Re: 60 vs. 30 ma - charging
From: Malcolm Watts[SMTP:MALCOLM-at-directorate.wnp.ac.nz]
Sent: Thursday, June 26, 1997 5:54 PM
Subject: Re: 60 vs. 30 ma - charging
Hi Skip, 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.
Since the neon leakage inductance and cap form an almost pure tuned
circuit (if you look at the neon resistance and inductance
vectorially, you see that inductance is by far the major determinant
of circuit action), the current in the circuit and voltage across the
components is almost in quadrature: i.e. v(t) = Asin(wt) and
i(t) = Bsin(wt + PI/4) = Bcos(wt). Taking the resistance into account
as well and modelling it in series with the inductance gives
Z = SQRT( R^2 + Xl^2 ). The phase angle comes somewhere between 80
and 90 degrees for a typical neon. For all three components,
Z = SQRT( R^2 + ( Xl - Xc )^2 ).