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Re: 60 vs. 30 ma - charging
From: John H. Couture[SMTP:couturejh-at-worldnet.att-dot-net]
Sent: Wednesday, June 25, 1997 3:36 PM
To: Tesla List
Subject: Re: 60 vs. 30 ma - charging
At 12:13 PM 6/25/97 +0000, you wrote:
>
>From: Skip Greiner[SMTP:sgreiner-at-wwnet-dot-com]
>Sent: Tuesday, June 24, 1997 11:31 PM
>To: Tesla List
>Subject: Re: 60 vs. 30 ma - charging
>
>Tesla List wrote:
>>
>> From: William Noble[SMTP:William_B_Noble-at-msn-dot-com]
>> Sent: Tuesday, June 24, 1997 1:50 AM
>> To: Tesla List
>> Subject: RE: 60 vs. 30 ma - charging
>>
>> the voltage on a capacitor is the integral of the current into the capacitor.
>> thus, if you hold constant current the voltage provides a ramp (dV/dT is a
>> constant). Conversely, the current into a capacitor is the deriviative of
>> voltage, eg i=c(dv/dt), so if you put a voltage step from an ideal source
into
>> a capacitor, you will have a single infinite current spike, thereafter 0.
>> Now, of course there is always resistance.
>>
>> anyway, the terse answer to your question is yes. However it's misleading.
>> If you look at a neon sign transformer as a constant current device, or at
>> least as a current limited device, and say that V=Vmax(cos(omega)) then you
>> will find that the current into the capacitor is probably at the transformer
>> max when omega is near 0, and the current may or may not reach 0 when
omega is
>> 90, depending on the value of C - if the integral of ImaxC for 1/240 of a
>> second is > Vmax then the capacitor will charge to Vmax, else it won't
>> (remember the voltage goes from 0 to max in 90 degrees = 1/240 sec at 60 hz)
>>
>> ----------
>big snip
>
>Hi All
>
>If the neon current goes to its maximum then the voltage goes to zero
>(almost). Try it!! I do not see how to write an equation which will show
>that the neon voltage is zero at max current and then increases as the
>current decreases. Perhaps you can show in an equation, not words, the
>relationship between current and voltage at the terminals of a neon.
>With that in hand we can go about integrating the current to get the
>final voltage on the cap.
>
>Skip
>
>-----------------------------------------------
Skip -
The neon transformer problem is not simple and has confused many coilers.
I like to look at the neon problem from a loading standpoint. The voltage
and wattage output will vary depending on the load. However, the current
wiil be constant only up to a design load then it will quickly reduce to
zero as the load goes to infinity (open circuit). The voltage and wattage
will be almost zero at almost zero load.
The design load is Z = V/I where V is the voltage rating of the neon and
I is the current rating. You will recognize this equation as the one that is
sometimes (incorrectly) used when selecting the TC primary capacitor size.
To find the voltage at different loads use
V = ZxI where Z is the load ohms and I is the rated current of
the neon transformer.
For example 12000 volt 30 ma 360 watt neon transformer
If load is 200 Kohms resistor
V = 200 000 x .03 = 6000 volts
W = V I = 6000 x .03 = 180 watts
The load resistor will get hot!
The voltage (Ohms law) and wattage is a linear relationship with the
load because the current is constant.
With a capacitor the loading on the neon is an exponential curve. The
voltage on the cap starts at zero (load almost zero) and builds up to the
voltage rating of the neon in an exponential curve (vs time) at constant
current. When the voltage on the cap equals the voltage rating of the neon
the current reduces to only the losses of the circuit (load is very high
impedance).
The big question is how does the TC pri cap load the neon when turned on
and off by the operating spark gap at different breaks per second? Do any
coilers on the List know how to find the optimum conditions? I have been
working on this problem for some time but with little success. Keep in mind
that the loading on the neon should be maximized for best results (an
iterative solution?).
John Couture