Re: What it takes to get big sparks?
The current flowing through this "air capacitor" is a large displacement
current. This is one reason why it seems to "float" around in the air and
doesn't always quickly strike the nearest ground. Sometimes the attachment
time is measured in full seconds vs. microseconds for DC and 60 Hz AC high
> From: Tesla List <tesla-at-pupman-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: What it takes to get big sparks?
> Date: Thursday, December 31, 1998 6:45 PM
> Original Poster: "Jim Lux" <jimlux-at-jpl.nasa.gov>
> I have been contemplating the whole issue of what factors most influence
> the length and size of the sparks from a tesla coil and have some ideas
> which I'm going to throw out for discussion.
> Starting with some assumptions:
> 1) The actual spark develops very quickly if there is a suitable low
> impedance source. The leader moves very very fast, essentially running
> entire length in a microsecond or thereabouts.
> 2) The charge for the spark comes almost exclusively from the top load.
> inductive impedance of the secondary coil is so high (when considered in
> time frame of a microsecond) that any sort of stored energy in the coil
> isn't going to get there in time (except of course, for the current
> flowing from the inductor into the capacitor as the top load voltage is
> 3) As the spark develops, it is essentially a capacitor that has
> capacitance (because it is getting bigger) and is charging at the same
> time, as well as heating the air up. This means that current is flowing
> through this "spark capacitor".
> I asked myself, where is the other connection of this capacitor?
> I think it is the ground (and the surroundings, etc.).
> And, because we are talking fast time scales here (microseconds, again),
> the inductance of any sort of wire connection from the bottom of the coil
> to the "ground" means that the coil can't really supply any appreciable
> current while the spark is forming.
> So then, what we have is an inductor (the secondary) which charges a
> capacitor (the top load working against the ground). That capacitor then
> breaks down (i.e. a leader forms) and discharges into the spark. The
> spark formation is almost independent of the secondary inductor because
> current required to develop the spark is much higher than that coming
> the inductor to the capacitor.
> This is why big top loads help make big sparks. And it is also why good
> grounds make big sparks. Interestingly, though, the nature of the ground
> is important. Of course, it has to be low impedance at the coil's basic
> operating frequency (otherwise, the secondary can't charge the top
> load/ground capacitor fast enough). Given the high inductance of most
> secondaries (milliHenries) compared to the inductance of even thin wires
> that are several feet long (microhenries), from this standpoint, it
> matter how you connect the base of the coil to the "ground".
> However, once that "capacitor" is charged, and it starts to discharge
> the spark), you want the capacitor to be low inductance. The top load is
> pretty low inductance (it is physically small) and resistance (it is
> metal), but what about that ground? How low impedance is that ground?
> Remember, at this time, what we are really looking at is one capacitor
> charging another through a resistance and inductance.
> So, I think what you want is a suitably conductive sheet to operate
> coil on top of. It doesn't have to be solid, but it does have to have low
> resistance per square, and, also important, low inductance per square
> I don't think one long wire running around would really do it). It can be
> insulated because it is part of a capacitor with mostly air dielectric,
> adding a thin layer of anything won't make any difference. It should be
> enough that it can supply enough charge to keep the spark going... I
> suggest that it should be big enough so that increasing its size by some
> increment doesn't increase the capacitance to the top load by more than
> 5-10%. This probably isn't all that big. For a big sheet which has low
> inductance, the propagation velocity of the charge is going to be a
> significant fraction of c (1 ft/nSec), and assuming you want your
> to be able to discharge in something on the order of 10-100 nSec, making
> bigger than 20-30 ft isn't going to buy you much.
> Thinking about how the field from the top load is distributed, in fact,
> hazard a guess that you should work against a disk of radius equal to the
> height of the top load above the disk, or maybe, twice that.
> Comments, thoughts, etc.....