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Re: Light Bulb Experiment (ala Brent Turner)

Tesla List wrote:
> 
> >From bert.hickman-at-aquila-dot-comTue Sep 24 22:38:25 1996
> Date: Tue, 24 Sep 1996 23:11:06 -0700
> From: Bert Hickman <bert.hickman-at-aquila-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Re: Light Bulb Experiment (ala Brent Turner)
> 
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> All:
> Thanks to everyone who responded re: the Light Bulb Experiment. Some
> specific feedback:
> 
> >From Brent Turner:
> > First off, what you are actually doing is using the light bulb as a
> > sort of power indicator, ala a hot-wire ammeter or watt-meter. Ironically,
> > I have seen a 100-watt bulb work better than a 40-watt bulb due to the
> > lowered filament resistance (snip>. A low wattage lamp has a higher filament
> > resistance, hence for a given.amount of power, will develop more voltage across it.
> 
> <SNIP>
> 
> I wasnt able to get a 100 Watt bulb to light, but I easily burned out a
> 15 Watt bulb.  I agree that  given a relatively fixed  amount of current
> flowing through the bulb to the corona, the power dissipated in the lamp
> will be a function of R*I^2. A smaller wattage light will develop more
> voltage across the filament, lighting it brighter. In my trials, large
> wattage bulbs were consistently dimmer.
> 
> >From Richard Hull:
> <SNIP>
> > The impedance of a Tesla coil's terminal is very high.  The impeadance of
> > pure air corona arc is even higher.  Thus the voltage differential
> > available across the bulb is high and the current is at a maximum for
> > this condition.  (It lights to a brillant glow)  When an arc strikes
> > ground, the current is maximum  as the coil goes to a very low impedance,
> > but the voltage across the coil and lamp are at a minimum (dim or
> > invisible glow).  To light a bulb, you need both current and voltage.
> > you will note that the arc current is brilliant on a ground hit. Most of
> > the energy is being fed to the air arc and not the lamp.
> 
> <SNIP>
> 
> Based upon an inductance of  73400 uH and an estimated self capacitance
> of
> about 15.8 pF, my secondary coils surge impedance should be about
> SQRT(L/C)=42k Ohms.  Based upon previous measurements of secondary Q
> with an
> air-only  discharge, the  impedance of  the air-only corona looks to be
> larger by at least an order of magnitude or more  (between 500k and 1
> Megohm). However, once the discharge to ground happens, I would expect
> the arc impedance to drop, and the instantaneous surge current to
> increase greatly. If this current were flowing through the filament, Id
> expect the bulb to increase in brightness unless the overall duty cycle
> decreased enough to more than offset the increased current flow.
> However. I now think these surge currents may be  bypassing the filament
> due to unseen arc-overs in the lampbase.
> I didnt quite follow the part about the coil going to a very low
> impedance during an arc strike to ground... Could you go into that a
> little bit more?? Thanks!


The base impedance of the coil to ground is the thing!  As this is the 
source of energy in a ringing coil.  This, if measured, will be in the 
few tens of ohms on a heavily top loaded system.  At resonance, ideally, 
the coil's DC resistance would be the only resistance in circuit. This is 
altered by the top loading capacitance. Regardless, the ground arcing 
coil is a very low source impedance.  Thus, the high currents.

Richard Hull, TCBOR