance. The change of tuning and also the change of the arc load resistance during a burst makes this difficult.

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