Resonant Voltage Formula Discussion
Date : Tue, 06 Jul 2004 16:45:19 -0600. Subject : Re: Resonant Voltage Formula
Original poster: "Jim L.
For simple RLC circuits (reasonable here), it's going to be of the general form
1-exp(-k*t)
----- Original Message -----
From: "Tesla list"
<tesla@pupman.com>
To:
<tesla@pupman.com>
Sent: Monday, July 05, 2004 8:31 PM
Subject: Resonant Voltage Formula
> Original poster: John
<fireba8104@yahoo.com>
>
> Hello all, As we all know a resonant circuit driven at it's resonant frequency
will have a final voltage of Vin * Q of the circuit. I wish to know if there is
a formula or method for finding the voltage at X number of cycles. This
> also might be useful in the up and coming DRSSTC field. Cheers, John.
Date : Tue, 06 Jul 2004 19:05:59 -0600. Subject : Re: Resonant Voltage Formula
Original poster: "Antonio Carlos M. de Queiroz"
<acmdq@uol.com.br>
Tesla list wrote:
>
> Original poster: John
<fireba8104@yahoo.com>
> As we all know a resonant circuit driven at it's resonant frequency will
> have a final voltage of Vin * Q of the circuit. I wish to know if there is
> a formula or method for finding the voltage at X number of cycles. This
> also might be useful in the up and coming DRSSTC field.
For a - single LCR circuit - the amplitude of the oscillations follows:
Vpeak = Vmax(1-exp(-t/T))
where T=2*Q/w0 and t is the time.
but t=n*2*pi/w0 where n is the number of cycles.
Vpeak = Vmax(1-exp(-n*pi/Q))
Q cycles: 0.96 Vmax
Q/2 cycles: 0.80 Vmax
Q/4 cycles: 0.54 Vmax
Antonio Carlos M. de Queiroz
Date : Tue, 06 Jul 2004 20:26:36 -0600. Subject : Re: Resonant Voltage Formula
Original poster: John
<fireba8104@yahoo.com>
Thanks Antonio, Others as well as I will find this very useful. Cheers,
John.
Date : Fri, 09 Jul 2004 07:32:08 -0600. Subject : Re: Resonant Voltage Formula
> As we all know a resonant circuit driven at it's resonant frequency will
have a final voltage of Vin * Q of the circuit. I wish to know if there is a
formula or method for finding the voltage at X number of cycles. This also
might be useful in the up and coming DRSSTC field.
For a - single LCR circuit - the amplitude of the oscillations follows:
Vpeak = Vmax(1-exp(-t/T))
where T=2*Q/w0 and t is the time.
but t=n*2*pi/w0 where n is the number of cycles.
Vpeak = Vmax(1-exp(-n*pi/Q))
Q cycles: 0.96 Vmax
Q/2 cycles: 0.80 Vmax
Q/4 cycles: 0.54 Vmax
Antonio Carlos M. de Queiroz"
The quantity [pi/Q] in "Vpeak = Vmax(1-exp(-n*pi/Q))" is known as the
"logarithmic decrement" of the waveform. The smaller the value the longer the
decay time. In "the good old days of spark gap transmitters"
the goal of all transmitter designers was to get the lowest possible value of
"decrement" since that resulted in the "narrowest wave" and least interference
to other services. The log decrement is just equal to the natural log of the
ratio of two successive voltage peaks which, for exponential decay, is constant
with time. Ed.
Date : Sat, 10 Jul 2004 21:54:09 -0600. Subject : Re: Resonant Voltage Formula
> Original poster: "Steve C.
>
The quantity [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 :(
This is exactly the formula for "ring up". The formula for "ring down" is Vpeak
= Vmax*exp(-n*pi/Q).
Antonio Carlos M. de Queiroz.
Date : Sat, 10 Jul 2004 21:54:37 -0600.
Subject : Re: Resonant Voltage Formula
Hi Steve, The logarithmic decay is because the energy loss per unit time decreases as the signal decays. I'm wondering on ramp up, if one can say the rate of energy pumped in is constant and deduce the envelope of the sine wave. Maybe you can scope the ring-up and make an approximation. Gerry R.
Date : Sat, 10 Jul 2004 21:54:16 -0600.
Subject : Re: Resonant Voltage Formula
> > Original poster: "Steve C.
> > >The quantity [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 :(
>
> This is exactly the formula for "ring up". The formula for "ring down" is
Vpeak = Vmax*exp(-n*pi/Q).
Hummm. Really, the expression for ring-up is more complex, because there is the
forced response to the input signal added to a transient response. But the
formula above gives a good approximation when Q is high and the excitation
frequency is close to the resonance frequency of the network.
Vmax=Q*Vin in the expression. Antonio Carlos M. de Queiroz.
Date : Sun, 11 Jul 2004 17:32:12 -0600
Subject : Re: Resonant Voltage Formula
Original poster: "Jim L.
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))