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Re: 60 vs. 30 ma - charging
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