Author Topic: Is it possible to go fully CW with a DRSSRC?  (Read 467 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.
K. 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.

Offline Power-Max

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #5 on: June 23, 2021, 06:56:19 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.

This is super enlightening, and seems to translate to my superficial observations! How was this equation derived, using the transformer model? The hard part would probably coming up with accurate figures for secondary Q factor and since the arc acts as a highly nonlinear load which is approximated to a variable shunt RC circuit.

So we know a high Qpri is bad. High Qpri conduction occurs when either term approaches infinity. The second term, (K2*Qsec)-1, grows large when the denominator approaches zero, when either the coupling approaches zero (running w/o a secondary) or when the Q factor of the secondary is very high itself (when it is short-circuited for instance, rejecting all magnetic flux) or if the topload does not permit breakout and acts as a high Q capacitor to the environment only.  The first term also approaches infinity when coupling approaches zero or Qsec gets too large, or if the term (1-f^2/fsec^2)^2 gets too high, which occurs at very high f. This last one is the least intuitive and is the most confusing. At high frequency, the inductive reactance dominates, the current will be triangular due to Xl dominating, and the current would be limited by Xl.

This term f confuses me even more because, from what I've observed, when 2 coils are coupled together the primary and the secondary resonant frequency both are impacted and change. As I understand from the transformer model, the impedance looking into the primary, you end up seeing both a magnetizing and leakage inductance, and the secondary RLC network is reflected to the primary via the turns ratio squared. (my knowledge on that model is a bit rusty though, but end result is a rather complex RLC network with multiple poles and zeros and LTspice plotting the voltage gain shows 2 resonant peaks and something I'll call an anti-resonance between them. The 2 resonant peaks tend to come together as K is reduced and spread apart as K is increased.

Which of these resonant peaks is preferred? I suppose the lower peak is more the result of the magnetizing inductance and the higher one more the result of the leakage inductance? Is is possible to construct the TC driver such it stays locked to only one of these peaks, is one of them more stable with arc loading? Is having the primary and secondary resonant frequency independently tuned to match ideal? (suppose secondary with large breakout runs 270kHz, tune the unloaded primary to also run at 270kHz?)

In the interest of saving my IGBTs (I bought a whole bunch of cheap 60N60 devices from LCSC, and HY1920W FETs to try out), is it worthwhile to use linear primary current feedback (resistor instead of back-to-back diodes) and set an inhibit pulse of fixed length (on the order of hundred milliseconds to a few seconds) using a LM339 comparator? Idea here is that when primary current exceeds a set limit, when the limit is exceeded the signal driving the gate drive IC is suppressed so the H bridge is disabled. Maybe through a flipflop with the RF feedback fed into the clock so that it disables and enables on a zero crossing. Not sure if that's really necessary though

Offline Uspring

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #6 on: June 24, 2021, 01:34:46 PM »
The equation was derived by calculating the primary current for a given input voltage and frequency f. The dissipative part of the primary current was then used to derive primary Q. Yes, getting a quantitative figure of the arc load is the hard part.

Quote
...or if the term (1-f^2/fsec^2)^2 gets too high, which occurs at very high f. This last one is the least intuitive and is the most confusing.
Usually f is quite close to fsec. The coil would be quite out of tune otherwise. Here is the place to confess, that the equation for Qpri is an approximation, which holds only if secondary resonance, primary resonance and driving frequency f aren't too far away from each other. This is normally the case.

Quote
Which of these resonant peaks is preferred?
It is possible to build drivers for either one of these resonant peaks, aka poles. A driver switching at zero current will start at the pole, which is closer to the primary resonance frequency. For a primary tuned lower than the secondary, that will be the lower pole. To run at the upper pole requires a few initial cycles at the upper pole frequency. After that it is possible to continue with zero current switching. The merits of upper vs lower pole are explained here:
https://highvoltageforum.net/index.php?topic=78.msg624#msg624

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Re: Is it possible to go fully CW with a DRSSRC?
« Reply #6 on: June 24, 2021, 01:34:46 PM »

 


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