---------- Forwarded message ----------
Date: Fri, 21 Sep 2007 11:25:30 -0500
From: Bert Hickman <bert.hickman@xxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: coupling losses ? (fwd)
Hi Chris,
It looks like you may have a bit of confusion here...
Tesla list wrote:
> ---------- Forwarded message ----------
> Date: Fri, 21 Sep 2007 10:23:00 +0100
> From: Chris Swinson <list@xxxxxxxxxxxxxxxxxxxxxxxxx>
> To: Tesla list <tesla@xxxxxxxxxx>
> Subject: coupling losses ?
>
> Hi all,
>
> If you start at 10KV on the primary, and the secondary is maybe 2" away,
> there is some distance there and coupling will be lower. As there is some
> distance, the 10KV line has to drop over the square of the distance ?
Coupling is purely a function of the respective geometries and spacing
between primary and secondary windings.
There are NO losses directly associated with coupling.
>
> Saying so, if stating out with 50% coupling, then it would be more like only
> 5KV actually gets onto the secondary coil ?
No. If there were no other losses in the system (such as spark gap,
resistance, radiation, etc.), then you could achieve 100% energy
transfer between the primary and secondary INDEPENDENT of the coupling
(for k>0). Lower coupling simply means that it will take longer to
transfer energy from one tuned LC circuit to the other. Looking at it
from a Conservation of Energy (COE) standpoint in a lossless system, all
of the initial energy stored within the primary tank cap would
eventually be transfered into the secondary LC system. However, the
transfer will occur over a number of cycles at the operating frequency -
the lower k, the longer the energy transfer process takes.
It's similar to pushing a "lossless" swing - if you give the swing a
larger push each time (via a higher k), the energy in th swing builds up
faster than if you used a smaller push (lower k). However, if the total
energy you provided (over time) is the same for the low k or high k
case, then the total energy transferred to the swing is also the same in
either case.
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
lower-k system takes longer to transfer energy to the secondary than a
higher k system, more energy gets lost per bang, leaving less ultimately
available at the secondary. However, there are limits to using high
coupling coefficients - racing sparks, flashovers. These tend to limit
practical coupling to about 0.2 - 0.25 for most two-coil systems unless
heroic measures are taken. Even so, a well-designed Tesla coil can
easily transfer over 85% of it's initial primary "bang" energy into the
secondary, even though the coupling coefficient may only be 0.1 - 0.2.
> After some of my own analysis of
> sparks gaps, its probable that my 10KV source will not turn off over the
> spark gap until about 8kv. So if we take 8KV as the actual primary voltage,
> loose 50% due to low coupling, then this leaves us with only 4KV actually
> driving the secondary coil ?
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).
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
>
> 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.
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).
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.
>
> cheers
> Chris
>
Best wishes,
Bert
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