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Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Wolfram wrote ...
There will be smoke, fire and burnt components long before the thermal resistance rise of the wiring becomes significant. Also, if wiring resistance was significant enough to limit performance with lower coupling, the dissipation in the wiring would lead to complete destruction in a short time.
Wolfram, Udo's calculations, as I understand them, suggest that one Ohm of resistance would be sufficient to limit further increase in primary current. I wouldn't expect one Ohm to result in total destruction, but what do I know?
I was just commenting on the way the capacitor bank is wired in, however I will stop posting in this thread if you think I'm on the wrong track.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
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Posts: 4245
Steve Conner wrote ...
Any info on the caps? What is the dielectric? Do they get hot?
This is the photo BSVi linked to:
EDIT: If that string of brown capacitors on the left is the MMC then I'm assuming that might have something to do with the problem.
I'm assuming the red wire going off towards the left is to the other end of the primary coil and that those toroidal inductors it passes through form some kind of current/frequency/phase sensor thing.
The only other reference I've found is "MMC=12.5nF"
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
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teravolt wrote ...
the way I see it the small electrolytics are in series and that bank is in parallel with the big cap and as for the wire I have seen worse
Those brown capacitors are definitely a series string, Teravolt. I'm pretty certain that is the MMC. The other end of the series string goes to the primary coil, I assume, and that red wire with the toroidal inductors around it is the return wire from the primary coil, but I'm only guessing.
Registered Member #1637
Joined: Sat Aug 16 2008, 04:47AM
Location: Kiev, Ukraine
Posts: 83
The boards has arrived. I now in process of soldering. This how bare board looks like:
What I want to look at specifically is the phase/frequency shift of the secondary current in relation to its self
Ok, I'll do that when I get coil running again.
Basically my running theory is that the frequency at the end of the burst should be much lower than when it started
With no streamer system resonates at 430kHz and with full streamer F drops down to 392kHz. So, frequency changes for about 10%.
Making the power ramp longer even by 5ms or so doesn’t make the sparks longer, rather chubbier. This principle of power per time is important
Agreed. I also think an optimal point exist in "power per time" vs "absolute power". Ramping power too fast explodes streamer into branches and pumping more power (more ramp time) stabilizes it.
Another my theory is that in optimal QCW operation we only compensate for streamer power loss. What I mean by this is that before we pump additional power, we have to wait before additional "chunk" of streamer is formed and consumes additional power. If we pump power faster, it will branch. This is consistent with the fact that linear ramp is optimal.
0.2 ohms, that I estimated by wire diameter, length and skin effect for BSVis coil.
it seems, that you need a loss resistance of about 1 Ohm to explain, what he observed. But that is a lot more than I get by a simple estimate from his wire thickness etc.
There is more to losses than resistive copper loss. The toriod is about as warm as primary wire. Secondary is warm too, once its wire even burned out. I think, huge amount of energt is wasted heating up all those things. MMC is cold, which is strange considering thin solder-coated leads. Bridge is cold.
Increasing coupling can be problematic due to flashovers, though.
Unlike drsstcs, voltage in QCW system is low, and you can prevent flashovers with conventional insulation materials.
BSVi, can we have some more photo's of the whole primary tank circuit, please?
I have to take more photos, but I'll do that after coil reconstruction.
Have you try adding more capacitance I.E. energy storage for your bridge to convert
No, unfortinutaley, I don't have access to another big capacitor. I have to get another one or two. I agree that 80v voltage drop is signifacant, throught. But voltage drop decrease logariphimcally with capacitance, so to decrease it much, you need really BIG primary bus capacitor.
I also afraid of big capacitors. With 1kJ energy stored, if circuit explode, IGBT's shrapnel could be dangerous. Eye protection required :)
Now, I think that the best (length-wise) solution to drive QCW is to build buck-boost converter, or some kind of insulated converter with stepup transformer. As we now know, higher voltage gives more bang than higher current. Bus capacitance becames not as important as in phase-shifting version, and you get softswitching everywere.
If this is for the large coupling case, I would assume the cap voltage drop during the burst to be considerably larger for the low coupling, since there is an increased current draw. It might be just large enough to stifle spark growth at some time during the burst.
It's good idea to take measuremet of this voltage drop with lower couping. I'll do this.
Any info on the caps? What is the dielectric? Do they get hot?
It's cheappy CBB-81. Metallized polypropylene dielectric. 8pcs of 2000v/100n caps in series. They are pretty cold.
Those brown capacitors are definitely a series string, Teravolt. I'm pretty certain that is the MMC. The other end of the series string goes to the primary coil, I assume, and that red wire with the toroidal inductors around it is the return wire from the primary coil, but I'm only guessing.
There will be smoke, fire and burnt components long before the thermal resistance rise of the wiring becomes significant. Also, if wiring resistance was significant enough to limit performance with lower coupling, the dissipation in the wiring would lead to complete destruction in a short time.
I tend to agree here. At 1 Ohm primary loss resistance, maybe a 100J or more per bang will get burned somewhere.
BVSi wrote:
No, unfortinutaley, I don't have access to another big capacitor. I have to get another one or two. I agree that 80v voltage drop is signifacant, throught. But voltage drop decrease logariphimcally with capacitance, so to decrease it much, you need really BIG primary bus capacitor.
It's not as bad as that. Doubling the caps will halve the voltage drop.
It's good idea to take measuremet of this voltage drop with lower couping. I'll do this.
Thank you in advance. I'm interested in what you get. If you increase primary current you won't get around beefing up the caps. At higher bus voltages, the drop will be lower in relation to it, so it will be less bothersome.
Registered Member #146
Joined: Sun Feb 12 2006, 04:21AM
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Posts: 1055
Now consider a lossy primary. Current will also increase here with lower coupling, increasing power input, but also energy is burned in the tank. If tank loss resistance is large enough, power output decreases, even though power input increases. That will be at the point, where about 50% of power input is burned up in the primary. This does not really make sense to me, since the coil performs too well to be that lossy. So I don't think, that less spark length with lower coupling can be explained by this alone.
I think the secondary resistance and impedance plays a role here too. The EMF driving the secondary sees more leakage inductance in the secondary, which has nearly as much impedance as the arc has, so wouldnt this cause a reduction in power as well and help explain the sensitivity to K?
Also, there has been some question about energy storage and i pose this hypothetical to maybe help think about the issue. Imagine you have 2 secondary coils, one with 20pF of effective capacitance and one with 40pF. Assume the Fres is identical. We know from measurements that it takes at least 40kV to begin spark production, so as i see it, the 40pF resonator requires twice the drive current (all else being equal) to achieve just 40kV (which is about half the peak voltage most QCWs will develop). If the coils had the same resistance, then significantly more power would be wasted in the 40pF resonator. However, the winding should only need half the inductance, which means less length and thicker wire. I suspect the resistance could go down sufficiently so that both secondaries have the same losses while producing 40kV. But still, there will be twice the primary amps required to drive the 40pF load. Whats the benefit? The same spark looks like a smaller load to the 40p coil and consequently the inverter can drive more power into the system and make bigger plasma because the source impedance is effectively smaller.
Bigger coils have higher energizing costs, and i think its really quite significant in terms of primary current during operation. Figuring out the right amount of energy storage for your tesla coil design i think is key to optimizing its efficiency. Seems obvious, and overlooked.
I think the secondary resistance and impedance plays a role here too. The EMF driving the secondary sees more leakage inductance in the secondary, which has nearly as much impedance as the arc has, so wouldnt this cause a reduction in power as well and help explain the sensitivity to K?
In a QCW setting, any obstruction of the power flow to the secondary, may it be from lower coupling or higher leakage inductance, will cause the primary to ramp up to higher currents, since it is less loaded. Leakage inductance is not consumptive, so the additional input power will go the secondary. For a short burst DRSSTC a significant part of this extra energy will be stored in the primary and may or may not be returned to the bus at the end of the burst.
...The same spark looks like a smaller load to the 40p coil and consequently the inverter can drive more power into the system and make bigger plasma because the source impedance is effectively smaller.
If the bridge can handle that, yes. Whether it is an advantage depends on whether the bridge has reached its current or voltage limit.
Bigger coils have higher energizing costs, and i think its really quite significant in terms of primary current during operation. Figuring out the right amount of energy storage for your tesla coil design i think is key to optimizing its efficiency. Seems obvious, and overlooked.
I believe, this applies in particular to coils with shorter bursts. Total bang energy and energy in the primary can be of similar magnitude. The energy in the tank does not have to be lost. One mode of operation can be to charge up the tank significantly by avoiding early breakout by detuning. When the arc finally starts, it will pull the secondary into tune, which will dump most of the primary energy into the secondary.
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