Re: 60 vs. 30 ma - charging

From: 	William Noble[SMTP:William_B_Noble-at-msn-dot-com]
Sent: 	Wednesday, June 25, 1997 11:51 PM
To: 	Tesla List
Subject: 	RE: 60 vs. 30 ma - charging

what you are describing is just a current limit - if the load increases beyond 
a certain point, the current is limited, which means that voltage drops 
(remember E=IR, so to hold I constant, you must increase R or decrease E).  I 
believe if you wanted to model the charging of the capacitor, you would use a 
current source, so Vc=IT.  At some point dV/dT of the input sinusoid will be 
below the current limit I, at which point you can then add in the remaining 
portion of the voltage using the standard cosine of the phase angle times the 
peak voltage equation.  The crossover point between the 2 equations is where 
the current calculated by using the cosine of the phase angle is equal to the 
current limit of the transformer.  If you really want one equation, it's just 
the sum of two integrals, one using constant current for T=0 to Tl, and one 
using cosine of the pnase angle for T=Tl to 90 degrees.  (yes, I know I'm 
mixing time and degrees, but they are equivalent, given (360*60) degrees per 

maybe to make this clearer, imagine that the current limit operates for the 
first 45 degrees.  In that case, you have a voltage curve that consists of a 
linear ramp for the first 45 degrees, and a sinusoidal leveling out for the 
next 45 degrees.  Since Sin is pretty linear near zero, this will look a lot 
like a sine function.  The current into the capacitor is CdV/dT, so just set 
that equal to the current limit, and solve to find the phase angle where the 
transformer comes out of current limit.


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.