Author Topic: Is it possible to go fully CW with a DRSSRC?  (Read 245 times)

Offline Power-Max

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Is it possible to go fully CW with a DRSSRC?
« on: June 03, 2021, 04:32:53 AM »
I almost managed it.

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I am using HY1920P MOSFETs (90A, 20mOhms, 200V) powered with rectified mains (video suggests 110VDC from 2 supplies, but most of the shots were with direct 170V mains) Problem is under certain arc loads the secondary pretty much rejects the magnetic flux and this seems to cause the reactive energy in the primary to build up to levels high enough to blow my MOSFETs.

I've since beefed up the H bridge by doubling down on the MOSFETs and using Class 1 ceramic capacitors in place of MKP film capacitors for the much lower dissipation factor and higher Q factor. This improved the performance considerably but results in blown capacitor! I'm not too surprised, using just 8 15nf 1210 size capacitors. Damn impressed they could handle it anyway. (I think I figured out the AC voltage was getting on the order of 1KV on the primary based on the calculated reactance)

The thing I don't understand about DRSSTC's is what the impedance looks like looking into the primary. In Spice, I can see that even if you take a primary and secondary tank tuned to the same frequency and then couple them together, you always end up with 2 resonant peaks, dependent on the coupling coefficient. the less coupled the closer the peaks.

Offline SalinsLV

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #1 on: June 04, 2021, 12:51:36 PM »
Does the reactance peak when You touch the coil or is it random? I remember i had a circuit that would short the mosfet when i touched the spark at high power.
Krisjanis Salins

Offline Uspring

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #2 on: June 04, 2021, 02:43:06 PM »
This post deals with the primary impedance of a DRSSTC: https://4hv.org/e107_plugins/forum/forum_viewtopic.php?128985
Primary impedance depends on almost everything, i.e. primary Z (=sqrt(Lpri/Cpri)), secondary impedance, tuning and also on arc loading of the secondary. A major difficulty with QCW coils is the change of arc load during arc growth. The arc changes the coils behaviour by its power draw and also by its capacitance, which modifies the secondary resonance frequency.

Offline Power-Max

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #3 on: June 09, 2021, 08:02:15 PM »
My very rudimentary understanding is that if perfectly in resonance (there are 2 resonant peaks with a DRSSTC and one anti-resonant peak which is quite annoying) they shift around with arc loading and how "deep" they go depends on the arc loading, as this can be modeled as a RC network as a secondary load. I need to figure out how to model a loosely coupled transformer like this better. The transformer model and reflecting impedance across a behavioral voltage/current source (used in the transformer model). I did this back in college with transformers and transistor models but I'm rusty lol)

But if I understand it right, what might be happening is with a worst-case load on the output (either no breakout at all or a dead short circuit on the secondary) what is happening is that the Q of the transformer network is only limited by the Q factor of the series resonant capacitor network and the effective resistance of the primary coil. MOSFETs or IGBTs are nearly always the limiting factor as their effective ohmic losses is about an order of magnitude or more higher that good quality tank circuit.

Offline Uspring

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #4 on: June 10, 2021, 06:31:40 PM »
Yes, too little or too much load will lead to large primary currents. This equation describes it:

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

Qpri is the effective Q of the primary tank due to its power loss to the secondary, k is the coupling, f the frequency the coil is running at, fsec is the secondary resonance frequency and Qsec the Q of the secondary caused by the arc load. A low Qpri implies an effective energy transfer to the secondary.

The equation is the sum of 2 terms, one having Qsec in the numerator and one in the denominator, so either Qsec being too large (=little arc load) or too small (=too much arc load) will cause a large Qpri, i.e. a large primary current. A large coupling k also reduces Qpri. The same is true for good tuning, i.e. f near to fsec, which makes the first term small.

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #4 on: June 10, 2021, 06:31:40 PM »

 


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