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What this means is that if your bridge’s max power handing is say already 12.448Kw then with 0.1k you have reached your limit. Were if you increased your coupling to 0.4K you now have an addition 5.53Kw of headroom, while still transferring the same amount of energy to the output in the same amount of time.
The figure of 5.53kW only applies to the first 1ms of the burst. After about 2ms the primary peak voltage has settled, so there is no additional energy anymore being stored in the tank. At that point input and output power will become equal (provided we omit copper losses). The drawback of lower coupling is, that the higher amount of stored energy in the primary will cause more losses in the copper resistance. But remember, that the OP wanted to have more current in the primary, which also means larger losses.
I'm still puzzled about BSVis observation, that decreasing coupling has such a bad effect on arc length. In the lossless case it should actually increase due to the higher power input. I've done some calculations for a lossy primary and 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.
Registered Member #2292
Joined: Fri Aug 14 2009, 05:33PM
Location: The Wild West AKA Arizona
Posts: 795
Uspring wrote ...
I pretty much agree up to this point:
What this means is that if your bridge’s max power handing is say already 12.448Kw then with 0.1k you have reached your limit. Were if you increased your coupling to 0.4K you now have an addition 5.53Kw of headroom, while still transferring the same amount of energy to the output in the same amount of time.
The figure of 5.53kW only applies to the first 1ms of the burst. After about 2ms the primary peak voltage has settled, so there is no additional energy anymore being stored in the tank. At that point input and output power will become equal (provided we omit copper losses). The drawback of lower coupling is, that the higher amount of stored energy in the primary will cause more losses in the copper resistance. But remember, that the OP wanted to have more current in the primary, which also means larger losses.
Udo,
You are correct in that with this simulation the primary current stabilizes after 2mS. However this wouldn’t be the case in a real system. In a DR the burst would almost always be shorter than 1mS. And in a QCW power would be added to the tank more controllably than my simulation, making the coil operate in a transient mode for much longer than 2mS.
So in the real world we would never see, operation where the tank current stabilized like it does in that simulation after 1mS. At least not in the way we operate most coils today.
But it brings up a good question! How would the coil operate if it was allowed to operate in the steady state with lower coupling? Personally I don’t know. I have always favored the ability to grow the current if needed, a perk of operating in transient mode.
You are correct in that with this simulation the primary current stabilizes after 2mS. However this wouldn’t be the case in a real system. In a DR the burst would almost always be shorter than 1mS. And in a QCW power would be added to the tank more controllably than my simulation, making the coil operate in a transient mode for much longer than 2mS.
So in the real world we would never see, operation where the tank current stabilized like it does in that simulation after 1mS. At least not in the way we operate most coils today.
I agree with your remarks on both short and longer burst operation.
But it brings up a good question! How would the coil operate if it was allowed to operate in the steady state with lower coupling? Personally I don’t know. I have always favored the ability to grow the current if needed, a perk of operating in transient mode.
Lower coupling will draw more current for a given input voltage as seen in the simulation. It will also make the coil more sensitive to detuning by the arc capacitance. If you can't obtain the desired amount of input current even at max input voltage, it is better to reduce primary inductance. If your current is too high, it is better to increase coupling rather than increase primary inductance. Increasing coupling can be problematic due to flashovers, though.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...
I'm still puzzled about BSVis observation, that decreasing coupling has such a bad effect on arc length. In the lossless case it should actually increase due to the higher power input. I've done some calculations for a lossy primary and 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.
It does make sense that if there is no increase in primary current when coupling is decreased then current in the primary is limited by resistance in the primary.
Maybe there are 'other resistances' in the primary circuit?
BSVi has linked to a photo on page two of this thread which he precedes with words something like 'My current setup is as messy as this'.
If those brown capacitors are the primary tank capacitors, then maybe the 'limiting resistance' is there, or maybe elsewhere in the primary tank circuit.
The primary coil itself is not the only resistance in the primary tank circuit.
EDIT: @BSVi, can we have some more photo's of the whole primary tank circuit, please?
Registered Member #195
Joined: Fri Feb 17 2006, 08:27PM
Location: Berkeley, ca.
Posts: 1111
BSVI, Ash I think has a point. Have you try adding more capacitance I.E. energy storage for your bridge to convert. If go to 20ms will there be enough gas at the end of the ramp up. How would it be pausible to calculate the amount of energy in joules that is needed for ramp up. if I was going to do it I would just keep adding caps till the spark stops growing or can it be pausible to go to far.
Registered Member #2292
Joined: Fri Aug 14 2009, 05:33PM
Location: The Wild West AKA Arizona
Posts: 795
Goodchild wrote ...
BSVi wrote ...
Excelent investigation, Goodchild! This is closely maches what i'm observing in real system. Seems that increasing bus voltage (and coupling) still the best way to increase streamer.
Maybe, it's even feasible to build buck-boost and use 1200v IGBTs to get ultralong streamers :)
As for tune, my experementation shows that you don't need good tune to have good streamers. That profs that secondary is almost resisitve load with low Q.
BSVi, I agree, this is what I observed with my QCW as well. As an alternative to making the bus voltage higher when raising coupling, the tank can be made lower impedance. This is useful if you are already running near the max bus voltage the system can handle. The trade off however is having to deal with larger currents in the primary LC.
Pardon quoting myself but, I pointed that out a while back. The tank impedance is one of the main factors that limits spark growth when you are running a fixed bus voltage.
And as such lower the tank Z (making the tank cap bigger) allows the current to ring up higher with the same driving voltage.
I believe, that an explanation of the effect of lower arc length with lower coupling may be a mixture of different issues:
First consider a lossless primary tank. Lowering coupling will increase current and therefore power input. Most of this goes to the output. There is also an effect, as Eric pointed out, that more energy will be stored in the primary tank. But this primary energy is about 0.25J for BSVis coil compared to the total bang energy of 364J. So this is a small effect and actually spark length should increase.
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.
BSVi wrote:
I have measured bus voltage before and after discharge. Before it is 379V and after - 300v. So I dont run out of voltage.
Considering capacitance of 13600uf, bang+losses energy is 364J.
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.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...
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.
If, as I suspect, the wiring to the capacitors is the 'limiting resistance' in a lossy primary tank circuit, then decreasing coupling will tend to increase current in the primary tank, which will in turn lead to extra heating, leading to increased resistance, leading to lower current.
My comments are purely based on the way the capacitors are wired up in the photo I mentioned above.
Current will only rise to a finite limit in any tank circuit, due to resistive (ohmic) and heating losses, even if completely uncoupled. Maybe this limit has been reached in BSVi's circuit?
Once this limit is reached, spark length depends entirely on coupling (and secondary losses), I think.
Registered Member #33
Joined: Sat Feb 04 2006, 01:31PM
Location: Norway
Posts: 971
Ash Small wrote ...
Uspring wrote ...
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.
If, as I suspect, the wiring to the capacitors is the 'limiting resistance' in a lossy primary tank circuit, then decreasing coupling will tend to increase current in the primary tank, which will in turn lead to extra heating, leading to increased resistance, leading to lower current.
My comments are purely based on the way the capacitors are wired up in the photo I mentioned above.
Current will only rise to a finite limit in any tank circuit, due to resistive (ohmic) and heating losses, even if completely uncoupled. Maybe this limit has been reached in BSVi's circuit?
Once this limit is reached, spark length depends entirely on coupling (and secondary losses), I think.
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.
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Increasing QCW streamer length
