Re: Resonant Voltage Formula

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jim Lux" <jimlux-at-earthlink-dot-net> 

It's actually exponential rise and decay, not logarithmic..

They're both for the same reason.. Take the decay case first.. it's easier.
The rate of energy loss is proportional to the amount of energy in the
system (that is, you lose a constant "percentage" each cycle).
This forms a simple differential equation f' = -kf (where the ' indicates a
derivative.. df/dt = -k *f)  and the f=exp(-kt) is a solution to this.

The increment/rise/charging is a bit different... the loss is the same, but
each cycle you are adding energy, until the energy lost = the energy added
(i.e. everything balances)... this makes it df/dt = k1 - k2f  (the k1 is the
amount you're adding, the -k2f is the amount lost each cycle..)

So, the solution for this one winds up being of the form ka *(1 - exp(-kat))


----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Saturday, July 10, 2004 8:54 PM
Subject: Re: Resonant Voltage Formula


 > Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net>
 >
 > Hi Steve,
 >
 > The logrithmic decay is be cause the energy loss per unit time decreases
as
 > the signal decays.  Im wondering on ramp up, if one can say the rate of
 > energy pumped in is constant and deduce the envelope of the sinewave.
Maybe
 > you can scope the ringup and make an approximation.
 >
 > Gerry R.
 >
 >  > Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com>
 >  >
 >  >  >The quantiti [pi/Q] in "Vpeak = Vmax(1-exp(-n*pi/Q))" is known as the
 >  >  >"logarithmic decrement" of the waveform.
 >  >
 >  > The original poster was not asking about how the waveform decays
 > naturally,
 >  > but how it rings _up_ when the resonance is forced by a SSTC-like
driver.
 > I
 >  > must admit that I don't know a simple formula for this one :(
 >  >
 >  > Steve C.
 >  >
 >  >
 >
 >