Re: Impedance Matching

From: 	Antonio Carlos M. de Queiroz[SMTP:acmq-at-compuland-dot-com.br]
Sent: 	Sunday, December 07, 1997 11:55 PM
To: 	Tesla List
Subject: 	Re: Impedance Matching

Alfred C. Erpel wrote:

>     I have a question that obviously shows that I don't have a total
> understanding of impedance.  Take the example of a neon sign transformer
> (NST) 60hz output of 10000V at .020 amps.
> Impedance Z = E / I     =    10,000 / .020   =   500,000

This is not properly impedance matching. You are only computing the 
impedance that will extract from the transformer the maximum power
that it can -normally- deliver. If you really want, it can deliver more
power than this.
>     If I am designing a Tesla Coil starting at the NST, the next thing I
> must do is pick a capacitor with a rating that will allow maximum energy
> transfer from the transformer.  Using the following formula (because I can
> follow instructions with the best of them):
> C = 1 / (2 * pi * Z * f)
> C = capacitance in picofarads
> Z = impedance of transformer (60 hz)
> f = frequency (hz)
> ergo, in this situation:
> C = .000000005 picofarads or .005 microfarads
>     OK. So the capacitor I build will have a capacitance of .005uF.

To this point, ok, but capacitances are measured in farads (F). 1 picofarad=1e-12 F.
The exact value for this capacitor is 1/(60*PI) microfarads (5.3 nF).
>     My Question :
>     *************
>     The energy that is in 1/2 cycle (180 degrees or 1/120th of a second) of
> output from the NST is 10,000V * .020A * 1/120sec. = 1.666 Watt-sec.

No. If the voltage waveform is sinusoidal, with a capacitive load the current
waveform is cosinusoidal. They are not in phase.

The power delivered to the capacitor from a zero crossing of the voltage
waveform to its next peak can be computed, by integrating the product of voltage
and current, as:

E=Integral from 0 to T/4 of 2*10000*0.02*sin(w*t)*cos(w*t)*dt, where w=2*pi*60
and T=1/60. This results in 5/3/PI=0.53 J (Joules).

> The
> energy that can be stored in a capacitor is E=1/2CV*2.  So, E = .5 *
> .000000005 * 10,000 * 2 = .25 Watt-sec.

Use the peak value of the voltage waveform, 10000*sqrt(2), and the voltage is 
squared, not multiplied by 2:

E=0.5*(1/(60*PI))*2*10000^2=5/3/PI = 0.53 J. As expected.

>     What is happening to these Watt-sec. (85% of every half cycle) of
> energy?

You mixed RMS values with peak values, and didn't compute the energies correctly.
Antonio Carlos M. de Queiroz