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Re: Secondary CURRENT
Tesla List wrote:
>
> Subscriber: sgd1-at-acpub.duke.edu Thu Feb 13 22:06:33 1997
> Date: Thu, 13 Feb 1997 16:25:31 -0500
> From: Stan <sgd1-at-acpub.duke.edu>
> To: tesla-at-pupman-dot-com
> Subject: Secondary CURRENT
>
> After following this list for several weeks, I have seen much discussion
> and debate about secondary voltage. However, I have seen no mention of
> secondary current. Anyone have any idea how much current can be
> expected at the secondary?
>
> [Part 2, Text/HTML 8 lines]
> [Unable to print this part]
Stan,
This turns out to be extremely interesting and complex! It depends when,
where, and how you measure it! Warning: the analysis below is NOT
intended to provide great rigorousness or precision, but only to provide
ballpark estimates. BTW, if you thought that estimating primary current
was frought with difficulty and controversy, its NOTHING compared to
trying to do the same on the secondary side! :^)
At the Coil Base:
================
The maximum current you'd expect to see coming off the base of your
secondary can be roughly approximated by invoking the conservation of
energy and using some simple math. The energy in the primary at the time
the gap first fires will be (1/2)CpVg^2 where Cp = the primary
capacitance (in Farads), and Vg = the voltage that the gap "fires" at. A
portion of this energy ultimately gets transferred to the secondary coil
(usually only 60% or less), where it sets up oscillations in the
secondary inducatance and the combination of the self-capacitance of the
secondary and the discharge terminal. If the secondary has an inductance
of Ls, it can be shown that the maximum base current wil be about:
Is(max) = X*Vg*SQRT(Cp/Ls) where X is typically 0.4 - 0.6.
and Cp and Ls are in uF and uH.
Example: 0.02 uF cap, discharged at 19000 Volts, and secondary
inductance of 73000 uH. Assume X = 0.6 (60%)
Is (max) = 0.6*19000*SQRT(0.02/73000)
= 0.61*19000*0.000523 ~ 6 Amps
This is why you want a nice beefy RF ground tied to the base of the
secondary. Larger coils may have base currents in the tens or even 100's
of amps.
At the Discharge Terminal:
=========================
This one's a lot tougher to estimate, particulary if there are streamers
to air coming off the top. Ypu'll find there's LOTS of differing
opinion here, and I'll probably take a few flames... but, what the
heck!! All cases assume good quenching.
There are actually three distinct cases:
1. No streamers:
If you've got a relatively large toroid at the top, the secondary coil
IS generating substantial current necessary to "swing" the toroid
voltage between positive and negative voltage extremes at the resonant
frequency. A "rough" approximation of this current is roughly the same
as the secondary base current (above). Note that, even without breakout,
there may be substantial "dark current" flow (mA-level)due to extensive
free-electron and ion production around the toroid.
2. Streamers to ground:
The peak currents coming off the toroid are surprisingly large!
Suppose you have a toroid of capacitance Ct and were able to drive it to
a peak voltage of Vsmax. At that instant, you'd have
(1/2)*CtVsmax^2 Joules (watt-seconds) of energy stored in the
electrostatic field surrounding the toroid and top portion of the
secondary. For example, a 30 pF toroid charged to 400 KV has about 2.4
Joules of electrostatic energy. If you now suddenly discharge this
energy to ground, the resulting discharge current can easily be
hundreds, or even thousands, of amps during the brief instant that the
discharge occurs! Heavy toroid-to-ground discharges typically last only
a few microseconds, and can discharge ALL of the electrostatic energy
residing in the secondary:toroid during this brief amount of
time.
Most coilers notice early on that flashovers to solid grounds are hot,
bright, and loud.. that's because they pack LOTS of current for a short
time. They are basically capacitive discharges! Imagine what these could
do if they went through you on the way to going to a solid ground...!
:^(
3. Streamers to air:
This is the toughest case of all. The streamers are dynamic and
inherently non-linear in their voltage versus current relationship. The
breakdown sequence can also be quite complex as newer streamers may
choose to follow in the footsteps of their predecessors, perhaps
"growing" overall streamer length in the process, or starting anew.
Finally, its not at all clear just what secondary voltage we've got at
any time. These streamers derive their energy from the secondary &
toroid, and they therefore alter the toroid voltage in the process of
heating and ionizing the air. In effect, the reactive currents seen in
case 1 above, are now augmented by an additional "lossy" component,
representing energy that's being removed from the secondary LC system
and dissipated as heat in the streamers.
The Q of the secondary system is a measure of its "lossiness". A number
of experimental measurements on operating coils have shown that the
secondary Q declines markedly once streamers are formed. One set of
experiments, for example, showed secondary Q dropping from about 150 to
the range of 10-20 depending upon the operational power level and how
"heavy" the streamers were. This was on a secondary & toroid with a
characteristic impedance at resonance of about 42000 Ohms. If we grossly
model this as leakage current due to the lossy dielectric (air) of the
secondary toroid cap, then this would represent an average "leakage"
resistance of about 420k Ohms under moderately heavy (Q=10) streamer
conditions. This implies that the "average" streamer current for this
coil was in the ballpark of about 1/10 of the reactive current covered
in the no-breakout case above or about:
Iave ~ 600 mA
However, streamer currents actually flow for only part of the total time
- being extinguished and reignited as the secondary output voltage
oscillates. The above may not fully reflect the fact that substantial
toroid capacitance may also provide short-duration, high-amplitude
current spikes during streamer propagation.
Another interesting experiment was done by forcing streamer currents to
flow through a 25 watt 120 Volt lightbulb. In this case, the lightbulb
could be partially lit, representing an "average" current of about 0.13
Amps (matched to an equal degree of illumination from a similar lamp
directly driven off a variac). A light bulb filament is a GREAT
integrator, capable of withstanding incredible abuse! The "on-time" for
streamer current was only about 100 uSec max, and the pulse repetition
rate was about 420 BPS, for a total "on-time" of about 420*100 = 42000
uS or 42 millisec. From this, a ballpark estimate of the peak current
can be back-figured:
Ipeak = 0.13/0.042 ~ 3 Amps (peak)
Again it must be emphasized that neither of the above measurements
constitute precise measurements - just an attempt to determine relative
magnitudes. It's also obvious that secondary measurements are one area
not very well instrumented. Hopefully, you've also got a better feel for
the complexities of the seemingly simple question you posed. :^)
As I prepare to duck from the oncoming flames and torrent of rotten
vegetable matter, I wish safe coiling to you, Stan!
-- Bert H --