Re: coupling losses ? (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Fri, 21 Sep 2007 21:24:01 +0100
From: Chris Swinson <list@xxxxxxxxxxxxxxxxxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: coupling losses ? (fwd)

HI Bert,

Thanks for your in dept analysis...


> If you include system losses, a portion of system's energy is lost over
> each cycle, leaving less available to make it to the secondary. Since a

Resistance, heat losses ?


> Wrong interpretation. The energy in the primary transfers to the
> secondary over a number of RF cycles. In most systems, the gases) in the
> gap(s) are still quite hot and conductive. As a result, the gap
> "reignites" at relatively low voltages - perhaps less than 10% of the
> initial ("cold") breakdown voltage. Re-ignition causes a failure to
> quench, which lets energy cycle back and forth between P->S, then S->P,
> many times. For most Tesla coils, this back and forth energy cycling
> process may continue many times before the gap finally stops conducting
> (quenches).


If we assume then that the spark gap is the main loss in the system, then if 
we replace with for example solid sate, then use a low K factor, we will 
need many cycles of course to transfer the energy, but what are the cycle by 
cycle losses then ?

In theory then at least, if we had zero losses, we can transfer energy in 
any number of cycles over any amount of coupling ?

As for spark gap conduction, If we have a 10KV source, the RSG will break 
down first at 10KV, now as the gaps move closer the voltage must drop ? In 
effect the spark gap may not actually turn off until the volts have droped 
to 8KV. Which would in effect mean we only put 8Kv into the primary ? Or did 
we pump 10KV as that was the inital fire voltage ? I have trouble with this, 
as the spark gap will not quench until the voltage has drops to maybe 8KV, 
in which case we only trapped 8KV in our primary system ?



> You may wish to study the extremely well written write-ups on Richie
> Burnett's page on TC Operation and Quenching to get a better
> understanding about spark gap coil operation, energy transfer, and the
> roles of coupling and quenching:
>
> http://www.richieburnett.co.uk/operation.html#operation
> http://www.richieburnett.co.uk/operatn2.html#quenching
>
>>

Thanks for that, I did read it a while ago, but It did not seen to answer a 
lot of things. I got confused as my small 12V coil testing shows that a low 
coupling of 10% translates to only 0.1V. On this basis there is a loss. So 
it is confusing to say the least!







>> On this basis the voltage gain of the secondary would have to be 50 to 
>> gain
>> 200KV output ? Does the "50" not mean the "Q" of the coil ?, or what 
>> tesla
>> called the "magnifying factor" ?
>
> Again, wrong interpretation. The Q of a secondary wound with high
> quality insulated wire is purely a function of its inductance versus its
> AC resistance at its operating frequency. The AC resistance of the
> winding includes the DC resistance plus additional resistive losses from
> skin and proximity effects, both of which act to reduce the "usable"
> conductor area of the wire. The Q of a close wound TC secondary is
> typically in the range of 100-300 - YMMV. The higher the Q of the
> secondary, the sharper it's resonant frequency.
>

Going by the archives, the general thinking Q factor does not matter. 
However as Q is divided by resistance, a typical coil has around 30 DC ohms, 
small dia wire will give higher AC resistance. So I would have thought using 
a resistance of less than 1 ohm would be a far superior design ?




> However, in spark gap coils (be they 2-coil classic or 3-coil
> magnifiers), it's actually COE (less losses) that ultimately governs the
>  output voltage. Since each bang has a fixed amount of energy, this is
> the maximum that can be transferred to the secondary if there were no
> system losses. Although having a high Q secondary is good design
> practice and helps to reduce overall system losses, system losses are
> actually dominated by the spark gap in the primary circuit and streamers
> (once you achieve breakout).
>


This is where it gets confusing. I ran a lot of simulations and found that 
while inductance is good for voltage gain, resistance goes up by almost a 
equal amount which makes more inductance actually bad. Tesla even states 
this in CSN book, Tesla then goes on to say that a large toroid is then 
needed to overcome the larger inductance.





> The Q of a coil is what Tesla sometimes called the "magnification
> factor". Tesla ultimately wanted to excite his largest systems from
> continuous (CW) RF sources, but unfortunately the technology to do so
> did not exist at the time. He knew that the higher the Q (of his
> secondary or tertiary coil), the higher the output voltage would be when
> excited from a constant amplitude CW source. A resonator with a Q of
> 300, when base excited from a low impedance 1,000 volt source, could
> develop 300 kV at the top (assumings no breakout). This is the "magic"
> of resonant systems. Sometimes called Q-multiplication, this process is
> how energy builds in the secondary of most vacuum tube and solid state
> coils. Even in these systems, the eventual output voltage is governed by
> a balance of energy: Energy into the power oscillator versus energy lost
> by system and streamer losses. The "effective" Q of the secondary
> plummets after breakout occurs, and so does the maximum output voltage.
>


http://www.future-technologies.co.uk/IMPULSE/graph/

My results show that a low resistance works better than a higher inductance 
and once you get below 1 ohm you really start to double up on the output 
voltage. While these are simulated tests, this does not include the actual 
practicalities in construction.

Cheers,
Chris