# Impedance Matching

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From: 	Alfred C. Erpel[SMTP:aerpel-at-op-dot-net]
Sent: 	Saturday, December 06, 1997 9:14 PM
To: 	Tesla List
Subject: 	Impedance Matching

Hello All,

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

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)

Z = impedance of transformer (60 hz)
f = frequency (hz)

ergo, in this situation:

OK. So the capacitor I build will have a capacitance of .005uF.

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.  The
energy that can be stored in a capacitor is E=1/2CV*2.  So, E = .5 *
.000000005 * 10,000 * 2 = .25 Watt-sec.
So each half cycle, the capacitor is presented with 1.666 Watt-sec. of
energy and gets fully charged (to its capacity:)of .25 Watt-sec., but that
leaves
(1.666 Watt-sec.  -   .25 Watt-sec.) or 1.416 Watt-sec. unaccounted for (to
me).
What is happening to these Watt-sec. (85% of every half cycle) of
energy?

Regards,
Alfred

```